Thread: "Physical 4D puzzle V2"

From: Melinda Green <melinda@superliminal.com>
Date: Wed, 26 Apr 2017 18:09:33 -0700
Subject: Physical 4D puzzle V2



Dear Cubists,

I've finished version 2 of my physical puzzle and uploaded a video of it here:
https://www.youtube.com/watch?v=zqftZ8kJKLo
Again, please don't share these videos outside this group as their purpose is just to get your feedback. I'll eventually replace them with a public video.

Here is an extra math puzzle that I bet you folks can answer: How many families of states does this puzzle have? In other words, if disassembled and reassembled in any random configuration the magnets allow, what are the odds that it can be solved? This has practical implications if all such configurations are solvable because it would provide a very easy way to fully scramble the puzzle.

And finally, a bit of fun: A relatively new friend of mine and new list member, Marc Ringuette, got excited enough to make his own version. He built it from EPP foam and colored tape, and used honey instead of magnets to hold it together. Check it out here: http://superliminal.com/cube/dessert_cube.jpg I don't know how practical a solution this is but it sure looks delicious! Welcome Marc!

-Melinda




From: Melinda Green <melinda@superliminal.com>
Date: Thu, 27 Apr 2017 18:00:05 -0700
Subject: Re: [MC4D] Physical 4D puzzle V2



--------------BB568385ED62C2A830A864EC
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: quoted-printable

The new arrangement of magnets allows every valid orientation of pieces. Th=
e only invalid ones are those where the diagonal lines cutting each cube's =
face cross each other rather than coincide. In other words, you can assembl=
e the puzzle in all ways that preserve the overall diamond/harlequin patter=
n. Just about every move you can think of on the whole puzzle is valid thou=
gh there are definitely invalid moves that the magnets allow. The most obvi=
ous invalid move is twisting of a single end cap.

I think your description of the center face is not correct though. Twists o=
f the outer faces cause twists "through" the center face, not "of" that fac=
e. Twists of the outer faces are twists of those faces themselves because t=
hey are the ones not changing, just like the center and outer faces of MC4D=
when you twist the center face. The only direct twist of the center face t=
hat this puzzle allows is a 90 degree twist about the outer axis. That happ=
ens when you simultaneously twist both end caps in the same direction.

Yes, it's quite straightforward reorienting the whole puzzle to put any of =
the four axes on the outside. This is a very nice improvement over the firs=
t version and should make it much easier to solve. You may be right that we=
just need to find the right way to think about the outside faces. I'll lea=
ve it to the math geniuses on the list to figure that out.

-Melinda

On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing] w=
rote:
>
> Hi Melinda,
>
> I do not agree with the criticism regarding the white and yellow stickers=
touching each other, this could simply be an effect of the different repre=
sentations of the puzzle. To really figure out if this indeed is a represen=
tation of a 2x2x2x2 we need to look at the possible moves (twists and rotat=
ions) and figure out the equivalent moves in the MC4D software. From the MC=
4D software, it's easy to understand that the only moves required are free =
twists of one of the faces (that is, only twisting the center face in the s=
tandard perspective projection in MC4D) and 4D rotations swapping which fac=
e is in the center (ctrl-clicking in MC4D). The first is possible in your p=
hysical puzzle by rotating the white and yellow subcubes (from here on I us=
e subcube to refer to the two halves of the puzzle and the colours of the s=
ubcubes to refer to the "outer colours"). The second is possible if it's po=
ssible to reach a solved state with any two colours on the subcubes that st=
ill allow you=20
> to perform the previously mentioned twists. This seems to be the case fro=
m your demonstration and is indeed true if the magnets allow the simple twi=
sts regardless of the colours of the subcubes. Thus, it is possible to let =
your puzzle be a representation of a 2x2x2x2, however, it might require tha=
t some moves that the magnets allow aren't used.
>
> Best regards,
> Joel
>
> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com melinda@superliminal.com> [4D_Cubing] <4D_Cubing@yahoogroups.com _Cubing@yahoogroups.com>>:
>
> Dear Cubists,
>
> I've finished version 2 of my physical puzzle and uploaded a video of=
it here:
> https://www.youtube.com/watch?v=3DzqftZ8kJKLo m/watch?v=3DzqftZ8kJKLo>
> Again, please don't share these videos outside this group as their pu=
rpose is just to get your feedback. I'll eventually replace them with a pub=
lic video.
>
> Here is an extra math puzzle that I bet you folks can answer: How man=
y families of states does this puzzle have? In other words, if disassembled=
and reassembled in any random configuration the magnets allow, what are th=
e odds that it can be solved? This has practical implications if all such c=
onfigurations are solvable because it would provide a very easy way to full=
y scramble the puzzle.
>
> And finally, a bit of fun: A relatively new friend of mine and new li=
st member, Marc Ringuette, got excited enough to make his own version. He b=
uilt it from EPP foam and colored tape, and used honey instead of magnets t=
o hold it together. Check it out here: http://superliminal.com/cube/dessert=
_cube.jpg I don't know how =
practical a solution this is but it sure looks delicious! Welcome Marc!
>
> -Melinda
>
>
>
>
>=20


--------------BB568385ED62C2A830A864EC
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable



">


The new arrangement of magnets allows every valid orientation of
pieces. The only invalid ones are those where the diagonal lines
cutting each cube's face cross each other rather than coincide. In
other words, you can assemble the puzzle in all ways that preserve
the overall diamond/harlequin pattern. Just about every move you can
think of on the whole puzzle is valid though there are definitely
invalid moves that the magnets allow. The most obvious invalid move
is twisting of a single end cap.



I think your description of the center face is not correct though.
Twists of the outer faces cause twists "through" the center face,
not "of" that face. Twists of the outer faces are twists of those
faces themselves because they are the ones not changing, just like
the center and outer faces of MC4D when you twist the center face.
The only direct twist of the center face that this puzzle allows is
a 90 degree twist about the outer axis. That happens when you
simultaneously twist both end caps in the same direction.



Yes, it's quite straightforward reorienting the whole puzzle to put
any of the four axes on the outside. This is a very nice improvement
over the first version and should make it much easier to solve. You
may be right that we just need to find the right way to think about
the outside faces. I'll leave it to the math geniuses on the list to
figure that out.



-Melinda



On 4/27/2017 10:31 AM, Joel Karlsson
mail.com">joelkarlsson97@gmail.com [4D_Cubing] wrote:

cite=3D"mid:CAEohJcG3N3hknw4fppRPtPW=3DpwWgj_Ux4yAO=3DoASS5Hp2Fcivg@mail.gm=
ail.com"
type=3D"cite">





Hi Melinda,




I do not agree with the criticism regarding the white and
yellow stickers touching each other, this could simply be an
effect of the different representations of the puzzle. To
really figure out if this indeed is a representation of a
2x2x2x2 we need to look at the possible moves (twists and
rotations) and figure out the equivalent moves in the MC4D
software. From the MC4D software, it's easy to understand
that the only moves required are free twists of one of the
faces (that is, only twisting the center face in the
standard perspective projection in MC4D) and 4D rotations
swapping which face is in the center (ctrl-clicking in
MC4D). The first is possible in your physical puzzle by
rotating the white and yellow subcubes (from here on I use
subcube to refer to the two halves of the puzzle and the
colours of the subcubes to refer to the "outer colours").
The second is possible if it's possible to reach a solved
state with any two colours on the subcubes that still allow
you to perform the previously mentioned twists. This seems
to be the case from your demonstration and is indeed true if
the magnets allow the simple twists regardless of the
colours of the subcubes. Thus, it is possible to let your
puzzle be a representation of a 2x2x2x2, however, it might
require that some moves that the magnets allow aren't used.




Best regards,


Joel




2017-04-27 3:09 GMT+02:00 Melinda Green
href=3D"mailto:melinda@superliminal.com">melinda@superliminal.c=
om

[4D_Cubing] < href=3D"mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4=
D_Cubing@yahoogroups.com
>
:

.8ex;border-left:1px #ccc solid;padding-left:1ex">

=C2=A0



Dear Cubists,



I've finished version 2 of my physical puzzle and
uploaded a video of it here:

href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJK=
Lo"
target=3D"_blank">https://www.youtube.com/watch?r>v=3DzqftZ8kJKLo


Again, please don't share these videos outside
this group as their purpose is just to get your
feedback. I'll eventually replace them with a
public video.



Here is an extra math puzzle that I bet you folks
can answer: How many families of states does this
puzzle have? In other words, if disassembled and
reassembled in any random configuration the
magnets allow, what are the odds that it can be
solved? This has practical implications if all
such configurations are solvable because it would
provide a very easy way to fully scramble the
puzzle.



And finally, a bit of fun: A relatively new friend
of mine and new list member, Marc Ringuette, got
excited enough to make his own version. He built
it from EPP foam and colored tape, and used honey
instead of magnets to hold it together. Check it
out here: href=3D"http://superliminal.com/cube/dessert_cube.j=
pg"
target=3D"_blank">http://superliminal.com/cube/>dessert_cube.jpg

I don't know how practical a solution this is but
it sure looks delicious! Welcome Marc!



-Melinda













=20=20=20=20=20=20







--------------BB568385ED62C2A830A864EC--




From: Joel Karlsson <joelkarlsson97@gmail.com>
Date: Fri, 28 Apr 2017 08:14:48 +0200
Subject: Re: [MC4D] Physical 4D puzzle V2



--94eb2c0d47a6ab5c4d054e33fd63
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Seems like there was a slight misunderstanding. I meant that you need to be
able to twist one of the faces and in MC4D the most natural choice is the
center face. In your physical puzzle you can achieve this type of twist by
twisting the two subcubes although this is indeed a twist of the subcubes
themselves and not the center face, however, this is still the same type of
twist just around another face.

If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
this puzzle. Hopefully the restrictions will be quite natural and only some
"strange" moves would be illegal. Regarding the "families of states" (aka
orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
twists preserves the parity of the pieces, meaning that only half of the
permutations you can achieve by disassembling and reassembling can be
reached through legal moves. Because of some geometrical properties of the
2x2x2x2 and its twists, which would take some time to discuss in detail
here, the orientation of the stickers mod 3 are preserved, meaning that the
last corner only can be oriented in one third of the number of orientations
for the other corners. This gives a total number of orbits of 2x3=3D6. To
check this result let's use this information to calculate all the possible
states of the 2x2x2x2; if there were no restrictions we would have 16! for
permuting the pieces (16 pieces) and 12^16 for orienting them (12
orientations for each corner). If we now take into account that there are 6
equally sized orbits this gets us to 12!16^12/6. However, we should also
note that the orientation of the puzzle as a hole is not set by some kind
of centerpieces and thus we need to devide with the number of orientations
of a 4D cube if we want all our states to be separated with twists and not
only rotations of the hole thing. The number of ways to orient a 4D cube in
space (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a
total of 12!16^12/(6*192) states which is indeed the same number that for
example David Smith arrived at during his calculations. Therefore, when
determining whether or not a twist on your puzzle is legal or not it is
sufficient and necessary to confirm that the twist is an even permutation
of the pieces and preserves the orientation of stickers mod 3.

Best regards,
Joel

Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
[4D_Cubing]" <4D_Cubing@yahoogroups.com>:



The new arrangement of magnets allows every valid orientation of pieces.
The only invalid ones are those where the diagonal lines cutting each
cube's face cross each other rather than coincide. In other words, you can
assemble the puzzle in all ways that preserve the overall diamond/harlequin
pattern. Just about every move you can think of on the whole puzzle is
valid though there are definitely invalid moves that the magnets allow. The
most obvious invalid move is twisting of a single end cap.

I think your description of the center face is not correct though. Twists
of the outer faces cause twists "through" the center face, not "of" that
face. Twists of the outer faces are twists of those faces themselves
because they are the ones not changing, just like the center and outer
faces of MC4D when you twist the center face. The only direct twist of the
center face that this puzzle allows is a 90 degree twist about the outer
axis. That happens when you simultaneously twist both end caps in the same
direction.

Yes, it's quite straightforward reorienting the whole puzzle to put any of
the four axes on the outside. This is a very nice improvement over the
first version and should make it much easier to solve. You may be right
that we just need to find the right way to think about the outside faces.
I'll leave it to the math geniuses on the list to figure that out.

-Melinda


On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
wrote:


Hi Melinda,

I do not agree with the criticism regarding the white and yellow stickers
touching each other, this could simply be an effect of the different
representations of the puzzle. To really figure out if this indeed is a
representation of a 2x2x2x2 we need to look at the possible moves (twists
and rotations) and figure out the equivalent moves in the MC4D software.
From the MC4D software, it's easy to understand that the only moves
required are free twists of one of the faces (that is, only twisting the
center face in the standard perspective projection in MC4D) and 4D
rotations swapping which face is in the center (ctrl-clicking in MC4D). The
first is possible in your physical puzzle by rotating the white and yellow
subcubes (from here on I use subcube to refer to the two halves of the
puzzle and the colours of the subcubes to refer to the "outer colours").
The second is possible if it's possible to reach a solved state with any
two colours on the subcubes that still allow you to perform the previously
mentioned twists. This seems to be the case from your demonstration and is
indeed true if the magnets allow the simple twists regardless of the
colours of the subcubes. Thus, it is possible to let your puzzle be a
representation of a 2x2x2x2, however, it might require that some moves that
the magnets allow aren't used.

Best regards,
Joel

2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
[4D_Cubing] <4D_Cubing@yahoogroups.com>:

>
>
> Dear Cubists,
>
> I've finished version 2 of my physical puzzle and uploaded a video of it
> here:
> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
> Again, please don't share these videos outside this group as their purpos=
e
> is just to get your feedback. I'll eventually replace them with a public
> video.
>
> Here is an extra math puzzle that I bet you folks can answer: How many
> families of states does this puzzle have? In other words, if disassembled
> and reassembled in any random configuration the magnets allow, what are t=
he
> odds that it can be solved? This has practical implications if all such
> configurations are solvable because it would provide a very easy way to
> fully scramble the puzzle.
>
> And finally, a bit of fun: A relatively new friend of mine and new list
> member, Marc Ringuette, got excited enough to make his own version. He
> built it from EPP foam and colored tape, and used honey instead of magnet=
s
> to hold it together. Check it out here: http://superliminal.com/cube/d
> essert_cube.jpg I don't know how practical a solution this is but it sure
> looks delicious! Welcome Marc!
>
> -Melinda
>
>



--94eb2c0d47a6ab5c4d054e33fd63
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Seems like there was a slight misunderstanding. I me=
ant that you need to be able to =C2=A0twist one of the faces and in MC4D th=
e most natural choice is the center face. In your physical puzzle you can a=
chieve this type of twist by twisting the two subcubes although this is ind=
eed a twist of the subcubes themselves and not the center face, however, th=
is is still the same type of twist just around another face.=C2=A0
v dir=3D"auto">
If the magnets are that allowing=
the 2x2x2x2 is obviously a subgroup of this puzzle. Hopefully the restrict=
ions will be quite natural and only some "strange" moves would be=
illegal. Regarding the "families of states" (aka orbits), the 2x=
2x2x2 has 6 orbits. As I mentioned earlier all allowed twists preserves the=
parity of the pieces, meaning that only half of the permutations you can a=
chieve by disassembling and reassembling can be reached through legal moves=
. Because of some geometrical properties of the 2x2x2x2 and its twists, whi=
ch would take some time to discuss in detail here, the orientation of the s=
tickers mod 3 are preserved, meaning that the last corner only can be orien=
ted in one third of the number of orientations for the other corners. This =
gives a total number of orbits of 2x3=3D6. To check this result let's u=
se this information to calculate all the possible states of the 2x2x2x2; if=
there were no restrictions we would have 16! for permuting the pieces (16 =
pieces) =C2=A0and 12^16 for orienting them (12 orientations for each corner=
). If we now take into account that there are 6 equally sized orbits this g=
ets us to 12!16^12/6. However, we should also note that the orientation of =
the puzzle as a hole is not set by some kind of centerpieces and thus we ne=
ed to devide with the number of orientations of a 4D cube if we want all ou=
r states to be separated with twists and not only rotations of the hole thi=
ng. The number of ways to orient a 4D cube in space (only allowing rotation=
s and not mirroring) is 8x6x4=3D192 giving a total of=C2=A0ont-family:sans-serif">12!16^12/(6*192) states which is indeed the same num=
ber that for example David Smith arrived at during his calculations. Theref=
ore, =C2=A0when determining whether or not a twist on your puzzle is legal =
or not it is sufficient and necessary to confirm that the twist is an even =
permutation of the pieces and preserves the orientation of stickers mod 3.<=
/span>

span>
Best re=
gards,=C2=A0
serif">Joel=C2=A0
dir=3D"auto">
Den 28 apr. 2017 3:02 fm skrev =
"Melinda Green melinda@sup=
erliminal.com
[4D_Cubing]" <oups.com">4D_Cubing@yahoogroups.com>:
quote class=3D"quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid=
;padding-left:1ex">












=20

=C2=A0







=20=20=20=20=20=20
=20=20=20=20=20=20


=20=20
=20=20
The new arrangement of magnets allows every valid orientation of
pieces. The only invalid ones are those where the diagonal lines
cutting each cube's face cross each other rather than coincide. In
other words, you can assemble the puzzle in all ways that preserve
the overall diamond/harlequin pattern. Just about every move you can
think of on the whole puzzle is valid though there are definitely
invalid moves that the magnets allow. The most obvious invalid move
is twisting of a single end cap.



I think your description of the center face is not correct though.
Twists of the outer faces cause twists "through" the center f=
ace,
not "of" that face. Twists of the outer faces are twists of t=
hose
faces themselves because they are the ones not changing, just like
the center and outer faces of MC4D when you twist the center face.
The only direct twist of the center face that this puzzle allows is
a 90 degree twist about the outer axis. That happens when you
simultaneously twist both end caps in the same direction.



Yes, it's quite straightforward reorienting the whole puzzle to put
any of the four axes on the outside. This is a very nice improvement
over the first version and should make it much easier to solve. You
may be right that we just need to find the right way to think about
the outside faces. I'll leave it to the math geniuses on the list t=
o
figure that out.



-Melinda








=20=20=20=20=20=20



Hi Melinda,




I do not agree with the criticism regarding the white and
yellow stickers touching each other, this could simply be an
effect of the different representations of the puzzle. To
really figure out if this indeed is a representation of a
2x2x2x2 we need to look at the possible moves (twists and
rotations) and figure out the equivalent moves in the MC4D
software. From the MC4D software, it's easy to understand
that the only moves required are free twists of one of the
faces (that is, only twisting the center face in the
standard perspective projection in MC4D) and 4D rotations
swapping which face is in the center (ctrl-clicking in
MC4D). The first is possible in your physical puzzle by
rotating the white and yellow subcubes (from here on I use
subcube to refer to the two halves of the puzzle and the
colours of the subcubes to refer to the "outer colours&quo=
t;).
The second is possible if it's possible to reach a solved
state with any two colours on the subcubes that still allow
you to perform the previously mentioned twists. This seems
to be the case from your demonstration and is indeed true if
the magnets allow the simple twists regardless of the
colours of the subcubes. Thus, it is possible to let your
puzzle be a representation of a 2x2x2x2, however, it might
require that some moves that the magnets allow aren't used.=





Best regards,


Joel



=20=20=20=20=20=20




=20=20




=20=20=20=20=20

=20=20=20=20







=20=20









--94eb2c0d47a6ab5c4d054e33fd63--




From: damienturtle@hotmail.co.uk
Date: 28 Apr 2017 20:50:24 +0000
Subject: Re: Physical 4D puzzle V2




From: damienturtle@hotmail.co.uk
Date: Fri, 28 Apr 2017 16:15:46 -0500
Subject: Re: Physical 4D puzzle V2



--001a113ed258bddb39054e409338
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

I agree with Don's arguments about adjacent sticker colors needing to be
next to each other. I think this can be turned into an accurate 2^4 with
coloring changes, so I agree with Joel too :)

To help me think about it, I started adding a new projection option for
spherical puzzles to MagicTile, which takes the two hemispheres of a puzzle
and maps them to two disks with identified boundaries connected at a point,
just like a physical "global chess
" game I have.
Melinda's puzzle is a lot like this up a dimension, so think about two
disjoint balls, each representing a hemisphere of the 2^4, each a "subcube"
of Melinda's puzzle. The two boundaries of the balls are identified with
each other and as you roll one around, the other half rolls around so that
identified points connect up. We need to have the same restriction on
Melinda's puzzle.

In the pristine state then, I think it'd be nice to have an internal
(hidden), solid colored octahedron on each half. The other 6 faces should
all have equal colors split between each hemisphere, 4 stickers on each
half. You should be able to reorient the two subcubes to make a half
octahedron of any color on each subcube. I just saw Matt's email and
picture, and it looks like we were going down the same thought path. I
think with recoloring (mirroring some of the current piece colorings)
though, the windmill's can be avoided (?)

Roice


On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
[4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:

>
>
> Seems like there was a slight misunderstanding. I meant that you need to
> be able to twist one of the faces and in MC4D the most natural choice is
> the center face. In your physical puzzle you can achieve this type of twi=
st
> by twisting the two subcubes although this is indeed a twist of the
> subcubes themselves and not the center face, however, this is still the
> same type of twist just around another face.
>
> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
> this puzzle. Hopefully the restrictions will be quite natural and only so=
me
> "strange" moves would be illegal. Regarding the "families of states" (aka
> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
> twists preserves the parity of the pieces, meaning that only half of the
> permutations you can achieve by disassembling and reassembling can be
> reached through legal moves. Because of some geometrical properties of th=
e
> 2x2x2x2 and its twists, which would take some time to discuss in detail
> here, the orientation of the stickers mod 3 are preserved, meaning that t=
he
> last corner only can be oriented in one third of the number of orientatio=
ns
> for the other corners. This gives a total number of orbits of 2x3=3D6. To
> check this result let's use this information to calculate all the possibl=
e
> states of the 2x2x2x2; if there were no restrictions we would have 16! fo=
r
> permuting the pieces (16 pieces) and 12^16 for orienting them (12
> orientations for each corner). If we now take into account that there are=
6
> equally sized orbits this gets us to 12!16^12/6. However, we should also
> note that the orientation of the puzzle as a hole is not set by some kind
> of centerpieces and thus we need to devide with the number of orientation=
s
> of a 4D cube if we want all our states to be separated with twists and no=
t
> only rotations of the hole thing. The number of ways to orient a 4D cube =
in
> space (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a
> total of 12!16^12/(6*192) states which is indeed the same number that for
> example David Smith arrived at during his calculations. Therefore, when
> determining whether or not a twist on your puzzle is legal or not it is
> sufficient and necessary to confirm that the twist is an even permutation
> of the pieces and preserves the orientation of stickers mod 3.
>
> Best regards,
> Joel
>
> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>
>
>
> The new arrangement of magnets allows every valid orientation of pieces.
> The only invalid ones are those where the diagonal lines cutting each
> cube's face cross each other rather than coincide. In other words, you ca=
n
> assemble the puzzle in all ways that preserve the overall diamond/harlequ=
in
> pattern. Just about every move you can think of on the whole puzzle is
> valid though there are definitely invalid moves that the magnets allow. T=
he
> most obvious invalid move is twisting of a single end cap.
>
> I think your description of the center face is not correct though. Twists
> of the outer faces cause twists "through" the center face, not "of" that
> face. Twists of the outer faces are twists of those faces themselves
> because they are the ones not changing, just like the center and outer
> faces of MC4D when you twist the center face. The only direct twist of th=
e
> center face that this puzzle allows is a 90 degree twist about the outer
> axis. That happens when you simultaneously twist both end caps in the sam=
e
> direction.
>
> Yes, it's quite straightforward reorienting the whole puzzle to put any o=
f
> the four axes on the outside. This is a very nice improvement over the
> first version and should make it much easier to solve. You may be right
> that we just need to find the right way to think about the outside faces.
> I'll leave it to the math geniuses on the list to figure that out.
>
> -Melinda
>
>
> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
> wrote:
>
>
> Hi Melinda,
>
> I do not agree with the criticism regarding the white and yellow stickers
> touching each other, this could simply be an effect of the different
> representations of the puzzle. To really figure out if this indeed is a
> representation of a 2x2x2x2 we need to look at the possible moves (twists
> and rotations) and figure out the equivalent moves in the MC4D software.
> From the MC4D software, it's easy to understand that the only moves
> required are free twists of one of the faces (that is, only twisting the
> center face in the standard perspective projection in MC4D) and 4D
> rotations swapping which face is in the center (ctrl-clicking in MC4D). T=
he
> first is possible in your physical puzzle by rotating the white and yello=
w
> subcubes (from here on I use subcube to refer to the two halves of the
> puzzle and the colours of the subcubes to refer to the "outer colours").
> The second is possible if it's possible to reach a solved state with any
> two colours on the subcubes that still allow you to perform the previousl=
y
> mentioned twists. This seems to be the case from your demonstration and i=
s
> indeed true if the magnets allow the simple twists regardless of the
> colours of the subcubes. Thus, it is possible to let your puzzle be a
> representation of a 2x2x2x2, however, it might require that some moves th=
at
> the magnets allow aren't used.
>
> Best regards,
> Joel
>
> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>
>>
>>
>> Dear Cubists,
>>
>> I've finished version 2 of my physical puzzle and uploaded a video of it
>> here:
>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>> Again, please don't share these videos outside this group as their
>> purpose is just to get your feedback. I'll eventually replace them with =
a
>> public video.
>>
>> Here is an extra math puzzle that I bet you folks can answer: How many
>> families of states does this puzzle have? In other words, if disassemble=
d
>> and reassembled in any random configuration the magnets allow, what are =
the
>> odds that it can be solved? This has practical implications if all such
>> configurations are solvable because it would provide a very easy way to
>> fully scramble the puzzle.
>>
>> And finally, a bit of fun: A relatively new friend of mine and new list
>> member, Marc Ringuette, got excited enough to make his own version. He
>> built it from EPP foam and colored tape, and used honey instead of magne=
ts
>> to hold it together. Check it out here: http://superliminal.com/cube/d
>> essert_cube.jpg I don't know how practical a solution this is but it
>> sure looks delicious! Welcome Marc!
>>
>> -Melinda
>>
>>
>
>
>
>
>=20
>

--001a113ed258bddb39054e409338
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

I agree with Don's arguments about adjacent sticker co=
lors needing to be next to each other.=C2=A0 I think this can be turned int=
o an accurate 2^4 with coloring changes, so I agree with Joel too :)
r>
To help me think about it, I started adding a new projection o=
ption for spherical puzzles to MagicTile, which takes the two hemispheres o=
f a puzzle and maps them to two disks with identified boundaries connected =
at a point, just like a physical "/global-chess/indexf.html" target=3D"_blank">global chess" game I =
have.=C2=A0 Melinda's puzzle is a lot like this up a dimension, so thin=
k about two disjoint balls, each representing a hemisphere of the 2^4, each=
a "subcube" of Melinda's puzzle.=C2=A0 The two boundaries of=
the balls are identified with each other and as you roll one around, the o=
ther half rolls around so that identified points connect up.=C2=A0 We need =
to have the same restriction on Melinda's puzzle.

<=
div>In the pristine state then, I think it'd be nice to have an interna=
l (hidden), solid colored octahedron on each half.=C2=A0 The other 6 faces =
should all have equal colors split between each hemisphere, 4 stickers on e=
ach half.=C2=A0 You should be able to reorient the two subcubes to make a h=
alf octahedron of any color on each subcube.=C2=A0 I just saw Matt's em=
ail and picture, and it looks like we were going down the same thought path=
.=C2=A0 I think with recoloring (mirroring some of the current piece colori=
ngs) though, the windmill's can be avoided (?)

>Roice


il_quote">On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com [4=
D_Cubing] <" target=3D"_blank">4D_Cubing@yahoogroups.com> wrote:
ckquote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #c=
cc solid;padding-left:1ex">






=20=20=20=20=20=20=20=20

















Seems like there was a slight misunderstanding. I me=
ant that you need to be able to =C2=A0twist one of the faces and in MC4D th=
e most natural choice is the center face. In your physical puzzle you can a=
chieve this type of twist by twisting the two subcubes although this is ind=
eed a twist of the subcubes themselves and not the center face, however, th=
is is still the same type of twist just around another face.=C2=A0
v dir=3D"auto">
If the magnets are that allowing=
the 2x2x2x2 is obviously a subgroup of this puzzle. Hopefully the restrict=
ions will be quite natural and only some "strange" moves would be=
illegal. Regarding the "families of states" (aka orbits), the 2x=
2x2x2 has 6 orbits. As I mentioned earlier all allowed twists preserves the=
parity of the pieces, meaning that only half of the permutations you can a=
chieve by disassembling and reassembling can be reached through legal moves=
. Because of some geometrical properties of the 2x2x2x2 and its twists, whi=
ch would take some time to discuss in detail here, the orientation of the s=
tickers mod 3 are preserved, meaning that the last corner only can be orien=
ted in one third of the number of orientations for the other corners. This =
gives a total number of orbits of 2x3=3D6. To check this result let's u=
se this information to calculate all the possible states of the 2x2x2x2; if=
there were no restrictions we would have 16! for permuting the pieces (16 =
pieces) =C2=A0and 12^16 for orienting them (12 orientations for each corner=
). If we now take into account that there are 6 equally sized orbits this g=
ets us to 12!16^12/6. However, we should also note that the orientation of =
the puzzle as a hole is not set by some kind of centerpieces and thus we ne=
ed to devide with the number of orientations of a 4D cube if we want all ou=
r states to be separated with twists and not only rotations of the hole thi=
ng. The number of ways to orient a 4D cube in space (only allowing rotation=
s and not mirroring) is 8x6x4=3D192 giving a total of=C2=A0ont-family:sans-serif">12!16^12/(6*192) states which is indeed the same num=
ber that for example David Smith arrived at during his calculations. Theref=
ore, =C2=A0when determining whether or not a twist on your puzzle is legal =
or not it is sufficient and necessary to confirm that the twist is an even =
permutation of the pieces and preserves the orientation of stickers mod 3.<=
/span>

span>
Best re=
gards,=C2=A0
serif">Joel=C2=A0
xtra" dir=3D"auto">
Den 28 apr. 2017 3:02 fm =
skrev "Melinda Green t=3D"_blank">melinda@superliminal.com [4D_Cubing]" <mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.c=
om
>:
882m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m_44319=
00632030069098quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;=
padding-left:1ex">












=20

=C2=A0


m_-7761579982835215532m_4431900632030069098m_6586884675826257012ygrp-mlmsg"=
>
38m_-7761579982835215532m_4431900632030069098m_6586884675826257012ygrp-msg"=
>


8538m_-7761579982835215532m_4431900632030069098m_6586884675826257012ygrp-te=
xt">
=20=20=20=20=20=20
=20=20=20=20=20=20


=20=20
=20=20
The new arrangement of magnets allows every valid orientation of
pieces. The only invalid ones are those where the diagonal lines
cutting each cube's face cross each other rather than coincide. In
other words, you can assemble the puzzle in all ways that preserve
the overall diamond/harlequin pattern. Just about every move you can
think of on the whole puzzle is valid though there are definitely
invalid moves that the magnets allow. The most obvious invalid move
is twisting of a single end cap.



I think your description of the center face is not correct though.
Twists of the outer faces cause twists "through" the center f=
ace,
not "of" that face. Twists of the outer faces are twists of t=
hose
faces themselves because they are the ones not changing, just like
the center and outer faces of MC4D when you twist the center face.
The only direct twist of the center face that this puzzle allows is
a 90 degree twist about the outer axis. That happens when you
simultaneously twist both end caps in the same direction.



Yes, it's quite straightforward reorienting the whole puzzle to put
any of the four axes on the outside. This is a very nice improvement
over the first version and should make it much easier to solve. You
may be right that we just need to find the right way to think about
the outside faces. I'll leave it to the math geniuses on the list t=
o
figure that out.



-Melinda

4244333051891278538m_-7761579982835215532m_4431900632030069098quoted-text">=




1278538m_-7761579982835215532m_4431900632030069098m_6586884675826257012moz-=
cite-prefix">On 4/27/2017 10:31 AM, Joel Karlsson
1278538m_-7761579982835215532m_4431900632030069098m_6586884675826257012moz-=
txt-link-abbreviated" href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_b=
lank">joelkarlsson97@gmail.com
[4D_Cubing] wrote:



=20=20=20=20=20=20



Hi Melinda,




I do not agree with the criticism regarding the white and
yellow stickers touching each other, this could simply be an
effect of the different representations of the puzzle. To
really figure out if this indeed is a representation of a
2x2x2x2 we need to look at the possible moves (twists and
rotations) and figure out the equivalent moves in the MC4D
software. From the MC4D software, it's easy to understand
that the only moves required are free twists of one of the
faces (that is, only twisting the center face in the
standard perspective projection in MC4D) and 4D rotations
swapping which face is in the center (ctrl-clicking in
MC4D). The first is possible in your physical puzzle by
rotating the white and yellow subcubes (from here on I use
subcube to refer to the two halves of the puzzle and the
colours of the subcubes to refer to the "outer colours&quo=
t;).
The second is possible if it's possible to reach a solved
state with any two colours on the subcubes that still allow
you to perform the previously mentioned twists. This seems
to be the case from your demonstration and is indeed true if
the magnets allow the simple twists regardless of the
colours of the subcubes. Thus, it is possible to let your
puzzle be a representation of a 2x2x2x2, however, it might
require that some moves that the magnets allow aren't used.=





Best regards,


Joel




2017-04-27 3:09 GMT+02:00 Melinda Green
mel=
inda@superliminal.com

[4D_Cubing] <oogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:r>
olid">

=C2=A0







=20=20=20=20=20=20




=20=20



3051891278538m_-7761579982835215532m_4431900632030069098quoted-text">
=20=20=20=20=20

=20=20=20=20







=20=20































--001a113ed258bddb39054e409338--




From: Roice Nelson <roice3@gmail.com>
Date: Fri, 28 Apr 2017 16:59:47 -0500
Subject: Re: [MC4D] Physical 4D puzzle V2



--94eb2c14a04428bce0054e413189
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

I sent my email a bit too fast. Ater staring/thinking a bit more, the
coloring Matt came up with is right-on if you want to put a solid color at
the center of each hemisphere. His comment about the "mirrored" pieces on
each side helped me understand better. 3 of the stickers are mirrored and
the 4th is the hidden color (different on each side for a given pair of
"mirrored" pieces). All faces behave identically as well, as they should.
It's a little bit of a bummer that it doesn't look very pristine in the
pristine state, but it does look like it should work as a 2^4.

I wonder if there might be some adjustments to be made when shapeways
allows printing translucent as a color :)

Roice


On Fri, Apr 28, 2017 at 4:15 PM, Roice Nelson wrote:

> I agree with Don's arguments about adjacent sticker colors needing to be
> next to each other. I think this can be turned into an accurate 2^4 with
> coloring changes, so I agree with Joel too :)
>
> To help me think about it, I started adding a new projection option for
> spherical puzzles to MagicTile, which takes the two hemispheres of a puzz=
le
> and maps them to two disks with identified boundaries connected at a poin=
t,
> just like a physical "global chess
> " game I have.
> Melinda's puzzle is a lot like this up a dimension, so think about two
> disjoint balls, each representing a hemisphere of the 2^4, each a "subcub=
e"
> of Melinda's puzzle. The two boundaries of the balls are identified with
> each other and as you roll one around, the other half rolls around so tha=
t
> identified points connect up. We need to have the same restriction on
> Melinda's puzzle.
>
> In the pristine state then, I think it'd be nice to have an internal
> (hidden), solid colored octahedron on each half. The other 6 faces shoul=
d
> all have equal colors split between each hemisphere, 4 stickers on each
> half. You should be able to reorient the two subcubes to make a half
> octahedron of any color on each subcube. I just saw Matt's email and
> picture, and it looks like we were going down the same thought path. I
> think with recoloring (mirroring some of the current piece colorings)
> though, the windmill's can be avoided (?)
>
> Roice
>
>
> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>
>>
>>
>> Seems like there was a slight misunderstanding. I meant that you need to
>> be able to twist one of the faces and in MC4D the most natural choice i=
s
>> the center face. In your physical puzzle you can achieve this type of tw=
ist
>> by twisting the two subcubes although this is indeed a twist of the
>> subcubes themselves and not the center face, however, this is still the
>> same type of twist just around another face.
>>
>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
>> this puzzle. Hopefully the restrictions will be quite natural and only s=
ome
>> "strange" moves would be illegal. Regarding the "families of states" (ak=
a
>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
>> twists preserves the parity of the pieces, meaning that only half of the
>> permutations you can achieve by disassembling and reassembling can be
>> reached through legal moves. Because of some geometrical properties of t=
he
>> 2x2x2x2 and its twists, which would take some time to discuss in detail
>> here, the orientation of the stickers mod 3 are preserved, meaning that =
the
>> last corner only can be oriented in one third of the number of orientati=
ons
>> for the other corners. This gives a total number of orbits of 2x3=3D6. T=
o
>> check this result let's use this information to calculate all the possib=
le
>> states of the 2x2x2x2; if there were no restrictions we would have 16! f=
or
>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>> orientations for each corner). If we now take into account that there ar=
e 6
>> equally sized orbits this gets us to 12!16^12/6. However, we should also
>> note that the orientation of the puzzle as a hole is not set by some kin=
d
>> of centerpieces and thus we need to devide with the number of orientatio=
ns
>> of a 4D cube if we want all our states to be separated with twists and n=
ot
>> only rotations of the hole thing. The number of ways to orient a 4D cube=
in
>> space (only allowing rotations and not mirroring) is 8x6x4=3D192 giving =
a
>> total of 12!16^12/(6*192) states which is indeed the same number that
>> for example David Smith arrived at during his calculations. Therefore,
>> when determining whether or not a twist on your puzzle is legal or not =
it
>> is sufficient and necessary to confirm that the twist is an even
>> permutation of the pieces and preserves the orientation of stickers mod =
3.
>>
>> Best regards,
>> Joel
>>
>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>
>>
>>
>> The new arrangement of magnets allows every valid orientation of pieces.
>> The only invalid ones are those where the diagonal lines cutting each
>> cube's face cross each other rather than coincide. In other words, you c=
an
>> assemble the puzzle in all ways that preserve the overall diamond/harleq=
uin
>> pattern. Just about every move you can think of on the whole puzzle is
>> valid though there are definitely invalid moves that the magnets allow. =
The
>> most obvious invalid move is twisting of a single end cap.
>>
>> I think your description of the center face is not correct though. Twist=
s
>> of the outer faces cause twists "through" the center face, not "of" that
>> face. Twists of the outer faces are twists of those faces themselves
>> because they are the ones not changing, just like the center and outer
>> faces of MC4D when you twist the center face. The only direct twist of t=
he
>> center face that this puzzle allows is a 90 degree twist about the outer
>> axis. That happens when you simultaneously twist both end caps in the sa=
me
>> direction.
>>
>> Yes, it's quite straightforward reorienting the whole puzzle to put any
>> of the four axes on the outside. This is a very nice improvement over th=
e
>> first version and should make it much easier to solve. You may be right
>> that we just need to find the right way to think about the outside faces=
.
>> I'll leave it to the math geniuses on the list to figure that out.
>>
>> -Melinda
>>
>>
>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com
>> [4D_Cubing] wrote:
>>
>>
>> Hi Melinda,
>>
>> I do not agree with the criticism regarding the white and yellow sticker=
s
>> touching each other, this could simply be an effect of the different
>> representations of the puzzle. To really figure out if this indeed is a
>> representation of a 2x2x2x2 we need to look at the possible moves (twist=
s
>> and rotations) and figure out the equivalent moves in the MC4D software.
>> From the MC4D software, it's easy to understand that the only moves
>> required are free twists of one of the faces (that is, only twisting the
>> center face in the standard perspective projection in MC4D) and 4D
>> rotations swapping which face is in the center (ctrl-clicking in MC4D). =
The
>> first is possible in your physical puzzle by rotating the white and yell=
ow
>> subcubes (from here on I use subcube to refer to the two halves of the
>> puzzle and the colours of the subcubes to refer to the "outer colours").
>> The second is possible if it's possible to reach a solved state with any
>> two colours on the subcubes that still allow you to perform the previous=
ly
>> mentioned twists. This seems to be the case from your demonstration and =
is
>> indeed true if the magnets allow the simple twists regardless of the
>> colours of the subcubes. Thus, it is possible to let your puzzle be a
>> representation of a 2x2x2x2, however, it might require that some moves t=
hat
>> the magnets allow aren't used.
>>
>> Best regards,
>> Joel
>>
>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>
>>>
>>>
>>> Dear Cubists,
>>>
>>> I've finished version 2 of my physical puzzle and uploaded a video of i=
t
>>> here:
>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>> Again, please don't share these videos outside this group as their
>>> purpose is just to get your feedback. I'll eventually replace them with=
a
>>> public video.
>>>
>>> Here is an extra math puzzle that I bet you folks can answer: How many
>>> families of states does this puzzle have? In other words, if disassembl=
ed
>>> and reassembled in any random configuration the magnets allow, what are=
the
>>> odds that it can be solved? This has practical implications if all such
>>> configurations are solvable because it would provide a very easy way to
>>> fully scramble the puzzle.
>>>
>>> And finally, a bit of fun: A relatively new friend of mine and new list
>>> member, Marc Ringuette, got excited enough to make his own version. He
>>> built it from EPP foam and colored tape, and used honey instead of magn=
ets
>>> to hold it together. Check it out here: http://superliminal.com/cube/d
>>> essert_cube.jpg I don't know how practical a solution this is but it
>>> sure looks delicious! Welcome Marc!
>>>
>>> -Melinda
>>>
>>>
>>
>>
>>
>>
>>=20
>>
>
>

--94eb2c14a04428bce0054e413189
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

I sent my email a bit too fast.=C2=A0 Ater staring/thinkin=
g a bit more, the coloring Matt came up with is right-on if you want to put=
a solid color at the center of each hemisphere.=C2=A0 His=C2=A0comment abo=
ut the "mirrored" pieces on each side helped me understand better=
. =C2=A03 of the stickers are mirrored and the 4th is the hidden color (dif=
ferent on each side for a given pair of "mirrored" pieces).=C2=A0=
All faces behave identically as well, as they should.=C2=A0 It's a lit=
tle bit of a bummer that it doesn't look very pristine in the pristine =
state, but it does look like it should work as a 2^4.

I =
wonder if there might be some adjustments to be made when shapeways allows =
printing translucent as a color :)

Roice
=


te">On Fri, Apr 28, 2017 at 4:15 PM, Roice Nelson <href=3D"mailto:roice3@gmail.com" target=3D"_blank">roice3@gmail.com>=
wrote:
8ex;border-left:1px #ccc solid;padding-left:1ex">
I agree w=
ith Don's arguments about adjacent sticker colors needing to be next to=
each other.=C2=A0 I think this can be turned into an accurate 2^4 with col=
oring changes, so I agree with Joel too :)

To help me th=
ink about it, I started adding a new projection option for spherical puzzle=
s to MagicTile, which takes the two hemispheres of a puzzle and maps them t=
o two disks with identified boundaries connected at a point, just like a ph=
ysical " target=3D"_blank">global chess" game I have.=C2=A0 Melinda's =
puzzle is a lot like this up a dimension, so think about two disjoint balls=
, each representing a hemisphere of the 2^4, each a "subcube" of =
Melinda's puzzle.=C2=A0 The two boundaries of the balls are identified =
with each other and as you roll one around, the other half rolls around so =
that identified points connect up.=C2=A0 We need to have the same restricti=
on on Melinda's puzzle.

In the pristine state =
then, I think it'd be nice to have an internal (hidden), solid colored =
octahedron on each half.=C2=A0 The other 6 faces should all have equal colo=
rs split between each hemisphere, 4 stickers on each half.=C2=A0 You should=
be able to reorient the two subcubes to make a half octahedron of any colo=
r on each subcube.=C2=A0 I just saw Matt's email and picture, and it lo=
oks like we were going down the same thought path.=C2=A0 I think with recol=
oring (mirroring some of the current piece colorings) though, the windmill&=
#39;s can be avoided (?)
">

Roice
=


On Fri,=
Apr 28, 2017 at 1:14 AM, Joel Karlsson ail.com" target=3D"_blank">joelkarlsson97@gmail.com [4D_Cubing] ir=3D"ltr"><k">4D_Cubing@yahoogroups.com> wrote:
gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-=
left:1ex">






=20=20=20=20=20=20=20=20

















Seems like there was a slight misunderstanding. I me=
ant that you need to be able to =C2=A0twist one of the faces and in MC4D th=
e most natural choice is the center face. In your physical puzzle you can a=
chieve this type of twist by twisting the two subcubes although this is ind=
eed a twist of the subcubes themselves and not the center face, however, th=
is is still the same type of twist just around another face.=C2=A0
v dir=3D"auto">
If the magnets are that allowing=
the 2x2x2x2 is obviously a subgroup of this puzzle. Hopefully the restrict=
ions will be quite natural and only some "strange" moves would be=
illegal. Regarding the "families of states" (aka orbits), the 2x=
2x2x2 has 6 orbits. As I mentioned earlier all allowed twists preserves the=
parity of the pieces, meaning that only half of the permutations you can a=
chieve by disassembling and reassembling can be reached through legal moves=
. Because of some geometrical properties of the 2x2x2x2 and its twists, whi=
ch would take some time to discuss in detail here, the orientation of the s=
tickers mod 3 are preserved, meaning that the last corner only can be orien=
ted in one third of the number of orientations for the other corners. This =
gives a total number of orbits of 2x3=3D6. To check this result let's u=
se this information to calculate all the possible states of the 2x2x2x2; if=
there were no restrictions we would have 16! for permuting the pieces (16 =
pieces) =C2=A0and 12^16 for orienting them (12 orientations for each corner=
). If we now take into account that there are 6 equally sized orbits this g=
ets us to 12!16^12/6. However, we should also note that the orientation of =
the puzzle as a hole is not set by some kind of centerpieces and thus we ne=
ed to devide with the number of orientations of a 4D cube if we want all ou=
r states to be separated with twists and not only rotations of the hole thi=
ng. The number of ways to orient a 4D cube in space (only allowing rotation=
s and not mirroring) is 8x6x4=3D192 giving a total of=C2=A0ont-family:sans-serif">12!16^12/(6*192) states which is indeed the same num=
ber that for example David Smith arrived at during his calculations. Theref=
ore, =C2=A0when determining whether or not a twist on your puzzle is legal =
or not it is sufficient and necessary to confirm that the twist is an even =
permutation of the pieces and preserves the orientation of stickers mod 3.<=
/span>

span>
Best re=
gards,=C2=A0
serif">Joel=C2=A0
xtra" dir=3D"auto">
Den 28 apr. 2017 3:02 fm =
skrev "Melinda Green t=3D"_blank">melinda@superliminal.com [4D_Cubing]" <mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.c=
om
>:
240m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77615=
79982835215532m_4431900632030069098quote" style=3D"margin:0 0 0 .8ex;border=
-left:1px #ccc solid;padding-left:1ex">












=20

=C2=A0


_-4244333051891278538m_-7761579982835215532m_4431900632030069098m_658688467=
5826257012ygrp-mlmsg">
4m_-4244333051891278538m_-7761579982835215532m_4431900632030069098m_6586884=
675826257012ygrp-msg">


054m_-4244333051891278538m_-7761579982835215532m_4431900632030069098m_65868=
84675826257012ygrp-text">
=20=20=20=20=20=20
=20=20=20=20=20=20


=20=20
=20=20
The new arrangement of magnets allows every valid orientation of
pieces. The only invalid ones are those where the diagonal lines
cutting each cube's face cross each other rather than coincide. In
other words, you can assemble the puzzle in all ways that preserve
the overall diamond/harlequin pattern. Just about every move you can
think of on the whole puzzle is valid though there are definitely
invalid moves that the magnets allow. The most obvious invalid move
is twisting of a single end cap.



I think your description of the center face is not correct though.
Twists of the outer faces cause twists "through" the center f=
ace,
not "of" that face. Twists of the outer faces are twists of t=
hose
faces themselves because they are the ones not changing, just like
the center and outer faces of MC4D when you twist the center face.
The only direct twist of the center face that this puzzle allows is
a 90 degree twist about the outer axis. That happens when you
simultaneously twist both end caps in the same direction.



Yes, it's quite straightforward reorienting the whole puzzle to put
any of the four axes on the outside. This is a very nice improvement
over the first version and should make it much easier to solve. You
may be right that we just need to find the right way to think about
the outside faces. I'll leave it to the math geniuses on the list t=
o
figure that out.



-Melinda

927561327128665054m_-4244333051891278538m_-7761579982835215532m_44319006320=
30069098quoted-text">



665054m_-4244333051891278538m_-7761579982835215532m_4431900632030069098m_65=
86884675826257012moz-cite-prefix">On 4/27/2017 10:31 AM, Joel Karlsson
665054m_-4244333051891278538m_-7761579982835215532m_4431900632030069098m_65=
86884675826257012moz-txt-link-abbreviated" href=3D"mailto:joelkarlsson97@gm=
ail.com" target=3D"_blank">joelkarlsson97@gmail.com
[4D_Cubing] wrote:<=
/div>


=20=20=20=20=20=20



Hi Melinda,




I do not agree with the criticism regarding the white and
yellow stickers touching each other, this could simply be an
effect of the different representations of the puzzle. To
really figure out if this indeed is a representation of a
2x2x2x2 we need to look at the possible moves (twists and
rotations) and figure out the equivalent moves in the MC4D
software. From the MC4D software, it's easy to understand
that the only moves required are free twists of one of the
faces (that is, only twisting the center face in the
standard perspective projection in MC4D) and 4D rotations
swapping which face is in the center (ctrl-clicking in
MC4D). The first is possible in your physical puzzle by
rotating the white and yellow subcubes (from here on I use
subcube to refer to the two halves of the puzzle and the
colours of the subcubes to refer to the "outer colours&quo=
t;).
The second is possible if it's possible to reach a solved
state with any two colours on the subcubes that still allow
you to perform the previously mentioned twists. This seems
to be the case from your demonstration and is indeed true if
the magnets allow the simple twists regardless of the
colours of the subcubes. Thus, it is possible to let your
puzzle be a representation of a 2x2x2x2, however, it might
require that some moves that the magnets allow aren't used.=





Best regards,


Joel




2017-04-27 3:09 GMT+02:00 Melinda Green
mel=
inda@superliminal.com

[4D_Cubing] <oogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:r>
olid">

=C2=A0
1327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030069=
098m_6586884675826257012m_3209267269419320979ygrp-mlmsg">








=20=20=20=20=20=20




=20=20



327128665054m_-4244333051891278538m_-7761579982835215532m_44319006320300690=
98quoted-text">
=20=20=20=20=20

=20=20=20=20







=20=20

































--94eb2c14a04428bce0054e413189--




From: Roice Nelson <roice3@gmail.com>
Date: Fri, 28 Apr 2017 17:44:04 -0500
Subject: Re: [MC4D] Physical 4D puzzle V2



--001a11459c067da9b8054e41cf7c
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Sorry for all the streaming, but I wanted to share one more thought. I now
completely agree with Joel/Matt about it behaving as a 2^4, even with the
original coloring. You just need to consider the corner colors of the two
subcubes (pink/purple near the end of the video) as being a window into the
interior of the piece. The other colors match up as desired. (Sorry if
folks already understood this after their emails and I'm just catching up!)

In fact, you could alter the coloring of the pieces slightly so that the
behavior was similar with the inverted coloring. At the corners where 3
colors meet on each piece, you could put a little circle of color of the
opposite 4th color. In Matt's windmill coloring then, you'd be able to see
all four colors of a piece, like you can with some of the pieces on
Melinda's original coloring. And again you'd consider the color circles a
window to the interior that did not require the same matching constraints
between the subcubes.

I'm looking forward to having one of these :)

Happy Friday everyone,
Roice


On Fri, Apr 28, 2017 at 4:59 PM, Roice Nelson wrote:

> I sent my email a bit too fast. Ater staring/thinking a bit more, the
> coloring Matt came up with is right-on if you want to put a solid color a=
t
> the center of each hemisphere. His comment about the "mirrored" pieces o=
n
> each side helped me understand better. 3 of the stickers are mirrored an=
d
> the 4th is the hidden color (different on each side for a given pair of
> "mirrored" pieces). All faces behave identically as well, as they should=
.
> It's a little bit of a bummer that it doesn't look very pristine in the
> pristine state, but it does look like it should work as a 2^4.
>
> I wonder if there might be some adjustments to be made when shapeways
> allows printing translucent as a color :)
>
> Roice
>
>
> On Fri, Apr 28, 2017 at 4:15 PM, Roice Nelson wrote:
>
>> I agree with Don's arguments about adjacent sticker colors needing to be
>> next to each other. I think this can be turned into an accurate 2^4 wit=
h
>> coloring changes, so I agree with Joel too :)
>>
>> To help me think about it, I started adding a new projection option for
>> spherical puzzles to MagicTile, which takes the two hemispheres of a puz=
zle
>> and maps them to two disks with identified boundaries connected at a poi=
nt,
>> just like a physical "global chess
>> " game I have.
>> Melinda's puzzle is a lot like this up a dimension, so think about two
>> disjoint balls, each representing a hemisphere of the 2^4, each a "subcu=
be"
>> of Melinda's puzzle. The two boundaries of the balls are identified wit=
h
>> each other and as you roll one around, the other half rolls around so th=
at
>> identified points connect up. We need to have the same restriction on
>> Melinda's puzzle.
>>
>> In the pristine state then, I think it'd be nice to have an internal
>> (hidden), solid colored octahedron on each half. The other 6 faces shou=
ld
>> all have equal colors split between each hemisphere, 4 stickers on each
>> half. You should be able to reorient the two subcubes to make a half
>> octahedron of any color on each subcube. I just saw Matt's email and
>> picture, and it looks like we were going down the same thought path. I
>> think with recoloring (mirroring some of the current piece colorings)
>> though, the windmill's can be avoided (?)
>>
>> Roice
>>
>>
>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
>> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>>
>>>
>>>
>>> Seems like there was a slight misunderstanding. I meant that you need t=
o
>>> be able to twist one of the faces and in MC4D the most natural choice =
is
>>> the center face. In your physical puzzle you can achieve this type of t=
wist
>>> by twisting the two subcubes although this is indeed a twist of the
>>> subcubes themselves and not the center face, however, this is still the
>>> same type of twist just around another face.
>>>
>>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
>>> this puzzle. Hopefully the restrictions will be quite natural and only =
some
>>> "strange" moves would be illegal. Regarding the "families of states" (a=
ka
>>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
>>> twists preserves the parity of the pieces, meaning that only half of th=
e
>>> permutations you can achieve by disassembling and reassembling can be
>>> reached through legal moves. Because of some geometrical properties of =
the
>>> 2x2x2x2 and its twists, which would take some time to discuss in detail
>>> here, the orientation of the stickers mod 3 are preserved, meaning that=
the
>>> last corner only can be oriented in one third of the number of orientat=
ions
>>> for the other corners. This gives a total number of orbits of 2x3=3D6. =
To
>>> check this result let's use this information to calculate all the possi=
ble
>>> states of the 2x2x2x2; if there were no restrictions we would have 16! =
for
>>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>>> orientations for each corner). If we now take into account that there a=
re 6
>>> equally sized orbits this gets us to 12!16^12/6. However, we should als=
o
>>> note that the orientation of the puzzle as a hole is not set by some ki=
nd
>>> of centerpieces and thus we need to devide with the number of orientati=
ons
>>> of a 4D cube if we want all our states to be separated with twists and =
not
>>> only rotations of the hole thing. The number of ways to orient a 4D cub=
e in
>>> space (only allowing rotations and not mirroring) is 8x6x4=3D192 giving=
a
>>> total of 12!16^12/(6*192) states which is indeed the same number that
>>> for example David Smith arrived at during his calculations. Therefore,
>>> when determining whether or not a twist on your puzzle is legal or not=
it
>>> is sufficient and necessary to confirm that the twist is an even
>>> permutation of the pieces and preserves the orientation of stickers mod=
3.
>>>
>>> Best regards,
>>> Joel
>>>
>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
>>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>>
>>>
>>>
>>> The new arrangement of magnets allows every valid orientation of pieces=
.
>>> The only invalid ones are those where the diagonal lines cutting each
>>> cube's face cross each other rather than coincide. In other words, you =
can
>>> assemble the puzzle in all ways that preserve the overall diamond/harle=
quin
>>> pattern. Just about every move you can think of on the whole puzzle is
>>> valid though there are definitely invalid moves that the magnets allow.=
The
>>> most obvious invalid move is twisting of a single end cap.
>>>
>>> I think your description of the center face is not correct though.
>>> Twists of the outer faces cause twists "through" the center face, not "=
of"
>>> that face. Twists of the outer faces are twists of those faces themselv=
es
>>> because they are the ones not changing, just like the center and outer
>>> faces of MC4D when you twist the center face. The only direct twist of =
the
>>> center face that this puzzle allows is a 90 degree twist about the oute=
r
>>> axis. That happens when you simultaneously twist both end caps in the s=
ame
>>> direction.
>>>
>>> Yes, it's quite straightforward reorienting the whole puzzle to put any
>>> of the four axes on the outside. This is a very nice improvement over t=
he
>>> first version and should make it much easier to solve. You may be right
>>> that we just need to find the right way to think about the outside face=
s.
>>> I'll leave it to the math geniuses on the list to figure that out.
>>>
>>> -Melinda
>>>
>>>
>>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com
>>> [4D_Cubing] wrote:
>>>
>>>
>>> Hi Melinda,
>>>
>>> I do not agree with the criticism regarding the white and yellow
>>> stickers touching each other, this could simply be an effect of the
>>> different representations of the puzzle. To really figure out if this
>>> indeed is a representation of a 2x2x2x2 we need to look at the possible
>>> moves (twists and rotations) and figure out the equivalent moves in the
>>> MC4D software. From the MC4D software, it's easy to understand that the
>>> only moves required are free twists of one of the faces (that is, only
>>> twisting the center face in the standard perspective projection in MC4D=
)
>>> and 4D rotations swapping which face is in the center (ctrl-clicking in
>>> MC4D). The first is possible in your physical puzzle by rotating the wh=
ite
>>> and yellow subcubes (from here on I use subcube to refer to the two hal=
ves
>>> of the puzzle and the colours of the subcubes to refer to the "outer
>>> colours"). The second is possible if it's possible to reach a solved st=
ate
>>> with any two colours on the subcubes that still allow you to perform th=
e
>>> previously mentioned twists. This seems to be the case from your
>>> demonstration and is indeed true if the magnets allow the simple twists
>>> regardless of the colours of the subcubes. Thus, it is possible to let =
your
>>> puzzle be a representation of a 2x2x2x2, however, it might require that
>>> some moves that the magnets allow aren't used.
>>>
>>> Best regards,
>>> Joel
>>>
>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>>
>>>>
>>>>
>>>> Dear Cubists,
>>>>
>>>> I've finished version 2 of my physical puzzle and uploaded a video of
>>>> it here:
>>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>>> Again, please don't share these videos outside this group as their
>>>> purpose is just to get your feedback. I'll eventually replace them wit=
h a
>>>> public video.
>>>>
>>>> Here is an extra math puzzle that I bet you folks can answer: How many
>>>> families of states does this puzzle have? In other words, if disassemb=
led
>>>> and reassembled in any random configuration the magnets allow, what ar=
e the
>>>> odds that it can be solved? This has practical implications if all suc=
h
>>>> configurations are solvable because it would provide a very easy way t=
o
>>>> fully scramble the puzzle.
>>>>
>>>> And finally, a bit of fun: A relatively new friend of mine and new lis=
t
>>>> member, Marc Ringuette, got excited enough to make his own version. He
>>>> built it from EPP foam and colored tape, and used honey instead of mag=
nets
>>>> to hold it together. Check it out here: http://superliminal.com/cube/d
>>>> essert_cube.jpg I don't know how practical a solution this is but it
>>>> sure looks delicious! Welcome Marc!
>>>>
>>>> -Melinda
>>>>
>>>>
>>>
>>>
>>>
>>>
>>>=20
>>>
>>
>>
>

--001a11459c067da9b8054e41cf7c
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Sorry for all the streaming, but I wanted to share one mor=
e thought.=C2=A0 I now completely agree with Joel/Matt about it behaving as=
a 2^4, even with the original coloring.=C2=A0 You just need to consider th=
e corner colors of the two subcubes (pink/purple near the end of the video)=
as being a window into the interior of the piece.=C2=A0 The other colors m=
atch up as desired. =C2=A0(Sorry if folks already understood this after the=
ir emails and I'm just catching up!)

In fact, you co=
uld alter the coloring of the pieces slightly so that the behavior was simi=
lar with the inverted coloring.=C2=A0 At the corners where 3 colors meet on=
each piece, you could put a little circle of color of the opposite 4th col=
or.=C2=A0 In Matt's windmill coloring then, you'd be able to see al=
l four colors of a piece, like you can with some of the pieces on Melinda&#=
39;s original coloring.=C2=A0 And again you'd consider the color circle=
s a window to the interior that did not require the same matching constrain=
ts between the subcubes.

I'm looking forward t=
o having one of these :)

Happy Friday everyone,iv>
Roice

"gmail_extra">
On Fri, Apr 28, 2017 at 4:59 P=
M, Roice Nelson <arget=3D"_blank">roice3@gmail.com> wrote:
s=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;pad=
ding-left:1ex">
I sent my email a bit too fast.=C2=A0 Ater =
staring/thinking a bit more, the coloring Matt came up with is right-on if =
you want to put a solid color at the center of each hemisphere.=C2=A0 His=
=C2=A0comment about the "mirrored" pieces on each side helped me =
understand better. =C2=A03 of the stickers are mirrored and the 4th is the =
hidden color (different on each side for a given pair of "mirrored&quo=
t; pieces).=C2=A0 All faces behave identically as well, as they should.=C2=
=A0 It's a little bit of a bummer that it doesn't look very pristin=
e in the pristine state, but it does look like it should work as a 2^4.>
I wonder if there might be some adjustments to be made when=
shapeways allows printing translucent as a color :)=


Roice
div>

=3D"gmail_extra">
On Fri, Apr 28, 2017 at 4:1=
5 PM, Roice Nelson <" target=3D"_blank">roice3@gmail.com> wrote:
lass=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;=
padding-left:1ex">
I agree with Don's arguments about a=
djacent sticker colors needing to be next to each other.=C2=A0 I think this=
can be turned into an accurate 2^4 with coloring changes, so I agree with =
Joel too :)

To help me think about it, I started adding =
a new projection option for spherical puzzles to MagicTile, which takes the=
two hemispheres of a puzzle and maps them to two disks with identified bou=
ndaries connected at a point, just like a physical "www.pa-network.com/global-chess/indexf.html" target=3D"_blank">global chess=
" game I have.=C2=A0 Melinda's puzzle is a lot like this up a =
dimension, so think about two disjoint balls, each representing a hemispher=
e of the 2^4, each a "subcube" of Melinda's puzzle.=C2=A0 The=
two boundaries of the balls are identified with each other and as you roll=
one around, the other half rolls around so that identified points connect =
up.=C2=A0 We need to have the same restriction on Melinda's puzzle.v>

In the pristine state then, I think it'd be nice =
to have an internal (hidden), solid colored octahedron on each half.=C2=A0 =
The other 6 faces should all have equal colors split between each hemispher=
e, 4 stickers on each half.=C2=A0 You should be able to reorient the two su=
bcubes to make a half octahedron of any color on each subcube.=C2=A0 I just=
saw Matt's email and picture, and it looks like we were going down the=
same thought path.=C2=A0 I think with recoloring (mirroring some of the cu=
rrent piece colorings) though, the windmill's can be avoided (?)
<=
span class=3D"m_-2858008468427403369HOEnZb">
r>
Roice
3369h5">


uote">On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson karlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com [4D_Cu=
bing] <rget=3D"_blank">4D_Cubing@yahoogroups.com> wrote:
ote class=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc s=
olid;padding-left:1ex">






=20=20=20=20=20=20=20=20

















Seems like there was a slight misunderstanding. I me=
ant that you need to be able to =C2=A0twist one of the faces and in MC4D th=
e most natural choice is the center face. In your physical puzzle you can a=
chieve this type of twist by twisting the two subcubes although this is ind=
eed a twist of the subcubes themselves and not the center face, however, th=
is is still the same type of twist just around another face.=C2=A0
v dir=3D"auto">
If the magnets are that allowing=
the 2x2x2x2 is obviously a subgroup of this puzzle. Hopefully the restrict=
ions will be quite natural and only some "strange" moves would be=
illegal. Regarding the "families of states" (aka orbits), the 2x=
2x2x2 has 6 orbits. As I mentioned earlier all allowed twists preserves the=
parity of the pieces, meaning that only half of the permutations you can a=
chieve by disassembling and reassembling can be reached through legal moves=
. Because of some geometrical properties of the 2x2x2x2 and its twists, whi=
ch would take some time to discuss in detail here, the orientation of the s=
tickers mod 3 are preserved, meaning that the last corner only can be orien=
ted in one third of the number of orientations for the other corners. This =
gives a total number of orbits of 2x3=3D6. To check this result let's u=
se this information to calculate all the possible states of the 2x2x2x2; if=
there were no restrictions we would have 16! for permuting the pieces (16 =
pieces) =C2=A0and 12^16 for orienting them (12 orientations for each corner=
). If we now take into account that there are 6 equally sized orbits this g=
ets us to 12!16^12/6. However, we should also note that the orientation of =
the puzzle as a hole is not set by some kind of centerpieces and thus we ne=
ed to devide with the number of orientations of a 4D cube if we want all ou=
r states to be separated with twists and not only rotations of the hole thi=
ng. The number of ways to orient a 4D cube in space (only allowing rotation=
s and not mirroring) is 8x6x4=3D192 giving a total of=C2=A0ont-family:sans-serif">12!16^12/(6*192) states which is indeed the same num=
ber that for example David Smith arrived at during his calculations. Theref=
ore, =C2=A0when determining whether or not a twist on your puzzle is legal =
or not it is sufficient and necessary to confirm that the twist is an even =
permutation of the pieces and preserves the orientation of stickers mod 3.<=
/span>

span>
Best re=
gards,=C2=A0
serif">Joel=C2=A0
xtra" dir=3D"auto">
Den 28 apr. 2017 3:02 fm =
skrev "Melinda Green t=3D"_blank">melinda@superliminal.com [4D_Cubing]" <mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.c=
om
>:
3369m_3421356551489074240m_6733134459512634882m_7927561327128665054m_-42443=
33051891278538m_-7761579982835215532m_4431900632030069098quote" style=3D"ma=
rgin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">












=20

=C2=A0


m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m_44319006=
32030069098m_6586884675826257012ygrp-mlmsg">
82m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m_443190=
0632030069098m_6586884675826257012ygrp-msg">


4882m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4431=
900632030069098m_6586884675826257012ygrp-text">
=20=20=20=20=20=20
=20=20=20=20=20=20


=20=20
=20=20
The new arrangement of magnets allows every valid orientation of
pieces. The only invalid ones are those where the diagonal lines
cutting each cube's face cross each other rather than coincide. In
other words, you can assemble the puzzle in all ways that preserve
the overall diamond/harlequin pattern. Just about every move you can
think of on the whole puzzle is valid though there are definitely
invalid moves that the magnets allow. The most obvious invalid move
is twisting of a single end cap.



I think your description of the center face is not correct though.
Twists of the outer faces cause twists "through" the center f=
ace,
not "of" that face. Twists of the outer faces are twists of t=
hose
faces themselves because they are the ones not changing, just like
the center and outer faces of MC4D when you twist the center face.
The only direct twist of the center face that this puzzle allows is
a 90 degree twist about the outer axis. That happens when you
simultaneously twist both end caps in the same direction.



Yes, it's quite straightforward reorienting the whole puzzle to put
any of the four axes on the outside. This is a very nice improvement
over the first version and should make it much easier to solve. You
may be right that we just need to find the right way to think about
the outside faces. I'll leave it to the math geniuses on the list t=
o
figure that out.



-Melinda

6733134459512634882m_7927561327128665054m_-4244333051891278538m_-7761579982=
835215532m_4431900632030069098quoted-text">



2634882m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4=
431900632030069098m_6586884675826257012moz-cite-prefix">On 4/27/2017 10:31 =
AM, Joel Karlsson
2634882m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4=
431900632030069098m_6586884675826257012moz-txt-link-abbreviated" href=3D"ma=
ilto:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.coma> [4D_Cubing] wrote:



=20=20=20=20=20=20



Hi Melinda,




I do not agree with the criticism regarding the white and
yellow stickers touching each other, this could simply be an
effect of the different representations of the puzzle. To
really figure out if this indeed is a representation of a
2x2x2x2 we need to look at the possible moves (twists and
rotations) and figure out the equivalent moves in the MC4D
software. From the MC4D software, it's easy to understand
that the only moves required are free twists of one of the
faces (that is, only twisting the center face in the
standard perspective projection in MC4D) and 4D rotations
swapping which face is in the center (ctrl-clicking in
MC4D). The first is possible in your physical puzzle by
rotating the white and yellow subcubes (from here on I use
subcube to refer to the two halves of the puzzle and the
colours of the subcubes to refer to the "outer colours&quo=
t;).
The second is possible if it's possible to reach a solved
state with any two colours on the subcubes that still allow
you to perform the previously mentioned twists. This seems
to be the case from your demonstration and is indeed true if
the magnets allow the simple twists regardless of the
colours of the subcubes. Thus, it is possible to let your
puzzle be a representation of a 2x2x2x2, however, it might
require that some moves that the magnets allow aren't used.=





Best regards,


Joel




2017-04-27 3:09 GMT+02:00 Melinda Green
mel=
inda@superliminal.com

[4D_Cubing] <oogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:r>
olid">

=C2=A0
34459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283521=
5532m_4431900632030069098m_6586884675826257012m_3209267269419320979ygrp-mlm=
sg">








=20=20=20=20=20=20




=20=20



4459512634882m_7927561327128665054m_-4244333051891278538m_-7761579982835215=
532m_4431900632030069098quoted-text">
=20=20=20=20=20

=20=20=20=20







=20=20



































--001a11459c067da9b8054e41cf7c--




From: Melinda Green <melinda@superliminal.com>
Date: Fri, 28 Apr 2017 16:04:58 -0700
Subject: Re: [MC4D] Physical 4D puzzle V2



--------------AC19E366C6DED7570859E7F9
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: quoted-printable

First off, thanks everyone for the helpful and encouraging feedback! Thanks=
Joel for showing us that there are 6 orbits in the 2^4 and for your rederi=
vation of the state count. And thanks Matt and Roice for pointing out the i=
mportance of the inverted views. It looks so strange in that configuration =
that I always want to get back to a normal view as quickly as possible, but=
it does seem equally valid, and as you've shown, it can be helpful for mor=
e than just finding short sequences.

I don't understand Matt's "pinwheel" configuration, but I will point out th=
at all that is needed to create your twin interior octahedra is a single ha=
lf-rotation like I showed in the video at 5:29 ch?v=3DzqftZ8kJKLo&t=3D5m29s>. The two main halves do end up being mirror i=
mages of each other on the visible outside like he described. Whether it's =
the pinwheel or the half-rotated version that's correct, I'm not sure that =
it's a bummer that the solved state is not at all obvious, so long as we ca=
n operate it in my original configuration and ignore the fact that the oute=
r faces touch. That would just mean that the "correct" view is evidence tha=
t that the more understandable view is legitimate.

I'm going to try to make a snapable V3 which should allow the pieces to be =
more easily taken apart and reassembled into other forms. Shapeways does of=
fer a single, clear translucent plastic that they call "Frosted Detail", an=
d another called "Transparent Acrylic", but I don't think that any sort of =
transparent stickers will help us, especially since this thing is chock ful=
l of magnets. The easiest way to let you see into the two hemispheres would=
be to simply truncate the pointy tips of the stickers. That already happen=
s a little bit due to the way I've rounded the edges. Here is a close-up ttp://superliminal.com/cube/inverted1.jpg> of a half-rotation in which you =
can see that the inner yellow and white faces are solved. Your suggestion o=
f little mapping dots on the corners also works, but just opening the exist=
ing window further would work more directly.

-Melinda

On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>
>
> I agree with Don's arguments about adjacent sticker colors needing to be =
next to each other. I think this can be turned into an accurate 2^4 with c=
oloring changes, so I agree with Joel too :)
>
> To help me think about it, I started adding a new projection option for s=
pherical puzzles to MagicTile, which takes the two hemispheres of a puzzle =
and maps them to two disks with identified boundaries connected at a point,=
just like a physical "global chess /indexf.html>" game I have. Melinda's puzzle is a lot like this up a dimen=
sion, so think about two disjoint balls, each representing a hemisphere of =
the 2^4, each a "subcube" of Melinda's puzzle. The two boundaries of the b=
alls are identified with each other and as you roll one around, the other h=
alf rolls around so that identified points connect up. We need to have the=
same restriction on Melinda's puzzle.
>
> In the pristine state then, I think it'd be nice to have an internal (hid=
den), solid colored octahedron on each half. The other 6 faces should all =
have equal colors split between each hemisphere, 4 stickers on each half. =
You should be able to reorient the two subcubes to make a half octahedron o=
f any color on each subcube. I just saw Matt's email and picture, and it l=
ooks like we were going down the same thought path. I think with recolorin=
g (mirroring some of the current piece colorings) though, the windmill's ca=
n be avoided (?)
>
> [...] After staring/thinking a bit more, the coloring Matt came up with i=
s right-on if you want to put a solid color at the center of each hemispher=
e. His comment about the "mirrored" pieces on each side helped me understa=
nd better. 3 of the stickers are mirrored and the 4th is the hidden color =
(different on each side for a given pair of "mirrored" pieces). All faces =
behave identically as well, as they should. It's a little bit of a bummer =
that it doesn't look very pristine in the pristine state, but it does look =
like it should work as a 2^4.
>
> I wonder if there might be some adjustments to be made when shapeways all=
ows printing translucent as a color :)
>
> [...] Sorry for all the streaming, but I wanted to share one more thought=
. I now completely agree with Joel/Matt about it behaving as a 2^4, even w=
ith the original coloring. You just need to consider the corner colors of =
the two subcubes (pink/purple near the end of the video) as being a window =
into the interior of the piece. The other colors match up as desired. (So=
rry if folks already understood this after their emails and I'm just catchi=
ng up!)
>
> In fact, you could alter the coloring of the pieces slightly so that the =
behavior was similar with the inverted coloring. At the corners where 3 co=
lors meet on each piece, you could put a little circle of color of the oppo=
site 4th color. In Matt's windmill coloring then, you'd be able to see all=
four colors of a piece, like you can with some of the pieces on Melinda's =
original coloring. And again you'd consider the color circles a window to =
the interior that did not require the same matching constraints between the=
subcubes.
>
> I'm looking forward to having one of these :)
>
> Happy Friday everyone,
> Roice
>
> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com <=
mailto:joelkarlsson97@gmail.com> [4D_Cubing] <4D_Cubing@yahoogroups.com ilto:4D_Cubing@yahoogroups.com>> wrote:
>
>
>
> Seems like there was a slight misunderstanding. I meant that you need=
to be able to twist one of the faces and in MC4D the most natural choice =
is the center face. In your physical puzzle you can achieve this type of tw=
ist by twisting the two subcubes although this is indeed a twist of the sub=
cubes themselves and not the center face, however, this is still the same t=
ype of twist just around another face.
>
> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup =
of this puzzle. Hopefully the restrictions will be quite natural and only s=
ome "strange" moves would be illegal. Regarding the "families of states" (a=
ka orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed tw=
ists preserves the parity of the pieces, meaning that only half of the perm=
utations you can achieve by disassembling and reassembling can be reached t=
hrough legal moves. Because of some geometrical properties of the 2x2x2x2 a=
nd its twists, which would take some time to discuss in detail here, the or=
ientation of the stickers mod 3 are preserved, meaning that the last corner=
only can be oriented in one third of the number of orientations for the ot=
her corners. This gives a total number of orbits of 2x3=3D6. To check this =
result let's use this information to calculate all the possible states of t=
he 2x2x2x2; if there were no restrictions we would have 16! for permuting t=
he pieces (16
> pieces) and 12^16 for orienting them (12 orientations for each corne=
r). If we now take into account that there are 6 equally sized orbits this =
gets us to 12!16^12/6. However, we should also note that the orientation of=
the puzzle as a hole is not set by some kind of centerpieces and thus we n=
eed to devide with the number of orientations of a 4D cube if we want all o=
ur states to be separated with twists and not only rotations of the hole th=
ing. The number of ways to orient a 4D cube in space (only allowing rotatio=
ns and not mirroring) is 8x6x4=3D192 giving a total of 12!16^12/(6*192) sta=
tes which is indeed the same number that for example David Smith arrived at=
during his calculations. Therefore, when determining whether or not a twi=
st on your puzzle is legal or not it is sufficient and necessary to confirm=
that the twist is an even permutation of the pieces and preserves the orie=
ntation of stickers mod 3.
>
> Best regards,
> Joel
>
> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.co=
m [4D_Cubing]" <4D_Cubing@yahoogroups.com=
>:
>
> The new arrangement of magnets allows every valid orientation of =
pieces. The only invalid ones are those where the diagonal lines cutting ea=
ch cube's face cross each other rather than coincide. In other words, you c=
an assemble the puzzle in all ways that preserve the overall diamond/harleq=
uin pattern. Just about every move you can think of on the whole puzzle is =
valid though there are definitely invalid moves that the magnets allow. The=
most obvious invalid move is twisting of a single end cap.
>
> I think your description of the center face is not correct though=
. Twists of the outer faces cause twists "through" the center face, not "of=
" that face. Twists of the outer faces are twists of those faces themselves=
because they are the ones not changing, just like the center and outer fac=
es of MC4D when you twist the center face. The only direct twist of the cen=
ter face that this puzzle allows is a 90 degree twist about the outer axis.=
That happens when you simultaneously twist both end caps in the same direc=
tion.
>
> Yes, it's quite straightforward reorienting the whole puzzle to p=
ut any of the four axes on the outside. This is a very nice improvement ove=
r the first version and should make it much easier to solve. You may be rig=
ht that we just need to find the right way to think about the outside faces=
. I'll leave it to the math geniuses on the list to figure that out.
>
> -Melinda
>
>
>
> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com ilto:joelkarlsson97@gmail.com> [4D_Cubing] wrote:
>>
>> Hi Melinda,
>>
>> I do not agree with the criticism regarding the white and yellow=
stickers touching each other, this could simply be an effect of the differ=
ent representations of the puzzle. To really figure out if this indeed is a=
representation of a 2x2x2x2 we need to look at the possible moves (twists =
and rotations) and figure out the equivalent moves in the MC4D software. Fr=
om the MC4D software, it's easy to understand that the only moves required =
are free twists of one of the faces (that is, only twisting the center face=
in the standard perspective projection in MC4D) and 4D rotations swapping =
which face is in the center (ctrl-clicking in MC4D). The first is possible =
in your physical puzzle by rotating the white and yellow subcubes (from her=
e on I use subcube to refer to the two halves of the puzzle and the colours=
of the subcubes to refer to the "outer colours"). The second is possible i=
f it's possible to reach a solved state with any two colours on the subcube=
s that still
>> allow you to perform the previously mentioned twists. This seems=
to be the case from your demonstration and is indeed true if the magnets a=
llow the simple twists regardless of the colours of the subcubes. Thus, it =
is possible to let your puzzle be a representation of a 2x2x2x2, however, i=
t might require that some moves that the magnets allow aren't used.
>>
>> Best regards,
>> Joel
>>
>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com=
[4D_Cubing] <4D_Cubing@yahoogroups.com <=
mailto:4D_Cubing@yahoogroups.com>>:
>>
>> Dear Cubists,
>>
>> I've finished version 2 of my physical puzzle and uploaded a=
video of it here:
>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo outube.com/watch?v=3DzqftZ8kJKLo>
>> Again, please don't share these videos outside this group as=
their purpose is just to get your feedback. I'll eventually replace them w=
ith a public video.
>>
>> Here is an extra math puzzle that I bet you folks can answer=
: How many families of states does this puzzle have? In other words, if dis=
assembled and reassembled in any random configuration the magnets allow, wh=
at are the odds that it can be solved? This has practical implications if a=
ll such configurations are solvable because it would provide a very easy wa=
y to fully scramble the puzzle.
>>
>> And finally, a bit of fun: A relatively new friend of mine a=
nd new list member, Marc Ringuette, got excited enough to make his own vers=
ion. He built it from EPP foam and colored tape, and used honey instead of =
magnets to hold it together. Check it out here: http://superliminal.com/cub=
e/dessert_cube.jpg I don't =
know how practical a solution this is but it sure looks delicious! Welcome =
Marc!
>>
>> -Melinda
>>
>>
>
>
>
>
>
>
>
>=20


--------------AC19E366C6DED7570859E7F9
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable



">


First off, thanks everyone for the helpful and encouraging feedback!
Thanks Joel for showing us that there are 6 orbits in the 2^4 and
for your rederivation of the state count. And thanks Matt and Roice
for pointing out the importance of the inverted views. It looks so
strange in that configuration that I always want to get back to a
normal view as quickly as possible, but it does seem equally valid,
and as you've shown, it can be helpful for more than just finding
short sequences.



I don't understand Matt's "pinwheel" configuration, but I will point
out that all that is needed to create your twin interior octahedra
is a single half-rotation like I showed in the video at href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJKLo&t=3D5m29s">=
5:29.
The two main halves do end up being mirror images of each other on
the visible outside like he described. Whether it's the pinwheel or
the half-rotated version that's correct, I'm not sure that it's a
bummer that the solved state is not at all obvious, so long as we
can operate it in my original configuration and ignore the fact that
the outer faces touch. That would just mean that the "correct" view
is evidence that that the more understandable view is legitimate.



I'm going to try to make a snapable V3 which should allow the pieces
to be more easily taken apart and reassembled into other forms.
Shapeways does offer a single, clear translucent plastic that they
call "Frosted Detail", and another called "Transparent Acrylic", but
I don't think that any sort of transparent stickers will help us,
especially since this thing is chock full of magnets. The easiest
way to let you see into the two hemispheres would be to simply
truncate the pointy tips of the stickers. That already happens a
little bit due to the way I've rounded the edges. Here is a href=3D"http://superliminal.com/cube/inverted1.jpg">close-up of
a half-rotation in which you can see that the inner yellow and white
faces are solved. Your suggestion of little mapping dots on the
corners also works, but just opening the existing window further
would work more directly.



-Melinda



On 4/28/2017 2:15 PM, Roice Nelson
">roice3@gmail.com [4D_Cubing] wrote:


cite=3D"mid:CAEMuGXrG8=3DsX--gWZFXj2Fqt9YoibEqihJfO4D2AMN9k0JLqYQ@mail.gmai=
l.com"
type=3D"cite">


I agree with Don's arguments about adjacent sticker
colors needing to be next to each other.=C2=A0 I think this can be
turned into an accurate 2^4 with coloring changes, so I agree
with Joel too :)



To help me think about it, I started adding a new
projection option for spherical puzzles to MagicTile, which
takes the two hemispheres of a puzzle and maps them to two
disks with identified boundaries connected at a point, just
like a physical " href=3D"http://www.pa-network.com/global-chess/indexf.html"
target=3D"_blank">global chess
" game I have.=C2=A0 Melinda'=
s
puzzle is a lot like this up a dimension, so think about two
disjoint balls, each representing a hemisphere of the 2^4,
each a "subcube" of Melinda's puzzle.=C2=A0 The two boundaries of
the balls are identified with each other and as you roll one
around, the other half rolls around so that identified points
connect up.=C2=A0 We need to have the same restriction on Melinda=
's
puzzle.




In the pristine state then, I think it'd be nice to have an
internal (hidden), solid colored octahedron on each half.=C2=A0 T=
he
other 6 faces should all have equal colors split between each
hemisphere, 4 stickers on each half.=C2=A0 You should be able to
reorient the two subcubes to make a half octahedron of any
color on each subcube.=C2=A0 I just saw Matt's email and picture,
and it looks like we were going down the same thought path.=C2=A0=
I
think with recoloring (mirroring some of the current piece
colorings) though, the windmill's can be avoided (?)




[...] After staring/thinking a bit more, the coloring Matt
came up with is right-on if you want to put a solid color at
the center of each hemisphere.=C2=A0 His=C2=A0comment about the
"mirrored" pieces on each side helped me understand better. =C2=
=A03
of the stickers are mirrored and the 4th is the hidden color
(different on each side for a given pair of "mirrored"
pieces).=C2=A0 All faces behave identically as well, as they
should.=C2=A0 It's a little bit of a bummer that it doesn't look
very pristine in the pristine state, but it does look like it
should work as a 2^4.



I wonder if there might be some adjustments to be made when
shapeways allows printing translucent as a color :)




[...] Sorry for all the streaming, but
I wanted to share one more thought.=C2=A0 I now completely agree
with Joel/Matt about it behaving as a 2^4, even with the
original coloring.=C2=A0 You just need to consider the corner
colors of the two subcubes (pink/purple near the end of the
video) as being a window into the interior of the piece.=C2=A0 Th=
e
other colors match up as desired. =C2=A0(Sorry if folks already
understood this after their emails and I'm just catching up!)



In fact, you could alter the coloring of the pieces
slightly so that the behavior was similar with the inverted
coloring.=C2=A0 At the corners where 3 colors meet on each piec=
e,
you could put a little circle of color of the opposite 4th
color.=C2=A0 In Matt's windmill coloring then, you'd be able to
see all four colors of a piece, like you can with some of
the pieces on Melinda's original coloring.=C2=A0 And again you'=
d
consider the color circles a window to the interior that did
not require the same matching constraints between the
subcubes.




I'm looking forward to having one of these :)




Happy Friday everyone,

Roice





On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson moz-do-not-send=3D"true"
href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_blank">jo=
elkarlsson97@gmail.com
[4D_Cubing] < href=3D"mailto:4D_Cubing@yahoogroups.com" target=3D"_blank"=
>4D_Cubing@yahoogroups.com
>

wrote:

.8ex;border-left:1px #ccc solid;padding-left:1ex">






Seems like there was a slight misunderstanding. I
meant that you need to be able to =C2=A0twist one of th=
e
faces and in MC4D the most natural choice is the
center face. In your physical puzzle you can achieve
this type of twist by twisting the two subcubes
although this is indeed a twist of the subcubes
themselves and not the center face, however, this is
still the same type of twist just around another
face.=C2=A0




If the magnets are that allowing the
2x2x2x2 is obviously a subgroup of this puzzle.
Hopefully the restrictions will be quite natural and
only some "strange" moves would be illegal.
Regarding the "families of states" (aka orbits), the
2x2x2x2 has 6 orbits. As I mentioned earlier all
allowed twists preserves the parity of the pieces,
meaning that only half of the permutations you can
achieve by disassembling and reassembling can be
reached through legal moves. Because of some
geometrical properties of the 2x2x2x2 and its
twists, which would take some time to discuss in
detail here, the orientation of the stickers mod 3
are preserved, meaning that the last corner only can
be oriented in one third of the number of
orientations for the other corners. This gives a
total number of orbits of 2x3=3D6. To check this
result let's use this information to calculate all
the possible states of the 2x2x2x2; if there were no
restrictions we would have 16! for permuting the
pieces (16 pieces) =C2=A0and 12^16 for orienting them (=
12
orientations for each corner). If we now take into
account that there are 6 equally sized orbits this
gets us to 12!16^12/6. However, we should also note
that the orientation of the puzzle as a hole is not
set by some kind of centerpieces and thus we need to
devide with the number of orientations of a 4D cube
if we want all our states to be separated with
twists and not only rotations of the hole thing. The
number of ways to orient a 4D cube in space (only
allowing rotations and not mirroring) is 8x6x4=3D192
giving a total of=C2=A0 style=3D"font-family:sans-serif">12!16^12/(6*192)
states which is indeed the same number that for
example David Smith arrived at during his
calculations. Therefore, =C2=A0when determining wheth=
er
or not a twist on your puzzle is legal or not it
is sufficient and necessary to confirm that the
twist is an even permutation of the pieces and
preserves the orientation of stickers mod 3.div>
=



=
Best
regards,=C2=A0

=
Joel=C2=A0





Den 28 apr. 2017 3:02
fm skrev "Melinda Green moz-do-not-send=3D"true"
href=3D"mailto:melinda@superliminal.com"
target=3D"_blank">melinda@superliminal.com
[4D_Cubing]" < href=3D"mailto:4D_Cubing@yahoogroups.com"
target=3D"_blank">4D_Cubing@yahoogroups.com
=
>: type=3D"attribution">
class=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_=
-7761579982835215532m_4431900632030069098quote"
style=3D"margin:0 0 0 .8ex;border-left:1px
#ccc solid;padding-left:1ex">

=C2=A0
id=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012ygrp-mlmsg">
id=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012ygrp-msg">
id=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012ygrp-text">

The new arrangement of magnets
allows every valid orientation of
pieces. The only invalid ones are
those where the diagonal lines
cutting each cube's face cross
each other rather than coincide.
In other words, you can assemble
the puzzle in all ways that
preserve the overall
diamond/harlequin pattern. Just
about every move you can think of
on the whole puzzle is valid
though there are definitely
invalid moves that the magnets
allow. The most obvious invalid
move is twisting of a single end
cap.



I think your description of the
center face is not correct though.
Twists of the outer faces cause
twists "through" the center face,
not "of" that face. Twists of the
outer faces are twists of those
faces themselves because they are
the ones not changing, just like
the center and outer faces of MC4D
when you twist the center face.
The only direct twist of the
center face that this puzzle
allows is a 90 degree twist about
the outer axis. That happens when
you simultaneously twist both end
caps in the same direction.



Yes, it's quite straightforward
reorienting the whole puzzle to
put any of the four axes on the
outside. This is a very nice
improvement over the first version
and should make it much easier to
solve. You may be right that we
just need to find the right way to
think about the outside faces.
I'll leave it to the math geniuses
on the list to figure that out.



-Melinda


class=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_=
-7761579982835215532m_4431900632030069098quoted-text">



class=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_=
-7761579982835215532m_4431900632030069098m_6586884675826257012moz-cite-pref=
ix">On
4/27/2017 10:31 AM, Joel
Karlsson moz-do-not-send=3D"true"
class=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_=
-7761579982835215532m_4431900632030069098m_6586884675826257012moz-txt-link-=
abbreviated"
href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@g=
mail.com
[4D_Cubing] wrote:






Hi Melinda,




I do not agree with the
criticism regarding the
white and yellow stickers
touching each other, this
could simply be an effect
of the different
representations of the
puzzle. To really figure
out if this indeed is a
representation of a
2x2x2x2 we need to look at
the possible moves (twists
and rotations) and figure
out the equivalent moves
in the MC4D software. From
the MC4D software, it's
easy to understand that
the only moves required
are free twists of one of
the faces (that is, only
twisting the center face
in the standard
perspective projection in
MC4D) and 4D rotations
swapping which face is in
the center (ctrl-clicking
in MC4D). The first is
possible in your physical
puzzle by rotating the
white and yellow subcubes
(from here on I use
subcube to refer to the
two halves of the puzzle
and the colours of the
subcubes to refer to the
"outer colours"). The
second is possible if it's
possible to reach a solved
state with any two colours
on the subcubes that still
allow you to perform the
previously mentioned
twists. This seems to be
the case from your
demonstration and is
indeed true if the magnets
allow the simple twists
regardless of the colours
of the subcubes. Thus, it
is possible to let your
puzzle be a representation
of a 2x2x2x2, however, it
might require that some
moves that the magnets
allow aren't used.




Best regards,


Joel




2017-0=
4-27
3:09 GMT+02:00 Melinda Green
href=3D"mailto:melinda@superl=
iminal.com"
target=3D"_blank">melinda@sup=
erliminal.com

[4D_Cubing] &=
lt; moz-do-not-send=3D"true"
href=3D"mailto:4D_Cubing@ya=
hoogroups.com"
target=3D"_blank">4D_Cubing=
@yahoogroups.com>
:

class=3D"gmail_quote"
style=3D"border-left:1px
#ccc solid">
style=3D"background-color:#=
fff">
=C2=A0
id=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012m_32092672694193=
20979ygrp-mlmsg">
id=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012m_32092672694193=
20979ygrp-msg">
id=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012m_32092672694193=
20979ygrp-text">

Dear Cubists,



I've finished
version 2 of my
physical puzzle
and uploaded a
video of it
here:

moz-do-not-send=
=3D"true"
href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJKLo" target=3D"_blank">ht=
tps://www.youtube.com/watch?v=3DzqftZ8kJKLo

Again, please
don't share
these videos
outside this
group as their
purpose is just
to get your
feedback. I'll
eventually
replace them
with a public
video.



Here is an extra
math puzzle that
I bet you folks
can answer: How
many families of
states does this
puzzle have? In
other words, if
disassembled and
reassembled in
any random
configuration
the magnets
allow, what are
the odds that it
can be solved?
This has
practical
implications if
all such
configurations
are solvable
because it would
provide a very
easy way to
fully scramble
the puzzle.



And finally, a
bit of fun: A
relatively new
friend of mine
and new list
member, Marc
Ringuette, got
excited enough
to make his own
version. He
built it from
EPP foam and
colored tape,
and used honey
instead of
magnets to hold
it together.
Check it out
here: moz-do-not-send=
=3D"true"
href=3D"http://superliminal.com/cube/dessert_cube.jpg" target=3D"_blank">ht=
tp://superliminal.com/cube/dessert_cube.jpg
I don't know how
practical a
solution this is
but it sure
looks delicious!
Welcome Marc!



-Melinda


















class=3D"m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_=
-7761579982835215532m_4431900632030069098quoted-text">






















=20=20=20=20=20=20







--------------AC19E366C6DED7570859E7F9--




From: Joel Karlsson <joelkarlsson97@gmail.com>
Date: Sun, 30 Apr 2017 22:51:41 +0200
Subject: Re: [MC4D] Physical 4D puzzle V2



--f403043896c8540888054e6879fe
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

I am no expert on group theory, so to better understand what twists are
legal I read through the part of Kamack and Keane's *The Rubik Tesseract *a=
bout
orienting the corners. Since all even permutations are allowed the easiest
way to check if a twist is legal might be to:
1. Check that the twist is an even permutation, that is: the same twist can
be done by performing an even number of piece swaps (2-cycles).
2. Check the periodicity of the twist. If A^k=3DI (A^k meaning performing t=
he
twist k times and I (the identity) representing the permutation of doing
nothing) and k is not divisible by 3 the twist A definitely doesn't violate
the restriction of the orientations since kx mod 3 =3D 0 and k mod 3 !=3D 0
implies x mod 3 =3D 0 meaning that the change of the total orientation x fo=
r
the twist A mod 3 is 0 (which precisely is the restriction of legal twists;
that they must preserve the orientation mod 3).

For instance, this implies that the restacking moves are legal 2x2x2x2
moves since both are composed of 8 2-cycles and both can be performed twice
(note that 2 is not divisible by 3) to obtain the identity.

Note that 1 and 2 are sufficient to check if a twist is legal but only 1 is
necessary; there can indeed exist a twist violating 2 that still is legal
and in that case, I believe that we might have to study the orientation
changes for that specific twist in more detail. However, if a twist can be
composed by other legal twists it is, of course, legal as well.

Best regards,
Joel

2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com
[4D_Cubing] <4D_Cubing@yahoogroups.com>:

>
>
> First off, thanks everyone for the helpful and encouraging feedback!
> Thanks Joel for showing us that there are 6 orbits in the 2^4 and for you=
r
> rederivation of the state count. And thanks Matt and Roice for pointing o=
ut
> the importance of the inverted views. It looks so strange in that
> configuration that I always want to get back to a normal view as quickly =
as
> possible, but it does seem equally valid, and as you've shown, it can be
> helpful for more than just finding short sequences.
>
> I don't understand Matt's "pinwheel" configuration, but I will point out
> that all that is needed to create your twin interior octahedra is a singl=
e
> half-rotation like I showed in the video at 5:29
> . The two main
> halves do end up being mirror images of each other on the visible outside
> like he described. Whether it's the pinwheel or the half-rotated version
> that's correct, I'm not sure that it's a bummer that the solved state is
> not at all obvious, so long as we can operate it in my original
> configuration and ignore the fact that the outer faces touch. That would
> just mean that the "correct" view is evidence that that the more
> understandable view is legitimate.
>
> I'm going to try to make a snapable V3 which should allow the pieces to b=
e
> more easily taken apart and reassembled into other forms. Shapeways does
> offer a single, clear translucent plastic that they call "Frosted Detail"=
,
> and another called "Transparent Acrylic", but I don't think that any sort
> of transparent stickers will help us, especially since this thing is choc=
k
> full of magnets. The easiest way to let you see into the two hemispheres
> would be to simply truncate the pointy tips of the stickers. That already
> happens a little bit due to the way I've rounded the edges. Here is a
> close-up of a half-rotation
> in which you can see that the inner yellow and white faces are solved. Yo=
ur
> suggestion of little mapping dots on the corners also works, but just
> opening the existing window further would work more directly.
>
> -Melinda
>
> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>
> I agree with Don's arguments about adjacent sticker colors needing to be
> next to each other. I think this can be turned into an accurate 2^4 with
> coloring changes, so I agree with Joel too :)
>
> To help me think about it, I started adding a new projection option for
> spherical puzzles to MagicTile, which takes the two hemispheres of a puzz=
le
> and maps them to two disks with identified boundaries connected at a poin=
t,
> just like a physical "global chess
> " game I have.
> Melinda's puzzle is a lot like this up a dimension, so think about two
> disjoint balls, each representing a hemisphere of the 2^4, each a "subcub=
e"
> of Melinda's puzzle. The two boundaries of the balls are identified with
> each other and as you roll one around, the other half rolls around so tha=
t
> identified points connect up. We need to have the same restriction on
> Melinda's puzzle.
>
> In the pristine state then, I think it'd be nice to have an internal
> (hidden), solid colored octahedron on each half. The other 6 faces shoul=
d
> all have equal colors split between each hemisphere, 4 stickers on each
> half. You should be able to reorient the two subcubes to make a half
> octahedron of any color on each subcube. I just saw Matt's email and
> picture, and it looks like we were going down the same thought path. I
> think with recoloring (mirroring some of the current piece colorings)
> though, the windmill's can be avoided (?)
>
> [...] After staring/thinking a bit more, the coloring Matt came up with i=
s
> right-on if you want to put a solid color at the center of each
> hemisphere. His comment about the "mirrored" pieces on each side helped =
me
> understand better. 3 of the stickers are mirrored and the 4th is the
> hidden color (different on each side for a given pair of "mirrored"
> pieces). All faces behave identically as well, as they should. It's a
> little bit of a bummer that it doesn't look very pristine in the pristine
> state, but it does look like it should work as a 2^4.
>
> I wonder if there might be some adjustments to be made when shapeways
> allows printing translucent as a color :)
>
> [...] Sorry for all the streaming, but I wanted to share one more
> thought. I now completely agree with Joel/Matt about it behaving as a 2^=
4,
> even with the original coloring. You just need to consider the corner
> colors of the two subcubes (pink/purple near the end of the video) as bei=
ng
> a window into the interior of the piece. The other colors match up as
> desired. (Sorry if folks already understood this after their emails and
> I'm just catching up!)
>
> In fact, you could alter the coloring of the pieces slightly so that the
> behavior was similar with the inverted coloring. At the corners where 3
> colors meet on each piece, you could put a little circle of color of the
> opposite 4th color. In Matt's windmill coloring then, you'd be able to s=
ee
> all four colors of a piece, like you can with some of the pieces on
> Melinda's original coloring. And again you'd consider the color circles =
a
> window to the interior that did not require the same matching constraints
> between the subcubes.
>
> I'm looking forward to having one of these :)
>
> Happy Friday everyone,
> Roice
>
> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>
>>
>>
>> Seems like there was a slight misunderstanding. I meant that you need to
>> be able to twist one of the faces and in MC4D the most natural choice i=
s
>> the center face. In your physical puzzle you can achieve this type of tw=
ist
>> by twisting the two subcubes although this is indeed a twist of the
>> subcubes themselves and not the center face, however, this is still the
>> same type of twist just around another face.
>>
>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
>> this puzzle. Hopefully the restrictions will be quite natural and only s=
ome
>> "strange" moves would be illegal. Regarding the "families of states" (ak=
a
>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
>> twists preserves the parity of the pieces, meaning that only half of the
>> permutations you can achieve by disassembling and reassembling can be
>> reached through legal moves. Because of some geometrical properties of t=
he
>> 2x2x2x2 and its twists, which would take some time to discuss in detail
>> here, the orientation of the stickers mod 3 are preserved, meaning that =
the
>> last corner only can be oriented in one third of the number of orientati=
ons
>> for the other corners. This gives a total number of orbits of 2x3=3D6. T=
o
>> check this result let's use this information to calculate all the possib=
le
>> states of the 2x2x2x2; if there were no restrictions we would have 16! f=
or
>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>> orientations for each corner). If we now take into account that there ar=
e 6
>> equally sized orbits this gets us to 12!16^12/6. However, we should also
>> note that the orientation of the puzzle as a hole is not set by some kin=
d
>> of centerpieces and thus we need to devide with the number of orientatio=
ns
>> of a 4D cube if we want all our states to be separated with twists and n=
ot
>> only rotations of the hole thing. The number of ways to orient a 4D cube=
in
>> space (only allowing rotations and not mirroring) is 8x6x4=3D192 giving =
a
>> total of 12!16^12/(6*192) states which is indeed the same number that
>> for example David Smith arrived at during his calculations. Therefore,
>> when determining whether or not a twist on your puzzle is legal or not =
it
>> is sufficient and necessary to confirm that the twist is an even
>> permutation of the pieces and preserves the orientation of stickers mod =
3.
>>
>> Best regards,
>> Joel
>>
>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>
>>
>>
>> The new arrangement of magnets allows every valid orientation of pieces.
>> The only invalid ones are those where the diagonal lines cutting each
>> cube's face cross each other rather than coincide. In other words, you c=
an
>> assemble the puzzle in all ways that preserve the overall diamond/harleq=
uin
>> pattern. Just about every move you can think of on the whole puzzle is
>> valid though there are definitely invalid moves that the magnets allow. =
The
>> most obvious invalid move is twisting of a single end cap.
>>
>> I think your description of the center face is not correct though. Twist=
s
>> of the outer faces cause twists "through" the center face, not "of" that
>> face. Twists of the outer faces are twists of those faces themselves
>> because they are the ones not changing, just like the center and outer
>> faces of MC4D when you twist the center face. The only direct twist of t=
he
>> center face that this puzzle allows is a 90 degree twist about the outer
>> axis. That happens when you simultaneously twist both end caps in the sa=
me
>> direction.
>>
>> Yes, it's quite straightforward reorienting the whole puzzle to put any
>> of the four axes on the outside. This is a very nice improvement over th=
e
>> first version and should make it much easier to solve. You may be right
>> that we just need to find the right way to think about the outside faces=
.
>> I'll leave it to the math geniuses on the list to figure that out.
>>
>> -Melinda
>>
>>
>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com
>> [4D_Cubing] wrote:
>>
>>
>> Hi Melinda,
>>
>> I do not agree with the criticism regarding the white and yellow sticker=
s
>> touching each other, this could simply be an effect of the different
>> representations of the puzzle. To really figure out if this indeed is a
>> representation of a 2x2x2x2 we need to look at the possible moves (twist=
s
>> and rotations) and figure out the equivalent moves in the MC4D software.
>> From the MC4D software, it's easy to understand that the only moves
>> required are free twists of one of the faces (that is, only twisting the
>> center face in the standard perspective projection in MC4D) and 4D
>> rotations swapping which face is in the center (ctrl-clicking in MC4D). =
The
>> first is possible in your physical puzzle by rotating the white and yell=
ow
>> subcubes (from here on I use subcube to refer to the two halves of the
>> puzzle and the colours of the subcubes to refer to the "outer colours").
>> The second is possible if it's possible to reach a solved state with any
>> two colours on the subcubes that still allow you to perform the previous=
ly
>> mentioned twists. This seems to be the case from your demonstration and =
is
>> indeed true if the magnets allow the simple twists regardless of the
>> colours of the subcubes. Thus, it is possible to let your puzzle be a
>> representation of a 2x2x2x2, however, it might require that some moves t=
hat
>> the magnets allow aren't used.
>>
>> Best regards,
>> Joel
>>
>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>
>>>
>>>
>>> Dear Cubists,
>>>
>>> I've finished version 2 of my physical puzzle and uploaded a video of i=
t
>>> here:
>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>> Again, please don't share these videos outside this group as their
>>> purpose is just to get your feedback. I'll eventually replace them with=
a
>>> public video.
>>>
>>> Here is an extra math puzzle that I bet you folks can answer: How many
>>> families of states does this puzzle have? In other words, if disassembl=
ed
>>> and reassembled in any random configuration the magnets allow, what are=
the
>>> odds that it can be solved? This has practical implications if all such
>>> configurations are solvable because it would provide a very easy way to
>>> fully scramble the puzzle.
>>>
>>> And finally, a bit of fun: A relatively new friend of mine and new list
>>> member, Marc Ringuette, got excited enough to make his own version. He
>>> built it from EPP foam and colored tape, and used honey instead of magn=
ets
>>> to hold it together. Check it out here: http://superliminal.com/cube/d
>>> essert_cube.jpg I don't know how practical a solution this is but it
>>> sure looks delicious! Welcome Marc!
>>>
>>> -Melinda
>>>
>>>
>>
>>
>>
>>
>>
>
>=20
>

--f403043896c8540888054e6879fe
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

I am no expert on group theory, s=
o to better understand what twists are legal I read through the part of Kam=
ack and Keane's The Rubik Tesseract about orienting the corners.=
Since all even permutations are allowed the easiest way to check if a twis=
t is legal might be to:
1. Check that the twist is an even permuta=
tion, that is: the same twist can be done by performing an even number of p=
iece swaps (2-cycles).
2. Check the periodicity of the twist. If A=
^k=3DI (A^k meaning performing the twist k times and I (the identity) repre=
senting the permutation of doing nothing) and k is not divisible by 3 the t=
wist A definitely doesn't violate the restriction of the orientations s=
ince kx mod 3 =3D 0 and k mod 3 !=3D 0 implies x mod 3 =3D 0 meaning that t=
he change of the total orientation x for the twist A mod 3 is 0 (which prec=
isely is the restriction of legal twists; that they must preserve the orien=
tation mod 3).

For instance, this implies that the restacking =
moves are legal 2x2x2x2 moves since both are composed of 8 2-cycles and bot=
h can be performed twice (note that 2 is not divisible by 3) to obtain the =
identity.

Note that 1 and 2 are sufficient to check if a =
twist is legal but only 1 is necessary; there can indeed exist a twist viol=
ating 2 that still is legal and in that case, I believe that we might have =
to study the orientation changes for that specific twist in more detail. Ho=
wever, if a twist can be composed by other legal twists it is, of course, l=
egal as well.

Best regards,
Joel
=

2017-04-29 1:04 G=
MT+02:00 Melinda Green melinda@=
superliminal.com
[4D_Cubing] <_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>=
;
:
order-left:1px #ccc solid;padding-left:1ex">












=20

=C2=A0







=20=20=20=20=20=20
=20=20=20=20=20=20


=20=20
=20=20
First off, thanks everyone for the helpful and encouraging feedback!
Thanks Joel for showing us that there are 6 orbits in the 2^4 and
for your rederivation of the state count. And thanks Matt and Roice
for pointing out the importance of the inverted views. It looks so
strange in that configuration that I always want to get back to a
normal view as quickly as possible, but it does seem equally valid,
and as you've shown, it can be helpful for more than just finding
short sequences.



I don't understand Matt's "pinwheel" configuration, b=
ut I will point
out that all that is needed to create your twin interior octahedra
is a single half-rotation like I showed in the video at s://www.youtube.com/watch?v=3DzqftZ8kJKLo&t=3D5m29s" target=3D"_blank">=
5:29
.
The two main halves do end up being mirror images of each other on
the visible outside like he described. Whether it's the pinwheel or
the half-rotated version that's correct, I'm not sure that it&#=
39;s a
bummer that the solved state is not at all obvious, so long as we
can operate it in my original configuration and ignore the fact that
the outer faces touch. That would just mean that the "correct"=
; view
is evidence that that the more understandable view is legitimate.



I'm going to try to make a snapable V3 which should allow the piece=
s
to be more easily taken apart and reassembled into other forms.
Shapeways does offer a single, clear translucent plastic that they
call "Frosted Detail", and another called "Transparent A=
crylic", but
I don't think that any sort of transparent stickers will help us,
especially since this thing is chock full of magnets. The easiest
way to let you see into the two hemispheres would be to simply
truncate the pointy tips of the stickers. That already happens a
little bit due to the way I've rounded the edges. Here is a =3D"http://superliminal.com/cube/inverted1.jpg" target=3D"_blank">close-up<=
/a> of
a half-rotation in which you can see that the inner yellow and white
faces are solved. Your suggestion of little mapping dots on the
corners also works, but just opening the existing window further
would work more directly.



-Melinda





=20=20=20=20=20=20
=20=20=20=20=20=20
I agree with Don's arguments ab=
out adjacent sticker
colors needing to be next to each other.=C2=A0 I think this can be
turned into an accurate 2^4 with coloring changes, so I agree
with Joel too :)



To help me think about it, I started adding a new
projection option for spherical puzzles to MagicTile, which
takes the two hemispheres of a puzzle and maps them to two
disks with identified boundaries connected at a point, just
like a physical "-chess/indexf.html" target=3D"_blank">global chess" game I have.=
=C2=A0 Melinda's
puzzle is a lot like this up a dimension, so think about two
disjoint balls, each representing a hemisphere of the 2^4,
each a "subcube" of Melinda's puzzle.=C2=A0 The two=
boundaries of
the balls are identified with each other and as you roll one
around, the other half rolls around so that identified points
connect up.=C2=A0 We need to have the same restriction on Melinda=
's
puzzle.




In the pristine state then, I think it'd be nice to have a=
n
internal (hidden), solid colored octahedron on each half.=C2=A0 T=
he
other 6 faces should all have equal colors split between each
hemisphere, 4 stickers on each half.=C2=A0 You should be able to
reorient the two subcubes to make a half octahedron of any
color on each subcube.=C2=A0 I just saw Matt's email and pict=
ure,
and it looks like we were going down the same thought path.=C2=A0=
I
think with recoloring (mirroring some of the current piece
colorings) though, the windmill's can be avoided (?)




[...] After staring/thinking a bit more, the coloring M=
att
came up with is right-on if you want to put a solid color at
the center of each hemisphere.=C2=A0 His=C2=A0comment about the
"mirrored" pieces on each side helped me understand bet=
ter. =C2=A03
of the stickers are mirrored and the 4th is the hidden color
(different on each side for a given pair of "mirrored"
pieces).=C2=A0 All faces behave identically as well, as they
should.=C2=A0 It's a little bit of a bummer that it doesn'=
;t look
very pristine in the pristine state, but it does look like it
should work as a 2^4.



I wonder if there might be some adjustments to be made when
shapeways allows printing translucent as a color :)




[...] Sorry for all the streaming, but
I wanted to share one more thought.=C2=A0 I now completely agree
with Joel/Matt about it behaving as a 2^4, even with the
original coloring.=C2=A0 You just need to consider the corner
colors of the two subcubes (pink/purple near the end of the
video) as being a window into the interior of the piece.=C2=A0 Th=
e
other colors match up as desired. =C2=A0(Sorry if folks already
understood this after their emails and I'm just catching up!)



In fact, you could alter the coloring of the pieces
slightly so that the behavior was similar with the inverted
coloring.=C2=A0 At the corners where 3 colors meet on each piec=
e,
you could put a little circle of color of the opposite 4th
color.=C2=A0 In Matt's windmill coloring then, you'd be=
able to
see all four colors of a piece, like you can with some of
the pieces on Melinda's original coloring.=C2=A0 And again =
you'd
consider the color circles a window to the interior that did
not require the same matching constraints between the
subcubes.




I'm looking forward to having one of these :)




Happy Friday everyone,

Roice





On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson o:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com
[4D_Cubing] <ahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>
wrote:

solid">






Seems like there was a slight misunderstanding. I
meant that you need to be able to =C2=A0twist one of th=
e
faces and in MC4D the most natural choice is the
center face. In your physical puzzle you can achieve
this type of twist by twisting the two subcubes
although this is indeed a twist of the subcubes
themselves and not the center face, however, this is
still the same type of twist just around another
face.=C2=A0




If the magnets are that allowing the
2x2x2x2 is obviously a subgroup of this puzzle.
Hopefully the restrictions will be quite natural and
only some "strange" moves would be illegal.
Regarding the "families of states" (aka orbit=
s), the
2x2x2x2 has 6 orbits. As I mentioned earlier all
allowed twists preserves the parity of the pieces,
meaning that only half of the permutations you can
achieve by disassembling and reassembling can be
reached through legal moves. Because of some
geometrical properties of the 2x2x2x2 and its
twists, which would take some time to discuss in
detail here, the orientation of the stickers mod 3
are preserved, meaning that the last corner only can
be oriented in one third of the number of
orientations for the other corners. This gives a
total number of orbits of 2x3=3D6. To check this
result let's use this information to calculate all
the possible states of the 2x2x2x2; if there were no
restrictions we would have 16! for permuting the
pieces (16 pieces) =C2=A0and 12^16 for orienting them (=
12
orientations for each corner). If we now take into
account that there are 6 equally sized orbits this
gets us to 12!16^12/6. However, we should also note
that the orientation of the puzzle as a hole is not
set by some kind of centerpieces and thus we need to
devide with the number of orientations of a 4D cube
if we want all our states to be separated with
twists and not only rotations of the hole thing. The
number of ways to orient a 4D cube in space (only
allowing rotations and not mirroring) is 8x6x4=3D192
giving a total of=C2=A0serif">12!16^12/(6*192)
states which is indeed the same number that for
example David Smith arrived at during his
calculations. Therefore, =C2=A0when determining wheth=
er
or not a twist on your puzzle is legal or not it
is sufficient and necessary to confirm that the
twist is an even permutation of the pieces and
preserves the orientation of stickers mod 3.
div>
=



=
Best
regards,=C2=A0

=
Joel=C2=A0





Den 28 apr. 2017 3:02
fm skrev "Melinda Green linda@superliminal.com" target=3D"_blank">melinda@superliminal.com
[4D_Cubing]" <@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:
ype=3D"attribution">
134459512634882m_7927561327128665054m_-4244333051891278538m_-77615799828352=
15532m_4431900632030069098quote" style=3D"border-left:1px #ccc solid">

=C2=A0
512634882m_7927561327128665054m_-4244333051891278538m_-7761579982835215532m=
_4431900632030069098m_6586884675826257012ygrp-mlmsg">
59512634882m_7927561327128665054m_-4244333051891278538m_-776157998283521553=
2m_4431900632030069098m_6586884675826257012ygrp-msg">
4459512634882m_7927561327128665054m_-4244333051891278538m_-7761579982835215=
532m_4431900632030069098m_6586884675826257012ygrp-text">

The new arrangement of magnets
allows every valid orientation of
pieces. The only invalid ones are
those where the diagonal lines
cutting each cube's face cross
each other rather than coincide.
In other words, you can assemble
the puzzle in all ways that
preserve the overall
diamond/harlequin pattern. Just
about every move you can think of
on the whole puzzle is valid
though there are definitely
invalid moves that the magnets
allow. The most obvious invalid
move is twisting of a single end
cap.



I think your description of the
center face is not correct though.
Twists of the outer faces cause
twists "through" the center=
face,
not "of" that face. Twists =
of the
outer faces are twists of those
faces themselves because they are
the ones not changing, just like
the center and outer faces of MC4D
when you twist the center face.
The only direct twist of the
center face that this puzzle
allows is a 90 degree twist about
the outer axis. That happens when
you simultaneously twist both end
caps in the same direction.



Yes, it's quite straightforward
reorienting the whole puzzle to
put any of the four axes on the
outside. This is a very nice
improvement over the first version
and should make it much easier to
solve. You may be right that we
just need to find the right way to
think about the outside faces.
I'll leave it to the math geniuse=
s
on the list to figure that out.



-Melinda


733134459512634882m_7927561327128665054m_-4244333051891278538m_-77615799828=
35215532m_4431900632030069098quoted-text">



_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998=
2835215532m_4431900632030069098m_6586884675826257012moz-cite-prefix">On
4/27/2017 10:31 AM, Joel
Karlsson 89261218m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-=
7761579982835215532m_4431900632030069098m_6586884675826257012moz-txt-link-a=
bbreviated" href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_blank">joel=
karlsson97@gmail.com

[4D_Cubing] wrote:






Hi Melinda,




I do not agree with the
criticism regarding the
white and yellow stickers
touching each other, this
could simply be an effect
of the different
representations of the
puzzle. To really figure
out if this indeed is a
representation of a
2x2x2x2 we need to look at
the possible moves (twists
and rotations) and figure
out the equivalent moves
in the MC4D software. From
the MC4D software, it's
easy to understand that
the only moves required
are free twists of one of
the faces (that is, only
twisting the center face
in the standard
perspective projection in
MC4D) and 4D rotations
swapping which face is in
the center (ctrl-clicking
in MC4D). The first is
possible in your physical
puzzle by rotating the
white and yellow subcubes
(from here on I use
subcube to refer to the
two halves of the puzzle
and the colours of the
subcubes to refer to the
"outer colours"). T=
he
second is possible if it'=
s
possible to reach a solved
state with any two colours
on the subcubes that still
allow you to perform the
previously mentioned
twists. This seems to be
the case from your
demonstration and is
indeed true if the magnets
allow the simple twists
regardless of the colours
of the subcubes. Thus, it
is possible to let your
puzzle be a representation
of a 2x2x2x2, however, it
might require that some
moves that the magnets
allow aren't used.




Best regards,


Joel




2017-0=
4-27
3:09 GMT+02:00 Melinda Green
liminal.com" target=3D"_blank">melinda@superliminal.com
[4D_Cubing] &=
lt;4D_Cubing=
@yahoogroups.com
>
:

e" style=3D"border-left:1px #ccc solid">
r:#fff">
=C2=A0
261218m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-77=
61579982835215532m_4431900632030069098m_6586884675826257012m_32092672694193=
20979ygrp-mlmsg">
89261218m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-=
7761579982835215532m_4431900632030069098m_6586884675826257012m_320926726941=
9320979ygrp-msg">
7589261218m_6733134459512634882m_7927561327128665054m_-4244333051891278538m=
_-7761579982835215532m_4431900632030069098m_6586884675826257012m_3209267269=
419320979ygrp-text">

Dear Cubists,



I've finished
version 2 of my
physical puzzle
and uploaded a
video of it
here:

www.youtube.com/watch?v=3DzqftZ8kJKLo" target=3D"_blank">https://www.youtub=
e.com/watch?v=3DzqftZ8kJKLo


Again, please
don't share
these videos
outside this
group as their
purpose is just
to get your
feedback. I'll
eventually
replace them
with a public
video.



Here is an extra
math puzzle that
I bet you folks
can answer: How
many families of
states does this
puzzle have? In
other words, if
disassembled and
reassembled in
any random
configuration
the magnets
allow, what are
the odds that it
can be solved?
This has
practical
implications if
all such
configurations
are solvable
because it would
provide a very
easy way to
fully scramble
the puzzle.



And finally, a
bit of fun: A
relatively new
friend of mine
and new list
member, Marc
Ringuette, got
excited enough
to make his own
version. He
built it from
EPP foam and
colored tape,
and used honey
instead of
magnets to hold
it together.
Check it out
here: tp://superliminal.com/cube/dessert_cube.jpg" target=3D"_blank">http://super=
liminal.com/cube/dessert_cube.jpg

I don't know ho=
w
practical a
solution this is
but it sure
looks delicious!
Welcome Marc!



-Melinda


















3134459512634882m_7927561327128665054m_-4244333051891278538m_-7761579982835=
215532m_4431900632030069098quoted-text">






















=20=20=20=20=20=20




=20=20




=20=20=20=20=20

=20=20=20=20







=20=20









--f403043896c8540888054e6879fe--




From: Joel Karlsson <joelkarlsson97@gmail.com>
Date: Wed, 3 May 2017 23:39:29 +0200
Subject: Re: [MC4D] Physical 4D puzzle V2



--f403045f4e60d81c2b054ea57d51
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Horrible typo... It seems like I made some typos in my email regarding the
state count. It should of course be 16!12^16/(6*192) and NOT
12!16^12/(6*192). However, I did calculate the correct number when
comparing with previous results so the actual derivation was correct.

Something of interest is that the physical pieces can be assembled in
16!24*12^15 wats since there are 16 pieces, the first one can be oriented
in 24 ways and the remaining can be oriented in 12 ways (since a corner
with 3 colours never touch a corner with just one colour). Dividing with 6
to get a single orbit still gives a factor 2*192 higher than the actual
count rather than 192. This shows that every state in the MC4D
representation has 2 representations in the physical puzzle. These two
representations must be the previously discussed, that the two halves
either have the same color on the outermost corners or the innermost
(forming an octahedron) when the puzzle is solved and thus both are
complete representations of the 2x2x2x2.

Best regards,
Joel Karlsson

Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" :

I am no expert on group theory, so to better understand what twists are
legal I read through the part of Kamack and Keane's *The Rubik Tesseract *a=
bout
orienting the corners. Since all even permutations are allowed the easiest
way to check if a twist is legal might be to:
1. Check that the twist is an even permutation, that is: the same twist can
be done by performing an even number of piece swaps (2-cycles).
2. Check the periodicity of the twist. If A^k=3DI (A^k meaning performing t=
he
twist k times and I (the identity) representing the permutation of doing
nothing) and k is not divisible by 3 the twist A definitely doesn't violate
the restriction of the orientations since kx mod 3 =3D 0 and k mod 3 !=3D 0
implies x mod 3 =3D 0 meaning that the change of the total orientation x fo=
r
the twist A mod 3 is 0 (which precisely is the restriction of legal twists;
that they must preserve the orientation mod 3).

For instance, this implies that the restacking moves are legal 2x2x2x2
moves since both are composed of 8 2-cycles and both can be performed twice
(note that 2 is not divisible by 3) to obtain the identity.

Note that 1 and 2 are sufficient to check if a twist is legal but only 1 is
necessary; there can indeed exist a twist violating 2 that still is legal
and in that case, I believe that we might have to study the orientation
changes for that specific twist in more detail. However, if a twist can be
composed by other legal twists it is, of course, legal as well.

Best regards,
Joel

2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com
[4D_Cubing] <4D_Cubing@yahoogroups.com>:

>
>
> First off, thanks everyone for the helpful and encouraging feedback!
> Thanks Joel for showing us that there are 6 orbits in the 2^4 and for you=
r
> rederivation of the state count. And thanks Matt and Roice for pointing o=
ut
> the importance of the inverted views. It looks so strange in that
> configuration that I always want to get back to a normal view as quickly =
as
> possible, but it does seem equally valid, and as you've shown, it can be
> helpful for more than just finding short sequences.
>
> I don't understand Matt's "pinwheel" configuration, but I will point out
> that all that is needed to create your twin interior octahedra is a singl=
e
> half-rotation like I showed in the video at 5:29
> . The two main
> halves do end up being mirror images of each other on the visible outside
> like he described. Whether it's the pinwheel or the half-rotated version
> that's correct, I'm not sure that it's a bummer that the solved state is
> not at all obvious, so long as we can operate it in my original
> configuration and ignore the fact that the outer faces touch. That would
> just mean that the "correct" view is evidence that that the more
> understandable view is legitimate.
>
> I'm going to try to make a snapable V3 which should allow the pieces to b=
e
> more easily taken apart and reassembled into other forms. Shapeways does
> offer a single, clear translucent plastic that they call "Frosted Detail"=
,
> and another called "Transparent Acrylic", but I don't think that any sort
> of transparent stickers will help us, especially since this thing is choc=
k
> full of magnets. The easiest way to let you see into the two hemispheres
> would be to simply truncate the pointy tips of the stickers. That already
> happens a little bit due to the way I've rounded the edges. Here is a
> close-up of a half-rotation
> in which you can see that the inner yellow and white faces are solved. Yo=
ur
> suggestion of little mapping dots on the corners also works, but just
> opening the existing window further would work more directly.
>
> -Melinda
>
> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>
> I agree with Don's arguments about adjacent sticker colors needing to be
> next to each other. I think this can be turned into an accurate 2^4 with
> coloring changes, so I agree with Joel too :)
>
> To help me think about it, I started adding a new projection option for
> spherical puzzles to MagicTile, which takes the two hemispheres of a puzz=
le
> and maps them to two disks with identified boundaries connected at a poin=
t,
> just like a physical "global chess
> " game I have.
> Melinda's puzzle is a lot like this up a dimension, so think about two
> disjoint balls, each representing a hemisphere of the 2^4, each a "subcub=
e"
> of Melinda's puzzle. The two boundaries of the balls are identified with
> each other and as you roll one around, the other half rolls around so tha=
t
> identified points connect up. We need to have the same restriction on
> Melinda's puzzle.
>
> In the pristine state then, I think it'd be nice to have an internal
> (hidden), solid colored octahedron on each half. The other 6 faces shoul=
d
> all have equal colors split between each hemisphere, 4 stickers on each
> half. You should be able to reorient the two subcubes to make a half
> octahedron of any color on each subcube. I just saw Matt's email and
> picture, and it looks like we were going down the same thought path. I
> think with recoloring (mirroring some of the current piece colorings)
> though, the windmill's can be avoided (?)
>
> [...] After staring/thinking a bit more, the coloring Matt came up with i=
s
> right-on if you want to put a solid color at the center of each
> hemisphere. His comment about the "mirrored" pieces on each side helped =
me
> understand better. 3 of the stickers are mirrored and the 4th is the
> hidden color (different on each side for a given pair of "mirrored"
> pieces). All faces behave identically as well, as they should. It's a
> little bit of a bummer that it doesn't look very pristine in the pristine
> state, but it does look like it should work as a 2^4.
>
> I wonder if there might be some adjustments to be made when shapeways
> allows printing translucent as a color :)
>
> [...] Sorry for all the streaming, but I wanted to share one more
> thought. I now completely agree with Joel/Matt about it behaving as a 2^=
4,
> even with the original coloring. You just need to consider the corner
> colors of the two subcubes (pink/purple near the end of the video) as bei=
ng
> a window into the interior of the piece. The other colors match up as
> desired. (Sorry if folks already understood this after their emails and
> I'm just catching up!)
>
> In fact, you could alter the coloring of the pieces slightly so that the
> behavior was similar with the inverted coloring. At the corners where 3
> colors meet on each piece, you could put a little circle of color of the
> opposite 4th color. In Matt's windmill coloring then, you'd be able to s=
ee
> all four colors of a piece, like you can with some of the pieces on
> Melinda's original coloring. And again you'd consider the color circles =
a
> window to the interior that did not require the same matching constraints
> between the subcubes.
>
> I'm looking forward to having one of these :)
>
> Happy Friday everyone,
> Roice
>
> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>
>>
>>
>> Seems like there was a slight misunderstanding. I meant that you need to
>> be able to twist one of the faces and in MC4D the most natural choice i=
s
>> the center face. In your physical puzzle you can achieve this type of tw=
ist
>> by twisting the two subcubes although this is indeed a twist of the
>> subcubes themselves and not the center face, however, this is still the
>> same type of twist just around another face.
>>
>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
>> this puzzle. Hopefully the restrictions will be quite natural and only s=
ome
>> "strange" moves would be illegal. Regarding the "families of states" (ak=
a
>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
>> twists preserves the parity of the pieces, meaning that only half of the
>> permutations you can achieve by disassembling and reassembling can be
>> reached through legal moves. Because of some geometrical properties of t=
he
>> 2x2x2x2 and its twists, which would take some time to discuss in detail
>> here, the orientation of the stickers mod 3 are preserved, meaning that =
the
>> last corner only can be oriented in one third of the number of orientati=
ons
>> for the other corners. This gives a total number of orbits of 2x3=3D6. T=
o
>> check this result let's use this information to calculate all the possib=
le
>> states of the 2x2x2x2; if there were no restrictions we would have 16! f=
or
>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>> orientations for each corner). If we now take into account that there ar=
e 6
>> equally sized orbits this gets us to 12!16^12/6. However, we should also
>> note that the orientation of the puzzle as a hole is not set by some kin=
d
>> of centerpieces and thus we need to devide with the number of orientatio=
ns
>> of a 4D cube if we want all our states to be separated with twists and n=
ot
>> only rotations of the hole thing. The number of ways to orient a 4D cube=
in
>> space (only allowing rotations and not mirroring) is 8x6x4=3D192 giving =
a
>> total of 12!16^12/(6*192) states which is indeed the same number that
>> for example David Smith arrived at during his calculations. Therefore,
>> when determining whether or not a twist on your puzzle is legal or not =
it
>> is sufficient and necessary to confirm that the twist is an even
>> permutation of the pieces and preserves the orientation of stickers mod =
3.
>>
>> Best regards,
>> Joel
>>
>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>
>>
>>
>> The new arrangement of magnets allows every valid orientation of pieces.
>> The only invalid ones are those where the diagonal lines cutting each
>> cube's face cross each other rather than coincide. In other words, you c=
an
>> assemble the puzzle in all ways that preserve the overall diamond/harleq=
uin
>> pattern. Just about every move you can think of on the whole puzzle is
>> valid though there are definitely invalid moves that the magnets allow. =
The
>> most obvious invalid move is twisting of a single end cap.
>>
>> I think your description of the center face is not correct though. Twist=
s
>> of the outer faces cause twists "through" the center face, not "of" that
>> face. Twists of the outer faces are twists of those faces themselves
>> because they are the ones not changing, just like the center and outer
>> faces of MC4D when you twist the center face. The only direct twist of t=
he
>> center face that this puzzle allows is a 90 degree twist about the outer
>> axis. That happens when you simultaneously twist both end caps in the sa=
me
>> direction.
>>
>> Yes, it's quite straightforward reorienting the whole puzzle to put any
>> of the four axes on the outside. This is a very nice improvement over th=
e
>> first version and should make it much easier to solve. You may be right
>> that we just need to find the right way to think about the outside faces=
.
>> I'll leave it to the math geniuses on the list to figure that out.
>>
>> -Melinda
>>
>>
>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com
>> [4D_Cubing] wrote:
>>
>>
>> Hi Melinda,
>>
>> I do not agree with the criticism regarding the white and yellow sticker=
s
>> touching each other, this could simply be an effect of the different
>> representations of the puzzle. To really figure out if this indeed is a
>> representation of a 2x2x2x2 we need to look at the possible moves (twist=
s
>> and rotations) and figure out the equivalent moves in the MC4D software.
>> From the MC4D software, it's easy to understand that the only moves
>> required are free twists of one of the faces (that is, only twisting the
>> center face in the standard perspective projection in MC4D) and 4D
>> rotations swapping which face is in the center (ctrl-clicking in MC4D). =
The
>> first is possible in your physical puzzle by rotating the white and yell=
ow
>> subcubes (from here on I use subcube to refer to the two halves of the
>> puzzle and the colours of the subcubes to refer to the "outer colours").
>> The second is possible if it's possible to reach a solved state with any
>> two colours on the subcubes that still allow you to perform the previous=
ly
>> mentioned twists. This seems to be the case from your demonstration and =
is
>> indeed true if the magnets allow the simple twists regardless of the
>> colours of the subcubes. Thus, it is possible to let your puzzle be a
>> representation of a 2x2x2x2, however, it might require that some moves t=
hat
>> the magnets allow aren't used.
>>
>> Best regards,
>> Joel
>>
>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>
>>>
>>>
>>> Dear Cubists,
>>>
>>> I've finished version 2 of my physical puzzle and uploaded a video of i=
t
>>> here:
>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>> Again, please don't share these videos outside this group as their
>>> purpose is just to get your feedback. I'll eventually replace them with=
a
>>> public video.
>>>
>>> Here is an extra math puzzle that I bet you folks can answer: How many
>>> families of states does this puzzle have? In other words, if disassembl=
ed
>>> and reassembled in any random configuration the magnets allow, what are=
the
>>> odds that it can be solved? This has practical implications if all such
>>> configurations are solvable because it would provide a very easy way to
>>> fully scramble the puzzle.
>>>
>>> And finally, a bit of fun: A relatively new friend of mine and new list
>>> member, Marc Ringuette, got excited enough to make his own version. He
>>> built it from EPP foam and colored tape, and used honey instead of magn=
ets
>>> to hold it together. Check it out here: http://superliminal.com/cube/d
>>> essert_cube.jpg I don't know how practical a solution this is but it
>>> sure looks delicious! Welcome Marc!
>>>
>>> -Melinda
>>>
>>>
>>
>>
>>
>>
>>
>
>=20
>

--f403045f4e60d81c2b054ea57d51
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable

Horrible typo... It seems like I made some typos in =
my email regarding the state count. It should of course be 16!12^16/(6*192)=
and NOT 12!16^12/(6*192). However, I did calculate the correct number when=
comparing with previous results so the actual derivation was correct.>

Something of interest is tha=
t the physical pieces can be assembled in 16!24*12^15 wats since there are =
16 pieces, the first one can be oriented in 24 ways and the remaining can b=
e oriented in 12 ways (since a corner with 3 colours never touch a corner w=
ith just one colour). Dividing with 6 to get a single orbit still gives a f=
actor 2*192 higher than the actual count rather than 192. This shows that e=
very state in the MC4D representation has 2 representations in the physical=
puzzle. These two representations must be the previously discussed, that t=
he two halves either have the same color on the outermost corners or the in=
nermost (forming an octahedron) when the puzzle is solved and thus both are=
complete representations of the 2x2x2x2.=C2=A0
class=3D"gmail_extra" dir=3D"auto">
r=3D"auto">Best regards,=C2=A0
>Joel Karlsson=C2=A0

class=3D"gmail_quote">Den 30 apr. 2017 10:51 em skrev "Joel Karlsson&q=
uot; <joelkarlsson97@gmail.c=
om
>:
rgin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
r">
I am no expert on group theory, so to better un=
derstand what twists are legal I read through the part of Kamack and Keane&=
#39;s The Rubik Tesseract about orienting the corners. Since all eve=
n permutations are allowed the easiest way to check if a twist is legal mig=
ht be to:
1. Check that the twist is an even permutation, that is:=
the same twist can be done by performing an even number of piece swaps (2-=
cycles).
2. Check the periodicity of the twist. If A^k=3DI (A^k me=
aning performing the twist k times and I (the identity) representing the pe=
rmutation of doing nothing) and k is not divisible by 3 the twist A definit=
ely doesn't violate the restriction of the orientations since kx mod 3 =
=3D 0 and k mod 3 !=3D 0 implies x mod 3 =3D 0 meaning that the change of t=
he total orientation x for the twist A mod 3 is 0 (which precisely is the r=
estriction of legal twists; that they must preserve the orientation mod 3).=


For instance, this implies that the restacking moves are lega=
l 2x2x2x2 moves since both are composed of 8 2-cycles and both can be perfo=
rmed twice (note that 2 is not divisible by 3) to obtain the identity.
<=
br>
Note that 1 and 2 are sufficient to check if a twist is legal=
but only 1 is necessary; there can indeed exist a twist violating 2 that s=
till is legal and in that case, I believe that we might have to study the o=
rientation changes for that specific twist in more detail. However, if a tw=
ist can be composed by other legal twists it is, of course, legal as well.<=
br>

Best regards,
Joel
elided-text">

2017=
-04-29 1:04 GMT+02:00 Melinda Green com" target=3D"_blank">melinda@superliminal.com [4D_Cubing] =3D"ltr"><>4D_Cubing@yahoogroups.com>:
ote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex=
">












=20

=C2=A0







=20=20=20=20=20=20
=20=20=20=20=20=20


=20=20
=20=20
First off, thanks everyone for the helpful and encouraging feedback!
Thanks Joel for showing us that there are 6 orbits in the 2^4 and
for your rederivation of the state count. And thanks Matt and Roice
for pointing out the importance of the inverted views. It looks so
strange in that configuration that I always want to get back to a
normal view as quickly as possible, but it does seem equally valid,
and as you've shown, it can be helpful for more than just finding
short sequences.



I don't understand Matt's "pinwheel" configuration, b=
ut I will point
out that all that is needed to create your twin interior octahedra
is a single half-rotation like I showed in the video at s://www.youtube.com/watch?v=3DzqftZ8kJKLo&t=3D5m29s" target=3D"_blank">=
5:29
.
The two main halves do end up being mirror images of each other on
the visible outside like he described. Whether it's the pinwheel or
the half-rotated version that's correct, I'm not sure that it&#=
39;s a
bummer that the solved state is not at all obvious, so long as we
can operate it in my original configuration and ignore the fact that
the outer faces touch. That would just mean that the "correct"=
; view
is evidence that that the more understandable view is legitimate.



I'm going to try to make a snapable V3 which should allow the piece=
s
to be more easily taken apart and reassembled into other forms.
Shapeways does offer a single, clear translucent plastic that they
call "Frosted Detail", and another called "Transparent A=
crylic", but
I don't think that any sort of transparent stickers will help us,
especially since this thing is chock full of magnets. The easiest
way to let you see into the two hemispheres would be to simply
truncate the pointy tips of the stickers. That already happens a
little bit due to the way I've rounded the edges. Here is a =3D"http://superliminal.com/cube/inverted1.jpg" target=3D"_blank">close-up<=
/a> of
a half-rotation in which you can see that the inner yellow and white
faces are solved. Your suggestion of little mapping dots on the
corners also works, but just opening the existing window further
would work more directly.



-Melinda





=20=20=20=20=20=20
=20=20=20=20=20=20
I agree with Don's arguments about adjacen=
t sticker
colors needing to be next to each other.=C2=A0 I think this can be
turned into an accurate 2^4 with coloring changes, so I agree
with Joel too :)



To help me think about it, I started adding a new
projection option for spherical puzzles to MagicTile, which
takes the two hemispheres of a puzzle and maps them to two
disks with identified boundaries connected at a point, just
like a physical "-chess/indexf.html" target=3D"_blank">global chess" game I have.=
=C2=A0 Melinda's
puzzle is a lot like this up a dimension, so think about two
disjoint balls, each representing a hemisphere of the 2^4,
each a "subcube" of Melinda's puzzle.=C2=A0 The two=
boundaries of
the balls are identified with each other and as you roll one
around, the other half rolls around so that identified points
connect up.=C2=A0 We need to have the same restriction on Melinda=
's
puzzle.




In the pristine state then, I think it'd be nice to have a=
n
internal (hidden), solid colored octahedron on each half.=C2=A0 T=
he
other 6 faces should all have equal colors split between each
hemisphere, 4 stickers on each half.=C2=A0 You should be able to
reorient the two subcubes to make a half octahedron of any
color on each subcube.=C2=A0 I just saw Matt's email and pict=
ure,
and it looks like we were going down the same thought path.=C2=A0=
I
think with recoloring (mirroring some of the current piece
colorings) though, the windmill's can be avoided (?)




[...] After staring/thinking a bit more, the coloring M=
att
came up with is right-on if you want to put a solid color at
the center of each hemisphere.=C2=A0 His=C2=A0comment about the
"mirrored" pieces on each side helped me understand bet=
ter. =C2=A03
of the stickers are mirrored and the 4th is the hidden color
(different on each side for a given pair of "mirrored"
pieces).=C2=A0 All faces behave identically as well, as they
should.=C2=A0 It's a little bit of a bummer that it doesn'=
;t look
very pristine in the pristine state, but it does look like it
should work as a 2^4.



I wonder if there might be some adjustments to be made when
shapeways allows printing translucent as a color :)




[...] Sorry for all the streaming, but
I wanted to share one more thought.=C2=A0 I now completely agree
with Joel/Matt about it behaving as a 2^4, even with the
original coloring.=C2=A0 You just need to consider the corner
colors of the two subcubes (pink/purple near the end of the
video) as being a window into the interior of the piece.=C2=A0 Th=
e
other colors match up as desired. =C2=A0(Sorry if folks already
understood this after their emails and I'm just catching up!)



In fact, you could alter the coloring of the pieces
slightly so that the behavior was similar with the inverted
coloring.=C2=A0 At the corners where 3 colors meet on each piec=
e,
you could put a little circle of color of the opposite 4th
color.=C2=A0 In Matt's windmill coloring then, you'd be=
able to
see all four colors of a piece, like you can with some of
the pieces on Melinda's original coloring.=C2=A0 And again =
you'd
consider the color circles a window to the interior that did
not require the same matching constraints between the
subcubes.




I'm looking forward to having one of these :)




Happy Friday everyone,

Roicev>




On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson o:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com
[4D_Cubing] <ahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>
wrote:

solid">






Seems like there was a slight misunderstanding. I
meant that you need to be able to =C2=A0twist one of th=
e
faces and in MC4D the most natural choice is the
center face. In your physical puzzle you can achieve
this type of twist by twisting the two subcubes
although this is indeed a twist of the subcubes
themselves and not the center face, however, this is
still the same type of twist just around another
face.=C2=A0




If the magnets are that allowing the
2x2x2x2 is obviously a subgroup of this puzzle.
Hopefully the restrictions will be quite natural and
only some "strange" moves would be illegal.
Regarding the "families of states" (aka orbit=
s), the
2x2x2x2 has 6 orbits. As I mentioned earlier all
allowed twists preserves the parity of the pieces,
meaning that only half of the permutations you can
achieve by disassembling and reassembling can be
reached through legal moves. Because of some
geometrical properties of the 2x2x2x2 and its
twists, which would take some time to discuss in
detail here, the orientation of the stickers mod 3
are preserved, meaning that the last corner only can
be oriented in one third of the number of
orientations for the other corners. This gives a
total number of orbits of 2x3=3D6. To check this
result let's use this information to calculate all
the possible states of the 2x2x2x2; if there were no
restrictions we would have 16! for permuting the
pieces (16 pieces) =C2=A0and 12^16 for orienting them (=
12
orientations for each corner). If we now take into
account that there are 6 equally sized orbits this
gets us to 12!16^12/6. However, we should also note
that the orientation of the puzzle as a hole is not
set by some kind of centerpieces and thus we need to
devide with the number of orientations of a 4D cube
if we want all our states to be separated with
twists and not only rotations of the hole thing. The
number of ways to orient a 4D cube in space (only
allowing rotations and not mirroring) is 8x6x4=3D192
giving a total of=C2=A0serif">12!16^12/(6*192)
states which is indeed the same number that for
example David Smith arrived at during his
calculations. Therefore, =C2=A0when determining wheth=
er
or not a twist on your puzzle is legal or not it
is sufficient and necessary to confirm that the
twist is an even permutation of the pieces and
preserves the orientation of stickers mod 3.
div>
=



=
Best
regards,=C2=A0

=
Joel=C2=A0





Den 28 apr. 2017 3:02
fm skrev "Melinda Green linda@superliminal.com" target=3D"_blank">melinda@superliminal.com
[4D_Cubing]" <@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:
ype=3D"attribution">
226697589261218m_6733134459512634882m_7927561327128665054m_-424433305189127=
8538m_-7761579982835215532m_4431900632030069098quote" style=3D"border-left:=
1px #ccc solid">

=C2=A0
589261218m_6733134459512634882m_7927561327128665054m_-4244333051891278538m_=
-7761579982835215532m_4431900632030069098m_6586884675826257012ygrp-mlmsg">
97589261218m_6733134459512634882m_7927561327128665054m_-4244333051891278538=
m_-7761579982835215532m_4431900632030069098m_6586884675826257012ygrp-msg">
6697589261218m_6733134459512634882m_7927561327128665054m_-42443330518912785=
38m_-7761579982835215532m_4431900632030069098m_6586884675826257012ygrp-text=
">

The new arrangement of magnets
allows every valid orientation of
pieces. The only invalid ones are
those where the diagonal lines
cutting each cube's face cross
each other rather than coincide.
In other words, you can assemble
the puzzle in all ways that
preserve the overall
diamond/harlequin pattern. Just
about every move you can think of
on the whole puzzle is valid
though there are definitely
invalid moves that the magnets
allow. The most obvious invalid
move is twisting of a single end
cap.



I think your description of the
center face is not correct though.
Twists of the outer faces cause
twists "through" the center=
face,
not "of" that face. Twists =
of the
outer faces are twists of those
faces themselves because they are
the ones not changing, just like
the center and outer faces of MC4D
when you twist the center face.
The only direct twist of the
center face that this puzzle
allows is a 90 degree twist about
the outer axis. That happens when
you simultaneously twist both end
caps in the same direction.



Yes, it's quite straightforward
reorienting the whole puzzle to
put any of the four axes on the
outside. This is a very nice
improvement over the first version
and should make it much easier to
solve. You may be right that we
just need to find the right way to
think about the outside faces.
I'll leave it to the math geniuse=
s
on the list to figure that out.



-Melinda


446226697589261218m_6733134459512634882m_7927561327128665054m_-424433305189=
1278538m_-7761579982835215532m_4431900632030069098quoted-text">



-4446226697589261218m_6733134459512634882m_7927561327128665054m_-4244333051=
891278538m_-7761579982835215532m_4431900632030069098m_6586884675826257012mo=
z-cite-prefix">On
4/27/2017 10:31 AM, Joel
Karlsson 3611828m_-4446226697589261218m_6733134459512634882m_7927561327128665054m_-4=
244333051891278538m_-7761579982835215532m_4431900632030069098m_658688467582=
6257012moz-txt-link-abbreviated" href=3D"mailto:joelkarlsson97@gmail.com" t=
arget=3D"_blank">joelkarlsson97@gmail.com

[4D_Cubing] wrote:






Hi Melinda,




I do not agree with the
criticism regarding the
white and yellow stickers
touching each other, this
could simply be an effect
of the different
representations of the
puzzle. To really figure
out if this indeed is a
representation of a
2x2x2x2 we need to look at
the possible moves (twists
and rotations) and figure
out the equivalent moves
in the MC4D software. From
the MC4D software, it's
easy to understand that
the only moves required
are free twists of one of
the faces (that is, only
twisting the center face
in the standard
perspective projection in
MC4D) and 4D rotations
swapping which face is in
the center (ctrl-clicking
in MC4D). The first is
possible in your physical
puzzle by rotating the
white and yellow subcubes
(from here on I use
subcube to refer to the
two halves of the puzzle
and the colours of the
subcubes to refer to the
"outer colours"). T=
he
second is possible if it'=
s
possible to reach a solved
state with any two colours
on the subcubes that still
allow you to perform the
previously mentioned
twists. This seems to be
the case from your
demonstration and is
indeed true if the magnets
allow the simple twists
regardless of the colours
of the subcubes. Thus, it
is possible to let your
puzzle be a representation
of a 2x2x2x2, however, it
might require that some
moves that the magnets
allow aren't used.




Best regards,


Joel




2017-0=
4-27
3:09 GMT+02:00 Melinda Green
liminal.com" target=3D"_blank">melinda@superliminal.com
[4D_Cubing] &=
lt;4D_Cubing=
@yahoogroups.com
>
:

e" style=3D"border-left:1px #ccc solid">
r:#fff">
=C2=A0
11828m_-4446226697589261218m_6733134459512634882m_7927561327128665054m_-424=
4333051891278538m_-7761579982835215532m_4431900632030069098m_65868846758262=
57012m_3209267269419320979ygrp-mlmsg">
3611828m_-4446226697589261218m_6733134459512634882m_7927561327128665054m_-4=
244333051891278538m_-7761579982835215532m_4431900632030069098m_658688467582=
6257012m_3209267269419320979ygrp-msg">
753611828m_-4446226697589261218m_6733134459512634882m_7927561327128665054m_=
-4244333051891278538m_-7761579982835215532m_4431900632030069098m_6586884675=
826257012m_3209267269419320979ygrp-text">

Dear Cubists,



I've finished
version 2 of my
physical puzzle
and uploaded a
video of it
here:

www.youtube.com/watch?v=3DzqftZ8kJKLo" target=3D"_blank">https://www.youtub=
e.com/watch?v=3DzqftZ8kJKLo


Again, please
don't share
these videos
outside this
group as their
purpose is just
to get your
feedback. I'll
eventually
replace them
with a public
video.



Here is an extra
math puzzle that
I bet you folks
can answer: How
many families of
states does this
puzzle have? In
other words, if
disassembled and
reassembled in
any random
configuration
the magnets
allow, what are
the odds that it
can be solved?
This has
practical
implications if
all such
configurations
are solvable
because it would
provide a very
easy way to
fully scramble
the puzzle.



And finally, a
bit of fun: A
relatively new
friend of mine
and new list
member, Marc
Ringuette, got
excited enough
to make his own
version. He
built it from
EPP foam and
colored tape,
and used honey
instead of
magnets to hold
it together.
Check it out
here: tp://superliminal.com/cube/dessert_cube.jpg" target=3D"_blank">http://super=
liminal.com/cube/dessert_cube.jpg

I don't know ho=
w
practical a
solution this is
but it sure
looks delicious!
Welcome Marc!



-Melinda


















6226697589261218m_6733134459512634882m_7927561327128665054m_-42443330518912=
78538m_-7761579982835215532m_4431900632030069098quoted-text">






















=20=20=20=20=20=20




=20=20




=20=20=20=20=20

=20=20=20=20







=20=20











--f403045f4e60d81c2b054ea57d51--




From: Melinda Green <melinda@superliminal.com>
Date: Wed, 3 May 2017 15:01:27 -0700
Subject: Re: [MC4D] Physical 4D puzzle V2



--------------0B3ABC49A067D6A587EBCEAB
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: quoted-printable

Thanks for the correction. A couple of things: First, when assembling one p=
iece at a time, I'd say there is only 1 way to place the first piece, not 2=
4. Otherwise you'd have to say that the 1x1x1 puzzle has 24 states. I under=
stand that this may be conventional, but to me, that just sounds silly.

Second, I have the feeling that the difference between the "two representat=
ions" you describe is simply one of those half-rotations I showed in the vi=
deo. In the normal solved state there is only one complete octahedron in th=
e very center, and in the half-rotated state there is one in the middle of =
each half of the "inverted" form. I consider them to be the same solved sta=
te.

-Melinda

On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing] wro=
te:
>
>
> Horrible typo... It seems like I made some typos in my email regarding th=
e state count. It should of course be 16!12^16/(6*192) and NOT 12!16^12/(6*=
192). However, I did calculate the correct number when comparing with previ=
ous results so the actual derivation was correct.
>
> Something of interest is that the physical pieces can be assembled in 16!=
24*12^15 ways since there are 16 pieces, the first one can be oriented in 2=
4 ways and the remaining can be oriented in 12 ways (since a corner with 3 =
colours never touch a corner with just one colour). Dividing with 6 to get =
a single orbit still gives a factor 2*192 higher than the actual count rath=
er than 192. This shows that every state in the MC4D representation has 2 r=
epresentations in the physical puzzle. These two representations must be th=
e previously discussed, that the two halves either have the same color on t=
he outermost corners or the innermost (forming an octahedron) when the puzz=
le is solved and thus both are complete representations of the 2x2x2x2.
>
> Best regards,
> Joel Karlsson
>
> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" >:
>
> I am no expert on group theory, so to better understand what twists a=
re legal I read through the part of Kamack and Keane's /The Rubik Tesseract=
/about orienting the corners. Since all even permutations are allowed the =
easiest way to check if a twist is legal might be to:
> 1. Check that the twist is an even permutation, that is: the same twi=
st can be done by performing an even number of piece swaps (2-cycles).
> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning perfor=
ming the twist k times and I (the identity) representing the permutation of=
doing nothing) and k is not divisible by 3 the twist A definitely doesn't =
violate the restriction of the orientations since kx mod 3 =3D 0 and k mod =
3 !=3D 0 implies x mod 3 =3D 0 meaning that the change of the total orienta=
tion x for the twist A mod 3 is 0 (which precisely is the restriction of le=
gal twists; that they must preserve the orientation mod 3).
>
> For instance, this implies that the restacking moves are legal 2x2x2x=
2 moves since both are composed of 8 2-cycles and both can be performed twi=
ce (note that 2 is not divisible by 3) to obtain the identity.
>
> Note that 1 and 2 are sufficient to check if a twist is legal but onl=
y 1 is necessary; there can indeed exist a twist violating 2 that still is =
legal and in that case, I believe that we might have to study the orientati=
on changes for that specific twist in more detail. However, if a twist can =
be composed by other legal twists it is, of course, legal as well.
>
> Best regards,
> Joel
>
> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com lto:melinda@superliminal.com> [4D_Cubing] <4D_Cubing@yahoogroups.com o:4D_Cubing@yahoogroups.com>>:
>
> First off, thanks everyone for the helpful and encouraging feedba=
ck! Thanks Joel for showing us that there are 6 orbits in the 2^4 and for y=
our rederivation of the state count. And thanks Matt and Roice for pointing=
out the importance of the inverted views. It looks so strange in that conf=
iguration that I always want to get back to a normal view as quickly as pos=
sible, but it does seem equally valid, and as you've shown, it can be helpf=
ul for more than just finding short sequences.
>
> I don't understand Matt's "pinwheel" configuration, but I will po=
int out that all that is needed to create your twin interior octahedra is a=
single half-rotation like I showed in the video at 5:29 be.com/watch?v=3DzqftZ8kJKLo&t=3D5m29s>. The two main halves do end up bein=
g mirror images of each other on the visible outside like he described. Whe=
ther it's the pinwheel or the half-rotated version that's correct, I'm not =
sure that it's a bummer that the solved state is not at all obvious, so lon=
g as we can operate it in my original configuration and ignore the fact tha=
t the outer faces touch. That would just mean that the "correct" view is ev=
idence that that the more understandable view is legitimate.
>
> I'm going to try to make a snapable V3 which should allow the pie=
ces to be more easily taken apart and reassembled into other forms. Shapewa=
ys does offer a single, clear translucent plastic that they call "Frosted D=
etail", and another called "Transparent Acrylic", but I don't think that an=
y sort of transparent stickers will help us, especially since this thing is=
chock full of magnets. The easiest way to let you see into the two hemisph=
eres would be to simply truncate the pointy tips of the stickers. That alre=
ady happens a little bit due to the way I've rounded the edges. Here is a c=
lose-up of a half-rotation in =
which you can see that the inner yellow and white faces are solved. Your su=
ggestion of little mapping dots on the corners also works, but just opening=
the existing window further would work more directly.
>
> -Melinda
>
> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com 3@gmail.com> [4D_Cubing] wrote:
>> I agree with Don's arguments about adjacent sticker colors needi=
ng to be next to each other. I think this can be turned into an accurate 2=
^4 with coloring changes, so I agree with Joel too :)
>>
>> To help me think about it, I started adding a new projection opt=
ion for spherical puzzles to MagicTile, which takes the two hemispheres of =
a puzzle and maps them to two disks with identified boundaries connected at=
a point, just like a physical "global chess bal-chess/indexf.html>" game I have. Melinda's puzzle is a lot like this u=
p a dimension, so think about two disjoint balls, each representing a hemis=
phere of the 2^4, each a "subcube" of Melinda's puzzle. The two boundaries=
of the balls are identified with each other and as you roll one around, th=
e other half rolls around so that identified points connect up. We need to =
have the same restriction on Melinda's puzzle.
>>
>> In the pristine state then, I think it'd be nice to have an inte=
rnal (hidden), solid colored octahedron on each half. The other 6 faces sh=
ould all have equal colors split between each hemisphere, 4 stickers on eac=
h half. You should be able to reorient the two subcubes to make a half oct=
ahedron of any color on each subcube. I just saw Matt's email and picture,=
and it looks like we were going down the same thought path. I think with =
recoloring (mirroring some of the current piece colorings) though, the wind=
mill's can be avoided (?)
>>
>> [...] After staring/thinking a bit more, the coloring Matt came =
up with is right-on if you want to put a solid color at the center of each =
hemisphere. His comment about the "mirrored" pieces on each side helped me=
understand better. 3 of the stickers are mirrored and the 4th is the hidd=
en color (different on each side for a given pair of "mirrored" pieces). A=
ll faces behave identically as well, as they should. It's a little bit of =
a bummer that it doesn't look very pristine in the pristine state, but it d=
oes look like it should work as a 2^4.
>>
>> I wonder if there might be some adjustments to be made when shap=
eways allows printing translucent as a color :)
>>
>> [...] Sorry for all the streaming, but I wanted to share one mor=
e thought. I now completely agree with Joel/Matt about it behaving as a 2^4=
, even with the original coloring. You just need to consider the corner co=
lors of the two subcubes (pink/purple near the end of the video) as being a=
window into the interior of the piece. The other colors match up as desir=
ed. (Sorry if folks already understood this after their emails and I'm jus=
t catching up!)
>>
>> In fact, you could alter the coloring of the pieces slightly so =
that the behavior was similar with the inverted coloring. At the corners w=
here 3 colors meet on each piece, you could put a little circle of color of=
the opposite 4th color. In Matt's windmill coloring then, you'd be able t=
o see all four colors of a piece, like you can with some of the pieces on M=
elinda's original coloring. And again you'd consider the color circles a wi=
ndow to the interior that did not require the same matching constraints bet=
ween the subcubes.
>>
>> I'm looking forward to having one of these :)
>>
>> Happy Friday everyone,
>> Roice
>>
>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gm=
ail.com [4D_Cubing] <4D_Cubing@yahoogroup=
s.com > wrote:
>>
>>
>>
>> Seems like there was a slight misunderstanding. I meant that=
you need to be able to twist one of the faces and in MC4D the most natura=
l choice is the center face. In your physical puzzle you can achieve this t=
ype of twist by twisting the two subcubes although this is indeed a twist o=
f the subcubes themselves and not the center face, however, this is still t=
he same type of twist just around another face.
>>
>> If the magnets are that allowing the 2x2x2x2 is obviously a =
subgroup of this puzzle. Hopefully the restrictions will be quite natural a=
nd only some "strange" moves would be illegal. Regarding the "families of s=
tates" (aka orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all a=
llowed twists preserves the parity of the pieces, meaning that only half of=
the permutations you can achieve by disassembling and reassembling can be =
reached through legal moves. Because of some geometrical properties of the =
2x2x2x2 and its twists, which would take some time to discuss in detail her=
e, the orientation of the stickers mod 3 are preserved, meaning that the la=
st corner only can be oriented in one third of the number of orientations f=
or the other corners. This gives a total number of orbits of 2x3=3D6. To ch=
eck this result let's use this information to calculate all the possible st=
ates of the 2x2x2x2; if there were no restrictions we would have 16! for pe=
rmuting the
>> pieces (16 pieces) and 12^16 for orienting them (12 orienta=
tions for each corner). If we now take into account that there are 6 equall=
y sized orbits this gets us to 12!16^12/6. However, we should also note tha=
t the orientation of the puzzle as a hole is not set by some kind of center=
pieces and thus we need to devide with the number of orientations of a 4D c=
ube if we want all our states to be separated with twists and not only rota=
tions of the hole thing. The number of ways to orient a 4D cube in space (o=
nly allowing rotations and not mirroring) is 8x6x4=3D192 giving a total of =
12!16^12/(6*192) states which is indeed the same number that for example Da=
vid Smith arrived at during his calculations. Therefore, when determining =
whether or not a twist on your puzzle is legal or not it is sufficient and =
necessary to confirm that the twist is an even permutation of the pieces an=
d preserves the orientation of stickers mod 3.
>>
>> Best regards,
>> Joel
>>
>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superl=
iminal.com [4D_Cubing]" <4D_Cubing@yahoog=
roups.com >:
>>
>> The new arrangement of magnets allows every valid orient=
ation of pieces. The only invalid ones are those where the diagonal lines c=
utting each cube's face cross each other rather than coincide. In other wor=
ds, you can assemble the puzzle in all ways that preserve the overall diamo=
nd/harlequin pattern. Just about every move you can think of on the whole p=
uzzle is valid though there are definitely invalid moves that the magnets a=
llow. The most obvious invalid move is twisting of a single end cap.
>>
>> I think your description of the center face is not corre=
ct though. Twists of the outer faces cause twists "through" the center face=
, not "of" that face. Twists of the outer faces are twists of those faces t=
hemselves because they are the ones not changing, just like the center and =
outer faces of MC4D when you twist the center face. The only direct twist o=
f the center face that this puzzle allows is a 90 degree twist about the ou=
ter axis. That happens when you simultaneously twist both end caps in the s=
ame direction.
>>
>> Yes, it's quite straightforward reorienting the whole pu=
zzle to put any of the four axes on the outside. This is a very nice improv=
ement over the first version and should make it much easier to solve. You m=
ay be right that we just need to find the right way to think about the outs=
ide faces. I'll leave it to the math geniuses on the list to figure that ou=
t.
>>
>> -Melinda
>>
>>
>>
>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmai=
l.com [4D_Cubing] wrote:
>>>
>>> Hi Melinda,
>>>
>>> I do not agree with the criticism regarding the white a=
nd yellow stickers touching each other, this could simply be an effect of t=
he different representations of the puzzle. To really figure out if this in=
deed is a representation of a 2x2x2x2 we need to look at the possible moves=
(twists and rotations) and figure out the equivalent moves in the MC4D sof=
tware. From the MC4D software, it's easy to understand that the only moves =
required are free twists of one of the faces (that is, only twisting the ce=
nter face in the standard perspective projection in MC4D) and 4D rotations =
swapping which face is in the center (ctrl-clicking in MC4D). The first is =
possible in your physical puzzle by rotating the white and yellow subcubes =
(from here on I use subcube to refer to the two halves of the puzzle and th=
e colours of the subcubes to refer to the "outer colours"). The second is p=
ossible if it's possible to reach a solved state with any two colours on th=
e subcubes that
>>> still allow you to perform the previously mentioned twi=
sts. This seems to be the case from your demonstration and is indeed true i=
f the magnets allow the simple twists regardless of the colours of the subc=
ubes. Thus, it is possible to let your puzzle be a representation of a 2x2x=
2x2, however, it might require that some moves that the magnets allow aren'=
t used.
>>>
>>> Best regards,
>>> Joel
>>>
>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superli=
minal.com [4D_Cubing] <4D_Cubing@yahoogro=
ups.com >:
>>>
>>> Dear Cubists,
>>>
>>> I've finished version 2 of my physical puzzle and u=
ploaded a video of it here:
>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo s://www.youtube.com/watch?v=3DzqftZ8kJKLo>
>>> Again, please don't share these videos outside this=
group as their purpose is just to get your feedback. I'll eventually repla=
ce them with a public video.
>>>
>>> Here is an extra math puzzle that I bet you folks c=
an answer: How many families of states does this puzzle have? In other word=
s, if disassembled and reassembled in any random configuration the magnets =
allow, what are the odds that it can be solved? This has practical implicat=
ions if all such configurations are solvable because it would provide a ver=
y easy way to fully scramble the puzzle.
>>>
>>> And finally, a bit of fun: A relatively new friend =
of mine and new list member, Marc Ringuette, got excited enough to make his=
own version. He built it from EPP foam and colored tape, and used honey in=
stead of magnets to hold it together. Check it out here: http://superlimina=
l.com/cube/dessert_cube.jpg =
I don't know how practical a solution this is but it sure looks delicious!=
Welcome Marc!
>>>
>>> -Melinda
>>>
>>>
>>
>>
>>
>>
>>
>
>
>
>
>
>=20


--------------0B3ABC49A067D6A587EBCEAB
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable



">


Thanks for the correction. A couple of things: First, when
assembling one piece at a time, I'd say there is only 1 way to place
the first piece, not 24. Otherwise you'd have to say that the 1x1x1
puzzle has 24 states. I understand that this may be conventional,
but to me, that just sounds silly.



Second, I have the feeling that the difference between the "two
representations" you describe is simply one of those half-rotations
I showed in the video. In the normal solved state there is only one
complete octahedron in the very center, and in the half-rotated
state there is one in the middle of each half of the "inverted"
form. I consider them to be the same solved state.



-Melinda



On 5/3/2017 2:39 PM, Joel Karlsson
mail.com">joelkarlsson97@gmail.com [4D_Cubing] wrote:


cite=3D"mid:CAEohJcEUFSdFQDrKpYNpHjkYUa54sA60E22Y7R4s6a4NC174RA@mail.gmail.=
com"
type=3D"cite">



Horrible typo... It seems like I made some typos in my
email regarding the state count. It should of course be
16!12^16/(6*192) and NOT 12!16^12/(6*192). However, I did
calculate the correct number when comparing with previous
results so the actual derivation was correct.




Something of interest is that the physical
pieces can be assembled in 16!24*12^15 ways since there are 16
pieces, the first one can be oriented in 24 ways and the
remaining can be oriented in 12 ways (since a corner with 3
colours never touch a corner with just one colour). Dividing
with 6 to get a single orbit still gives a factor 2*192 higher
than the actual count rather than 192. This shows that every
state in the MC4D representation has 2 representations in the
physical puzzle. These two representations must be the
previously discussed, that the two halves either have the same
color on the outermost corners or the innermost (forming an
octahedron) when the puzzle is solved and thus both are
complete representations of the 2x2x2x2.=C2=A0





Best regards,=C2=A0

Joel Karlsson=C2=A0



Den 30 apr. 2017 10:51 em skrev
"Joel Karlsson" < href=3D"mailto:joelkarlsson97@gmail.com">joelkarlsson97@gma=
il.com
>: type=3D"attribution">
.8ex;border-left:1px #ccc solid;padding-left:1ex">





I am no expert on group theory, so to
better understand what twists are legal I
read through the part of Kamack and Keane's
The Rubik Tesseract about orienting
the corners. Since all even permutations are
allowed the easiest way to check if a twist
is legal might be to:


1. Check that the twist is an even
permutation, that is: the same twist can be
done by performing an even number of piece
swaps (2-cycles).


2. Check the periodicity of the twist. If A^k=3DI
(A^k meaning performing the twist k times and I
(the identity) representing the permutation of
doing nothing) and k is not divisible by 3 the
twist A definitely doesn't violate the
restriction of the orientations since kx mod 3 =3D
0 and k mod 3 !=3D 0 implies x mod 3 =3D 0 meaning
that the change of the total orientation x for
the twist A mod 3 is 0 (which precisely is the
restriction of legal twists; that they must
preserve the orientation mod 3).




For instance, this implies that the restacking
moves are legal 2x2x2x2 moves since both are
composed of 8 2-cycles and both can be performed
twice (note that 2 is not divisible by 3) to
obtain the identity.




Note that 1 and 2 are sufficient to check if a
twist is legal but only 1 is necessary; there can
indeed exist a twist violating 2 that still is
legal and in that case, I believe that we might
have to study the orientation changes for that
specific twist in more detail. However, if a twist
can be composed by other legal twists it is, of
course, legal as well.





Best regards,


Joel





2017-04-29 1:04 GMT+02:00
Melinda Green href=3D"mailto:melinda@superliminal.com"
target=3D"_blank">melinda@superliminal.com

[4D_Cubing] < moz-do-not-send=3D"true"
href=3D"mailto:4D_Cubing@yahoogroups.com"
target=3D"_blank">4D_Cubing@yahoogroups.com&g=
t;
:

0 .8ex;border-left:1px #ccc
solid;padding-left:1ex">

=C2=A0
id=3D"m_2995809254753611828m_-44462266975892612=
18ygrp-mlmsg">
id=3D"m_2995809254753611828m_-444622669758926=
1218ygrp-msg">
id=3D"m_2995809254753611828m_-4446226697589=
261218ygrp-text">

First off, thanks everyone for the
helpful and encouraging feedback!
Thanks Joel for showing us that there
are 6 orbits in the 2^4 and for your
rederivation of the state count. And
thanks Matt and Roice for pointing out
the importance of the inverted views.
It looks so strange in that
configuration that I always want to
get back to a normal view as quickly
as possible, but it does seem equally
valid, and as you've shown, it can be
helpful for more than just finding
short sequences.



I don't understand Matt's "pinwheel"
configuration, but I will point out
that all that is needed to create your
twin interior octahedra is a single
half-rotation like I showed in the
video at href=3D"https://www.youtube.com/watch?v=
=3DzqftZ8kJKLo&t=3D5m29s"
target=3D"_blank">5:29
. The two
main halves do end up being mirror
images of each other on the visible
outside like he described. Whether
it's the pinwheel or the half-rotated
version that's correct, I'm not sure
that it's a bummer that the solved
state is not at all obvious, so long
as we can operate it in my original
configuration and ignore the fact that
the outer faces touch. That would just
mean that the "correct" view is
evidence that that the more
understandable view is legitimate.



I'm going to try to make a snapable V3
which should allow the pieces to be
more easily taken apart and
reassembled into other forms.
Shapeways does offer a single, clear
translucent plastic that they call
"Frosted Detail", and another called
"Transparent Acrylic", but I don't
think that any sort of transparent
stickers will help us, especially
since this thing is chock full of
magnets. The easiest way to let you
see into the two hemispheres would be
to simply truncate the pointy tips of
the stickers. That already happens a
little bit due to the way I've rounded
the edges. Here is a moz-do-not-send=3D"true"
href=3D"http://superliminal.com/cube/in=
verted1.jpg"
target=3D"_blank">close-up of a
half-rotation in which you can see
that the inner yellow and white faces
are solved. Your suggestion of little
mapping dots on the corners also
works, but just opening the existing
window further would work more
directly.



-Melinda




class=3D"m_2995809254753611828m_-44462266=
97589261218moz-cite-prefix">On
4/28/2017 2:15 PM, Roice Nelson moz-do-not-send=3D"true"
class=3D"m_2995809254753611828m_-4446226697589261218moz-txt-link-abbreviate=
d"
href=3D"mailto:roice3@gmail.com"
target=3D"_blank">roice3@gmail.com
[4D_Cubing] wrote:



I agree with
Don's arguments about adjacent
sticker colors needing to be next
to each other.=C2=A0 I think this can
be turned into an accurate 2^4
with coloring changes, so I agree
with Joel too :)



To help me think about it, I
started adding a new projection
option for spherical puzzles to
MagicTile, which takes the two
hemispheres of a puzzle and maps
them to two disks with
identified boundaries connected
at a point, just like a physical
" href=3D"http://www.pa-network.com=
/global-chess/indexf.html"
target=3D"_blank">global chess>"
game I have.=C2=A0 Melinda's puzzle
is a lot like this up a
dimension, so think about two
disjoint balls, each
representing a hemisphere of the
2^4, each a "subcube" of
Melinda's puzzle.=C2=A0 The two
boundaries of the balls are
identified with each other and
as you roll one around, the
other half rolls around so that
identified points connect up.=C2=A0
We need to have the same
restriction on Melinda's puzzle.iv>



In the pristine state then, I
think it'd be nice to have an
internal (hidden), solid colored
octahedron on each half.=C2=A0 The
other 6 faces should all have
equal colors split between each
hemisphere, 4 stickers on each
half.=C2=A0 You should be able to
reorient the two subcubes to
make a half octahedron of any
color on each subcube.=C2=A0 I just
saw Matt's email and picture,
and it looks like we were going
down the same thought path.=C2=A0 I
think with recoloring (mirroring
some of the current piece
colorings) though, the
windmill's can be avoided (?)





[...] After staring/thinking a
bit more, the coloring Matt came
up with is right-on if you want to
put a solid color at the center of
each hemisphere.=C2=A0 His=C2=A0comme=
nt
about the "mirrored" pieces on
each side helped me understand
better. =C2=A03 of the stickers are
mirrored and the 4th is the hidden
color (different on each side for
a given pair of "mirrored"
pieces).=C2=A0 All faces behave
identically as well, as they
should.=C2=A0 It's a little bit of a
bummer that it doesn't look very
pristine in the pristine state,
but it does look like it should
work as a 2^4.



I wonder if there might be some
adjustments to be made when
shapeways allows printing
translucent as a color :)
iv>



[...] Sorry
for all the streaming, but I
wanted to share one more thought.=C2=
=A0
I now completely agree with
Joel/Matt about it behaving as a
2^4, even with the original
coloring.=C2=A0 You just need to
consider the corner colors of the
two subcubes (pink/purple near the
end of the video) as being a
window into the interior of the
piece.=C2=A0 The other colors match u=
p
as desired. =C2=A0(Sorry if folks
already understood this after
their emails and I'm just catching
up!)



In fact, you could alter
the coloring of the pieces
slightly so that the behavior
was similar with the inverted
coloring.=C2=A0 At the corners
where 3 colors meet on each
piece, you could put a little
circle of color of the
opposite 4th color.=C2=A0 In Matt=
's
windmill coloring then, you'd
be able to see all four colors
of a piece, like you can with
some of the pieces on
Melinda's original coloring.=C2=
=A0
And again you'd consider the
color circles a window to the
interior that did not require
the same matching constraints
between the subcubes.




I'm looking forward to
having one of these :)




Happy Friday everyone,



class=3D"m_2995809254753611828h5"=
>
Roice





On Fri, Apr 28, 2017 at 1:14
AM, Joel Karlsson moz-do-not-send=3D"true"
href=3D"mailto:joelkarlsson97=
@gmail.com"
target=3D"_blank">joelkarlsso=
n97@gmail.com

[4D_Cubing] &=
lt; moz-do-not-send=3D"true"
href=3D"mailto:4D_Cubing@ya=
hoogroups.com"
target=3D"_blank">4D_Cubing=
@yahoogroups.com>

wrote:

class=3D"gmail_quote"
style=3D"border-left:1px
#ccc solid">





Seems like there
was a slight
misunderstanding. I
meant that you need
to be able to =C2=A0twi=
st
one of the faces and
in MC4D the most
natural choice is
the center face. In
your physical puzzle
you can achieve this
type of twist by
twisting the two
subcubes although
this is indeed a
twist of the
subcubes themselves
and not the center
face, however, this
is still the same
type of twist just
around another
face.=C2=A0




If the
magnets are that
allowing the 2x2x2x2
is obviously a
subgroup of this
puzzle. Hopefully
the restrictions
will be quite
natural and only
some "strange" moves
would be illegal.
Regarding the
"families of states"
(aka orbits), the
2x2x2x2 has 6
orbits. As I
mentioned earlier
all allowed twists
preserves the parity
of the pieces,
meaning that only
half of the
permutations you can
achieve by
disassembling and
reassembling can be
reached through
legal moves. Because
of some geometrical
properties of the
2x2x2x2 and its
twists, which would
take some time to
discuss in detail
here, the
orientation of the
stickers mod 3 are
preserved, meaning
that the last corner
only can be oriented
in one third of the
number of
orientations for the
other corners. This
gives a total number
of orbits of 2x3=3D6.
To check this result
let's use this
information to
calculate all the
possible states of
the 2x2x2x2; if
there were no
restrictions we
would have 16! for
permuting the pieces
(16 pieces) =C2=A0and
12^16 for orienting
them (12
orientations for
each corner). If we
now take into
account that there
are 6 equally sized
orbits this gets us
to 12!16^12/6.
However, we should
also note that the
orientation of the
puzzle as a hole is
not set by some kind
of centerpieces and
thus we need to
devide with the
number of
orientations of a 4D
cube if we want all
our states to be
separated with
twists and not only
rotations of the
hole thing. The
number of ways to
orient a 4D cube in
space (only allowing
rotations and not
mirroring) is
8x6x4=3D192 giving a
total of=C2=A0 style=3D"font-family:=
sans-serif">12!16^12/(6*192)
states which is
indeed the same
number that for
example David
Smith arrived at
during his
calculations.
Therefore, =C2=A0when
determining
whether or not a
twist on your
puzzle is legal or
not it is
sufficient and
necessary to
confirm that the
twist is an even
permutation of the
pieces and
preserves the
orientation of
stickers mod 3.>

style=3D"font-family:sans-serif">


style=3D"font-family:sans-serif">Best regards,=C2=A0

style=3D"font-family:sans-serif">Joel=C2=A0



class=3D"gmail_extr=
a"
dir=3D"auto">

class=3D"gmail_qu=
ote">Den
28 apr. 2017
3:02 fm skrev
"Melinda Green
moz-do-not-send=
=3D"true"
href=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlim=
inal.com
[4D_Cubing]"
< moz-do-not-send=
=3D"true"
href=3D"mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoo=
groups.com>:type=3D"attribution">
class=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_=
7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632=
030069098quote"
style=3D"border-left:1px #ccc solid">
style=3D"backgrou=
nd-color:#fff">
=C2=A0n>
id=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_792=
7561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030=
069098m_6586884675826257012ygrp-mlmsg">
id=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_792=
7561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030=
069098m_6586884675826257012ygrp-msg">
id=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_792=
7561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030=
069098m_6586884675826257012ygrp-text">

The new
arrangement of
magnets allows
every valid
orientation of
pieces. The
only invalid
ones are those
where the
diagonal lines
cutting each
cube's face
cross each
other rather
than coincide.
In other
words, you can
assemble the
puzzle in all
ways that
preserve the
overall
diamond/harlequin
pattern. Just
about every
move you can
think of on
the whole
puzzle is
valid though
there are
definitely
invalid moves
that the
magnets allow.
The most
obvious
invalid move
is twisting of
a single end
cap.



I think your
description of
the center
face is not
correct
though. Twists
of the outer
faces cause
twists
"through" the
center face,
not "of" that
face. Twists
of the outer
faces are
twists of
those faces
themselves
because they
are the ones
not changing,
just like the
center and
outer faces of
MC4D when you
twist the
center face.
The only
direct twist
of the center
face that this
puzzle allows
is a 90 degree
twist about
the outer
axis. That
happens when
you
simultaneously
twist both end
caps in the
same
direction.



Yes, it's
quite
straightforward
reorienting
the whole
puzzle to put
any of the
four axes on
the outside.
This is a very
nice
improvement
over the first
version and
should make it
much easier to
solve. You may
be right that
we just need
to find the
right way to
think about
the outside
faces. I'll
leave it to
the math
geniuses on
the list to
figure that
out.



-Melinda


class=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_=
7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632=
030069098quoted-text">



class=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_=
7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632=
030069098m_6586884675826257012moz-cite-prefix">On
4/27/2017
10:31 AM, Joel
Karlsson moz-do-not-send=
=3D"true"
class=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_=
7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632=
030069098m_6586884675826257012moz-txt-link-abbreviated"
href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@g=
mail.com
[4D_Cubing]
wrote:

type=3D"cite">>



Hi
Melinda,




I do not agree
with the
criticism
regarding the
white and
yellow
stickers
touching each
other, this
could simply
be an effect
of the
different
representations
of the puzzle.
To really
figure out if
this indeed is
a
representation
of a 2x2x2x2
we need to
look at the
possible moves
(twists and
rotations) and
figure out the
equivalent
moves in the
MC4D software.
From the MC4D
software, it's
easy to
understand
that the only
moves required
are free
twists of one
of the faces
(that is, only
twisting the
center face in
the standard
perspective
projection in
MC4D) and 4D
rotations
swapping which
face is in the
center
(ctrl-clicking
in MC4D). The
first is
possible in
your physical
puzzle by
rotating the
white and
yellow
subcubes (from
here on I use
subcube to
refer to the
two halves of
the puzzle and
the colours of
the subcubes
to refer to
the "outer
colours"). The
second is
possible if
it's possible
to reach a
solved state
with any two
colours on the
subcubes that
still allow
you to perform
the previously
mentioned
twists. This
seems to be
the case from
your
demonstration
and is indeed
true if the
magnets allow
the simple
twists
regardless of
the colours of
the subcubes.
Thus, it is
possible to
let your
puzzle be a
representation
of a 2x2x2x2,
however, it
might require
that some
moves that the
magnets allow
aren't used.




Best regards,


Joel


class=3D"gmail_ex=
tra">

class=3D"gmail_qu=
ote">2017-04-27
3:09 GMT+02:00
Melinda Green
moz-do-not-send=
=3D"true"
href=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlim=
inal.com
[4D_Cubing] dir=3D"ltr"><<=
a
moz-do-not-send=3D"true" href=3D"mailto:4D_Cubing@yahoogroups.com"
target=3D"_blank"=
>4D_Cubing@yahoogroups.com>
:

class=3D"gmail_qu=
ote"
style=3D"border-left:1px #ccc solid">
style=3D"backgrou=
nd-color:#fff">
=C2=A0n>
id=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_792=
7561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030=
069098m_6586884675826257012m_3209267269419320979ygrp-mlmsg">
id=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_792=
7561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030=
069098m_6586884675826257012m_3209267269419320979ygrp-msg">
id=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_792=
7561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632030=
069098m_6586884675826257012m_3209267269419320979ygrp-text">

Dear
Cubists,



I've finished
version 2 of
my physical
puzzle and
uploaded a
video of it
here:

moz-do-not-send=
=3D"true"
href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJKLo" target=3D"_blank">ht=
tps://www.youtube.com/watch?v=3DzqftZ8kJKLo

Again, please
don't share
these videos
outside this
group as their
purpose is
just to get
your feedback.
I'll
eventually
replace them
with a public
video.



Here is an
extra math
puzzle that I
bet you folks
can answer:
How many
families of
states does
this puzzle
have? In other
words, if
disassembled
and
reassembled in
any random
configuration
the magnets
allow, what
are the odds
that it can be
solved? This
has practical
implications
if all such
configurations
are solvable
because it
would provide
a very easy
way to fully
scramble the
puzzle.



And finally, a
bit of fun: A
relatively new
friend of mine
and new list
member, Marc
Ringuette, got
excited enough
to make his
own version.
He built it
from EPP foam
and colored
tape, and used
honey instead
of magnets to
hold it
together.
Check it out
here: moz-do-not-send=
=3D"true"
href=3D"http://superliminal.com/cube/dessert_cube.jpg" target=3D"_blank">ht=
tp://superliminal.com/cube/dessert_cube.jpg
I don't know
how practical
a solution
this is but it
sure looks
delicious!
Welcome Marc!



-Melinda


















class=3D"m_2995809254753611828m_-4446226697589261218m_6733134459512634882m_=
7927561327128665054m_-4244333051891278538m_-7761579982835215532m_4431900632=
030069098quoted-text">












































=20=20=20=20=20=20







--------------0B3ABC49A067D6A587EBCEAB--




From: Melinda Green <melinda@superliminal.com>
Date: Thu, 18 May 2017 18:33:05 -0700
Subject: Re: [MC4D] Physical 4D puzzle V2



--------------DC3048547D20144DBFC49280
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: quoted-printable

Hello Joel,

Thanks for drilling into this puzzle. Finding good ways to discuss and thin=
k about moves and representations will be key. I'll comment on some details=
in-line.

On 5/14/2017 6:16 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing] wr=
ote:
>
> Yes, that is correct and in fact, you should divide not only with 24 for =
the orientation but also with 16 for the placement if you want to calculate=
unique states (since the 2x2x2x2 doesn't have fixed centerpieces). The poi=
nt, however, was that if you don=E2=80=99t take that into account you get a=
factor of 24*16=3D384 (meaning that the puzzle has 384 representations of =
every unique state) instead of the factor of 192 which you get when calcula=
ting the states from the virtual puzzle and hence every state of the virtua=
l puzzle has two representations in the physical puzzle. Yes exactly, they =
are indeed the same solved (or other) state and you are correct that the ha=
lf rotation (taking off a 2x2 layer and placing it at the other end of the =
puzzle) takes you from one representation to the same state with the other =
representation. This means that the restacking move (taking off the front 2=
x4 layer and placing it behind the other 2x4 layer) can be expressed with h=
alf-rotations and=20
> ordinary twists and rotations (which you might have pointed out already).

Yes, I made that claim in the video but didn't show it because I have yet t=
o record such a sequence. I've only stumbled through it a few times. I talk=
ed about it at 5:53 s> though I mistakenly called it a twist, when I should have called it a se=
quence.

>
> I think I've found six moves including ordinary twists and a restacking m=
ove that is identical to a half-rotation and thus it's easy to compose a re=
stacking move with one half-rotation and five ordinary twists. There might =
be an error since I've only played with the puzzle in my mind so it would b=
e great if you, Melinda, could confirm this (the sequence is described late=
r in this email).

You mean "RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-"? Yes, that works. There =
do seem to be easier ways to do that beginning with an ordinary rolling rot=
ation. I don't see those in your notation, but the equivalent using a pair =
of twists would be Rx Lx' Sx Vy if I got that right.

>
> To be able to communicate move sequences properly we need notation for re=
presenting twists, rotations, half-rotations, restacking moves and folds. F=
eel free to come with other suggestion but you can find mine below. Please =
read the following thoroughly (maybe twice) to make sure that you understan=
d everything since misinterpreted notation could potentially become a night=
mare and feel free to ask questions if there is something that needs clarif=
ication.
> *
> Coordinate system and labelling:*
>
> Let's introduce a global coordinate system. In whatever state the puzzle =
is let the positive x-axis point upwards, the positive y-axis towards you a=
nd the positive z-axis to the right (note that this is a right-hand system)=
.

I see the utility of a global coordinate system, but this one seems rather =
non-standard. I suggest that X be to the right, and Y up since these are ne=
ar-universal standards. Z can be in or out. I have no opinion. If there is =
any convention in the twisty puzzle community, I'd go with that.

Note also that the wiki may be a good place to document and iterate on term=
inology, descriptions and diagrams. Ray added a "notation" section to the 3=
^4 page here , and I know that one=
other member was thinking of collecting a set of moves on another wiki pag=
e.

>
> Now let's name the 8 faces of the puzzle. The right face is denoted with =
R, the left with L, the top U (up), the bottom D (down), the front F, the b=
ack B, the center K (kata) and the last one A (ana). The R and L faces are =
either the outer corners of the right and left halves respectively or the i=
nner corners of these halves (forming octahedra) depending on the represent=
ation of the puzzle. The U, D, F and B faces are either two diamond shapes =
(looking something like this: <><>) on the corresponding side of the puzzle=
or one whole and two half diamond shapes (><><). The K face is either an o=
ctahedron in the center of the puzzle or the outer corners of the center 2x=
2x2 block. Lastly, the A face is either two diamond shapes, one on the righ=
t and one on the left side of the puzzle, or another shape that=E2=80=99s a=
little bit hard to describe with just a few words (the white stickers at 5=
:10 in the latest video, after the half-rotation but before the restacking =
move).
>

I think it's more correct to say that the K face is either an octahedron at=
the origin (A<>K<>A) or in the center of one of the main halves, with the =
A face inside the other half(>A<>K<). This was what I was getting at in my =
previous message. You do later talk about octahedral faces being in either =
the center or the two main halves, so this is just terminology. But about "=
the outer corners of the center 2x2x2 block", this cannot be the A or K fac=
e as you've labeled them. You've been calling these the L/R faces, but the =
left-right distinction disappears in the half-rotated state, so maybe "left=
" and "right" aren't the best names. To me, they are always the "outside" f=
aces, regardless. You can distinguish them as the left and right outside fa=
ces in one representation, or as the center and end outside faces in the ot=
her. (Or perhaps "end" versus "ind" if we want to be cute.)

I'm also a little torn about naming the interior faces ana and kata, not be=
cause of the names themselves which I like, but because the mysterious face=
s to me are the outermost ones you're calling R and L. It only requires a s=
imple rotation to move faces in and out of the interior (octahedral) positi=
ons, but it's much more difficult to move another axis into the L/R/outer d=
irection.

So maybe the directions can be

* Up-Down
* Front-Back
* Ind-End
* Ana-Kata

I'm not in love with it and will be happy with anything that works. Thought=
s anyone?

>
> We also need a name for normal 3D-rotations, restacking moves and folds (=
note that a half-rotation is a kind of restacking move). Let O be the name =
for a rOtation (note that the origin O doesn't move during a rotation, by t=
he way, these are 3D rotations of the physical puzzle), let S represent a r=
eStacking move and V a folding/clamshell move (you can remember this by thi=
nking of V as a folded line).
>

I think it's fine to call the clamshell move a fold or denote it as V. I ju=
st wouldn't consider it to be a basic move since it's a simple composite of=
3 basic twists as shown here A&t=3D19m40s>. In general, I think there are so many useful composite moves=
that we need to be able to easily make them up ad hoc with substitutions l=
ike *Let **=E2=86=93 =3D **Rx Lx'*. These are really macro moves which can =
be nested. That said, it's a particularly useful move so it's probably wort=
h describing in some formal way like you do in detail below.

>
>
> Further, let I (capital i) be the identity, preserving the state and rota=
tion of the puzzle. We cannot use I to indicate what moves should be perfor=
med but it's still useful as we will see later. Since we also want to be ab=
le to express if a sequence of moves is a rotation, preserving the state of=
the puzzle but possibly representing it in a different way, we can introdu=
ce mod(rot) (modulo rotation). So, if a move sequence P satisfies P mod(rot=
) =3D I, that means that the state of the puzzle is the same before and aft=
er P is performed although the rotation and representation of the puzzle ar=
e allowed to change. I do also want to introduce mod(3rot) (modulo 3D-rotat=
ion) and P mod(3rot) =3D I means that if the right 3D-rotation (a combinati=
on of O moves as we will see later) is applied to the puzzle after P you ge=
t the identity I. Moreover, let the standard rotation of the puzzle be any =
rotation such that the longer side is parallel with the z-axis, that is the=
puzzle forms a=20
> 2x2x4 (2 pieces thick in the x-direction, 2 in the y-direction and 4 in t=
he z-direction), and the K face is an octahedron.
> *
> Rotations and twists:
>
> *Now we can move on to name actual moves. The notation of a move is a com=
bination of a capital letter and a lowercase letter. O followed by x, y or =
z is a rotation of the whole puzzle around the corresponding axis in the ma=
thematical positive direction (counterclockwise/the way your right-hand fin=
gers curl if you point in the direction of the axis with your thumb). For e=
xample, Ox is a rotation around the x-axis that turns a 2x2x4 into a 2x4x2.=
A name of a face (U, D, F, B, R, L, K or A) followed by x, y or z means: d=
etach the 8 pieces that have a sticker belonging to the face and then turn =
those pieces around the global axis. For example, if the longer side of the=
puzzle is parallel to the z-axis (the standard rotation), Rx means: take t=
he right 2x2x2 block and turn it around the global x-axis in the mathematic=
al positive direction. Note that what moves are physically possible and all=
owed is determined by the rotation of the puzzle (I will come back to this =
later). Further=20
> note that Ox mod(rot) =3D Oy mod(rot) =3D Oz mod(rot) =3D I, meaning that=
the 3D-rotations of the physical puzzle corresponds to a 4D-rotation of th=
e represented 2x2x2x2.
>

There are several, distinct types of rotations, none of which change the st=
ate of the puzzle, and I think we need a way to be unambiguous about them. =
The types I see are

1. Simple reorientation of the physical puzzle in the hand, no magnets inv=
olved. IE your 'O' moves and maybe analogous to mouse-dragging in MC4D?
2. Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-cl=
ick?
3. Half-rotations. Maybe analogous to a "face first" view? (ctrl-click on =
a 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by Cubie")
4. Whole-puzzle reorientations that move an arbitrary axis into the "outer=
" 2 faces. No MC4D analog.


> *Inverses and performing a move more than once*
>
> To mark that a move should be performed n times let's put ^n after it. Fo=
r convenience when writing and speaking let ' (prime) represent ^-1 (the in=
verse) and n represent ^n. The inverse P' of some permutation P is the perm=
utation that satisfies P P' =3D P' P =3D I (the identity). For example (Rx)=
' means: do Rx backwards, which corresponds to rotating the right 2x2x2 blo=
ck in the mathematical negative direction (clockwise) around the x-axis and=
(Rx)2 means: perform Rx twice. However, we can also define powers of just =
the lowercase letters, for example, Rx2 =3D Rx^2 =3D Rxx =3D Rx Rx. So x2=
=3Dx^2 means: do whatever the capital letter specifies two times with respe=
ct to the x-axis. We can see that the capital letter naturally is distribut=
ed over the two lowercase letters. Rx' =3D Rx^-1 means: do whatever the cap=
ital letter specifies but in the other direction than you would have if the=
prime wouldn't have been there (note that thus, x'=3Dx^-1 can be seen as t=
he negative x-axis). Note that=20
> (Rx)^2 =3D Rx^2 and (Rx)' =3D Rx' which is true for all twists and rotati=
ons but that doesn't have to be the case for other types of moves (restacks=
and folds).
> *
> Restacking moves*
>
> A restacking move is an S followed by either x, y or z. Here the lowercas=
e letter specifies in what direction to restack. For example, Sy (from the =
standard rotation) means: take the front 8 pieces and put them at the back,=
whereas Sx means: take the top 8 pieces and put them at the bottom. Note t=
hat Sx is equivalent to taking the bottom 8 pieces and putting them at the =
top. However, if we want to be able to make half-rotations we sometimes nee=
d to restack through a plane that doesn't go through the origin. In the sta=
ndard rotation, let Sz be the normal restack (taking the 8 right pieces, th=
e right 2x2x2 block, and putting them on the left), Sz+ be the restack wher=
e you split the puzzle in the plane further in the positive z-direction (ta=
king the right 2x2x1 cap of 4 pieces and putting it at the left end of the =
puzzle) and Sz- the restack where you split the puzzle in the plane further=
in the negative z-direction (taking the left 2x2x1 cap of 4 pieces and put=
ting it at the=20
> right end of the puzzle). If the longer side of the puzzle is parallel to=
the x-axis instead, Sx+ would take the top 1x2x2 cap and put it on the bot=
tom. Note that in the standard rotation Sz mod(rot) =3D I. For restacks we =
see that (Sx)' =3D Sx' =3D Sx (true for y and z too of course), that (Sz+)'=
=3D Sz- and that (Sz+)2 =3D (Sz-)2 =3D Sz. We can define Sz'+ to have mean=
ing by thinking of z=E2=80=99 as the negative z-axis and with that in mind =
it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus, (Sz)=E2=80=99 =3D =
Sz=E2=80=99 and (Sz+)=E2=80=99 =3D Sz=E2=80=99+.
>
> *Fold moves*
>
> A fold might be a little bit harder to describe in an intuitive way. Firs=
t, let's think about what folds are interesting moves. The folds that canno=
t be expressed as rotations and restacks are unfolding the puzzle to a 4x4 =
and then folding it back along another axis. If we start with the standard =
rotation and unfold the puzzle into a 1x4x4 (making it look like a 4x4 from=
above) the only folds that will achieve something you can't do with a rest=
ack mod(rot) is folding it to a 2x4x2 so that the longer side is parallel w=
ith the y-axis after the fold. Thus, there are 8 interesting fold moves for=
any given rotation of the puzzle since there are 4 ways to unfold it to a =
4x4 and then 2 ways of folding it back that make the move different from a =
restack move mod(rot).

Assuming you complete a folding move in the same representation (<><> or ><=
><), then there are only two interesting choices. That's because it doesn't=
matter which end of a chosen cutting plane you open it from, the end resul=
t will be the same. That also means that any two consecutive clamshell move=
s along the same cutting plane will undo each other. It further suggests th=
at any interesting sequence of clamshell moves must alternate between the t=
wo possible long cut directions, meaning there is no choice involved. 12 cl=
amshell moves will cycle back to the initial state.

There is one other weird folding move where you open it in one direction an=
d then fold the two halves back-to-back in a different direction. If you si=
mply kept folding along the initial hinge, you'd simply have a restacking. =
When completed the other way, it's equivalent to a restacking plus a clamsh=
ell, so I don't think it's useful though it is somewhat interesting.

>
> Let's call these 8 folds interesting fold moves. Note that an interesting=
fold move always changes which axis the longer side of the puzzle is paral=
lel with. Further note that both during unfold and fold all pieces are move=
d; it would be possible to have 8 of the pieces fixed during an unfold and =
folding the other half 180 degrees but I think that it=E2=80=99s more intui=
tive that these moves fold both halves 90 degrees and performing them with =
180-degree folds might therefore lead to errors since the puzzle might get =
rotated differently. To illustrate a correct unfold without a puzzle: Put y=
our palms together such that your thumbs point upward and your fingers forw=
ard. Now turn your right hand 90 degrees clockwise and your left hand 90 de=
grees counterclockwise such that the normal to your palms point up, your fi=
ngers point forward, your right thumb to the right and your left thumb to t=
he left. That was what will later be called a Vx unfold and the folds are s=
imply reversed unfolds.=20
> (I might have used the word =E2=80=9Cfold=E2=80=9D in two different ways =
but will try to use the term =E2=80=9Cfold move=E2=80=9D when referring to =
the move composed by an unfold and a fold rather than simply calling these =
moves =E2=80=9Cfolds=E2=80=9D.)
>
> To specify the unfold let's use V followed by one of x, y, z, x', y' and =
z'. The lowercase letter describes in which direction to unfold. Vx means u=
nfold in the direction of the positive x-axis and Vx' in the direction of t=
he negative x-axis, if that makes any sense. I will try to explain more pre=
cisely what I mean with the example Vx from the standard rotation (it might=
also help to read the last sentences in the previous paragraph again). So,=
the puzzle is in the standard rotation and thus have the form 2x2x4 (x-, y=
- and z-thickness respectively). The first part (the unfolding) of the move=
specified with Vx is to unfold the puzzle in the x-direction, making it a =
1x4x4 (note that the thickness in the x-direction is 1 after the Vx unfold,=
which is no coincidence). There are two ways to do that; either the sides =
of the pieces that are initially touching another piece (inside of the puzz=
le in the x,z-plane and your palms in the hand example) are facing up or do=
wn after the=20
> unfold. Let Vx be the unfold where these sides point in the direction of =
the positive x-axis (up) and Vx' the other one where these sides point in t=
he direction of the negative x-axis (down) after the unfold. Note that if t=
he longer side of the puzzle is parallel to the z-axis only Vx, Vx', Vy and=
Vy' are possible. Now we need to specify how to fold the puzzle back to co=
mplete the folding move. Given an unfold, say Vx, there are only two ways t=
o fold that are interesting (not turning the fold move into a restack mod(r=
ot)) and you have to fold it perpendicular to the unfold to create an inter=
esting fold move. So, if you start with the standard rotation and do Vx you=
have a 1x4x4 that you have to fold into a 2x4x2. To distinguish the two po=
ssibilities, use + or - after the Vx. Let Vx+ be the unfold Vx followed by =
the interesting fold that makes the sides that are initially touching anoth=
er piece (before the unfold) touch another piece after the fold move is com=
pleted and let=20
> Vx- be the other interesting fold move that starts with the unfold Vx. (T=
hus, continuing with the hand example, if you want to do a Vx+ first do the=
Vx unfold described in the end of the previous paragraph and then fold you=
r hands such that your fingers point up, the normal to your palms point for=
ward, the right palm is touching the right-hand fingers, the left palm is t=
ouching the left-hand fingers, the right thumb is pointing to the right and=
the left thumb is pointing to the left). Note that the two halves of the p=
uzzle always should be folded 90 degrees each and you should never make a f=
old or unfold where you fold just one half 180-degrees (if you want to use =
my notation, that is). Further note that Vx+ Sx mod(3rot) =3D Vx- and that =
Vx+ Vx+ =3D I which is equivalent to (Vx+)=E2=80=99 =3D Vx+ and this is tru=
e for all fold moves (note that after a Vx+ another Vx+ is always possible)=
.
> *
> The 2x2x2x2 in the MC4D software*
>
> The notation above can also be applied to the 2x2x2x2 in the MC4D program=
. There, you are not allowed to do S or V moves but instead, you are allowe=
d to do the [crtl]+[left-click] moves. This can easily be represented with =
notation similar to the above. Let=E2=80=99s use C (as in Centering) and on=
e of x, y and z. For example, Cx would be to rotate the face in the positiv=
e x-direction aka the U face to the center. Thus, Cz' is simply [ctrl]+[lef=
t-click] on the L face and similarly for the other C moves. The O, U, D, F,=
B, R, L, K and A moves are performed in the same way as above so, for exam=
ple Rx would be a [left-click] on the top-side of the right face. In this r=
epresentation of the puzzle almost all moves are allowed; all U, D, F, B, R=
, L, K, O and C moves are possible regardless of rotation and only A moves =
(and of course the rightfully forbidden S and V moves) are impossible regar=
dless of rotation. Note that R and L moves in the software correspond to th=
e same moves of the=20
> physical puzzle but this is not generally true (I will come back to this =
later).
> *
> Possible moves (so far) in the standard rotation*
>
> In the standard rotation, the possible/allowed moves with the definitions=
above are:
>
> O moves, all of these are always possible in any state and rotation of th=
e puzzle since they are simply 3D-rotations.
>
> R, L moves, all of these as well since the puzzle has a right and left 2x=
2x2 block in the standard rotation.
>
> U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x4 blocks wi=
th less symmetry than a 2x2x2 block.
>
> F, B moves, just Fy2 and By2 since the front and back are 2x1x4 blocks wi=
th less symmetry than a 2x2x2 block.
>
> K moves, all possible since this is a rotation of the center 2x2x2 block.

I think the only legal 90 degree twists of the K face are those about the l=
ong axis. I believe this is what Christopher Locke was saying in this messa=
ge 45>. To see why there is no straightforward way to perform other 90 degree =
twists, you only need to perform a 90 degree twist on an outer (L/R) face a=
nd then reorient the whole puzzle along a different outer axis. If the orig=
inal twist was not about the new long axis, then there is clearly no straig=
htforward way to undo that twist.

>
> A moves, only Az moves since this is two 2x2x1 blocks that have to be rot=
ated together.
>
> S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined in the stan=
dard rotation.
>
> V moves, not Vz+ or Vz- since the definition doesn't give these meaning w=
hen the long side of the puzzle is parallel with the z-axis.
>
> Note that for example Fz2 is not allowed since this won't take you to a s=
tate of the puzzle. To allow more moves we need to extend the definitions (=
after the extension in the next paragraph all rotations and twists (O, R, L=
, U, D, F, B, K, A) are possible in any rotation and only which S and V mov=
es are possible depend on the state and rotation of the puzzle).
>
>
> *Extension of some definitions*
>
> It's possible to make an extension that allows all O, R, L, U, D, F, B, K=
and A moves in any state. I will explain how this can be done in the stand=
ard rotation but it applies analogously to any other rotation where the K f=
ace is an octahedron. First let's focus on U, D, F and B and because of the=
symmetry of the puzzle all of these are analogous so I will only explain o=
ne. The extension that makes all U moves possible (note that the U face is =
in the positive x-direction) is as follows: when making an U move first det=
ach the 8 top pieces which gives you a 1x2x4 block, fold this block into a =
2x2x2 block in the positive x-direction (similar to the later part of a Vx+=
move from the standard rotation) such that the U face form an octahedron, =
rotate this 2x2x2 block around the specified axis (for example around the z=
-axis if you are doing an Uz move), reverse the fold you just did creating =
a 1x2x4 block again and reattach the block.

I noticed something like this the other day but realized that it only seems=
to work for rotations along the long dimension (z in your example). These =
are already easily accomplished by a simple rotation to put the face in que=
stion on the end caps, followed by a double end-cap twist.

This is as far as I'm going to comment for the moment because the informati=
on gets very dense and I've been mulling and picking over your message for =
several days already. In short, I really like your attempt to provide a com=
plete system of notation for discussing this puzzle and will be curious to =
hear your thoughts on my comments so far. I hope others will chime in too.

One final thought is that a real "acid test" of any notation system for thi=
s puzzle will be attempt to translate some algorithms from MC4D. I would mo=
st like to see a sequence that flips a single piece, like the second 4-colo=
r series on this page or.htm> of Roice's solution, or his pair of twirled corners at the end of t=
his page . On=
e trick will be to minimize the number of whole-puzzle reorientations neede=
d, but really any sequence that works will be great evidence that the puzzl=
es are equivalent. I suspect that this sort of exercise will never be pract=
ical because it will require too many reorientations, and that entirely new=
methods will be needed to actually solve this puzzle.

Best,
-Melinda

> The A moves can be done very similarly but after you have detached the tw=
o 2x2x1 blocks you don't fold them but instead you stack them similar to a =
Sz move, creating a 2x2x2 block with the A face as an octahedron in the mid=
dle and then reverse the process after you have rotated the block as specif=
ied (for example around the negative y-axis if you are doing an Ay' move). =
Note that these extended moves are closely related to the normal moves and =
for example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99 in the standard rotatio=
n and note that (Ry Ly=E2=80=99) mod(rot) =3D (Ly Ry=E2=80=99) mod(rot) =3D=
I (this applies to the other extended moves as well).
>
> If the cube is in the half-rotated state, where both the R and L faces ar=
e octahedra, you can extend the definitions very similarly. The only thing =
you have to change is how you fold the 2x4 blocks when performing a U, D, F=
or B move. Instead of folding the two 2x2 halves of the 2x4 into a 2x2 you=
have to fold the end 1x2 block 180 degrees such that the face forms an oct=
ahedron.
>
> These moves might be a little bit harder to perform, to me especially the=
A moves seems a bit awkward, so I don't know if it's good to use them or n=
ot. However, the A moves are not necessary if you allow Sz in the standard =
rotation (which you really should since Sz mod(rot) =3D I in the standard r=
otation) and thus it might not be too bad to use this extended version. The=
notation supports both variants so if you don=E2=80=99t want to use these =
extended moves that shouldn=E2=80=99t be a problem. Note that, however, for=
example Ux (physical puzzle) !=3D Ux (virtual puzzle) where !=3D means =E2=
=80=9Cnot equal to=E2=80=9D (more about legal/illegal moves later).
>
> *Generalisation of the notation*
>
> Let=E2=80=99s generalise the notation to make it easier to use and to mak=
e it work for any n^4 cube in the MC4D software. Previously, we saw that (R=
x)2 =3D Rx Rx and if we allow ourselves to rewrite this as (Rx)2 =3D Rx2 =
=3D Rxx =3D Rx Rx we see that the capital letter naturally can be distribut=
ed over the lowercase letters. We can make this more general and say that a=
ny capital letter followed by several lowercase letters means the same thin=
g as the capital letter distributed over the lowercase letters. Like Rxyz =
=3D Rx Ry Rz and here R can be exchanged with any capital letter and xyz ca=
n be exchanged with any sequence of lowercase letters. We can also allow se=
veral capital letters and one lowercase letter, for example RLx and let=E2=
=80=99s define this as RLx =3D Rx Lx so that the lowercase letter can be di=
stributed over the capital letters. We can also define a capital letter fol=
lowed by =E2=80=98 (prime) like R=E2=80=99x =3D Rx=E2=80=99 and R=E2=80=99x=
y =3D Rx=E2=80=99y=E2=80=99 so the =E2=80=98 (prime) is distributed over th=
e lowercase letters. Note that=20
> we don=E2=80=99t define a capital to any other power than -1 like this si=
nce for example R2x =3D RRx might seem like a good idea at first but it isn=
=E2=80=99t very useful since R2 and RR are the same lengths (and powers gre=
ater than two are seldom used) and we will see that we can define R2 in ano=
ther way that generalises the notation to all n^4 cubes.
>
> Okay, let=E2=80=99s define R2 and similar moves now and have in mind what=
moves we want to be possible for a n^4 cube. The moves that we cannot achi=
eve with the notation this far is twisting deeper slices. To match the nota=
tion with the controls of the MC4D software let R2x be the move similar to =
Rx but twisting the 2^nd layer instead of the top one and similarly for oth=
er capital letters, numbers (up to n) and lowercase letters. Thus, R2x is p=
erformed as Rx but holding down the number 2 key. Just as in the program, w=
hen no number is specified 1 is assumed and you can combine several numbers=
like R12x to twist both the first and second layer. This notation does not=
apply to rotations (O) folding moves (V) and restacking moves (S) (I suppo=
se you could redefine the S move using this deeper-slice-notation and use S=
1z as Sz+, S2z as Sz and S3z as Sz- but since these moves are only allowed =
for the physical 2x2x2x2 I think that the notation with + and =E2=80=93 is =
better since S followed by a=20
> lowercase letter without +/- always means splitting the cube in a coordin=
ate plane that way, not sure though so input would be great). The direction=
of the twist R3x should be the same as Rx meaning that if Rx takes sticker=
s belonging to K and move them to F, so should R3x, in accordance with the =
controls of the MC4D software. Note that for a 3x3x3x3 it=E2=80=99s true th=
at R3z =3D Lz whereas R3x =3D Lx=E2=80=99 (note that R and L are the faces =
in the z-directions so because of the symmetry of the cube it will also be =
true that for example U3x =3D Bx whereas U3y=3DBy=E2=80=99).
>
> What about the case with several capital letters and several lowercase le=
tters, for instance, RLxy? I see two natural definitions of this. Either, w=
e could have RLxy =3D Rxy Lxy or we could define it as RLxy =3D RLx RLy. Th=
ese are generally not the same (if you exchange R and L with any allowed ca=
pital letter and similarly for x and y). I don=E2=80=99t know what is best,=
what do you think? The situations I find this most useful in are RL=E2=80=
=99xy to do a rotation and RLxy as a twist. However, since R and L are oppo=
site faces their operations commute which imply RL=E2=80=99xy(1^st definiti=
on) =3D Rxy L=E2=80=99xy =3D Rx Ry Lx=E2=80=99 Ly=E2=80=99 =3D Rx Lx=E2=80=
=99 Ry Ly=E2=80=99 =3D RL=E2=80=99x RL=E2=80=99y =3D RL=E2=80=99xy (2^nd de=
finition) and similarly for the other case with RLxy. Hopefully, we can fin=
d another useful sequence of moves where this notation can be used with onl=
y one of the definitions and can thereby decide which definition to use. Pe=
rsonally, I feel like RLxy =3D RLx RLy is the more intuitive definition but=
I don=E2=80=99t have any good argument for this so=20
> I=E2=80=99ll leave the question open.
>
> For convenience, it might be good to be able to separate moves like Rxy a=
nd RLx from the basic moves Rx, Oy etc when speaking and writing. Let=E2=80=
=99s call the basic moves that only contain one capital letter and one lowe=
rcase letter (possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or a numbe=
r) _simple moves_ (like Rx, L'y, Ux2 and D3y=E2=80=99) and the moves that c=
ontain more than one capital letter or more than one lowercase letter _comp=
osed moves_.
>
> *More about inverses*
>
> This list can obviously be made longer but here are some identities that =
are good to know and understand. Note that R, L, U, x, y and z below just a=
re examples, the following is true in general for non-folds (however, S mov=
es are fine).
>
> (P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99 P1=E2=80=99=
(Pi is an arbitrary permutation for i=3D1,2,=E2=80=A6n)
> (Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx
> (RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x
> (Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+(just as an example with rest=
acking moves, note that the inverse doesn=E2=80=99t change the + or -)
> RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=80=99xyz(tru=
e for both definitions)
> (RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=80=99yx(tru=
e for both definitions)
>
> For V moves we have that: (Vx+)=E2=80=99 =3D Vx+!=3D Vx=E2=80=99+(!=3D me=
ans =E2=80=9Cnot equal to=E2=80=9D)
>
> *Some important notes on legal/illegal moves
>
> *Although there are a lot of moves possible with this notation we might n=
ot want to use them all. If we really want a 2x2x2x2 and not something else=
I think that we should try to stick to moves that are legal 2x2x2x2 moves =
as far as possible (note that I said legal moves and not permutations (a le=
gal permutation can be made up of one or more legal moves)). Clarification:=
cycling three of the edge-pieces of a Rubik=E2=80=99s cube is a legal perm=
utation but not a legal move, a legal move is a rotation of the cube or a t=
wist of one of the layers. In this section I will only address simple moves=
and simply refer to them as moves (legal composed moves are moves composed=
by legal simple moves).
>
> I do believe that all moves allowed by my notation are legal permutations=
based on their periodicity (they have a period of 2 or 4 and are all even =
permutations of the pieces). So, which of them correspond to legal 2x2x2x2 =
moves? The O moves are obviously legal moves since they are equal to the id=
entity mod(rot). The same goes for restacking (S) (with or without +/-) in =
the direction of the longest side of the puzzle (Sz, Sz+ and Sz- in the sta=
ndard rotation) since these are rotations and half-rotations that don=E2=80=
=99t change the state of the puzzle. Restacking in the other directions and=
fold moves (V) are however not legal moves since they are made of 8 2-cycl=
es and change the state of the puzzle (note that they, however, are legal p=
ermutations). The rest of the moves (R, L, F, B, U, D, K and A) can be divi=
ded into two sets: (1) the moves where you rotate a 2x2x2 block with an oct=
ahedron inside and (2) the moves where you rotate a 2x2x2 block without an =
octahedron inside. A=20
> move belonging to (2) is always legal. We can see this by observing what =
a Rx does with the pieces in the standard rotation with just K forming an o=
ctahedron. The stickers move in 6 4-cycles and if the puzzle is solved the =
U and D faces still _looks_ solved after the move. A move belonging to set =
(1) is legal either if it=E2=80=99s an 180-degree twist or if it=E2=80=99s =
a rotation around the axis parallel with the longest side of the puzzle (th=
e z-axis in the standard rotation). Quite interestingly these are exactly t=
he moves that don=E2=80=99t mix up the R and L stickers with the rest in th=
e standard rotation. I think I know a way to prove that no legal 2x2x2x2 mo=
ve can mix up these stickers with the rest and this has to do with the fact=
that these stickers form an inverted octahedron (with the corners pointing=
outward) instead of a normal octahedron (let=E2=80=99s call this hypothesi=
s * for now). Note that all legal twists (R, L, F, B, U, D, K and A) of the=
physical puzzle correspond to the same twist=20
> in the MC4D software.
>
> So, what moves should we add to the set of legal moves be able to get to =
every state of the 2x2x2x2? I think that we should add the restacking moves=
and folding moves since Melinda has already found a pretty short sequence =
of those moves to make a rotation that changes which colours are on the R a=
nd L faces. That sequence, starting from the standard rotation, is: Oy Sx+z=
Vy+ Ozy=E2=80=99 Vy+ Ozy=E2=80=99 Vy+ =3D Oy Sx+z(Vy+ Ozy=E2=80=99)3 mod(3=
rot) =3D I mod(rot) (hopefully I got that right). What I have found (which =
I mentioned previously) is that (from the standard rotation): RLx Ly2 Sy By=
2 Ly2 RLx=E2=80=99 =3D Sz- and since this is equivalent with Sy =3D Ly2 RLx=
=E2=80=99 Sz- RLx Ly2 By2 the restacking moves that are not legal moves are=
not very complicated permutations and therefore I think that we can accept=
them since they help us mix up the R and L stickers with other faces. In a=
sense, the folding moves are =E2=80=9Cmore illegal=E2=80=9D since they can=
not be composed by the legal moves (according to hypothesis *). This is als=
o=20
> true for the illegal moves belonging to set (1) discussed above. However,=
since the folding moves is probably easier to perform and is enough to rea=
ch every state of the 2x2x2x2 I think that we should use them and not the i=
llegal moves belonging to (1). Note, once again, that all moves described b=
y the notation are legal permutations (even the ones that I just a few word=
s ago referred to as illegal moves) so if you wish you can use all of them =
and still only reach legal 2x2x2x2 states. However (in a strict sense) one =
could argue that you are not solving the 2x2x2x2 if you use illegal moves. =
If you only use illegal moves to compose rotations (that is, create a permu=
tation including illegal moves that are equal to I mod(rot)) and not actual=
ly using the illegal moves as twists I would classify that as solving a 2x2=
x2x2. What do you think about this?
>
> *What moves to use?
>
> *Here=E2=80=99s a short list of the simple moves that I think should be u=
sed for the physical 2x2x2x2. Note that this is just my thoughts and you ma=
y use the notation to describe any move that it can describe if you wish to=
. The following list assumes that the puzzle is in the standard rotation bu=
t is analogous for other representations where the K face is an octahedron.
>
> O, all since they are I mod(rot),
> R, L, all since they are legal (note Rx (physical puzzle) =3D Rx (virtual=
2x2x2x2)),
> U, D, only x2 since these are the only legal easy-to-perform moves,
> F, B, only y2 since these are the only legal easy-to-perform moves,
> K, A, only z, z=E2=80=99 and z2 since these are the only legal easy-to-pe=
rform moves,
> S, at least z, z+ and z- since these are equal to I mod(rot),
> S, possibly x and y since these help us perform rotations and is easy to =
compose (not necessary to reach all states and not legal though),
> V, all 8 allowed by the rotation of the puzzle (at least one is necessary=
to reach all states and if you allow one the others are easy to achieve an=
yway).
>
> If you start with the standard rotation and then perform Sz+ the followin=
g applies instead (this applies analogously to any other rotation where the=
R and L faces are octahedra).
>
> O, all since they are I mod(rot),
> R, L, only z, z=E2=80=99 and z2 since these are the only legal easy-to-pe=
rform moves,
> U, D, only x2 since these are the only legal easy-to-perform moves,
> F, B, only y2 since these are the only legal easy-to-perform moves,
> K, A, at least z, z=E2=80=99 and z2, possibly all (since they are legal) =
although some might be hard to perform.
> S, V, same as above.
>
> Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D I mod(rot) (=
!=3D for not equal) which implies that the Ux2 move when R and L are octahe=
dra is different from the Ux2 move when K is an octahedron. (Actually, the =
sequence above is equal to Uy2).
>
> Regarding virtual n^4 cubes I think that all O, C, R, L, U, D, F, B, K an=
d A moves should be used since they are all legal and really the only thing=
you need (left-clicking on an edge or corner piece in the computer program=
can be described quite easily with the notation, for example, Kzy2 is left=
-clicking on the top-front edge piece on the K face).
>
> I hope this was possible to follow and understand. Feel free to ask quest=
ions about the notation if you find anything ambiguous.
>
> Best regards,
> Joel Karlsson
>
> Den 4 maj 2017 12:01 fm skrev "Melinda Green melinda@superliminal.com ilto:melinda@superliminal.com> [4D_Cubing]" <4D_Cubing@yahoogroups.com lto:4D_Cubing@yahoogroups.com>>:
>
>
> Thanks for the correction. A couple of things: First, when assembling=
one piece at a time, I'd say there is only 1 way to place the first piece,=
not 24. Otherwise you'd have to say that the 1x1x1 puzzle has 24 states. I=
understand that this may be conventional, but to me, that just sounds sill=
y.
>
> Second, I have the feeling that the difference between the "two repre=
sentations" you describe is simply one of those half-rotations I showed in =
the video. In the normal solved state there is only one complete octahedron=
in the very center, and in the half-rotated state there is one in the midd=
le of each half of the "inverted" form. I consider them to be the same solv=
ed state.
>
> -Melinda
>
>
> On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com oelkarlsson97@gmail.com> [4D_Cubing] wrote:
>> Horrible typo... It seems like I made some typos in my email regardi=
ng the state count. It should of course be 16!12^16/(6*192) and NOT 12!16^1=
2/(6*192). However, I did calculate the correct number when comparing with =
previous results so the actual derivation was correct.
>>
>> Something of interest is that the physical pieces can be assembled i=
n 16!24*12^15 ways since there are 16 pieces, the first one can be oriented=
in 24 ways and the remaining can be oriented in 12 ways (since a corner wi=
th 3 colours never touch a corner with just one colour). Dividing with 6 to=
get a single orbit still gives a factor 2*192 higher than the actual count=
rather than 192. This shows that every state in the MC4D representation ha=
s 2 representations in the physical puzzle. These two representations must =
be the previously discussed, that the two halves either have the same color=
on the outermost corners or the innermost (forming an octahedron) when the=
puzzle is solved and thus both are complete representations of the 2x2x2x2=
.
>>
>> Best regards,
>> Joel Karlsson
>>
>> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" l.com >:
>>
>> I am no expert on group theory, so to better understand what twi=
sts are legal I read through the part of Kamack and Keane's /The Rubik Tess=
eract /about orienting the corners. Since all even permutations are allowed=
the easiest way to check if a twist is legal might be to:
>> 1. Check that the twist is an even permutation, that is: the sam=
e twist can be done by performing an even number of piece swaps (2-cycles).
>> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning p=
erforming the twist k times and I (the identity) representing the permutati=
on of doing nothing) and k is not divisible by 3 the twist A definitely doe=
sn't violate the restriction of the orientations since kx mod 3 =3D 0 and k=
mod 3 !=3D 0 implies x mod 3 =3D 0 meaning that the change of the total or=
ientation x for the twist A mod 3 is 0 (which precisely is the restriction =
of legal twists; that they must preserve the orientation mod 3).
>>
>> For instance, this implies that the restacking moves are legal 2=
x2x2x2 moves since both are composed of 8 2-cycles and both can be performe=
d twice (note that 2 is not divisible by 3) to obtain the identity.
>>
>> Note that 1 and 2 are sufficient to check if a twist is legal bu=
t only 1 is necessary; there can indeed exist a twist violating 2 that stil=
l is legal and in that case, I believe that we might have to study the orie=
ntation changes for that specific twist in more detail. However, if a twist=
can be composed by other legal twists it is, of course, legal as well.
>>
>> Best regards,
>> Joel
>>
>> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com=
[4D_Cubing] <4D_Cubing@yahoogroups.com <=
mailto:4D_Cubing@yahoogroups.com>>:
>>
>>
>> First off, thanks everyone for the helpful and encouraging f=
eedback! Thanks Joel for showing us that there are 6 orbits in the 2^4 and =
for your rederivation of the state count. And thanks Matt and Roice for poi=
nting out the importance of the inverted views. It looks so strange in that=
configuration that I always want to get back to a normal view as quickly a=
s possible, but it does seem equally valid, and as you've shown, it can be =
helpful for more than just finding short sequences.
>>
>> I don't understand Matt's "pinwheel" configuration, but I wi=
ll point out that all that is needed to create your twin interior octahedra=
is a single half-rotation like I showed in the video at 5:29 youtube.com/watch?v=3DzqftZ8kJKLo&t=3D5m29s>. The two main halves do end up=
being mirror images of each other on the visible outside like he described=
. Whether it's the pinwheel or the half-rotated version that's correct, I'm=
not sure that it's a bummer that the solved state is not at all obvious, s=
o long as we can operate it in my original configuration and ignore the fac=
t that the outer faces touch. That would just mean that the "correct" view =
is evidence that that the more understandable view is legitimate.
>>
>> I'm going to try to make a snapable V3 which should allow th=
e pieces to be more easily taken apart and reassembled into other forms. Sh=
apeways does offer a single, clear translucent plastic that they call "Fros=
ted Detail", and another called "Transparent Acrylic", but I don't think th=
at any sort of transparent stickers will help us, especially since this thi=
ng is chock full of magnets. The easiest way to let you see into the two he=
mispheres would be to simply truncate the pointy tips of the stickers. That=
already happens a little bit due to the way I've rounded the edges. Here i=
s a close-up of a half-rotatio=
n in which you can see that the inner yellow and white faces are solved. Yo=
ur suggestion of little mapping dots on the corners also works, but just op=
ening the existing window further would work more directly.
>>
>> -Melinda
>>
>>
>> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com roice3@gmail.com> [4D_Cubing] wrote:
>>> I agree with Don's arguments about adjacent sticker colors =
needing to be next to each other. I think this can be turned into an accur=
ate 2^4 with coloring changes, so I agree with Joel too :)
>>>
>>> To help me think about it, I started adding a new projectio=
n option for spherical puzzles to MagicTile, which takes the two hemisphere=
s of a puzzle and maps them to two disks with identified boundaries connect=
ed at a point, just like a physical "global chess m/global-chess/indexf.html>" game I have. Melinda's puzzle is a lot like th=
is up a dimension, so think about two disjoint balls, each representing a h=
emisphere of the 2^4, each a "subcube" of Melinda's puzzle. The two bounda=
ries of the balls are identified with each other and as you roll one around=
, the other half rolls around so that identified points connect up. We nee=
d to have the same restriction on Melinda's puzzle.
>>>
>>> In the pristine state then, I think it'd be nice to have an=
internal (hidden), solid colored octahedron on each half. The other 6 face=
s should all have equal colors split between each hemisphere, 4 stickers on=
each half. You should be able to reorient the two subcubes to make a half =
octahedron of any color on each subcube. I just saw Matt's email and pictur=
e, and it looks like we were going down the same thought path. I think wit=
h recoloring (mirroring some of the current piece colorings) though, the wi=
ndmill's can be avoided (?)
>>>
>>> [...] After staring/thinking a bit more, the coloring Matt =
came up with is right-on if you want to put a solid color at the center of =
each hemisphere. His comment about the "mirrored" pieces on each side helpe=
d me understand better. 3 of the stickers are mirrored and the 4th is the =
hidden color (different on each side for a given pair of "mirrored" pieces)=
. All faces behave identically as well, as they should. It's a little bit=
of a bummer that it doesn't look very pristine in the pristine state, but =
it does look like it should work as a 2^4.
>>>
>>> I wonder if there might be some adjustments to be made when=
shapeways allows printing translucent as a color :)
>>>
>>> [...] Sorry for all the streaming, but I wanted to share on=
e more thought. I now completely agree with Joel/Matt about it behaving as =
a 2^4, even with the original coloring. You just need to consider the corn=
er colors of the two subcubes (pink/purple near the end of the video) as be=
ing a window into the interior of the piece. The other colors match up as d=
esired. (Sorry if folks already understood this after their emails and I'm=
just catching up!)
>>>
>>> In fact, you could alter the coloring of the pieces slightl=
y so that the behavior was similar with the inverted coloring. At the corn=
ers where 3 colors meet on each piece, you could put a little circle of col=
or of the opposite 4th color. In Matt's windmill coloring then, you'd be a=
ble to see all four colors of a piece, like you can with some of the pieces=
on Melinda's original coloring. And again you'd consider the color circle=
s a window to the interior that did not require the same matching constrain=
ts between the subcubes.
>>>
>>> I'm looking forward to having one of these :)
>>>
>>> Happy Friday everyone,
>>> Roice
>>>
>>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson=
97@gmail.com [4D_Cubing] <4D_Cubing@yahoo=
groups.com > wrote:
>>>
>>>
>>>
>>> Seems like there was a slight misunderstanding. I meant=
that you need to be able to twist one of the faces and in MC4D the most n=
atural choice is the center face. In your physical puzzle you can achieve t=
his type of twist by twisting the two subcubes although this is indeed a tw=
ist of the subcubes themselves and not the center face, however, this is st=
ill the same type of twist just around another face.
>>>
>>> If the magnets are that allowing the 2x2x2x2 is obvious=
ly a subgroup of this puzzle. Hopefully the restrictions will be quite natu=
ral and only some "strange" moves would be illegal. Regarding the "families=
of states" (aka orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier =
all allowed twists preserves the parity of the pieces, meaning that only ha=
lf of the permutations you can achieve by disassembling and reassembling ca=
n be reached through legal moves. Because of some geometrical properties of=
the 2x2x2x2 and its twists, which would take some time to discuss in detai=
l here, the orientation of the stickers mod 3 are preserved, meaning that t=
he last corner only can be oriented in one third of the number of orientati=
ons for the other corners. This gives a total number of orbits of 2x3=3D6. =
To check this result let's use this information to calculate all the possib=
le states of the 2x2x2x2; if there were no restrictions we would have 16! f=
or permuting the
>>> pieces (16 pieces) and 12^16 for orienting them (12 or=
ientations for each corner). If we now take into account that there are 6 e=
qually sized orbits this gets us to 12!16^12/6. However, we should also not=
e that the orientation of the puzzle as a hole is not set by some kind of c=
enterpieces and thus we need to devide with the number of orientations of a=
4D cube if we want all our states to be separated with twists and not only=
rotations of the hole thing. The number of ways to orient a 4D cube in spa=
ce (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a tota=
l of 12!16^12/(6*192) states which is indeed the same number that for examp=
le David Smith arrived at during his calculations. Therefore, when determi=
ning whether or not a twist on your puzzle is legal or not it is sufficient=
and necessary to confirm that the twist is an even permutation of the piec=
es and preserves the orientation of stickers mod 3.
>>>
>>> Best regards,
>>> Joel
>>>
>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@s=
uperliminal.com [4D_Cubing]" <4D_Cubing@y=
ahoogroups.com >:
>>>
>>> The new arrangement of magnets allows every valid o=
rientation of pieces. The only invalid ones are those where the diagonal li=
nes cutting each cube's face cross each other rather than coincide. In othe=
r words, you can assemble the puzzle in all ways that preserve the overall =
diamond/harlequin pattern. Just about every move you can think of on the wh=
ole puzzle is valid though there are definitely invalid moves that the magn=
ets allow. The most obvious invalid move is twisting of a single end cap.
>>>
>>> I think your description of the center face is not =
correct though. Twists of the outer faces cause twists "through" the center=
face, not "of" that face. Twists of the outer faces are twists of those fa=
ces themselves because they are the ones not changing, just like the center=
and outer faces of MC4D when you twist the center face. The only direct tw=
ist of the center face that this puzzle allows is a 90 degree twist about t=
he outer axis. That happens when you simultaneously twist both end caps in =
the same direction.
>>>
>>> Yes, it's quite straightforward reorienting the who=
le puzzle to put any of the four axes on the outside. This is a very nice i=
mprovement over the first version and should make it much easier to solve. =
You may be right that we just need to find the right way to think about the=
outside faces. I'll leave it to the math geniuses on the list to figure th=
at out.
>>>
>>> -Melinda
>>>
>>>
>>>
>>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97=
@gmail.com [4D_Cubing] wrote:
>>>>
>>>> Hi Melinda,
>>>>
>>>> I do not agree with the criticism regarding the wh=
ite and yellow stickers touching each other, this could simply be an effect=
of the different representations of the puzzle. To really figure out if th=
is indeed is a representation of a 2x2x2x2 we need to look at the possible =
moves (twists and rotations) and figure out the equivalent moves in the MC4=
D software. From the MC4D software, it's easy to understand that the only m=
oves required are free twists of one of the faces (that is, only twisting t=
he center face in the standard perspective projection in MC4D) and 4D rotat=
ions swapping which face is in the center (ctrl-clicking in MC4D). The firs=
t is possible in your physical puzzle by rotating the white and yellow subc=
ubes (from here on I use subcube to refer to the two halves of the puzzle a=
nd the colours of the subcubes to refer to the "outer colours"). The second=
is possible if it's possible to reach a solved state with any two colours =
on the subcubes
>>>> that still allow you to perform the previously men=
tioned twists. This seems to be the case from your demonstration and is ind=
eed true if the magnets allow the simple twists regardless of the colours o=
f the subcubes. Thus, it is possible to let your puzzle be a representation=
of a 2x2x2x2, however, it might require that some moves that the magnets a=
llow aren't used.
>>>>
>>>> Best regards,
>>>> Joel
>>>>
>>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@su=
perliminal.com [4D_Cubing] <4D_Cubing@yah=
oogroups.com >:
>>>>
>>>> Dear Cubists,
>>>>
>>>> I've finished version 2 of my physical puzzle =
and uploaded a video of it here:
>>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo =

>>>> Again, please don't share these videos outside=
this group as their purpose is just to get your feedback. I'll eventually =
replace them with a public video.
>>>>
>>>> Here is an extra math puzzle that I bet you fo=
lks can answer: How many families of states does this puzzle have? In other=
words, if disassembled and reassembled in any random configuration the mag=
nets allow, what are the odds that it can be solved? This has practical imp=
lications if all such configurations are solvable because it would provide =
a very easy way to fully scramble the puzzle.
>>>>
>>>> And finally, a bit of fun: A relatively new fr=
iend of mine and new list member, Marc Ringuette, got excited enough to mak=
e his own version. He built it from EPP foam and colored tape, and used hon=
ey instead of magnets to hold it together. Check it out here: http://superl=
iminal.com/cube/dessert_cube.jpg .jpg> I don't know how practical a solution this is but it sure looks delic=
ious! Welcome Marc!
>>>>
>>>> -Melinda
>>>>
>>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>>


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Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable



">


Hello Joel,



Thanks for drilling into this puzzle. Finding good ways to discuss
and think about moves and representations will be key. I'll
comment on some details in-line.




On 5/14/2017 6:16 AM,
Joel Karlsson href=3D"mailto:joelkarlsson97@gmail.com">joelkarlsson97@gmail.com=

[4D_Cubing] wrote:



cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">





Yes, that is
correct and in fact, you should divide not only with 24
for the orientation but also with 16 for the placement if
you want to calculate unique states (since the 2x2x2x2
doesn't have fixed centerpieces). The point, however, was
that if you don=E2=80=99t take that into account you get a fa=
ctor
of 24*16=3D384 (meaning that the puzzle has 384
representations of every unique state) instead of the
factor of 192 which you get when calculating the states
from the virtual puzzle and hence every state of the
virtual puzzle has two representations in the physical
puzzle. Yes exactly, they are indeed the same solved (or
other) state and you are correct that the half rotation
(taking off a 2x2 layer and placing it at the other end of
the puzzle) takes you from one representation to the same
state with the other representation. This means that the n>restacking

move (taking off the front 2x4 layer and placing it behind
the other 2x4 layer) can be expressed with half-rotations
and ordinary twists and rotations (which you might have
pointed out already).






Yes, I made that claim in the video but didn't show it because I
have yet to record such a sequence. I've only stumbled through it
a few times. I talked about it at href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJKLo&t=3D5m53s=
">5:53
though I mistakenly called it a twist, when I should have called
it a sequence.




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">



I think I've
found six moves including ordinary twists and a restack=
ing

move that is identical to a half-rotation and thus it's
easy to compose a restacking move with one
half-rotation and five ordinary twists. There might be an
error since I've only played with the puzzle in my mind so
it would be great if you, Melinda, could confirm this (the
sequence is described later in this email).






You mean " style=3D"line-height:107%" lang=3D"EN-GB">RLx Ly2 Sy By2 Ly2 RLx=E2=
=80=99 =3D
Sz-"? Yes, that works. There do seem to be easier ways to do
that beginning with an ordinary rolling rotation. I don't see
those in your notation, but the equivalent using a pair of
twists would be Rx Lx' Sx Vy if I got that right.




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">

"
lang=3D"EN-GB">

To be able to communicate move sequences properly we need
notation for representing twists, rotations,
half-rotations, restacking moves and folds. Feel free to
come with other suggestion but you can find mine below.
Please read the following thoroughly (maybe twice) to make
sure that you understand everything since misinterpreted
notation could potentially become a nightmare and feel
free to ask questions if there is something that needs
clarification.



Coordinate system and labelling:
style=3D"line-height:107%" lang=3D"EN-GB">



Let's introduce a global coordinate system. In whatever
state the puzzle is let the positive x-axis point upwards,
the positive y-axis towards you and the positive z-axis to
the right (note that this is a right-hand system). >





I see the utility of a global coordinate system, but this one
seems rather non-standard. I suggest that X be to the right, and Y
up since these are near-universal standards. Z can be in or out. I
have no opinion. If there is any convention in the twisty puzzle
community, I'd go with that.



Note also that the wiki may be a good place to document and
iterate on terminology, descriptions and diagrams. Ray added a
"notation" section to the 3^4 page href=3D"http://wiki.superliminal.com/wiki/3%5E4">here, and I
know that one other member was thinking of collecting a set of
moves on another wiki page.




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">



Now let's name
the 8 faces of the puzzle. The right face is denoted with
R, the left with L, the top U (up), the bottom D (down),
the front F, the back B, the center K (kata) and the last
one A (ana). The R and L faces are either the
outer corners of the right and left halves respectively or
the inner corners of these halves (forming octahedra)
depending on the representation of the puzzle. The U, D, F
and B faces are either two diamond shapes (looking
something like this: <><>) on the
corresponding side of the puzzle or one whole and two half
diamond shapes (><><). The K face is either an
octahedron in the center of the puzzle or the outer
corners of the center 2x2x2 block. Lastly, the A face is
either two diamond shapes, one on the right and one on the
left side of the puzzle, or another shape that=E2=80=99s a li=
ttle
bit hard to describe with just a few words (the white
stickers at 5:10 in the latest video, after the
half-rotation but before the restacking move).








I think it's more correct to say that the K face is either an
octahedron at the origin (A<>K<>A) or in the center of
one of the main halves, with the A face inside the other half style=3D"line-height:107%" lang=3D"EN-GB"> (>A<>K<).
This was what I was getting at in my previous message. You do
later talk about octahedral faces being in either the center or
the two main halves, so this is just terminology. But about "> style=3D"line-height:107%" lang=3D"EN-GB"> style=3D"line-height:107%" lang=3D"EN-GB">the outer corners of th=
e
center 2x2x2 block", this cannot be the A or K face as you've
labeled them. You've been calling these the L/R faces, but the
left-right distinction disappears in the half-rotated state,
so maybe "left" and "right" aren't the best names. To me, they
are always the "outside" faces, regardless. You can
distinguish them as the left and right outside faces in one
representation, or as the center and end outside faces in the
other. (Or perhaps "end" versus "ind" if we want to be cute.)an>

style=3D"line-height:107%" lang=3D"EN-GB">=C2=A0>
I'm also a little
torn about naming the interior faces ana and kata, not because
of the names themselves which I like, but because the mysterious
faces to me are the outermost ones you're calling R and L. It
only requires a simple rotation to move faces in and out of the
interior (octahedral) positions, but it's much more difficult to
move another axis into the L/R/outer direction.



So maybe the directions can be



  • >Up-Down

  • >Front-Back

  • >Ind-End

  • >Ana-Kata


I'm n=
ot
in love with it and will be happy with anything that works.
Thoughts anyone?





cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">



We also need a
name for normal 3D-rotations, restacking moves and folds
(note that a half-rotation is a kind of restackingn>
move). Let O be the name for a rOtation (note that the
origin O doesn't move during a rotation, by the way, these
are 3D rotations of the physical puzzle), let S represent
a reStacking move and V a folding/clamshell move (you can
remember this by thinking of V as a folded line).








I think it's fine to call the clamshell move a fold or denote it
as V. I just wouldn't consider it to be a basic move since it's a
simple composite of 3 basic twists as shown href=3D"https://www.youtube.com/watch?v=3DAsx653BGDWA&t=3D19m40=
s">here.
In general, I think there are so many useful composite moves that
we need to be able to easily make them up ad hoc with
substitutions like Let
=E2=86=93 =
=3D
size=3D"-1"> lang=3D"EN-GB">Rx Lx'. These are really
macro moves which can be nested. That said, it's a particularly
useful move so it's probably worth describing in some formal way
like you do in detail below.




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">





Further, let I (capital i) be the identity, preserving the
state and rotation of the puzzle. We cannot use I to
indicate what moves should be performed but it's still
useful as we will see later. Since we also want to be able
to express if a sequence of moves is a rotation,
preserving the state of the puzzle but possibly
representing it in a different way, we can introduce
mod(rot) (modulo rotation). So, if a move sequence P
satisfies P mod(rot) =3D I, that means that the state of the
puzzle is the same before and after P is performed
although the rotation and representation of the puzzle are
allowed to change. I do also want to introduce mod(3rot)
(modulo 3D-rotation) and P mod(3rot) =3D I
means that if the right 3D-rotation (a combination of O
moves as we will see later) is applied to the puzzle after
P you get the identity I. Moreover, let the standard
rotation of the puzzle be any rotation such that the
longer side is parallel with the z-axis, that is the
puzzle forms a 2x2x4 (2 pieces thick in the x-direction, 2
in the y-direction and 4 in the z-direction), and the K
face is an octahedron.



Rotations and twists:



Now we can move on to name actual moves. The notation
of a move is a combination of a capital letter and a
lowercase letter. O followed by x, y or z is a rotation of
the whole puzzle around the corresponding axis in the
mathematical positive direction (counterclockwise/the way
your right-hand fingers curl if you point in the direction
of the axis with your thumb). For example, Ox is a
rotation around the x-axis that turns a 2x2x4 into a
2x4x2. A name of a face (U, D, F, B, R, L, K or A)
followed by x, y or z means: detach the 8 pieces that have
a sticker belonging to the face and then turn those pieces
around the global axis. For example, if the longer side of
the puzzle is parallel to the z-axis (the standard
rotation), Rx means: take the right 2x2x2 block and turn
it around the global x-axis in the mathematical positive
direction. Note that what moves are physically possible
and allowed is determined by the rotation of the puzzle (I
will come back to this later). Further note that Ox
mod(rot) =3D Oy mod(rot) =3D Oz mod(rot) =3D I, meaning that =
the
3D-rotations of the physical puzzle corresponds to a
4D-rotation of the represented 2x2x2x2.








There are several, distinct types of rotations, none of which
change the state of the puzzle, and I think we need a way to be
unambiguous about them. The types I see are



  1. Simple reorientation of the physical puzzle
    in the hand, no magnets involved. IE your 'O' moves and maybe
    analogous to mouse-dragging in MC4D?

  2. Rolling of one 2x2x2 half against the other.
    Maybe analogous to ctrl-click?

  3. Half-rotations. Maybe analogous to a "face
    first" view? (ctrl-click on a 2-color piece of the 3^4 with
    the setting "Ctrl-Click Rotates: by Cubie")

  4. Whole-puzzle reorientations that move an
    arbitrary axis into the "outer" 2 faces. No MC4D analog.
    li>

style=3D"line-height:107%" lang=3D"EN-GB">

n>
cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">

"
lang=3D"EN-GB"> Inverses and performing a move more than
once




To mark that a move should be performed n times let's put
^n after it. For convenience when writing and speaking let
' (prime) represent ^-1 (the inverse) and n represent ^n.
The inverse P' of some permutation P is the permutation
that satisfies P P' =3D P' P =3D I (the identity). For exampl=
e
(Rx)' means: do Rx backwards, which corresponds to
rotating the right 2x2x2 block in the mathematical
negative direction (clockwise) around the x-axis and (Rx)2
means: perform Rx twice. However, we can also define
powers of just the lowercase letters, for example, Rx2 =3D
Rx^2 =3D Rxx =3D Rx Rx. So x2=3Dx^2 means: do whatever the
capital letter specifies two times with respect to the
x-axis. We can see that the capital letter naturally is
distributed over the two lowercase letters. Rx' =3D Rx^-1
means: do whatever the capital letter specifies but in the
other direction than you would have if the prime wouldn't
have been there (note that thus, x'=3Dx^-1 can be seen as
the negative x-axis). Note that (Rx)^2 =3D Rx^2 and (Rx)' =3D
Rx' which is true for all twists and rotations but that
doesn't have to be the case for other types of moves
(restacks and folds).



Restacking moves




A restacking move is an S followed by either x, y or z.
Here the lowercase letter specifies in what direction to
restack. For example, Sy (from the standard rotation)
means: take the front 8 pieces and put them at the back,
whereas Sx means: take the top 8 pieces and put them at
the bottom. Note that Sx is equivalent to taking the
bottom 8 pieces and putting them at the top. However, if
we want to be able to make half-rotations we sometimes
need to restack through a plane that doesn't go through
the origin. In the standard rotation, let Sz be the normal
restack (taking the 8 right pieces, the right 2x2x2 block,
and putting them on the left), Sz+ be the restack where
you split the puzzle in the plane further in the positive
z-direction (taking the right 2x2x1 cap of 4 pieces and
putting it at the left end of the puzzle) and Sz- the
restack where you split the puzzle in the plane further in
the negative z-direction (taking the left 2x2x1 cap of 4
pieces and putting it at the right end of the puzzle). If
the longer side of the puzzle is parallel to the x-axis
instead, Sx+ would take the top 1x2x2 cap and put it on
the bottom. Note that in the standard rotation Sz mod(rot)
=3D I. For restacks we see that (Sx)' =3D Sx' =3D Sx (true fo=
r y
and z too of course), that (Sz+)' =3D Sz- and that (Sz+)2 =3D
(Sz-)2 =3D Sz. We can define Sz'+ to have meaning by
thinking of z=E2=80=99 as the negative z-axis and with that i=
n
mind it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus,=
(Sz)=E2=80=99 =3D Sz=E2=80=99
and (Sz+)=E2=80=99 =3D Sz=E2=80=99+.



Fold
moves
lang=3D"EN-GB">



A fold might be a little bit harder to describe in an
intuitive way. First, let's think about what folds are
interesting moves. The folds that cannot be expressed as
rotations and restacks are unfolding the puzzle to a 4x4
and then folding it back along another axis. If we start
with the standard rotation and unfold the puzzle into a
1x4x4 (making it look like a 4x4 from above) the only
folds that will achieve something you can't do with a
restack mod(rot) is folding it to a 2x4x2 so that the
longer side is parallel with the y-axis after the fold.
Thus, there are 8 interesting fold moves for any given
rotation of the puzzle since there are 4 ways to unfold it
to a 4x4 and then 2 ways of folding it back that make the
move different from a restack move mod(rot).







Assuming you complete a folding move in the same representation
(<><> or ><><), then there are only two
interesting choices. That's because it doesn't matter which end of
a chosen cutting plane you open it from, the end result will be
the same. That also means that any two consecutive clamshell moves
along the same cutting plane will undo each other. It further
suggests that any interesting sequence of clamshell moves must
alternate between the two possible long cut directions, meaning
there is no choice involved. 12 clamshell moves will cycle back to
the initial state.



There is one other weird folding move where you open it in one
direction and then fold the two halves back-to-back in a different
direction. If you simply kept folding along the initial hinge,
you'd simply have a restacking. When completed the other way, it's
equivalent to a restacking plus a clamshell, so I don't think it's
useful though it is somewhat interesting.




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">



Let's call thes=
e
8 folds interesting fold moves. Note that an interesting
fold move always changes which axis the longer side of the
puzzle is parallel with. Further note that both during
unfold and fold all pieces are moved; it would be possible
to have 8 of the pieces fixed during an unfold and folding
the other half 180 degrees but I think that it=E2=80=99s more
intuitive that these moves fold both halves 90 degrees and
performing them with 180-degree folds might therefore lead
to errors since the puzzle might get rotated differently.
To illustrate a correct unfold without a puzzle: Put your
palms together such that your thumbs point upward and your
fingers forward. Now turn your right hand 90 degrees
clockwise and your left hand 90 degrees counterclockwise
such that the normal to your palms point up, your fingers
point forward, your right thumb to the right and your left
thumb to the left. That was what will later be called a Vx
unfold and the folds are simply reversed unfolds. (I might
have used the word =E2=80=9Cfold=E2=80=9D in two different wa=
ys but will
try to use the term =E2=80=9Cfold move=E2=80=9D when referrin=
g to the move
composed by an unfold and a fold rather than simply
calling these moves =E2=80=9Cfolds=E2=80=9D.)
g=3D"EN-GB">



To
specify the unfold let's use V followed by one of x, y, z,
x', y' and z'. The lowercase letter describes in which
direction to unfold. Vx means unfold in the direction of
the positive x-axis and Vx' in the direction of the
negative x-axis, if that makes any sense. I will try to
explain more precisely what I mean with the example Vx
from the standard rotation (it might also help to read the
last sentences in the previous paragraph again). So, the
puzzle is in the standard rotation and thus have the form
2x2x4 (x-, y- and z-thickness respectively). The first
part (the unfolding) of the move specified with Vx is to
unfold the puzzle in the x-direction, making it a 1x4x4
(note that the thickness in the x-direction is 1 after the
Vx unfold, which is no coincidence). There are two ways to
do that; either the sides of the pieces that are initially
touching another piece (inside of the puzzle in the
x,z-plane and your palms in the hand example) are facing
up or down after the unfold. Let Vx be the unfold where
these sides point in the direction of the positive x-axis
(up) and Vx' the other one where these sides point in the
direction of the negative x-axis (down) after the unfold.
Note that if the longer side of the puzzle is parallel to
the z-axis only Vx, Vx', Vy and Vy' are possible. Now we
need to specify how to fold the puzzle back to complete
the folding move. Given an unfold, say Vx, there are only
two ways to fold that are interesting (not turning the
fold move into a restack mod(rot)) and you have to fold it
perpendicular to the unfold to create an interesting fold
move. So, if you start with the standard rotation and do
Vx you have a 1x4x4 that you have to fold into a 2x4x2. To
distinguish the two possibilities, use + or - after the
Vx. Let Vx+ be the unfold Vx followed by the interesting
fold that makes the sides that are initially touching
another piece (before the unfold) touch another piece
after the fold move is completed and let Vx- be the other
interesting fold move that starts with the unfold Vx.
(Thus, continuing with the hand example, if you want to do
a Vx+ first do the Vx unfold described in the end of the
previous paragraph and then fold your hands such that your
fingers point up, the normal to your palms point forward,
the right palm is touching the right-hand fingers, the
left palm is touching the left-hand fingers, the right
thumb is pointing to the right and the left thumb is
pointing to the left). Note that the two halves of the
puzzle always should be folded 90 degrees each and you
should never make a fold or unfold where you fold just one
half 180-degrees (if you want to use my notation, that
is). Further note that Vx+ Sx mod(3rot) =3D Vx- and that Vx+
Vx+ =3D I which is equivalent to (Vx+)=E2=80=99 =3D Vx+ and t=
his is
true for all fold moves (note that after a Vx+ another Vx+
is always possible).



The 2x2x2x2 in the MC4D software
style=3D"line-height:107%" lang=3D"EN-GB">



The notation above can also be applied to the 2x2x2x2 in
the MC4D program. There, you are not allowed to do S or V
moves but instead, you are allowed to do the
[crtl]+[left-click] moves. This can easily be represented
with notation similar to the above. Let=E2=80=99s use C (as i=
n
Centering) and one of x, y and z. For example, Cx would be
to rotate the face in the positive x-direction aka the U
face to the center. Thus, Cz' is simply
[ctrl]+[left-click] on the L face and similarly for the
other C moves. The O, U, D, F, B, R, L, K and A moves are
performed in the same way as above so, for example Rx
would be a [left-click] on the top-side of the right face.
In this representation of the puzzle almost all moves are
allowed; all U, D, F, B, R, L, K, O and C moves are
possible regardless of rotation and only A moves (and of
course the rightfully forbidden S and V moves) are
impossible regardless of rotation. Note that R and L moves
in the software correspond to the same moves of the
physical puzzle but this is not generally true (I will
come back to this later).



Possible moves (so far) in the standard rotation




In the standard rotation, the possible/allowed moves with
the definitions above are:



O moves, all of these are always possible
in any state and rotation of the puzzle since they are
simply 3D-rotations.




R, L moves, all of these as well since
the puzzle has a right and left 2x2x2 block in the
standard rotation.




U. D moves, just Ux2 and Dx2 since the
top and bottom are 1x2x4 blocks with less symmetry than a
2x2x2 block.




F, B moves, just Fy2 and By2 since the
front and back are 2x1x4 blocks with less symmetry than a
2x2x2 block.




K moves, all possible since this is a
rotation of the center 2x2x2 block.





I think the only legal 90 degree twists of the K face are those
about the long axis. I believe this is what Christopher Locke was
saying in href=3D"https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/message=
s/3645">this
message. To see why there is no straightforward way to
perform other 90 degree twists, you only need to perform a 90
degree twist on an outer (L/R) face and then reorient the whole
puzzle along a different outer axis. If the original twist was not
about the new long axis, then there is clearly no straightforward
way to undo that twist.




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">



A moves, only Az moves since this is two
2x2x1 blocks that have to be rotated together.




S moves, not Sx+, Sx-, Sy+ or Sy- since
those are not defined in the standard rotation.




V moves, not Vz+ or Vz- since the
definition doesn't give these meaning when the long side
of the puzzle is parallel with the z-axis.




Note that for example Fz2 is not allowed
since this won't take you to a state of the puzzle. To
allow more moves we need to extend the definitions (after
the extension in the next paragraph all rotations and
twists (O, R, L, U, D, F, B, K, A) are possible in any
rotation and only which S and V moves are possible depend
on the state and rotation of the puzzle).






Extension of some definitions



It's possible to make an extension that allows all O, R,
L, U, D, F, B, K and A moves in any state. I will explain
how this can be done in the standard rotation but it
applies analogously to any other rotation where the K face
is an octahedron. First let's focus on U, D, F and B and
because of the symmetry of the puzzle all of these are
analogous so I will only explain one. The extension that
makes all U moves possible (note that the U face is in the
positive x-direction) is as follows: when making an U move
first detach the 8 top pieces which gives you a 1x2x4
block, fold this block into a 2x2x2 block in the positive
x-direction (similar to the later part of a Vx+ move from
the standard rotation) such that the U face form an
octahedron, rotate this 2x2x2 block around the specified
axis (for example around the z-axis if you are doing an Uz
move), reverse the fold you just did creating a 1x2x4
block again and reattach the block.





I noticed something like this the other day but realized that it
only seems to work for rotations along the long dimension (z in
your example). These are already easily accomplished by a simple
rotation to put the face in question on the end caps, followed by
a double end-cap twist.



This is as far as I'm going to comment for the moment because the
information gets very dense and I've been mulling and picking over
your message for several days already. In short, I really like
your attempt to provide a complete system of notation for
discussing this puzzle and will be curious to hear your thoughts
on my comments so far. I hope others will chime in too.



One final thought is that a real "acid test" of any notation
system for this puzzle will be attempt to translate some
algorithms from MC4D. I would most like to see a sequence that
flips a single piece, like the second 4-color series on href=3D"http://superliminal.com/cube/solution/pages/four_color.htm"=
>this
page of Roice's solution, or his pair of twirled corners at
the end of href=3D"http://superliminal.com/cube/solution/pages/series_hints.ht=
m">this
page. One trick will be to minimize the number of
whole-puzzle reorientations needed, but really any sequence that
works will be great evidence that the puzzles are equivalent. I
suspect that this sort of exercise will never be practical because
it will require too many reorientations, and that entirely new
methods will be needed to actually solve this puzzle.



Best,

-Melinda




cite=3D"mid:CAEohJcGnRf6+Vthd9OmTisqWDZ+npcc=3DFVvk3Avck728Sa1K4w@mail.gmai=
l.com"
type=3D"cite">

"
lang=3D"EN-GB">The A moves can be done very similarly but
after you have detached the two 2x2x1 blocks you don't
fold them but instead you stack them similar to a Sz move,
creating a 2x2x2 block with the A face as an octahedron in
the middle and then reverse the process after you have
rotated the block as specified (for example around the
negative y-axis if you are doing an Ay' move). Note that
these extended moves are closely related to the normal
moves and for example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99=
in the
standard rotation and note that (Ry Ly=E2=80=99) mod(rot) =3D=
(Ly
Ry=E2=80=99) mod(rot) =3D I (this applies to the other extend=
ed
moves as well).



If the cube is in the half-rotated state, where both the R
and L faces are octahedra, you can extend the definitions
very similarly. The only thing you have to change is how
you fold the 2x4 blocks when performing a U, D, F or B
move. Instead of folding the two 2x2 halves of the 2x4
into a 2x2 you have to fold the end 1x2 block 180 degrees
such that the face forms an octahedron.



These moves might be a little bit harder to perform, to me
especially the A moves seems a bit awkward, so I don't
know if it's good to use them or not. However, the A moves
are not necessary if you allow Sz in the standard rotation
(which you really should since Sz mod(rot) =3D I in the
standard rotation) and thus it might not be too bad to use
this extended version. The notation supports both variants
so if you don=E2=80=99t want to use these extended moves that
shouldn=E2=80=99t be a problem. Note that, however, for examp=
le Ux
(physical puzzle) !=3D Ux (virtual puzzle) where !=3D means
=E2=80=9Cnot equal to=E2=80=9D (more about legal/illegal move=
s later).



style=3D"line-height:107%;color:black" lang=3D"EN-GB">Gener=
alisation
of the notation
style=3D"line-height:107%;color:black" lang=3D"EN-GB">



Let=E2=80=99s generalise the notation to make it easier to us=
e and
to make it work for any n^4 cube in the MC4D software.
Previously, we saw that (Rx)2 =3D Rx Rx and if we allow
ourselves to rewrite this as (Rx)2 =3D Rx2 =3D Rxx =3D Rx Rx =
we
see that the capital letter naturally can be distributed
over the lowercase letters. We can make this more general
and say that any capital letter followed by several
lowercase letters means the same thing as the capital
letter distributed over the lowercase letters. Like Rxyz =3D
Rx Ry Rz and here R can be exchanged with any capital
letter and xyz can be exchanged with any sequence of
lowercase letters. We can also allow several capital
letters and one lowercase letter, for example RLx and
let=E2=80=99s define this as RLx =3D Rx Lx so that the lowerc=
ase
letter can be distributed over the capital letters. We can
also define a capital letter followed by =E2=80=98 (prime) li=
ke
R=E2=80=99x =3D Rx=E2=80=99 and R=E2=80=99xy =3D Rx=E2=80=99y=
=E2=80=99 so the =E2=80=98 (prime) is distributed
over the lowercase letters. Note that we don=E2=80=99t define=
a
capital to any other power than -1 like this since for
example R2x =3D RRx might seem like a good idea at first but
it isn=E2=80=99t very useful since R2 and RR are the same len=
gths
(and powers greater than two are seldom used) and we will
see that we can define R2 in another way that generalises
the notation to all n^4 cubes.



Okay, let=E2=80=99s define R2 and similar moves now and have =
in
mind what moves we want to be possible for a n^4 cube. The
moves that we cannot achieve with the notation this far is
twisting deeper slices. To match the notation with the
controls of the MC4D software let R2x be the move similar
to Rx but twisting the 2nd layer instead of the
top one and similarly for other capital letters, numbers
(up to n) and lowercase letters. Thus, R2x is performed as
Rx but holding down the number 2 key. Just as in the
program, when no number is specified 1 is assumed and you
can combine several numbers like R12x to twist both the
first and second layer. This notation does not apply to
rotations (O) folding moves (V) and restacking moves (S)
(I suppose you could redefine the S move using this
deeper-slice-notation and use S1z as Sz+, S2z as Sz and
S3z as Sz- but since these moves are only allowed for the
physical 2x2x2x2 I think that the notation with + and =E2=80=
=93 is
better since S followed by a lowercase letter without +/-
always means splitting the cube in a coordinate plane that
way, not sure though so input would be great). The
direction of the twist R3x should be the same as Rx
meaning that if Rx takes stickers belonging to K and move
them to F, so should R3x, in accordance with the controls
of the MC4D software. Note that for a 3x3x3x3 it=E2=80=99s tr=
ue
that R3z =3D Lz whereas R3x =3D Lx=E2=80=99 (note that R and =
L are the
faces in the z-directions so because of the symmetry of
the cube it will also be true that for example U3x =3D Bx
whereas U3y=3DBy=E2=80=99).



What about the case with several capital letters and
several lowercase letters, for instance, RLxy? I see two
natural definitions of this. Either, we could have RLxy =3D
Rxy Lxy or we could define it as RLxy =3D RLx RLy. These are
generally not the same (if you exchange R and L with any
allowed capital letter and similarly for x and y). I don=E2=
=80=99t
know what is best, what do you think? The situations I
find this most useful in are RL=E2=80=99xy to do a rotation a=
nd
RLxy as a twist. However, since R and L are opposite faces
their operations commute which imply RL=E2=80=99xy=C2=
=A0
(1st
definition) =3D Rxy L=E2=80=99xy =3D Rx Ry Lx=E2=80=99 Ly=E2=
=80=99 =3D Rx Lx=E2=80=99 Ry Ly=E2=80=99 =3D
RL=E2=80=99x RL=E2=80=99y =3D RL=E2=80=99xy (2nd d=
efinition) and
similarly for the other case with RLxy. Hopefully, we can
find another useful sequence of moves where this notation
can be used with only one of the definitions and can
thereby decide which definition to use. Personally, I feel
like RLxy =3D RLx RLy is the more intuitive definition but I
don=E2=80=99t have any good argument for this so I=E2=80=99ll=
leave the
question open.



For convenience, it might be good to be able to separate
moves like Rxy and RLx from the basic moves Rx, Oy etc
when speaking and writing. Let=E2=80=99s call the basic moves=
that
only contain one capital letter and one lowercase letter
(possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or a numb=
er) simple
moves
(like Rx, L'y, Ux2 and D3y=E2=80=99) and the move=
s
that contain more than one capital letter or more than one
lowercase letter composed moves. style=3D"line-height:107%" lang=3D"EN-GB">



More about inverses



This list can obviously be made longer but here are some
identities that are good to know and understand. Note that
R, L, U, x, y and z below just are examples, the following
is true in general for non-folds (however, S moves are
fine).



(P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99=
P1=E2=80=99 (Pi is an arbitrary
permutation for i=3D1,2,=E2=80=A6n)

(Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx

(RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x

(Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+=C2=A0=C2=
=A0
(just as an example
with restacking moves, note that the inverse doesn=E2=80=99t
change the + or -)

RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=
=80=99xyz=C2=A0=C2=A0 (true for both
definitions)

(RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=
=80=99yx=C2=A0=C2=A0 (true for both
definitions)



For V moves we have that: (Vx+)=E2=80=99 =3D Vx+=C2=A0=
=C2=A0
!=3D
Vx=E2=80=99+=C2=A0=C2=A0 (!=3D means =E2=80=9Cno=
t equal to=E2=80=9D)



Some important notes on legal/illegal moves



Although there are a lot of moves possible with this
notation we might not want to use them all. If we really
want a 2x2x2x2 and not something else I think that we
should try to stick to moves that are legal 2x2x2x2 moves
as far as possible (note that I said legal moves and not
permutations (a legal permutation can be made up of one or
more legal moves)). Clarification: cycling three of the
edge-pieces of a Rubik=E2=80=99s cube is a legal permutation =
but
not a legal move, a legal move is a rotation of the cube
or a twist of one of the layers. In this section I will
only address simple moves and simply refer to them as
moves (legal composed moves are moves composed by legal
simple moves).



I do believe that all moves allowed by my notation are
legal permutations based on their periodicity (they have a
period of 2 or 4 and are all even permutations of the
pieces). So, which of them correspond to legal 2x2x2x2
moves? The O moves are obviously legal moves since they
are equal to the identity mod(rot). The same goes for
restacking (S) (with or without +/-) in the direction of
the longest side of the puzzle (Sz, Sz+ and Sz- in the
standard rotation) since these are rotations and
half-rotations that don=E2=80=99t change the state of the puz=
zle.
Restacking in the other directions and fold moves (V) are
however not legal moves since they are made of 8 2-cycles
and change the state of the puzzle (note that they,
however, are legal permutations). The rest of the moves
(R, L, F, B, U, D, K and A) can be divided into two sets:
(1) the moves where you rotate a 2x2x2 block with an
octahedron inside and (2) the moves where you rotate a
2x2x2 block without an octahedron inside. A move belonging
to (2) is always legal. We can see this by observing what
a Rx does with the pieces in the standard rotation with
just K forming an octahedron. The stickers move in 6
4-cycles and if the puzzle is solved the U and D faces
still looks solved after the move. A move
belonging to set (1) is legal either if it=E2=80=99s an 180-d=
egree
twist or if it=E2=80=99s a rotation around the axis parallel =
with
the longest side of the puzzle (the z-axis in the standard
rotation). Quite interestingly these are exactly the moves
that don=E2=80=99t mix up the R and L stickers with the rest =
in
the standard rotation. I think I know a way to prove that
no legal 2x2x2x2 move can mix up these stickers with the
rest and this has to do with the fact that these stickers
form an inverted octahedron (with the corners pointing
outward) instead of a normal octahedron (let=E2=80=99s call t=
his
hypothesis * for now). Note that all legal twists (R, L,
F, B, U, D, K and A) of the physical puzzle correspond to
the same twist in the MC4D software.



So, what moves should we add to the set of legal moves be
able to get to every state of the 2x2x2x2? I think that we
should add the restacking moves and folding moves since
Melinda has already found a pretty short sequence of those
moves to make a rotation that changes which colours are on
the R and L faces. That sequence, starting from the
standard rotation, is: Oy Sx+z Vy+ Ozy=E2=80=99 Vy+ Ozy=E2=80=
=99 Vy+ =3D Oy
Sx+z=C2=A0 (Vy+ Ozy=E2=80=99)3 mod(3rot) =3D I m=
od(rot)
(hopefully I got that right). What I have found (which I
mentioned previously) is that (from the standard
rotation): RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz- and since =
this is
equivalent with Sy =3D Ly2 RLx=E2=80=99 Sz- RLx Ly2 By2 the
restacking moves that are not legal moves are not very
complicated permutations and therefore I think that we can
accept them since they help us mix up the R and L stickers
with other faces. In a sense, the folding moves are =E2=80=9C=
more
illegal=E2=80=9D since they cannot be composed by the legal m=
oves
(according to hypothesis *). This is also true for the
illegal moves belonging to set (1) discussed above.
However, since the folding moves is probably easier to
perform and is enough to reach every state of the 2x2x2x2
I think that we should use them and not the illegal moves
belonging to (1). Note, once again, that all moves
described by the notation are legal permutations (even the
ones that I just a few words ago referred to as illegal
moves) so if you wish you can use all of them and still
only reach legal 2x2x2x2 states. However (in a strict
sense) one could argue that you are not solving the
2x2x2x2 if you use illegal moves. If you only use illegal
moves to compose rotations (that is, create a permutation
including illegal moves that are equal to I mod(rot)) and
not actually using the illegal moves as twists I would
classify that as solving a 2x2x2x2. What do you think
about this?



What moves to use?



Here=E2=80=99s a short list of the simple moves that I th=
ink
should be used for the physical 2x2x2x2. Note that this is
just my thoughts and you may use the notation to describe
any move that it can describe if you wish to. The
following list assumes that the puzzle is in the standard
rotation but is analogous for other representations where
the K face is an octahedron.



O, all since they are I mod(rot),

R, L, all since they are legal (note Rx (physical puzzle)
=3D Rx (virtual 2x2x2x2)),

U, D, only x2 since these are the only legal
easy-to-perform moves,

F, B, only y2 since these are the only legal
easy-to-perform moves,

K, A, only z, z=E2=80=99 and z2 since these are the only lega=
l
easy-to-perform moves,

S, at least z, z+ and z- since these are equal to I
mod(rot),

S, possibly x and y since these help us perform rotations
and is easy to compose (not necessary to reach all states
and not legal though),

V, all 8 allowed by the rotation of the puzzle (at least
one is necessary to reach all states and if you allow one
the others are easy to achieve anyway).



If you start with the standard rotation and then perform
Sz+ the following applies instead (this applies
analogously to any other rotation where the R and L faces
are octahedra).



O, all since they are I mod(rot),

R, L, only z, z=E2=80=99 and z2 since these are the only lega=
l
easy-to-perform moves,

U, D, only x2 since these are the only legal
easy-to-perform moves,

F, B, only y2 since these are the only legal
easy-to-perform moves,

K, A, at least z, z=E2=80=99 and z2, possibly all (since they=
are
legal) although some might be hard to perform.

S, V, same as above.



Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D
I mod(rot) (!=3D for not equal) which implies that the Ux2
move when R and L are octahedra is different from the Ux2
move when K is an octahedron. (Actually, the sequence
above is equal to Uy2).



Regarding virtual n^4 cubes I think that all O, C, R, L,
U, D, F, B, K and A moves should be used since they are
all legal and really the only thing you need
(left-clicking on an edge or corner piece in the computer
program can be described quite easily with the notation,
for example, Kzy2 is left-clicking on the top-front edge
piece on the K face).



I hope this was possible to follow and understand. Feel
free to ask questions about the notation if you find
anything ambiguous.an
style=3D"line-height:107%" lang=3D"EN">



Best regards,

Joel Karlsson
line-height: 107%;" lang=3D"EN-GB">






Den 4 maj 2017
12:01 fm skrev "Melinda Green moz-do-not-send=3D"true"
href=3D"mailto:melinda@superliminal.com"
target=3D"_blank">melinda@superliminal.com
[4D_Cubing]" < href=3D"mailto:4D_Cubing@yahoogroups.com"
target=3D"_blank">4D_Cubing@yahoogroups.com
>: type=3D"attribution">

class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254quote"
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rgb(204,204,204);padding-left:1ex">
size=3D"-1"> =C2=
=A0

id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151ygrp-mlmsg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151ygrp-msg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151ygrp-text">


Thanks for the correction. A couple of
things: First, when assembling one piece at
a time, I'd say there is only 1 way to place
the first piece, not 24. Otherwise you'd
have to say that the 1x1x1 puzzle has 24
states. I understand that this may be
conventional, but to me, that just sounds
silly.



Second, I have the feeling that the
difference between the "two representations"
you describe is simply one of those
half-rotations I showed in the video. In the
normal solved state there is only one
complete octahedron in the very center, and
in the half-rotated state there is one in
the middle of each half of the "inverted"
form. I consider them to be the same solved
state.



-Melinda


class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254quoted-text"> size=3D"-1">




class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151moz-cite-prefix"> size=3D"-1">On 5/3/2017 2:39 PM, Joel
Karlsson class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151moz-txt-link-abbreviated"
href=3D"mailto:joelkarlsson97@gmail.com"
target=3D"_blank">joelkarlsson97@gmail.co=
m

[4D_Cubing] wrote:





class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254quoted-text">
Horrible typo... It
seems like I made some typos in my
email regarding the state count. It
should of course be 16!12^16/(6*192)
and NOT 12!16^12/(6*192). However, I
did calculate the correct number
when comparing with previous results
so the actual derivation was
correct.





Something
of interest is that the physical
pieces can be assembled in 16!24*12^15
ways since there are 16 pieces, the
first one can be oriented in 24 ways
and the remaining can be oriented in
12 ways (since a corner with 3 colours
never touch a corner with just one
colour). Dividing with 6 to get a
single orbit still gives a factor
2*192 higher than the actual count
rather than 192. This shows that every
state in the MC4D representation has 2
representations in the physical
puzzle. These two representations must
be the previously discussed, that the
two halves either have the same color
on the outermost corners or the
innermost (forming an octahedron) when
the puzzle is solved and thus both are
complete representations of the
2x2x2x2.=C2=A0

class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254elided-text">

<=
font
size=3D"-1">


<=
font
size=3D"-1">Best regards,=C2=A0>

<=
font
size=3D"-1">Joel Karlsson=C2=A0>

<=
font
size=3D"-1">


size=3D"-1">Den 30 apr. 2017 10:51
em skrev "Joel Karlsson" < moz-do-not-send=3D"true"
href=3D"mailto:joelkarlsson97@gma=
il.com"
target=3D"_blank">joelkarlsson97@=
gmail.com>: type=3D"attribution">

class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151quote"
style=3D"border-left:1px solid
rgb(204,204,204)">





I
am no expert on
group theory, so
to better
understand what
twists are legal I
read through the
part of Kamack and
Keane's The
Rubik Tesseract
about
orienting the
corners. Since all
even permutations
are allowed the
easiest way to
check if a twist
is legal might be
to:


1.
Check that the twist
is an even
permutation, that
is: the same twist
can be done by
performing an even
number of piece
swaps (2-cycles).


2.
Check the periodicity
of the twist. If A^k=3DI
(A^k meaning
performing the twist k
times and I (the
identity) representing
the permutation of
doing nothing) and k
is not divisible by 3
the twist A definitely
doesn't violate the
restriction of the
orientations since kx
mod 3 =3D 0 and k mod 3
!=3D 0 implies x mod 3 =
=3D
0 meaning that the
change of the total
orientation x for the
twist A mod 3 is 0
(which precisely is
the restriction of
legal twists; that
they must preserve the
orientation mod 3).

<=
br>

For
instance, this implies
that the restacking
moves are legal 2x2x2x2
moves since both are
composed of 8 2-cycles
and both can be
performed twice (note
that 2 is not divisible
by 3) to obtain the
identity.

>

Note
that 1 and 2 are
sufficient to check if a
twist is legal but only
1 is necessary; there
can indeed exist a twist
violating 2 that still
is legal and in that
case, I believe that we
might have to study the
orientation changes for
that specific twist in
more detail. However, if
a twist can be composed
by other legal twists it
is, of course, legal as
well.





Best
regards,


Joel


class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151elided-text">
size=3D"-1">


t
size=3D"-1">2017-04-29
1:04 GMT+02:00 Melinda
Green moz-do-not-send=3D"true"
href=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlim=
inal.com
[4D_Cubing] dir=3D"ltr">< moz-do-not-send=3D"true=
"
href=3D"mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoo=
groups.com>:


class=3D"gmail_quote"
style=3D"border-left:1px
solid rgb(204,204,204)">
style=3D"background-color=
:rgb(255,255,255)"> size=3D"-1"> =C2=
=A0


id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218ygrp=
-mlmsg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218ygrp=
-msg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218ygrp=
-text">
<=
br>
First off,
thanks
everyone for
the helpful
and
encouraging
feedback!
Thanks Joel
for showing us
that there are
6 orbits in
the 2^4 and
for your
rederivation
of the state
count. And
thanks Matt
and Roice for
pointing out
the importance
of the
inverted
views. It
looks so
strange in
that
configuration
that I always
want to get
back to a
normal view as
quickly as
possible, but
it does seem
equally valid,
and as you've
shown, it can
be helpful for
more than just
finding short
sequences.

size=3D"-1">

I don't
understand
Matt's
"pinwheel"
configuration,
but I will
point out that
all that is
needed to
create your
twin interior
octahedra is a
single
half-rotation
like I showed
in the video
at moz-do-not-send=
=3D"true"
href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJKLo&t=3D5m29s"
target=3D"_blank"=
>5:29.
The two main
halves do end
up being
mirror images
of each other
on the visible
outside like
he described.
Whether it's
the pinwheel
or the
half-rotated
version that's
correct, I'm
not sure that
it's a bummer
that the
solved state
is not at all
obvious, so
long as we can
operate it in
my original
configuration
and ignore the
fact that the
outer faces
touch. That
would just
mean that the
"correct" view
is evidence
that that the
more
understandable
view is
legitimate.

size=3D"-1">

I'm going to
try to make a
snapable V3
which should
allow the
pieces to be
more easily
taken apart
and
reassembled
into other
forms.
Shapeways does
offer a
single, clear
translucent
plastic that
they call
"Frosted
Detail", and
another called
"Transparent
Acrylic", but
I don't think
that any sort
of transparent
stickers will
help us,
especially
since this
thing is chock
full of
magnets. The
easiest way to
let you see
into the two
hemispheres
would be to
simply
truncate the
pointy tips of
the stickers.
That already
happens a
little bit due
to the way
I've rounded
the edges.
Here is a moz-do-not-send=3D"true" href=3D"http://superliminal.com/cube/inverted1.jpg=
"
target=3D"_blank">close-up of a half-rotation in which you can see
that the inner
yellow and
white faces
are solved.
Your
suggestion of
little mapping
dots on the
corners also
works, but
just opening
the existing
window further
would work
more directly.>
size=3D"-1">

-Melinda>





class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
oz-cite-prefix"> size=3D"-1">On
4/28/2017 2:15
PM, Roice
Nelson moz-do-not-send=
=3D"true"
class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
oz-txt-link-abbreviated"
href=3D"mailto:roice3@gmail.com" target=3D"_blank">roice3@gmail.com
[4D_Cubing]
wrote:


type=3D"cite">
=
size=3D"-1">>I
agree with
Don's
arguments
about adjacent
sticker colors
needing to be
next to each
other.=C2=A0 I
think this can
be turned into
an accurate
2^4 with
coloring
changes, so I
agree with
Joel too :)



To help
me think about
it, I started
adding a new
projection
option for
spherical
puzzles to
MagicTile,
which takes
the two
hemispheres of
a puzzle and
maps them to
two disks with
identified
boundaries
connected at a
point, just
like a
physical " moz-do-not-send=
=3D"true"
href=3D"http://www.pa-network.com/global-chess/indexf.html"
target=3D"_blank"=
>global
chess"
game I have.=C2=
=A0
Melinda's
puzzle is a
lot like this
up a
dimension, so
think about
two disjoint
balls, each
representing a
hemisphere of
the 2^4, each
a "subcube" of
Melinda's
puzzle.=C2=A0 The
two boundaries
of the balls
are identified
with each
other and as
you roll one
around, the
other half
rolls around
so that
identified
points connect
up.=C2=A0 We need
to have the
same
restriction on
Melinda's
puzzle.




In the
pristine state
then, I think
it'd be nice
to have an
internal
(hidden),
solid colored
octahedron on
each half.=C2=A0
The other 6
faces should
all have equal
colors split
between each
hemisphere, 4
stickers on
each half.=C2=A0
You should be
able to
reorient the
two subcubes
to make a half
octahedron of
any color on
each subcube.=C2=
=A0
I just saw
Matt's email
and picture,
and it looks
like we were
going down the
same thought
path.=C2=A0 I thi=
nk
with
recoloring
(mirroring
some of the
current piece
colorings)
though, the
windmill's can
be avoided (?)iv>




size=3D"-1">[...]
After
staring/thinking
a bit more,
the coloring
Matt came up
with is
right-on if
you want to
put a solid
color at the
center of each
hemisphere.=C2=A0
His=C2=A0comment
about the
"mirrored"
pieces on each
side helped me
understand
better. =C2=A03 o=
f
the stickers
are mirrored
and the 4th is
the hidden
color
(different on
each side for
a given pair
of "mirrored"
pieces).=C2=A0 Al=
l
faces behave
identically as
well, as they
should.=C2=A0 It'=
s
a little bit
of a bummer
that it
doesn't look
very pristine
in the
pristine
state, but it
does look like
it should work
as a 2^4. =




I wonder if
there might be
some
adjustments to
be made when
shapeways
allows
printing
translucent as
a color :)
=




class=3D"gmail_ex=
tra"> size=3D"-1">[...]
Sorry for all
the streaming,
but I wanted
to share one
more thought.=C2=
=A0
I now
completely
agree with
Joel/Matt
about it
behaving as a
2^4, even with
the original
coloring.=C2=A0 Y=
ou
just need to
consider the
corner colors
of the two
subcubes
(pink/purple
near the end
of the video)
as being a
window into
the interior
of the piece.=C2=
=A0
The other
colors match
up as desired.
=C2=A0(Sorry if
folks already
understood
this after
their emails
and I'm just
catching up!)




In fact,
you could
alter the
coloring of
the pieces
slightly so
that the
behavior was
similar with
the inverted
coloring.=C2=A0 A=
t
the corners
where 3 colors
meet on each
piece, you
could put a
little circle
of color of
the opposite
4th color.=C2=A0 =
In
Matt's
windmill
coloring then,
you'd be able
to see all
four colors of
a piece, like
you can with
some of the
pieces on
Melinda's
original
coloring.=C2=A0 A=
nd
again you'd
consider the
color circles
a window to
the interior
that did not
require the
same matching
constraints
between the
subcubes.




I'm
looking
forward to
having one of
these :)




Happy
Friday
everyone,



class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828h5">
Roice

class=3D"gmail_qu=
ote">



On Fri, Apr
28, 2017 at
1:14 AM, Joel
Karlsson moz-do-not-send=
=3D"true"
href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@g=
mail.com
[4D_Cubing] dir=3D"ltr"><<=
a
moz-do-not-send=3D"true" href=3D"mailto:4D_Cubing@yahoogroups.com"
target=3D"_blank"=
>4D_Cubing@yahoogroups.com>
wrote:

class=3D"gmail_qu=
ote"
style=3D"border-left:1px solid rgb(204,204,204)">




dir=3D"auto">
Seems
like there was
a slight
misunderstanding.
I meant that
you need to be
able to =C2=A0twi=
st
one of the
faces and in
MC4D the most
natural choice
is the center
face. In your
physical
puzzle you can
achieve this
type of twist
by twisting
the two
subcubes
although this
is indeed a
twist of the
subcubes
themselves and
not the center
face, however,
this is still
the same type
of twist just
around another
face.=C2=A0

dir=3D"auto">


dir=3D"auto">If
the magnets
are that
allowing the
2x2x2x2 is
obviously a
subgroup of
this puzzle.
Hopefully the
restrictions
will be quite
natural and
only some
"strange"
moves would be
illegal.
Regarding the
"families of
states" (aka
orbits), the
2x2x2x2 has 6
orbits. As I
mentioned
earlier all
allowed twists
preserves the
parity of the
pieces,
meaning that
only half of
the
permutations
you can
achieve by
disassembling
and
reassembling
can be reached
through legal
moves. Because
of some
geometrical
properties of
the 2x2x2x2
and its
twists, which
would take
some time to
discuss in
detail here,
the
orientation of
the stickers
mod 3 are
preserved,
meaning that
the last
corner only
can be
oriented in
one third of
the number of
orientations
for the other
corners. This
gives a total
number of
orbits of
2x3=3D6. To
check this
result let's
use this
information to
calculate all
the possible
states of the
2x2x2x2; if
there were no
restrictions
we would have
16! for
permuting the
pieces (16
pieces) =C2=A0and
12^16 for
orienting them
(12
orientations
for each
corner). If we
now take into
account that
there are 6
equally sized
orbits this
gets us to
12!16^12/6.
However, we
should also
note that the
orientation of
the puzzle as
a hole is not
set by some
kind of
centerpieces
and thus we
need to devide
with the
number of
orientations
of a 4D cube
if we want all
our states to
be separated
with twists
and not only
rotations of
the hole
thing. The
number of ways
to orient a 4D
cube in space
(only allowing
rotations and
not mirroring)
is 8x6x4=3D192
giving a total
of=C2=A012!16^12/(6*192) states which is indeed the same number that for
example David
Smith arrived
at during his
calculations.
Therefore,
=C2=A0when
determining
whether or not
a twist on
your puzzle is
legal or not
it is
sufficient and
necessary to
confirm that
the twist is
an even
permutation of
the pieces and
preserves the
orientation of
stickers mod
3.

dir=3D"auto">


dir=3D"auto">Best
regards,=C2=A0iv>
dir=3D"auto">Joel=
=C2=A0


dir=3D"auto">
class=3D"gmail_ex=
tra"
dir=3D"auto">

class=3D"gmail_qu=
ote">Den
28 apr. 2017
3:02 fm skrev
"Melinda Green
moz-do-not-send=
=3D"true"
href=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlim=
inal.com
[4D_Cubing]"
< moz-do-not-send=
=3D"true"
href=3D"mailto:4D_Cubing@yahoogroups.com" target=3D"_blank">4D_Cubing@yahoo=
groups.com>:type=3D"attribution">
class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998=
2835215532m_4431900632030069098quote"
style=3D"border-left:1px solid rgb(204,204,204)">
style=3D"backgrou=
nd-color:rgb(255,255,255)">
=C2=A0n>
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m_67=
33134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283=
5215532m_4431900632030069098m_6586884675826257012ygrp-mlmsg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m_67=
33134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283=
5215532m_4431900632030069098m_6586884675826257012ygrp-msg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m_67=
33134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283=
5215532m_4431900632030069098m_6586884675826257012ygrp-text">

The new
arrangement of
magnets allows
every valid
orientation of
pieces. The
only invalid
ones are those
where the
diagonal lines
cutting each
cube's face
cross each
other rather
than coincide.
In other
words, you can
assemble the
puzzle in all
ways that
preserve the
overall
diamond/harlequin
pattern. Just
about every
move you can
think of on
the whole
puzzle is
valid though
there are
definitely
invalid moves
that the
magnets allow.
The most
obvious
invalid move
is twisting of
a single end
cap.



I think your
description of
the center
face is not
correct
though. Twists
of the outer
faces cause
twists
"through" the
center face,
not "of" that
face. Twists
of the outer
faces are
twists of
those faces
themselves
because they
are the ones
not changing,
just like the
center and
outer faces of
MC4D when you
twist the
center face.
The only
direct twist
of the center
face that this
puzzle allows
is a 90 degree
twist about
the outer
axis. That
happens when
you
simultaneously
twist both end
caps in the
same
direction.



Yes, it's
quite
straightforward
reorienting
the whole
puzzle to put
any of the
four axes on
the outside.
This is a very
nice
improvement
over the first
version and
should make it
much easier to
solve. You may
be right that
we just need
to find the
right way to
think about
the outside
faces. I'll
leave it to
the math
geniuses on
the list to
figure that
out.



-Melinda


class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998=
2835215532m_4431900632030069098quoted-text">



class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998=
2835215532m_4431900632030069098m_6586884675826257012moz-cite-prefix">On
4/27/2017
10:31 AM, Joel
Karlsson moz-do-not-send=
=3D"true"
class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998=
2835215532m_4431900632030069098m_6586884675826257012moz-txt-link-abbreviate=
d"
href=3D"mailto:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@g=
mail.com
[4D_Cubing]
wrote:

type=3D"cite">>



Hi
Melinda,




I do not agree
with the
criticism
regarding the
white and
yellow
stickers
touching each
other, this
could simply
be an effect
of the
different
representations
of the puzzle.
To really
figure out if
this indeed is
a
representation
of a 2x2x2x2
we need to
look at the
possible moves
(twists and
rotations) and
figure out the
equivalent
moves in the
MC4D software.
From the MC4D
software, it's
easy to
understand
that the only
moves required
are free
twists of one
of the faces
(that is, only
twisting the
center face in
the standard
perspective
projection in
MC4D) and 4D
rotations
swapping which
face is in the
center
(ctrl-clicking
in MC4D). The
first is
possible in
your physical
puzzle by
rotating the
white and
yellow
subcubes (from
here on I use
subcube to
refer to the
two halves of
the puzzle and
the colours of
the subcubes
to refer to
the "outer
colours"). The
second is
possible if
it's possible
to reach a
solved state
with any two
colours on the
subcubes that
still allow
you to perform
the previously
mentioned
twists. This
seems to be
the case from
your
demonstration
and is indeed
true if the
magnets allow
the simple
twists
regardless of
the colours of
the subcubes.
Thus, it is
possible to
let your
puzzle be a
representation
of a 2x2x2x2,
however, it
might require
that some
moves that the
magnets allow
aren't used.




Best regards,


Joel


class=3D"gmail_ex=
tra">

class=3D"gmail_qu=
ote">2017-04-27
3:09 GMT+02:00
Melinda Green
moz-do-not-send=
=3D"true"
href=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlim=
inal.com
[4D_Cubing] dir=3D"ltr"><<=
a
moz-do-not-send=3D"true" href=3D"mailto:4D_Cubing@yahoogroups.com"
target=3D"_blank"=
>4D_Cubing@yahoogroups.com>:

class=3D"gmail_qu=
ote"
style=3D"border-left:1px solid rgb(204,204,204)">
style=3D"backgrou=
nd-color:rgb(255,255,255)">
=C2=A0n>
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m_67=
33134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283=
5215532m_4431900632030069098m_6586884675826257012m_3209267269419320979ygrp-=
mlmsg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m_67=
33134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283=
5215532m_4431900632030069098m_6586884675826257012m_3209267269419320979ygrp-=
msg">
id=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968950=
6161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m_67=
33134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998283=
5215532m_4431900632030069098m_6586884675826257012m_3209267269419320979ygrp-=
text">

Dear
Cubists,



I've finished
version 2 of
my physical
puzzle and
uploaded a
video of it
here:

moz-do-not-send=
=3D"true"
href=3D"https://www.youtube.com/watch?v=3DzqftZ8kJKLo" target=3D"_blank">ht=
tps://www.youtube.com/watch?v=3DzqftZ8kJKLo

Again, please
don't share
these videos
outside this
group as their
purpose is
just to get
your feedback.
I'll
eventually
replace them
with a public
video.



Here is an
extra math
puzzle that I
bet you folks
can answer:
How many
families of
states does
this puzzle
have? In other
words, if
disassembled
and
reassembled in
any random
configuration
the magnets
allow, what
are the odds
that it can be
solved? This
has practical
implications
if all such
configurations
are solvable
because it
would provide
a very easy
way to fully
scramble the
puzzle.



And finally, a
bit of fun: A
relatively new
friend of mine
and new list
member, Marc
Ringuette, got
excited enough
to make his
own version.
He built it
from EPP foam
and colored
tape, and used
honey instead
of magnets to
hold it
together.
Check it out
here: moz-do-not-send=
=3D"true"
href=3D"http://superliminal.com/cube/dessert_cube.jpg" target=3D"_blank">ht=
tp://superliminal.com/cube/dessert_cube.jpg
I don't know
how practical
a solution
this is but it
sure looks
delicious!
Welcome Marc!



-Melinda


















class=3D"gmail-m_6473756560834928383gmail-m_7083180584533685750m_-594094968=
9506161254m_2386783995576461151m_2995809254753611828m_-4446226697589261218m=
_6733134459512634882m_7927561327128665054m_-4244333051891278538m_-776157998=
2835215532m_4431900632030069098quoted-text">

























<=
br>




































--------------DC3048547D20144DBFC49280--




From: Joel Karlsson <joelkarlsson97@gmail.com>
Date: Mon, 22 May 2017 22:32:21 +0200
Subject: Re: [MC4D] Physical 4D puzzle V2



Hi Melinda,

Thank you for the feedback. Regarding the coordinate system, it=E2=80=99s j=
ust
a matter of preference. I thought it would be nice to have a
xy-symmetry but understand that it might be more practical to follow
conventions. So, adapting your suggestion, let's redefine the axes as
x pointing right, y up and z towards you. Note that this means that
the longer side of the puzzle is parallel with the x-axis in the
standard rotation. From here on, I will use this new coordinate
system.

Regarding the name of the faces. Since which faces are the "outer
ones" changes with how you rotate the puzzle and what state the puzzle
is in, I think that the labelling of the faces should be independent
of which faces are the outer faces (forming what will be referred to
as inverted octahedra). Since the puzzle is a representation of a
4d-cube in 3d-space it will (with our coordinate system) always have
two faces "belonging" to every axis and two faces that don't belong to
any axis. Therefore, it makes sense to label the faces in such a way
that the face in for example the positive x direction is R, always.
So, the K face (which is one of the faces not belonging to a
particular axis) is always the face only belonging to the center 2x2x2
block (and this can be either an octahedron or an inverted octahedron
(the outer corners) depending on the representation of the puzzle).
The distinction between left and right never disappears; the R face is
always the face only belonging to the right half of the puzzle and can
be an octahedron, an inverted octahedron, two half octahedra (<><>) or
one half and two quarter octahedra (><><). Note that the
half-rotations are 90-degree rotations of the puzzle; for example, Sx+
(physical puzzle) =3D Cx' (virtual puzzle) is the rotation that does L
-> K -> R -> A -> L so the face that previously was the L face is now
the K face. From the standard rotation (where R and L are the inverted
octahedra before the rotation), this would mean that the K and A faces
are inverted octahedra after a Sx+ rotation. What faces are the
"mysterious" (or more precisely symmetry breaking, forming inverted
octahedra instead of regular octahedra) depend on the rotation of the
puzzle and can be any two opposite faces (R and L, U and D, F and B or
A and K). It might be useful to be able to describe what faces are
inverted octahedra since this determines what moves are legal so let's
say that the puzzle is in an RL representation if R and L are inverted
octahedra and similarly for other states. Thus, Sx+ can take you from
an RL representation to an AK representation (what a long word let's
use rep for short). Note that the AK rep doesn't specify which axis
the longer side should be parallel with so let's just add a lowercase
letter to indicate this (an AKx rep is thus a state where the A and K
faces are inverted octahedra and the longer side is parallel with the
x-axis). In conclusion, I would like to keep the names of the faces as
I first defined them and hope that it's clearer what I mean with them
and that the names of the faces are not related to what faces form
inverted octahedra.

You also wrote:
"There are several, distinct types of rotations, none of which change
the state of the puzzle, and I think we need a way to be unambiguous
about them. The types I see are

1. Simple reorientation of the physical puzzle in the hand, no magnets
involved. IE your 'O' moves and maybe analogous to mouse-dragging in
MC4D?
2. Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-cli=
ck?
3. Half-rotations. Maybe analogous to a "face first" view? (ctrl-click
on a 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by
Cubie")
4. Whole-puzzle reorientations that move an arbitrary axis into the
"outer" 2 faces. No MC4D analog."

The rotations (as far as I know) are the O moves (type one rotation),
which are indeed analogous to mouse-dragging in MC4D, rolling the
2x2x2 halves (type two rotation) (these are easily described as for
example RL'y in an RL rep) which (in a non-AK rep) is a ctrl-click on
a non-inverted face (that is ctrl-click on a face that is currently
represented with an octahedron), half-rotations (also type two
rotations) which is a ctrl-click on an inverted face and sequences
that for example from an RL rep can take the R face to K without
turning the puzzle into an AK rep (type three rotation). Type one and
two rotations are legal moves but type three contain illegal 2x2x2x2
moves according to hypothesis * in my previous email (however, they
are very important and needed if we wish to be able to reach all
states).

Regarding fold moves, from the standard rotation (RL rep) are you
saying that Vy+ =3D Vy'+ or something like Vy+ =3D Vy'+ FB'x2 =3D Vy'+
mod(rot)? The former seems not to be correct but I believe the latter
is, please correct me if I'm wrong. I don't assume that you fold the
puzzle back to the same representation. This is what + and -
indicates, + preserve the representation mod(rot) (i.e AK rep both
before and after or neither before nor after) and - changes it (going
from AK rep to non-AK rep or vice versa).

"I think the only legal 90 degree twists of the K face are those about
the long axis. I believe this is what Christopher Locke was saying in
this message. To see why there is no straightforward way to perform
other 90 degree twists, you only need to perform a 90 degree twist on
an outer (L/R) face and then reorient the whole puzzle along a
different outer axis. If the original twist was not about the new long
axis, then there is clearly no straightforward way to undo that
twist."

Yes, as pointed out further down in my previous email. The notation
allows all 90-degree rotations after the section "extensions of some
definitions" although only 180-degree rotations are legal for
octahedral faces around axes not parallel with the longer side of the
puzzle (see the section =E2=80=9Csome important notes on legal/illegal
moves=E2=80=9D).

"I noticed something like this the other day but realized that it only
seems to work for rotations along the long dimension (z in your
example). These are already easily accomplished by a simple rotation
to put the face in question on the end caps, followed by a double
end-cap twist."

It works for the other rotations as well although these are not legal
moves. They could possibly be used instead of the illegal S moves
(along axes not parallel with the longer side) and the illegal V moves
(V- (minus) moves which are closely related to the illegal S moves,
example from RL rep: Vy- =3D Vy+ Sx Oz2) but as I mentioned in my
previous email I do believe that it's better to use the S and V moves
since you have found a relatively short way to perform a type three
rotation with those.

Let:
P1 =3D Sy+ Uyz2 Sy-
P1=E2=80=99 =3D P1 (P1 is its own inverse)
P2 =3D (P1 UD=E2=80=99z P1 UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z=E2=80=
=99 P1 UD=E2=80=99z)2 (P2=3DP_2 not P^2
whereas the two at the end means: perform twice)
P2=E2=80=99 =3D (UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z P1 UD=E2=80=99z =
P1 UD=E2=80=99z=E2=80=99 P1)2
P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x =3D I mod(r=
ot) (type three rotation)
P3=E2=80=99 =3D Oy=E2=80=99 P3 Oy=E2=80=99
P4 =3D P2 P3 P2 P3=E2=80=99 P2=E2=80=99 P3 P2=E2=80=99 P3=E2=80=99

P4 from UD rep is a 164 move sequence rotating only one corner in its
place. The sequence is inspired by Roice =E2=80=9Csecond four-color series=
=E2=80=9D
but I have changed the =E2=80=9CTop 9=E2=80=9D moves to pure rotations sinc=
e it=E2=80=99s all
that=E2=80=99s necessary and a bit shorter to perform. P3 is your type thre=
e
rotation (with a rotation added at the end) and P2 is Roice =E2=80=9Cthird
three-color series=E2=80=9D. Written with my notation but for the virtual
puzzle, the sequences (Roice original since it=E2=80=99s easier to perform =
K
twists than O rotations in MC4D) are:
Q1 =3D (Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=99 Rz2y=E2=80=99)2 (ana=
logue to P2)
Q1=E2=80=99 =3D (Rz2y=E2=80=99 Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=99=
)2
Q2 =3D Q1 Kxy=E2=80=99 Q1 Kyx=E2=80=99 Q1=E2=80=99 Kxy=E2=80=99 Q1=E2=
=80=99 Kyx=E2=80=99 (analogue to P4 but
only 36 moves)

How to read faster (the example applies to a 3x3x3x3): moves like
Kz2y=E2=80=99 are clicking on 3C-pieces in MC4D. If the move is written in
this way, [uppercase letter] [lowercase letter]2 [lowercase letter
possibly with =E2=80=98 (prime)], there=E2=80=99s a quite quick way to real=
ize which
piece this is. The uppercase letter specifies which face the piece to
press is on and the first lowercase letter (followed by 2) specifies
one of the sides of that face that the piece belong to. There are then
4 possible pieces. Sadly, the piece do not lie in the direction of the
last letter from the center of the side of the face but you have to
move one edge clockwise from this. So, Kz2y=E2=80=99 is a click on the edge=
on
the front side of the K face one step clockwise from the negative
y-axis (thus, the front left 3C-piece on the K face). It=E2=80=99s a bit
unfortunate that this =E2=80=9Crule=E2=80=9D isn=E2=80=99t even simpler but=
it's at least true
for all of these moves (as far as I know). Moves like Kxy=E2=80=99 are
left/right-clicking on a corner piece but currently I don=E2=80=99t know a
fast way to determine which corner. Any ideas?

Best regards,
Joel Karlsson

2017-05-19 3:33 GMT+02:00 Melinda Green melinda@superliminal.com
[4D_Cubing] <4D_Cubing@yahoogroups.com>:
>
>
> Hello Joel,
>
> Thanks for drilling into this puzzle. Finding good ways to discuss and th=
ink
> about moves and representations will be key. I'll comment on some details
> in-line.
>
> On 5/14/2017 6:16 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
> wrote:
>
>
> Yes, that is correct and in fact, you should divide not only with 24 for =
the
> orientation but also with 16 for the placement if you want to calculate
> unique states (since the 2x2x2x2 doesn't have fixed centerpieces). The
> point, however, was that if you don=E2=80=99t take that into account you =
get a
> factor of 24*16=3D384 (meaning that the puzzle has 384 representations of
> every unique state) instead of the factor of 192 which you get when
> calculating the states from the virtual puzzle and hence every state of t=
he
> virtual puzzle has two representations in the physical puzzle. Yes exactl=
y,
> they are indeed the same solved (or other) state and you are correct that
> the half rotation (taking off a 2x2 layer and placing it at the other end=
of
> the puzzle) takes you from one representation to the same state with the
> other representation. This means that the restacking move (taking off the
> front 2x4 layer and placing it behind the other 2x4 layer) can be express=
ed
> with half-rotations and ordinary twists and rotations (which you might ha=
ve
> pointed out already).
>
>
> Yes, I made that claim in the video but didn't show it because I have yet=
to
> record such a sequence. I've only stumbled through it a few times. I talk=
ed
> about it at 5:53 though I mistakenly called it a twist, when I should hav=
e
> called it a sequence.
>
>
> I think I've found six moves including ordinary twists and a restacking m=
ove
> that is identical to a half-rotation and thus it's easy to compose a
> restacking move with one half-rotation and five ordinary twists. There mi=
ght
> be an error since I've only played with the puzzle in my mind so it would=
be
> great if you, Melinda, could confirm this (the sequence is described late=
r
> in this email).
>
>
> You mean "RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-"? Yes, that works. Ther=
e do seem to
> be easier ways to do that beginning with an ordinary rolling rotation. I
> don't see those in your notation, but the equivalent using a pair of twis=
ts
> would be Rx Lx' Sx Vy if I got that right.
>
>
> To be able to communicate move sequences properly we need notation for
> representing twists, rotations, half-rotations, restacking moves and fold=
s.
> Feel free to come with other suggestion but you can find mine below. Plea=
se
> read the following thoroughly (maybe twice) to make sure that you underst=
and
> everything since misinterpreted notation could potentially become a
> nightmare and feel free to ask questions if there is something that needs
> clarification.
>
> Coordinate system and labelling:
>
> Let's introduce a global coordinate system. In whatever state the puzzle =
is
> let the positive x-axis point upwards, the positive y-axis towards you an=
d
> the positive z-axis to the right (note that this is a right-hand system).
>
>
> I see the utility of a global coordinate system, but this one seems rathe=
r
> non-standard. I suggest that X be to the right, and Y up since these are
> near-universal standards. Z can be in or out. I have no opinion. If there=
is
> any convention in the twisty puzzle community, I'd go with that.
>
> Note also that the wiki may be a good place to document and iterate on
> terminology, descriptions and diagrams. Ray added a "notation" section to
> the 3^4 page here, and I know that one other member was thinking of
> collecting a set of moves on another wiki page.
>
>
> Now let's name the 8 faces of the puzzle. The right face is denoted with =
R,
> the left with L, the top U (up), the bottom D (down), the front F, the ba=
ck
> B, the center K (kata) and the last one A (ana). The R and L faces are
> either the outer corners of the right and left halves respectively or the
> inner corners of these halves (forming octahedra) depending on the
> representation of the puzzle. The U, D, F and B faces are either two diam=
ond
> shapes (looking something like this: <><>) on the corresponding side of t=
he
> puzzle or one whole and two half diamond shapes (><><). The K face is eit=
her
> an octahedron in the center of the puzzle or the outer corners of the cen=
ter
> 2x2x2 block. Lastly, the A face is either two diamond shapes, one on the
> right and one on the left side of the puzzle, or another shape that=E2=80=
=99s a
> little bit hard to describe with just a few words (the white stickers at
> 5:10 in the latest video, after the half-rotation but before the restacki=
ng
> move).
>
>
> I think it's more correct to say that the K face is either an octahedron =
at
> the origin (A<>K<>A) or in the center of one of the main halves, with the=
A
> face inside the other half (>A<>K<). This was what I was getting at in my
> previous message. You do later talk about octahedral faces being in eithe=
r
> the center or the two main halves, so this is just terminology. But about
> "the outer corners of the center 2x2x2 block", this cannot be the A or K
> face as you've labeled them. You've been calling these the L/R faces, but
> the left-right distinction disappears in the half-rotated state, so maybe
> "left" and "right" aren't the best names. To me, they are always the
> "outside" faces, regardless. You can distinguish them as the left and rig=
ht
> outside faces in one representation, or as the center and end outside fac=
es
> in the other. (Or perhaps "end" versus "ind" if we want to be cute.)
>
> I'm also a little torn about naming the interior faces ana and kata, not
> because of the names themselves which I like, but because the mysterious
> faces to me are the outermost ones you're calling R and L. It only requir=
es
> a simple rotation to move faces in and out of the interior (octahedral)
> positions, but it's much more difficult to move another axis into the
> L/R/outer direction.
>
> So maybe the directions can be
>
> Up-Down
> Front-Back
> Ind-End
> Ana-Kata
>
> I'm not in love with it and will be happy with anything that works. Thoug=
hts
> anyone?
>
>
> We also need a name for normal 3D-rotations, restacking moves and folds
> (note that a half-rotation is a kind of restacking move). Let O be the na=
me
> for a rOtation (note that the origin O doesn't move during a rotation, by
> the way, these are 3D rotations of the physical puzzle), let S represent =
a
> reStacking move and V a folding/clamshell move (you can remember this by
> thinking of V as a folded line).
>
>
> I think it's fine to call the clamshell move a fold or denote it as V. I
> just wouldn't consider it to be a basic move since it's a simple composit=
e
> of 3 basic twists as shown here. In general, I think there are so many
> useful composite moves that we need to be able to easily make them up ad =
hoc
> with substitutions like Let =E2=86=93 =3D Rx Lx'. These are really macro =
moves which
> can be nested. That said, it's a particularly useful move so it's probabl=
y
> worth describing in some formal way like you do in detail below.
>
>
>
> Further, let I (capital i) be the identity, preserving the state and
> rotation of the puzzle. We cannot use I to indicate what moves should be
> performed but it's still useful as we will see later. Since we also want =
to
> be able to express if a sequence of moves is a rotation, preserving the
> state of the puzzle but possibly representing it in a different way, we c=
an
> introduce mod(rot) (modulo rotation). So, if a move sequence P satisfies =
P
> mod(rot) =3D I, that means that the state of the puzzle is the same befor=
e and
> after P is performed although the rotation and representation of the puzz=
le
> are allowed to change. I do also want to introduce mod(3rot) (modulo
> 3D-rotation) and P mod(3rot) =3D I means that if the right 3D-rotation (a
> combination of O moves as we will see later) is applied to the puzzle aft=
er
> P you get the identity I. Moreover, let the standard rotation of the puzz=
le
> be any rotation such that the longer side is parallel with the z-axis, th=
at
> is the puzzle forms a 2x2x4 (2 pieces thick in the x-direction, 2 in the
> y-direction and 4 in the z-direction), and the K face is an octahedron.
>
> Rotations and twists:
>
> Now we can move on to name actual moves. The notation of a move is a
> combination of a capital letter and a lowercase letter. O followed by x, =
y
> or z is a rotation of the whole puzzle around the corresponding axis in t=
he
> mathematical positive direction (counterclockwise/the way your right-hand
> fingers curl if you point in the direction of the axis with your thumb). =
For
> example, Ox is a rotation around the x-axis that turns a 2x2x4 into a 2x4=
x2.
> A name of a face (U, D, F, B, R, L, K or A) followed by x, y or z means:
> detach the 8 pieces that have a sticker belonging to the face and then tu=
rn
> those pieces around the global axis. For example, if the longer side of t=
he
> puzzle is parallel to the z-axis (the standard rotation), Rx means: take =
the
> right 2x2x2 block and turn it around the global x-axis in the mathematica=
l
> positive direction. Note that what moves are physically possible and allo=
wed
> is determined by the rotation of the puzzle (I will come back to this
> later). Further note that Ox mod(rot) =3D Oy mod(rot) =3D Oz mod(rot) =3D=
I,
> meaning that the 3D-rotations of the physical puzzle corresponds to a
> 4D-rotation of the represented 2x2x2x2.
>
>
> There are several, distinct types of rotations, none of which change the
> state of the puzzle, and I think we need a way to be unambiguous about th=
em.
> The types I see are
>
> Simple reorientation of the physical puzzle in the hand, no magnets
> involved. IE your 'O' moves and maybe analogous to mouse-dragging in MC4D=
?
> Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-clic=
k?
> Half-rotations. Maybe analogous to a "face first" view? (ctrl-click on a
> 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by Cubie")
> Whole-puzzle reorientations that move an arbitrary axis into the "outer" =
2
> faces. No MC4D analog.
>
>
> Inverses and performing a move more than once
>
> To mark that a move should be performed n times let's put ^n after it. Fo=
r
> convenience when writing and speaking let ' (prime) represent ^-1 (the
> inverse) and n represent ^n. The inverse P' of some permutation P is the
> permutation that satisfies P P' =3D P' P =3D I (the identity). For exampl=
e (Rx)'
> means: do Rx backwards, which corresponds to rotating the right 2x2x2 blo=
ck
> in the mathematical negative direction (clockwise) around the x-axis and
> (Rx)2 means: perform Rx twice. However, we can also define powers of just
> the lowercase letters, for example, Rx2 =3D Rx^2 =3D Rxx =3D Rx Rx. So x2=
=3Dx^2
> means: do whatever the capital letter specifies two times with respect to
> the x-axis. We can see that the capital letter naturally is distributed o=
ver
> the two lowercase letters. Rx' =3D Rx^-1 means: do whatever the capital l=
etter
> specifies but in the other direction than you would have if the prime
> wouldn't have been there (note that thus, x'=3Dx^-1 can be seen as the
> negative x-axis). Note that (Rx)^2 =3D Rx^2 and (Rx)' =3D Rx' which is tr=
ue for
> all twists and rotations but that doesn't have to be the case for other
> types of moves (restacks and folds).
>
> Restacking moves
>
> A restacking move is an S followed by either x, y or z. Here the lowercas=
e
> letter specifies in what direction to restack. For example, Sy (from the
> standard rotation) means: take the front 8 pieces and put them at the bac=
k,
> whereas Sx means: take the top 8 pieces and put them at the bottom. Note
> that Sx is equivalent to taking the bottom 8 pieces and putting them at t=
he
> top. However, if we want to be able to make half-rotations we sometimes n=
eed
> to restack through a plane that doesn't go through the origin. In the
> standard rotation, let Sz be the normal restack (taking the 8 right piece=
s,
> the right 2x2x2 block, and putting them on the left), Sz+ be the restack
> where you split the puzzle in the plane further in the positive z-directi=
on
> (taking the right 2x2x1 cap of 4 pieces and putting it at the left end of
> the puzzle) and Sz- the restack where you split the puzzle in the plane
> further in the negative z-direction (taking the left 2x2x1 cap of 4 piece=
s
> and putting it at the right end of the puzzle). If the longer side of the
> puzzle is parallel to the x-axis instead, Sx+ would take the top 1x2x2 ca=
p
> and put it on the bottom. Note that in the standard rotation Sz mod(rot) =
=3D
> I. For restacks we see that (Sx)' =3D Sx' =3D Sx (true for y and z too of
> course), that (Sz+)' =3D Sz- and that (Sz+)2 =3D (Sz-)2 =3D Sz. We can de=
fine Sz'+
> to have meaning by thinking of z=E2=80=99 as the negative z-axis and with=
that in
> mind it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus, (Sz)=E2=80=
=99 =3D Sz=E2=80=99 and (Sz+)=E2=80=99 =3D Sz=E2=80=99+.
>
> Fold moves
>
> A fold might be a little bit harder to describe in an intuitive way. Firs=
t,
> let's think about what folds are interesting moves. The folds that cannot=
be
> expressed as rotations and restacks are unfolding the puzzle to a 4x4 and
> then folding it back along another axis. If we start with the standard
> rotation and unfold the puzzle into a 1x4x4 (making it look like a 4x4 fr=
om
> above) the only folds that will achieve something you can't do with a
> restack mod(rot) is folding it to a 2x4x2 so that the longer side is
> parallel with the y-axis after the fold. Thus, there are 8 interesting fo=
ld
> moves for any given rotation of the puzzle since there are 4 ways to unfo=
ld
> it to a 4x4 and then 2 ways of folding it back that make the move differe=
nt
> from a restack move mod(rot).
>
>
> Assuming you complete a folding move in the same representation (<><> or
>><><), then there are only two interesting choices. That's because it
> doesn't matter which end of a chosen cutting plane you open it from, the =
end
> result will be the same. That also means that any two consecutive clamshe=
ll
> moves along the same cutting plane will undo each other. It further sugge=
sts
> that any interesting sequence of clamshell moves must alternate between t=
he
> two possible long cut directions, meaning there is no choice involved. 12
> clamshell moves will cycle back to the initial state.
>
> There is one other weird folding move where you open it in one direction =
and
> then fold the two halves back-to-back in a different direction. If you
> simply kept folding along the initial hinge, you'd simply have a restacki=
ng.
> When completed the other way, it's equivalent to a restacking plus a
> clamshell, so I don't think it's useful though it is somewhat interesting=
.
>
>
> Let's call these 8 folds interesting fold moves. Note that an interesting
> fold move always changes which axis the longer side of the puzzle is
> parallel with. Further note that both during unfold and fold all pieces a=
re
> moved; it would be possible to have 8 of the pieces fixed during an unfol=
d
> and folding the other half 180 degrees but I think that it=E2=80=99s more=
intuitive
> that these moves fold both halves 90 degrees and performing them with
> 180-degree folds might therefore lead to errors since the puzzle might ge=
t
> rotated differently. To illustrate a correct unfold without a puzzle: Put
> your palms together such that your thumbs point upward and your fingers
> forward. Now turn your right hand 90 degrees clockwise and your left hand=
90
> degrees counterclockwise such that the normal to your palms point up, you=
r
> fingers point forward, your right thumb to the right and your left thumb =
to
> the left. That was what will later be called a Vx unfold and the folds ar=
e
> simply reversed unfolds. (I might have used the word =E2=80=9Cfold=E2=80=
=9D in two different
> ways but will try to use the term =E2=80=9Cfold move=E2=80=9D when referr=
ing to the move
> composed by an unfold and a fold rather than simply calling these moves
> =E2=80=9Cfolds=E2=80=9D.)
>
> To specify the unfold let's use V followed by one of x, y, z, x', y' and =
z'.
> The lowercase letter describes in which direction to unfold. Vx means unf=
old
> in the direction of the positive x-axis and Vx' in the direction of the
> negative x-axis, if that makes any sense. I will try to explain more
> precisely what I mean with the example Vx from the standard rotation (it
> might also help to read the last sentences in the previous paragraph agai=
n).
> So, the puzzle is in the standard rotation and thus have the form 2x2x4 (=
x-,
> y- and z-thickness respectively). The first part (the unfolding) of the m=
ove
> specified with Vx is to unfold the puzzle in the x-direction, making it a
> 1x4x4 (note that the thickness in the x-direction is 1 after the Vx unfol=
d,
> which is no coincidence). There are two ways to do that; either the sides=
of
> the pieces that are initially touching another piece (inside of the puzzl=
e
> in the x,z-plane and your palms in the hand example) are facing up or dow=
n
> after the unfold. Let Vx be the unfold where these sides point in the
> direction of the positive x-axis (up) and Vx' the other one where these
> sides point in the direction of the negative x-axis (down) after the unfo=
ld.
> Note that if the longer side of the puzzle is parallel to the z-axis only
> Vx, Vx', Vy and Vy' are possible. Now we need to specify how to fold the
> puzzle back to complete the folding move. Given an unfold, say Vx, there =
are
> only two ways to fold that are interesting (not turning the fold move int=
o a
> restack mod(rot)) and you have to fold it perpendicular to the unfold to
> create an interesting fold move. So, if you start with the standard rotat=
ion
> and do Vx you have a 1x4x4 that you have to fold into a 2x4x2. To
> distinguish the two possibilities, use + or - after the Vx. Let Vx+ be th=
e
> unfold Vx followed by the interesting fold that makes the sides that are
> initially touching another piece (before the unfold) touch another piece
> after the fold move is completed and let Vx- be the other interesting fol=
d
> move that starts with the unfold Vx. (Thus, continuing with the hand
> example, if you want to do a Vx+ first do the Vx unfold described in the =
end
> of the previous paragraph and then fold your hands such that your fingers
> point up, the normal to your palms point forward, the right palm is touch=
ing
> the right-hand fingers, the left palm is touching the left-hand fingers, =
the
> right thumb is pointing to the right and the left thumb is pointing to th=
e
> left). Note that the two halves of the puzzle always should be folded 90
> degrees each and you should never make a fold or unfold where you fold ju=
st
> one half 180-degrees (if you want to use my notation, that is). Further n=
ote
> that Vx+ Sx mod(3rot) =3D Vx- and that Vx+ Vx+ =3D I which is equivalent =
to
> (Vx+)=E2=80=99 =3D Vx+ and this is true for all fold moves (note that aft=
er a Vx+
> another Vx+ is always possible).
>
> The 2x2x2x2 in the MC4D software
>
> The notation above can also be applied to the 2x2x2x2 in the MC4D program=
.
> There, you are not allowed to do S or V moves but instead, you are allowe=
d
> to do the [crtl]+[left-click] moves. This can easily be represented with
> notation similar to the above. Let=E2=80=99s use C (as in Centering) and =
one of x, y
> and z. For example, Cx would be to rotate the face in the positive
> x-direction aka the U face to the center. Thus, Cz' is simply
> [ctrl]+[left-click] on the L face and similarly for the other C moves. Th=
e
> O, U, D, F, B, R, L, K and A moves are performed in the same way as above
> so, for example Rx would be a [left-click] on the top-side of the right
> face. In this representation of the puzzle almost all moves are allowed; =
all
> U, D, F, B, R, L, K, O and C moves are possible regardless of rotation an=
d
> only A moves (and of course the rightfully forbidden S and V moves) are
> impossible regardless of rotation. Note that R and L moves in the softwar=
e
> correspond to the same moves of the physical puzzle but this is not
> generally true (I will come back to this later).
>
> Possible moves (so far) in the standard rotation
>
> In the standard rotation, the possible/allowed moves with the definitions
> above are:
>
> O moves, all of these are always possible in any state and rotation of th=
e
> puzzle since they are simply 3D-rotations.
>
> R, L moves, all of these as well since the puzzle has a right and left 2x=
2x2
> block in the standard rotation.
>
> U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x4 blocks wi=
th
> less symmetry than a 2x2x2 block.
>
> F, B moves, just Fy2 and By2 since the front and back are 2x1x4 blocks wi=
th
> less symmetry than a 2x2x2 block.
>
> K moves, all possible since this is a rotation of the center 2x2x2 block.
>
>
> I think the only legal 90 degree twists of the K face are those about the
> long axis. I believe this is what Christopher Locke was saying in this
> message. To see why there is no straightforward way to perform other 90
> degree twists, you only need to perform a 90 degree twist on an outer (L/=
R)
> face and then reorient the whole puzzle along a different outer axis. If =
the
> original twist was not about the new long axis, then there is clearly no
> straightforward way to undo that twist.
>
>
> A moves, only Az moves since this is two 2x2x1 blocks that have to be
> rotated together.
>
> S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined in the
> standard rotation.
>
> V moves, not Vz+ or Vz- since the definition doesn't give these meaning w=
hen
> the long side of the puzzle is parallel with the z-axis.
>
> Note that for example Fz2 is not allowed since this won't take you to a
> state of the puzzle. To allow more moves we need to extend the definition=
s
> (after the extension in the next paragraph all rotations and twists (O, R=
,
> L, U, D, F, B, K, A) are possible in any rotation and only which S and V
> moves are possible depend on the state and rotation of the puzzle).
>
>
> Extension of some definitions
>
> It's possible to make an extension that allows all O, R, L, U, D, F, B, K
> and A moves in any state. I will explain how this can be done in the
> standard rotation but it applies analogously to any other rotation where =
the
> K face is an octahedron. First let's focus on U, D, F and B and because o=
f
> the symmetry of the puzzle all of these are analogous so I will only expl=
ain
> one. The extension that makes all U moves possible (note that the U face =
is
> in the positive x-direction) is as follows: when making an U move first
> detach the 8 top pieces which gives you a 1x2x4 block, fold this block in=
to
> a 2x2x2 block in the positive x-direction (similar to the later part of a
> Vx+ move from the standard rotation) such that the U face form an
> octahedron, rotate this 2x2x2 block around the specified axis (for exampl=
e
> around the z-axis if you are doing an Uz move), reverse the fold you just
> did creating a 1x2x4 block again and reattach the block.
>
>
> I noticed something like this the other day but realized that it only see=
ms
> to work for rotations along the long dimension (z in your example). These
> are already easily accomplished by a simple rotation to put the face in
> question on the end caps, followed by a double end-cap twist.
>
> This is as far as I'm going to comment for the moment because the
> information gets very dense and I've been mulling and picking over your
> message for several days already. In short, I really like your attempt to
> provide a complete system of notation for discussing this puzzle and will=
be
> curious to hear your thoughts on my comments so far. I hope others will
> chime in too.
>
> One final thought is that a real "acid test" of any notation system for t=
his
> puzzle will be attempt to translate some algorithms from MC4D. I would mo=
st
> like to see a sequence that flips a single piece, like the second 4-color
> series on this page of Roice's solution, or his pair of twirled corners a=
t
> the end of this page. One trick will be to minimize the number of
> whole-puzzle reorientations needed, but really any sequence that works wi=
ll
> be great evidence that the puzzles are equivalent. I suspect that this so=
rt
> of exercise will never be practical because it will require too many
> reorientations, and that entirely new methods will be needed to actually
> solve this puzzle.
>
> Best,
> -Melinda
>
> The A moves can be done very similarly but after you have detached the tw=
o
> 2x2x1 blocks you don't fold them but instead you stack them similar to a =
Sz
> move, creating a 2x2x2 block with the A face as an octahedron in the midd=
le
> and then reverse the process after you have rotated the block as specifie=
d
> (for example around the negative y-axis if you are doing an Ay' move). No=
te
> that these extended moves are closely related to the normal moves and for
> example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99 in the standard rotation =
and note that (Ry
> Ly=E2=80=99) mod(rot) =3D (Ly Ry=E2=80=99) mod(rot) =3D I (this applies t=
o the other extended
> moves as well).
>
> If the cube is in the half-rotated state, where both the R and L faces ar=
e
> octahedra, you can extend the definitions very similarly. The only thing =
you
> have to change is how you fold the 2x4 blocks when performing a U, D, F o=
r B
> move. Instead of folding the two 2x2 halves of the 2x4 into a 2x2 you hav=
e
> to fold the end 1x2 block 180 degrees such that the face forms an
> octahedron.
>
> These moves might be a little bit harder to perform, to me especially the=
A
> moves seems a bit awkward, so I don't know if it's good to use them or no=
t.
> However, the A moves are not necessary if you allow Sz in the standard
> rotation (which you really should since Sz mod(rot) =3D I in the standard
> rotation) and thus it might not be too bad to use this extended version. =
The
> notation supports both variants so if you don=E2=80=99t want to use these=
extended
> moves that shouldn=E2=80=99t be a problem. Note that, however, for exampl=
e Ux
> (physical puzzle) !=3D Ux (virtual puzzle) where !=3D means =E2=80=9Cnot =
equal to=E2=80=9D (more
> about legal/illegal moves later).
>
> Generalisation of the notation
>
> Let=E2=80=99s generalise the notation to make it easier to use and to mak=
e it work
> for any n^4 cube in the MC4D software. Previously, we saw that (Rx)2 =3D =
Rx Rx
> and if we allow ourselves to rewrite this as (Rx)2 =3D Rx2 =3D Rxx =3D Rx=
Rx we
> see that the capital letter naturally can be distributed over the lowerca=
se
> letters. We can make this more general and say that any capital letter
> followed by several lowercase letters means the same thing as the capital
> letter distributed over the lowercase letters. Like Rxyz =3D Rx Ry Rz and=
here
> R can be exchanged with any capital letter and xyz can be exchanged with =
any
> sequence of lowercase letters. We can also allow several capital letters =
and
> one lowercase letter, for example RLx and let=E2=80=99s define this as RL=
x =3D Rx Lx
> so that the lowercase letter can be distributed over the capital letters.=
We
> can also define a capital letter followed by =E2=80=98 (prime) like R=E2=
=80=99x =3D Rx=E2=80=99 and
> R=E2=80=99xy =3D Rx=E2=80=99y=E2=80=99 so the =E2=80=98 (prime) is distri=
buted over the lowercase letters.
> Note that we don=E2=80=99t define a capital to any other power than -1 li=
ke this
> since for example R2x =3D RRx might seem like a good idea at first but it
> isn=E2=80=99t very useful since R2 and RR are the same lengths (and power=
s greater
> than two are seldom used) and we will see that we can define R2 in anothe=
r
> way that generalises the notation to all n^4 cubes.
>
> Okay, let=E2=80=99s define R2 and similar moves now and have in mind what=
moves we
> want to be possible for a n^4 cube. The moves that we cannot achieve with
> the notation this far is twisting deeper slices. To match the notation wi=
th
> the controls of the MC4D software let R2x be the move similar to Rx but
> twisting the 2nd layer instead of the top one and similarly for other
> capital letters, numbers (up to n) and lowercase letters. Thus, R2x is
> performed as Rx but holding down the number 2 key. Just as in the program=
,
> when no number is specified 1 is assumed and you can combine several numb=
ers
> like R12x to twist both the first and second layer. This notation does no=
t
> apply to rotations (O) folding moves (V) and restacking moves (S) (I supp=
ose
> you could redefine the S move using this deeper-slice-notation and use S1=
z
> as Sz+, S2z as Sz and S3z as Sz- but since these moves are only allowed f=
or
> the physical 2x2x2x2 I think that the notation with + and =E2=80=93 is be=
tter since
> S followed by a lowercase letter without +/- always means splitting the c=
ube
> in a coordinate plane that way, not sure though so input would be great).
> The direction of the twist R3x should be the same as Rx meaning that if R=
x
> takes stickers belonging to K and move them to F, so should R3x, in
> accordance with the controls of the MC4D software. Note that for a 3x3x3x=
3
> it=E2=80=99s true that R3z =3D Lz whereas R3x =3D Lx=E2=80=99 (note that =
R and L are the faces
> in the z-directions so because of the symmetry of the cube it will also b=
e
> true that for example U3x =3D Bx whereas U3y=3DBy=E2=80=99).
>
> What about the case with several capital letters and several lowercase
> letters, for instance, RLxy? I see two natural definitions of this. Eithe=
r,
> we could have RLxy =3D Rxy Lxy or we could define it as RLxy =3D RLx RLy.=
These
> are generally not the same (if you exchange R and L with any allowed capi=
tal
> letter and similarly for x and y). I don=E2=80=99t know what is best, wha=
t do you
> think? The situations I find this most useful in are RL=E2=80=99xy to do =
a rotation
> and RLxy as a twist. However, since R and L are opposite faces their
> operations commute which imply RL=E2=80=99xy (1st definition) =3D Rxy L=
=E2=80=99xy =3D Rx Ry
> Lx=E2=80=99 Ly=E2=80=99 =3D Rx Lx=E2=80=99 Ry Ly=E2=80=99 =3D RL=E2=80=99=
x RL=E2=80=99y =3D RL=E2=80=99xy (2nd definition) and similarly
> for the other case with RLxy. Hopefully, we can find another useful seque=
nce
> of moves where this notation can be used with only one of the definitions
> and can thereby decide which definition to use. Personally, I feel like R=
Lxy
> =3D RLx RLy is the more intuitive definition but I don=E2=80=99t have any=
good
> argument for this so I=E2=80=99ll leave the question open.
>
> For convenience, it might be good to be able to separate moves like Rxy a=
nd
> RLx from the basic moves Rx, Oy etc when speaking and writing. Let=E2=80=
=99s call
> the basic moves that only contain one capital letter and one lowercase
> letter (possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or a number) s=
imple moves (like
> Rx, L'y, Ux2 and D3y=E2=80=99) and the moves that contain more than one c=
apital
> letter or more than one lowercase letter composed moves.
>
> More about inverses
>
> This list can obviously be made longer but here are some identities that =
are
> good to know and understand. Note that R, L, U, x, y and z below just are
> examples, the following is true in general for non-folds (however, S move=
s
> are fine).
>
> (P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99 P1=E2=80=99=
(Pi is an arbitrary permutation for
> i=3D1,2,=E2=80=A6n)
> (Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx
> (RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x
> (Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+ (just as an example with r=
estacking moves, note that
> the inverse doesn=E2=80=99t change the + or -)
> RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=80=99xyz (=
true for both definitions)
> (RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=80=99yx (=
true for both definitions)
>
> For V moves we have that: (Vx+)=E2=80=99 =3D Vx+ !=3D Vx=E2=80=99+ (!=
=3D means =E2=80=9Cnot equal to=E2=80=9D)
>
> Some important notes on legal/illegal moves
>
> Although there are a lot of moves possible with this notation we might no=
t
> want to use them all. If we really want a 2x2x2x2 and not something else =
I
> think that we should try to stick to moves that are legal 2x2x2x2 moves a=
s
> far as possible (note that I said legal moves and not permutations (a leg=
al
> permutation can be made up of one or more legal moves)). Clarification:
> cycling three of the edge-pieces of a Rubik=E2=80=99s cube is a legal per=
mutation
> but not a legal move, a legal move is a rotation of the cube or a twist o=
f
> one of the layers. In this section I will only address simple moves and
> simply refer to them as moves (legal composed moves are moves composed by
> legal simple moves).
>
> I do believe that all moves allowed by my notation are legal permutations
> based on their periodicity (they have a period of 2 or 4 and are all even
> permutations of the pieces). So, which of them correspond to legal 2x2x2x=
2
> moves? The O moves are obviously legal moves since they are equal to the
> identity mod(rot). The same goes for restacking (S) (with or without +/-)=
in
> the direction of the longest side of the puzzle (Sz, Sz+ and Sz- in the
> standard rotation) since these are rotations and half-rotations that don=
=E2=80=99t
> change the state of the puzzle. Restacking in the other directions and fo=
ld
> moves (V) are however not legal moves since they are made of 8 2-cycles a=
nd
> change the state of the puzzle (note that they, however, are legal
> permutations). The rest of the moves (R, L, F, B, U, D, K and A) can be
> divided into two sets: (1) the moves where you rotate a 2x2x2 block with =
an
> octahedron inside and (2) the moves where you rotate a 2x2x2 block withou=
t
> an octahedron inside. A move belonging to (2) is always legal. We can see
> this by observing what a Rx does with the pieces in the standard rotation
> with just K forming an octahedron. The stickers move in 6 4-cycles and if
> the puzzle is solved the U and D faces still looks solved after the move.=
A
> move belonging to set (1) is legal either if it=E2=80=99s an 180-degree t=
wist or if
> it=E2=80=99s a rotation around the axis parallel with the longest side of=
the puzzle
> (the z-axis in the standard rotation). Quite interestingly these are exac=
tly
> the moves that don=E2=80=99t mix up the R and L stickers with the rest in=
the
> standard rotation. I think I know a way to prove that no legal 2x2x2x2 mo=
ve
> can mix up these stickers with the rest and this has to do with the fact
> that these stickers form an inverted octahedron (with the corners pointin=
g
> outward) instead of a normal octahedron (let=E2=80=99s call this hypothes=
is * for
> now). Note that all legal twists (R, L, F, B, U, D, K and A) of the physi=
cal
> puzzle correspond to the same twist in the MC4D software.
>
> So, what moves should we add to the set of legal moves be able to get to
> every state of the 2x2x2x2? I think that we should add the restacking mov=
es
> and folding moves since Melinda has already found a pretty short sequence=
of
> those moves to make a rotation that changes which colours are on the R an=
d L
> faces. That sequence, starting from the standard rotation, is: Oy Sx+z Vy=
+
> Ozy=E2=80=99 Vy+ Ozy=E2=80=99 Vy+ =3D Oy Sx+z (Vy+ Ozy=E2=80=99)3 mod(3r=
ot) =3D I mod(rot) (hopefully I
> got that right). What I have found (which I mentioned previously) is that
> (from the standard rotation): RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz- and=
since this
> is equivalent with Sy =3D Ly2 RLx=E2=80=99 Sz- RLx Ly2 By2 the restacking=
moves that
> are not legal moves are not very complicated permutations and therefore I
> think that we can accept them since they help us mix up the R and L stick=
ers
> with other faces. In a sense, the folding moves are =E2=80=9Cmore illegal=
=E2=80=9D since
> they cannot be composed by the legal moves (according to hypothesis *). T=
his
> is also true for the illegal moves belonging to set (1) discussed above.
> However, since the folding moves is probably easier to perform and is eno=
ugh
> to reach every state of the 2x2x2x2 I think that we should use them and n=
ot
> the illegal moves belonging to (1). Note, once again, that all moves
> described by the notation are legal permutations (even the ones that I ju=
st
> a few words ago referred to as illegal moves) so if you wish you can use =
all
> of them and still only reach legal 2x2x2x2 states. However (in a strict
> sense) one could argue that you are not solving the 2x2x2x2 if you use
> illegal moves. If you only use illegal moves to compose rotations (that i=
s,
> create a permutation including illegal moves that are equal to I mod(rot)=
)
> and not actually using the illegal moves as twists I would classify that =
as
> solving a 2x2x2x2. What do you think about this?
>
> What moves to use?
>
> Here=E2=80=99s a short list of the simple moves that I think should be us=
ed for the
> physical 2x2x2x2. Note that this is just my thoughts and you may use the
> notation to describe any move that it can describe if you wish to. The
> following list assumes that the puzzle is in the standard rotation but is
> analogous for other representations where the K face is an octahedron.
>
> O, all since they are I mod(rot),
> R, L, all since they are legal (note Rx (physical puzzle) =3D Rx (virtual
> 2x2x2x2)),
> U, D, only x2 since these are the only legal easy-to-perform moves,
> F, B, only y2 since these are the only legal easy-to-perform moves,
> K, A, only z, z=E2=80=99 and z2 since these are the only legal easy-to-pe=
rform
> moves,
> S, at least z, z+ and z- since these are equal to I mod(rot),
> S, possibly x and y since these help us perform rotations and is easy to
> compose (not necessary to reach all states and not legal though),
> V, all 8 allowed by the rotation of the puzzle (at least one is necessary=
to
> reach all states and if you allow one the others are easy to achieve
> anyway).
>
> If you start with the standard rotation and then perform Sz+ the followin=
g
> applies instead (this applies analogously to any other rotation where the=
R
> and L faces are octahedra).
>
> O, all since they are I mod(rot),
> R, L, only z, z=E2=80=99 and z2 since these are the only legal easy-to-pe=
rform
> moves,
> U, D, only x2 since these are the only legal easy-to-perform moves,
> F, B, only y2 since these are the only legal easy-to-perform moves,
> K, A, at least z, z=E2=80=99 and z2, possibly all (since they are legal) =
although
> some might be hard to perform.
> S, V, same as above.
>
> Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D I mod(rot) (=
!=3D
> for not equal) which implies that the Ux2 move when R and L are octahedra=
is
> different from the Ux2 move when K is an octahedron. (Actually, the seque=
nce
> above is equal to Uy2).
>
> Regarding virtual n^4 cubes I think that all O, C, R, L, U, D, F, B, K an=
d A
> moves should be used since they are all legal and really the only thing y=
ou
> need (left-clicking on an edge or corner piece in the computer program ca=
n
> be described quite easily with the notation, for example, Kzy2 is
> left-clicking on the top-front edge piece on the K face).
>
> I hope this was possible to follow and understand. Feel free to ask
> questions about the notation if you find anything ambiguous.
>
> Best regards,
> Joel Karlsson
>
> Den 4 maj 2017 12:01 fm skrev "Melinda Green melinda@superliminal.com
> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>
>
>
> Thanks for the correction. A couple of things: First, when assembling one
> piece at a time, I'd say there is only 1 way to place the first piece, no=
t
> 24. Otherwise you'd have to say that the 1x1x1 puzzle has 24 states. I
> understand that this may be conventional, but to me, that just sounds sil=
ly.
>
> Second, I have the feeling that the difference between the "two
> representations" you describe is simply one of those half-rotations I sho=
wed
> in the video. In the normal solved state there is only one complete
> octahedron in the very center, and in the half-rotated state there is one=
in
> the middle of each half of the "inverted" form. I consider them to be the
> same solved state.
>
> -Melinda
>
>
> On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
> wrote:
>
> Horrible typo... It seems like I made some typos in my email regarding th=
e
> state count. It should of course be 16!12^16/(6*192) and NOT
> 12!16^12/(6*192). However, I did calculate the correct number when compar=
ing
> with previous results so the actual derivation was correct.
>
> Something of interest is that the physical pieces can be assembled in
> 16!24*12^15 ways since there are 16 pieces, the first one can be oriented=
in
> 24 ways and the remaining can be oriented in 12 ways (since a corner with=
3
> colours never touch a corner with just one colour). Dividing with 6 to ge=
t a
> single orbit still gives a factor 2*192 higher than the actual count rath=
er
> than 192. This shows that every state in the MC4D representation has 2
> representations in the physical puzzle. These two representations must be
> the previously discussed, that the two halves either have the same color =
on
> the outermost corners or the innermost (forming an octahedron) when the
> puzzle is solved and thus both are complete representations of the 2x2x2x=
2.
>
> Best regards,
> Joel Karlsson
>
> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" >:
>
> I am no expert on group theory, so to better understand what twists are
> legal I read through the part of Kamack and Keane's The Rubik Tesseract
> about orienting the corners. Since all even permutations are allowed the
> easiest way to check if a twist is legal might be to:
> 1. Check that the twist is an even permutation, that is: the same twist c=
an
> be done by performing an even number of piece swaps (2-cycles).
> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning performing=
the
> twist k times and I (the identity) representing the permutation of doing
> nothing) and k is not divisible by 3 the twist A definitely doesn't viola=
te
> the restriction of the orientations since kx mod 3 =3D 0 and k mod 3 !=3D=
0
> implies x mod 3 =3D 0 meaning that the change of the total orientation x =
for
> the twist A mod 3 is 0 (which precisely is the restriction of legal twist=
s;
> that they must preserve the orientation mod 3).
>
> For instance, this implies that the restacking moves are legal 2x2x2x2 mo=
ves
> since both are composed of 8 2-cycles and both can be performed twice (no=
te
> that 2 is not divisible by 3) to obtain the identity.
>
> Note that 1 and 2 are sufficient to check if a twist is legal but only 1 =
is
> necessary; there can indeed exist a twist violating 2 that still is legal
> and in that case, I believe that we might have to study the orientation
> changes for that specific twist in more detail. However, if a twist can b=
e
> composed by other legal twists it is, of course, legal as well.
>
> Best regards,
> Joel
>
> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com [4D_Cubi=
ng]
> <4D_Cubing@yahoogroups.com>:
>>
>>
>>
>> First off, thanks everyone for the helpful and encouraging feedback!
>> Thanks Joel for showing us that there are 6 orbits in the 2^4 and for yo=
ur
>> rederivation of the state count. And thanks Matt and Roice for pointing =
out
>> the importance of the inverted views. It looks so strange in that
>> configuration that I always want to get back to a normal view as quickly=
as
>> possible, but it does seem equally valid, and as you've shown, it can be
>> helpful for more than just finding short sequences.
>>
>> I don't understand Matt's "pinwheel" configuration, but I will point out
>> that all that is needed to create your twin interior octahedra is a sing=
le
>> half-rotation like I showed in the video at 5:29. The two main halves do=
end
>> up being mirror images of each other on the visible outside like he
>> described. Whether it's the pinwheel or the half-rotated version that's
>> correct, I'm not sure that it's a bummer that the solved state is not at=
all
>> obvious, so long as we can operate it in my original configuration and
>> ignore the fact that the outer faces touch. That would just mean that th=
e
>> "correct" view is evidence that that the more understandable view is
>> legitimate.
>>
>> I'm going to try to make a snapable V3 which should allow the pieces to =
be
>> more easily taken apart and reassembled into other forms. Shapeways does
>> offer a single, clear translucent plastic that they call "Frosted Detail=
",
>> and another called "Transparent Acrylic", but I don't think that any sor=
t of
>> transparent stickers will help us, especially since this thing is chock =
full
>> of magnets. The easiest way to let you see into the two hemispheres woul=
d be
>> to simply truncate the pointy tips of the stickers. That already happens=
a
>> little bit due to the way I've rounded the edges. Here is a close-up of =
a
>> half-rotation in which you can see that the inner yellow and white faces=
are
>> solved. Your suggestion of little mapping dots on the corners also works=
,
>> but just opening the existing window further would work more directly.
>>
>> -Melinda
>>
>>
>> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>>
>> I agree with Don's arguments about adjacent sticker colors needing to be
>> next to each other. I think this can be turned into an accurate 2^4 wit=
h
>> coloring changes, so I agree with Joel too :)
>>
>> To help me think about it, I started adding a new projection option for
>> spherical puzzles to MagicTile, which takes the two hemispheres of a puz=
zle
>> and maps them to two disks with identified boundaries connected at a poi=
nt,
>> just like a physical "global chess" game I have. Melinda's puzzle is a =
lot
>> like this up a dimension, so think about two disjoint balls, each
>> representing a hemisphere of the 2^4, each a "subcube" of Melinda's puzz=
le.
>> The two boundaries of the balls are identified with each other and as yo=
u
>> roll one around, the other half rolls around so that identified points
>> connect up. We need to have the same restriction on Melinda's puzzle.
>>
>> In the pristine state then, I think it'd be nice to have an internal
>> (hidden), solid colored octahedron on each half. The other 6 faces shou=
ld
>> all have equal colors split between each hemisphere, 4 stickers on each
>> half. You should be able to reorient the two subcubes to make a half
>> octahedron of any color on each subcube. I just saw Matt's email and
>> picture, and it looks like we were going down the same thought path. I
>> think with recoloring (mirroring some of the current piece colorings)
>> though, the windmill's can be avoided (?)
>>
>> [...] After staring/thinking a bit more, the coloring Matt came up with =
is
>> right-on if you want to put a solid color at the center of each hemisphe=
re.
>> His comment about the "mirrored" pieces on each side helped me understan=
d
>> better. 3 of the stickers are mirrored and the 4th is the hidden color
>> (different on each side for a given pair of "mirrored" pieces). All fac=
es
>> behave identically as well, as they should. It's a little bit of a bumm=
er
>> that it doesn't look very pristine in the pristine state, but it does lo=
ok
>> like it should work as a 2^4.
>>
>> I wonder if there might be some adjustments to be made when shapeways
>> allows printing translucent as a color :)
>>
>> [...] Sorry for all the streaming, but I wanted to share one more though=
t.
>> I now completely agree with Joel/Matt about it behaving as a 2^4, even w=
ith
>> the original coloring. You just need to consider the corner colors of t=
he
>> two subcubes (pink/purple near the end of the video) as being a window i=
nto
>> the interior of the piece. The other colors match up as desired. (Sorr=
y if
>> folks already understood this after their emails and I'm just catching u=
p!)
>>
>> In fact, you could alter the coloring of the pieces slightly so that the
>> behavior was similar with the inverted coloring. At the corners where 3
>> colors meet on each piece, you could put a little circle of color of the
>> opposite 4th color. In Matt's windmill coloring then, you'd be able to =
see
>> all four colors of a piece, like you can with some of the pieces on
>> Melinda's original coloring. And again you'd consider the color circles=
a
>> window to the interior that did not require the same matching constraint=
s
>> between the subcubes.
>>
>> I'm looking forward to having one of these :)
>>
>> Happy Friday everyone,
>> Roice
>>
>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
>> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>>>
>>>
>>>
>>> Seems like there was a slight misunderstanding. I meant that you need t=
o
>>> be able to twist one of the faces and in MC4D the most natural choice =
is
>>> the center face. In your physical puzzle you can achieve this type of t=
wist
>>> by twisting the two subcubes although this is indeed a twist of the sub=
cubes
>>> themselves and not the center face, however, this is still the same typ=
e of
>>> twist just around another face.
>>>
>>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
>>> this puzzle. Hopefully the restrictions will be quite natural and only =
some
>>> "strange" moves would be illegal. Regarding the "families of states" (a=
ka
>>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed t=
wists
>>> preserves the parity of the pieces, meaning that only half of the
>>> permutations you can achieve by disassembling and reassembling can be
>>> reached through legal moves. Because of some geometrical properties of =
the
>>> 2x2x2x2 and its twists, which would take some time to discuss in detail
>>> here, the orientation of the stickers mod 3 are preserved, meaning that=
the
>>> last corner only can be oriented in one third of the number of orientat=
ions
>>> for the other corners. This gives a total number of orbits of 2x3=3D6. =
To
>>> check this result let's use this information to calculate all the possi=
ble
>>> states of the 2x2x2x2; if there were no restrictions we would have 16! =
for
>>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>>> orientations for each corner). If we now take into account that there a=
re 6
>>> equally sized orbits this gets us to 12!16^12/6. However, we should als=
o
>>> note that the orientation of the puzzle as a hole is not set by some ki=
nd of
>>> centerpieces and thus we need to devide with the number of orientations=
of a
>>> 4D cube if we want all our states to be separated with twists and not o=
nly
>>> rotations of the hole thing. The number of ways to orient a 4D cube in =
space
>>> (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a tot=
al of
>>> 12!16^12/(6*192) states which is indeed the same number that for exampl=
e
>>> David Smith arrived at during his calculations. Therefore, when determ=
ining
>>> whether or not a twist on your puzzle is legal or not it is sufficient =
and
>>> necessary to confirm that the twist is an even permutation of the piece=
s and
>>> preserves the orientation of stickers mod 3.
>>>
>>> Best regards,
>>> Joel
>>>
>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
>>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>>
>>>
>>>
>>> The new arrangement of magnets allows every valid orientation of pieces=
.
>>> The only invalid ones are those where the diagonal lines cutting each c=
ube's
>>> face cross each other rather than coincide. In other words, you can ass=
emble
>>> the puzzle in all ways that preserve the overall diamond/harlequin patt=
ern.
>>> Just about every move you can think of on the whole puzzle is valid tho=
ugh
>>> there are definitely invalid moves that the magnets allow. The most obv=
ious
>>> invalid move is twisting of a single end cap.
>>>
>>> I think your description of the center face is not correct though. Twis=
ts
>>> of the outer faces cause twists "through" the center face, not "of" tha=
t
>>> face. Twists of the outer faces are twists of those faces themselves be=
cause
>>> they are the ones not changing, just like the center and outer faces of=
MC4D
>>> when you twist the center face. The only direct twist of the center fac=
e
>>> that this puzzle allows is a 90 degree twist about the outer axis. That
>>> happens when you simultaneously twist both end caps in the same directi=
on.
>>>
>>> Yes, it's quite straightforward reorienting the whole puzzle to put any
>>> of the four axes on the outside. This is a very nice improvement over t=
he
>>> first version and should make it much easier to solve. You may be right=
that
>>> we just need to find the right way to think about the outside faces. I'=
ll
>>> leave it to the math geniuses on the list to figure that out.
>>>
>>> -Melinda
>>>
>>>
>>>
>>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubin=
g]
>>> wrote:
>>>
>>>
>>> Hi Melinda,
>>>
>>> I do not agree with the criticism regarding the white and yellow sticke=
rs
>>> touching each other, this could simply be an effect of the different
>>> representations of the puzzle. To really figure out if this indeed is a
>>> representation of a 2x2x2x2 we need to look at the possible moves (twis=
ts
>>> and rotations) and figure out the equivalent moves in the MC4D software=
.
>>> From the MC4D software, it's easy to understand that the only moves req=
uired
>>> are free twists of one of the faces (that is, only twisting the center =
face
>>> in the standard perspective projection in MC4D) and 4D rotations swappi=
ng
>>> which face is in the center (ctrl-clicking in MC4D). The first is possi=
ble
>>> in your physical puzzle by rotating the white and yellow subcubes (from=
here
>>> on I use subcube to refer to the two halves of the puzzle and the colou=
rs of
>>> the subcubes to refer to the "outer colours"). The second is possible i=
f
>>> it's possible to reach a solved state with any two colours on the subcu=
bes
>>> that still allow you to perform the previously mentioned twists. This s=
eems
>>> to be the case from your demonstration and is indeed true if the magnet=
s
>>> allow the simple twists regardless of the colours of the subcubes. Thus=
, it
>>> is possible to let your puzzle be a representation of a 2x2x2x2, howeve=
r, it
>>> might require that some moves that the magnets allow aren't used.
>>>
>>> Best regards,
>>> Joel
>>>
>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>>>
>>>>
>>>>
>>>> Dear Cubists,
>>>>
>>>> I've finished version 2 of my physical puzzle and uploaded a video of =
it
>>>> here:
>>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>>> Again, please don't share these videos outside this group as their
>>>> purpose is just to get your feedback. I'll eventually replace them wit=
h a
>>>> public video.
>>>>
>>>> Here is an extra math puzzle that I bet you folks can answer: How many
>>>> families of states does this puzzle have? In other words, if disassemb=
led
>>>> and reassembled in any random configuration the magnets allow, what ar=
e the
>>>> odds that it can be solved? This has practical implications if all suc=
h
>>>> configurations are solvable because it would provide a very easy way t=
o
>>>> fully scramble the puzzle.
>>>>
>>>> And finally, a bit of fun: A relatively new friend of mine and new lis=
t
>>>> member, Marc Ringuette, got excited enough to make his own version. He=
built
>>>> it from EPP foam and colored tape, and used honey instead of magnets t=
o hold
>>>> it together. Check it out here:
>>>> http://superliminal.com/cube/dessert_cube.jpg I don't know how practic=
al a
>>>> solution this is but it sure looks delicious! Welcome Marc!
>>>>
>>>> -Melinda
>>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>
>
>
>=20




From: Joel Karlsson <joelkarlsson97@gmail.com>
Date: Mon, 29 May 2017 11:57:18 +0200
Subject: Re: [MC4D] Physical 4D puzzle V2



Hello,

Just a quick correction regarding a previous statement. From the
calculation of the states of the puzzle, we can see that if we choose
a state and rotation there are still two representations of that state
and rotation in the physical puzzle. Previously, we also said that
these states are separated by a half-rotation such as Rx+ from an RL
rep or AKx rep. This is not the case. When I calculated the states of
the puzzle I assumed that the longer side of the puzzle should be
parallel with the x-axis. If we don't make that assumption we find
that the number of states (not distinct states, some of these are just
separated by a rotation) are 3*16!*24*12^15 since there are three
possible choices for which axis should be parallel with the longer
side. This means that, in fact, each distinct state has 6
representations in the physical puzzle. This is actually what we
should expect since it should be possible to represent every state in
an RL rep, an UD rep, an FB rep, an AKx rep, an AKy rep and an AKz
rep. In the solved state, the rotation of the puzzle is determined by
which colour belongs to each face and given such a rotation there are
indeed six representations: first, choose an axis that the longer side
should be parallel with (3 alternatives) and then choose either the AK
rep or the non-AK rep (2 alternatives). However, the so-called
"half-rotation" changes which colour belongs to which face so this is
not a move that gets you from one representation of a state to
another. From an UD rep with:
P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x ,
P3 Ox'z' takes you to an FB rep of the same state
Oy P3 Oz' takes you to an RL rep of the same state
UD'x P3 UD'z' Sy- Oy takes you to an AKy rep of the same state
UD'x P3 Oyx' Sz+ takes you to an AKz rep of the same state and
UD'x P3 UD'x' Oz' Sx- takes you to an AKx rep of the same state
From this, we can see that the AK states are closely related and it's
possible to change between them without illegal moves. For example
going from an AKy rep to an AKz rep of the same state could be done
with:
(UD'x P3 UD'z' Sy- Oy)' UD'x P3 Oyx' Sz+ =3D Oy' Sy+ UD'z
P3' UD'x' UD'x P3 Oyx' Sz+ =3D Oy' Sy+ UD'z Oyx' Sz+

Note that these sequences don't change the state nor the rotation of
the puzzle but they do change the representation. Let's call these
types of sequences J moves. Note that what I have previously called a
type three rotation is actually a J move and a rotation. Further note
that J =3D I mod(rep) (modulo representation). The J moves changes which
pair of faces are symmetry breaking and this is the type of moves that
needs to be added to the set of legal moves to make the physical
puzzle an actual 2x2x2x2. A J move is a move that is equal to I
mod(rep, rot) but that isn't equal to I mod(rot) (which means that it
has to preserve the state, is allowed to change the rotation and has
to change the representation/what faces (not in the left/right sense
but rather in the sense of colours) are symmetry-breaking ).

Best regards,
Joel Karlsson

2017-05-22 22:32 GMT+02:00 Joel Karlsson :
> Hi Melinda,
>
> Thank you for the feedback. Regarding the coordinate system, it=E2=80=99s=
just
> a matter of preference. I thought it would be nice to have a
> xy-symmetry but understand that it might be more practical to follow
> conventions. So, adapting your suggestion, let's redefine the axes as
> x pointing right, y up and z towards you. Note that this means that
> the longer side of the puzzle is parallel with the x-axis in the
> standard rotation. From here on, I will use this new coordinate
> system.
>
> Regarding the name of the faces. Since which faces are the "outer
> ones" changes with how you rotate the puzzle and what state the puzzle
> is in, I think that the labelling of the faces should be independent
> of which faces are the outer faces (forming what will be referred to
> as inverted octahedra). Since the puzzle is a representation of a
> 4d-cube in 3d-space it will (with our coordinate system) always have
> two faces "belonging" to every axis and two faces that don't belong to
> any axis. Therefore, it makes sense to label the faces in such a way
> that the face in for example the positive x direction is R, always.
> So, the K face (which is one of the faces not belonging to a
> particular axis) is always the face only belonging to the center 2x2x2
> block (and this can be either an octahedron or an inverted octahedron
> (the outer corners) depending on the representation of the puzzle).
> The distinction between left and right never disappears; the R face is
> always the face only belonging to the right half of the puzzle and can
> be an octahedron, an inverted octahedron, two half octahedra (<><>) or
> one half and two quarter octahedra (><><). Note that the
> half-rotations are 90-degree rotations of the puzzle; for example, Sx+
> (physical puzzle) =3D Cx' (virtual puzzle) is the rotation that does L
> -> K -> R -> A -> L so the face that previously was the L face is now
> the K face. From the standard rotation (where R and L are the inverted
> octahedra before the rotation), this would mean that the K and A faces
> are inverted octahedra after a Sx+ rotation. What faces are the
> "mysterious" (or more precisely symmetry breaking, forming inverted
> octahedra instead of regular octahedra) depend on the rotation of the
> puzzle and can be any two opposite faces (R and L, U and D, F and B or
> A and K). It might be useful to be able to describe what faces are
> inverted octahedra since this determines what moves are legal so let's
> say that the puzzle is in an RL representation if R and L are inverted
> octahedra and similarly for other states. Thus, Sx+ can take you from
> an RL representation to an AK representation (what a long word let's
> use rep for short). Note that the AK rep doesn't specify which axis
> the longer side should be parallel with so let's just add a lowercase
> letter to indicate this (an AKx rep is thus a state where the A and K
> faces are inverted octahedra and the longer side is parallel with the
> x-axis). In conclusion, I would like to keep the names of the faces as
> I first defined them and hope that it's clearer what I mean with them
> and that the names of the faces are not related to what faces form
> inverted octahedra.
>
> You also wrote:
> "There are several, distinct types of rotations, none of which change
> the state of the puzzle, and I think we need a way to be unambiguous
> about them. The types I see are
>
> 1. Simple reorientation of the physical puzzle in the hand, no magnets
> involved. IE your 'O' moves and maybe analogous to mouse-dragging in
> MC4D?
> 2. Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-c=
lick?
> 3. Half-rotations. Maybe analogous to a "face first" view? (ctrl-click
> on a 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by
> Cubie")
> 4. Whole-puzzle reorientations that move an arbitrary axis into the
> "outer" 2 faces. No MC4D analog."
>
> The rotations (as far as I know) are the O moves (type one rotation),
> which are indeed analogous to mouse-dragging in MC4D, rolling the
> 2x2x2 halves (type two rotation) (these are easily described as for
> example RL'y in an RL rep) which (in a non-AK rep) is a ctrl-click on
> a non-inverted face (that is ctrl-click on a face that is currently
> represented with an octahedron), half-rotations (also type two
> rotations) which is a ctrl-click on an inverted face and sequences
> that for example from an RL rep can take the R face to K without
> turning the puzzle into an AK rep (type three rotation). Type one and
> two rotations are legal moves but type three contain illegal 2x2x2x2
> moves according to hypothesis * in my previous email (however, they
> are very important and needed if we wish to be able to reach all
> states).
>
> Regarding fold moves, from the standard rotation (RL rep) are you
> saying that Vy+ =3D Vy'+ or something like Vy+ =3D Vy'+ FB'x2 =3D Vy'+
> mod(rot)? The former seems not to be correct but I believe the latter
> is, please correct me if I'm wrong. I don't assume that you fold the
> puzzle back to the same representation. This is what + and -
> indicates, + preserve the representation mod(rot) (i.e AK rep both
> before and after or neither before nor after) and - changes it (going
> from AK rep to non-AK rep or vice versa).
>
> "I think the only legal 90 degree twists of the K face are those about
> the long axis. I believe this is what Christopher Locke was saying in
> this message. To see why there is no straightforward way to perform
> other 90 degree twists, you only need to perform a 90 degree twist on
> an outer (L/R) face and then reorient the whole puzzle along a
> different outer axis. If the original twist was not about the new long
> axis, then there is clearly no straightforward way to undo that
> twist."
>
> Yes, as pointed out further down in my previous email. The notation
> allows all 90-degree rotations after the section "extensions of some
> definitions" although only 180-degree rotations are legal for
> octahedral faces around axes not parallel with the longer side of the
> puzzle (see the section =E2=80=9Csome important notes on legal/illegal
> moves=E2=80=9D).
>
> "I noticed something like this the other day but realized that it only
> seems to work for rotations along the long dimension (z in your
> example). These are already easily accomplished by a simple rotation
> to put the face in question on the end caps, followed by a double
> end-cap twist."
>
> It works for the other rotations as well although these are not legal
> moves. They could possibly be used instead of the illegal S moves
> (along axes not parallel with the longer side) and the illegal V moves
> (V- (minus) moves which are closely related to the illegal S moves,
> example from RL rep: Vy- =3D Vy+ Sx Oz2) but as I mentioned in my
> previous email I do believe that it's better to use the S and V moves
> since you have found a relatively short way to perform a type three
> rotation with those.
>
> Let:
> P1 =3D Sy+ Uyz2 Sy-
> P1=E2=80=99 =3D P1 (P1 is its own inverse)
> P2 =3D (P1 UD=E2=80=99z P1 UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z=E2=
=80=99 P1 UD=E2=80=99z)2 (P2=3DP_2 not P^2
> whereas the two at the end means: perform twice)
> P2=E2=80=99 =3D (UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z P1 UD=E2=80=99=
z P1 UD=E2=80=99z=E2=80=99 P1)2
> P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x =3D I mod=
(rot) (type three rotation)
> P3=E2=80=99 =3D Oy=E2=80=99 P3 Oy=E2=80=99
> P4 =3D P2 P3 P2 P3=E2=80=99 P2=E2=80=99 P3 P2=E2=80=99 P3=E2=80=99
>
> P4 from UD rep is a 164 move sequence rotating only one corner in its
> place. The sequence is inspired by Roice =E2=80=9Csecond four-color serie=
s=E2=80=9D
> but I have changed the =E2=80=9CTop 9=E2=80=9D moves to pure rotations si=
nce it=E2=80=99s all
> that=E2=80=99s necessary and a bit shorter to perform. P3 is your type th=
ree
> rotation (with a rotation added at the end) and P2 is Roice =E2=80=9Cthir=
d
> three-color series=E2=80=9D. Written with my notation but for the virtual
> puzzle, the sequences (Roice original since it=E2=80=99s easier to perfor=
m K
> twists than O rotations in MC4D) are:
> Q1 =3D (Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=99 Rz2y=E2=80=99)2 (a=
nalogue to P2)
> Q1=E2=80=99 =3D (Rz2y=E2=80=99 Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=
=99)2
> Q2 =3D Q1 Kxy=E2=80=99 Q1 Kyx=E2=80=99 Q1=E2=80=99 Kxy=E2=80=99 Q1=
=E2=80=99 Kyx=E2=80=99 (analogue to P4 but
> only 36 moves)
>
> How to read faster (the example applies to a 3x3x3x3): moves like
> Kz2y=E2=80=99 are clicking on 3C-pieces in MC4D. If the move is written i=
n
> this way, [uppercase letter] [lowercase letter]2 [lowercase letter
> possibly with =E2=80=98 (prime)], there=E2=80=99s a quite quick way to re=
alize which
> piece this is. The uppercase letter specifies which face the piece to
> press is on and the first lowercase letter (followed by 2) specifies
> one of the sides of that face that the piece belong to. There are then
> 4 possible pieces. Sadly, the piece do not lie in the direction of the
> last letter from the center of the side of the face but you have to
> move one edge clockwise from this. So, Kz2y=E2=80=99 is a click on the ed=
ge on
> the front side of the K face one step clockwise from the negative
> y-axis (thus, the front left 3C-piece on the K face). It=E2=80=99s a bit
> unfortunate that this =E2=80=9Crule=E2=80=9D isn=E2=80=99t even simpler b=
ut it's at least true
> for all of these moves (as far as I know). Moves like Kxy=E2=80=99 are
> left/right-clicking on a corner piece but currently I don=E2=80=99t know =
a
> fast way to determine which corner. Any ideas?
>
> Best regards,
> Joel Karlsson
>
> 2017-05-19 3:33 GMT+02:00 Melinda Green melinda@superliminal.com
> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>
>>
>> Hello Joel,
>>
>> Thanks for drilling into this puzzle. Finding good ways to discuss and t=
hink
>> about moves and representations will be key. I'll comment on some detail=
s
>> in-line.
>>
>> On 5/14/2017 6:16 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
>> wrote:
>>
>>
>> Yes, that is correct and in fact, you should divide not only with 24 for=
the
>> orientation but also with 16 for the placement if you want to calculate
>> unique states (since the 2x2x2x2 doesn't have fixed centerpieces). The
>> point, however, was that if you don=E2=80=99t take that into account you=
get a
>> factor of 24*16=3D384 (meaning that the puzzle has 384 representations o=
f
>> every unique state) instead of the factor of 192 which you get when
>> calculating the states from the virtual puzzle and hence every state of =
the
>> virtual puzzle has two representations in the physical puzzle. Yes exact=
ly,
>> they are indeed the same solved (or other) state and you are correct tha=
t
>> the half rotation (taking off a 2x2 layer and placing it at the other en=
d of
>> the puzzle) takes you from one representation to the same state with the
>> other representation. This means that the restacking move (taking off th=
e
>> front 2x4 layer and placing it behind the other 2x4 layer) can be expres=
sed
>> with half-rotations and ordinary twists and rotations (which you might h=
ave
>> pointed out already).
>>
>>
>> Yes, I made that claim in the video but didn't show it because I have ye=
t to
>> record such a sequence. I've only stumbled through it a few times. I tal=
ked
>> about it at 5:53 though I mistakenly called it a twist, when I should ha=
ve
>> called it a sequence.
>>
>>
>> I think I've found six moves including ordinary twists and a restacking =
move
>> that is identical to a half-rotation and thus it's easy to compose a
>> restacking move with one half-rotation and five ordinary twists. There m=
ight
>> be an error since I've only played with the puzzle in my mind so it woul=
d be
>> great if you, Melinda, could confirm this (the sequence is described lat=
er
>> in this email).
>>
>>
>> You mean "RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-"? Yes, that works. The=
re do seem to
>> be easier ways to do that beginning with an ordinary rolling rotation. I
>> don't see those in your notation, but the equivalent using a pair of twi=
sts
>> would be Rx Lx' Sx Vy if I got that right.
>>
>>
>> To be able to communicate move sequences properly we need notation for
>> representing twists, rotations, half-rotations, restacking moves and fol=
ds.
>> Feel free to come with other suggestion but you can find mine below. Ple=
ase
>> read the following thoroughly (maybe twice) to make sure that you unders=
tand
>> everything since misinterpreted notation could potentially become a
>> nightmare and feel free to ask questions if there is something that need=
s
>> clarification.
>>
>> Coordinate system and labelling:
>>
>> Let's introduce a global coordinate system. In whatever state the puzzle=
is
>> let the positive x-axis point upwards, the positive y-axis towards you a=
nd
>> the positive z-axis to the right (note that this is a right-hand system)=
.
>>
>>
>> I see the utility of a global coordinate system, but this one seems rath=
er
>> non-standard. I suggest that X be to the right, and Y up since these are
>> near-universal standards. Z can be in or out. I have no opinion. If ther=
e is
>> any convention in the twisty puzzle community, I'd go with that.
>>
>> Note also that the wiki may be a good place to document and iterate on
>> terminology, descriptions and diagrams. Ray added a "notation" section t=
o
>> the 3^4 page here, and I know that one other member was thinking of
>> collecting a set of moves on another wiki page.
>>
>>
>> Now let's name the 8 faces of the puzzle. The right face is denoted with=
R,
>> the left with L, the top U (up), the bottom D (down), the front F, the b=
ack
>> B, the center K (kata) and the last one A (ana). The R and L faces are
>> either the outer corners of the right and left halves respectively or th=
e
>> inner corners of these halves (forming octahedra) depending on the
>> representation of the puzzle. The U, D, F and B faces are either two dia=
mond
>> shapes (looking something like this: <><>) on the corresponding side of =
the
>> puzzle or one whole and two half diamond shapes (><><). The K face is ei=
ther
>> an octahedron in the center of the puzzle or the outer corners of the ce=
nter
>> 2x2x2 block. Lastly, the A face is either two diamond shapes, one on the
>> right and one on the left side of the puzzle, or another shape that=E2=
=80=99s a
>> little bit hard to describe with just a few words (the white stickers at
>> 5:10 in the latest video, after the half-rotation but before the restack=
ing
>> move).
>>
>>
>> I think it's more correct to say that the K face is either an octahedron=
at
>> the origin (A<>K<>A) or in the center of one of the main halves, with th=
e A
>> face inside the other half (>A<>K<). This was what I was getting at in m=
y
>> previous message. You do later talk about octahedral faces being in eith=
er
>> the center or the two main halves, so this is just terminology. But abou=
t
>> "the outer corners of the center 2x2x2 block", this cannot be the A or K
>> face as you've labeled them. You've been calling these the L/R faces, bu=
t
>> the left-right distinction disappears in the half-rotated state, so mayb=
e
>> "left" and "right" aren't the best names. To me, they are always the
>> "outside" faces, regardless. You can distinguish them as the left and ri=
ght
>> outside faces in one representation, or as the center and end outside fa=
ces
>> in the other. (Or perhaps "end" versus "ind" if we want to be cute.)
>>
>> I'm also a little torn about naming the interior faces ana and kata, not
>> because of the names themselves which I like, but because the mysterious
>> faces to me are the outermost ones you're calling R and L. It only requi=
res
>> a simple rotation to move faces in and out of the interior (octahedral)
>> positions, but it's much more difficult to move another axis into the
>> L/R/outer direction.
>>
>> So maybe the directions can be
>>
>> Up-Down
>> Front-Back
>> Ind-End
>> Ana-Kata
>>
>> I'm not in love with it and will be happy with anything that works. Thou=
ghts
>> anyone?
>>
>>
>> We also need a name for normal 3D-rotations, restacking moves and folds
>> (note that a half-rotation is a kind of restacking move). Let O be the n=
ame
>> for a rOtation (note that the origin O doesn't move during a rotation, b=
y
>> the way, these are 3D rotations of the physical puzzle), let S represent=
a
>> reStacking move and V a folding/clamshell move (you can remember this by
>> thinking of V as a folded line).
>>
>>
>> I think it's fine to call the clamshell move a fold or denote it as V. I
>> just wouldn't consider it to be a basic move since it's a simple composi=
te
>> of 3 basic twists as shown here. In general, I think there are so many
>> useful composite moves that we need to be able to easily make them up ad=
hoc
>> with substitutions like Let =E2=86=93 =3D Rx Lx'. These are really macro=
moves which
>> can be nested. That said, it's a particularly useful move so it's probab=
ly
>> worth describing in some formal way like you do in detail below.
>>
>>
>>
>> Further, let I (capital i) be the identity, preserving the state and
>> rotation of the puzzle. We cannot use I to indicate what moves should be
>> performed but it's still useful as we will see later. Since we also want=
to
>> be able to express if a sequence of moves is a rotation, preserving the
>> state of the puzzle but possibly representing it in a different way, we =
can
>> introduce mod(rot) (modulo rotation). So, if a move sequence P satisfies=
P
>> mod(rot) =3D I, that means that the state of the puzzle is the same befo=
re and
>> after P is performed although the rotation and representation of the puz=
zle
>> are allowed to change. I do also want to introduce mod(3rot) (modulo
>> 3D-rotation) and P mod(3rot) =3D I means that if the right 3D-rotation (=
a
>> combination of O moves as we will see later) is applied to the puzzle af=
ter
>> P you get the identity I. Moreover, let the standard rotation of the puz=
zle
>> be any rotation such that the longer side is parallel with the z-axis, t=
hat
>> is the puzzle forms a 2x2x4 (2 pieces thick in the x-direction, 2 in the
>> y-direction and 4 in the z-direction), and the K face is an octahedron.
>>
>> Rotations and twists:
>>
>> Now we can move on to name actual moves. The notation of a move is a
>> combination of a capital letter and a lowercase letter. O followed by x,=
y
>> or z is a rotation of the whole puzzle around the corresponding axis in =
the
>> mathematical positive direction (counterclockwise/the way your right-han=
d
>> fingers curl if you point in the direction of the axis with your thumb).=
For
>> example, Ox is a rotation around the x-axis that turns a 2x2x4 into a 2x=
4x2.
>> A name of a face (U, D, F, B, R, L, K or A) followed by x, y or z means:
>> detach the 8 pieces that have a sticker belonging to the face and then t=
urn
>> those pieces around the global axis. For example, if the longer side of =
the
>> puzzle is parallel to the z-axis (the standard rotation), Rx means: take=
the
>> right 2x2x2 block and turn it around the global x-axis in the mathematic=
al
>> positive direction. Note that what moves are physically possible and all=
owed
>> is determined by the rotation of the puzzle (I will come back to this
>> later). Further note that Ox mod(rot) =3D Oy mod(rot) =3D Oz mod(rot) =
=3D I,
>> meaning that the 3D-rotations of the physical puzzle corresponds to a
>> 4D-rotation of the represented 2x2x2x2.
>>
>>
>> There are several, distinct types of rotations, none of which change the
>> state of the puzzle, and I think we need a way to be unambiguous about t=
hem.
>> The types I see are
>>
>> Simple reorientation of the physical puzzle in the hand, no magnets
>> involved. IE your 'O' moves and maybe analogous to mouse-dragging in MC4=
D?
>> Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-cli=
ck?
>> Half-rotations. Maybe analogous to a "face first" view? (ctrl-click on a
>> 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by Cubie"=
)
>> Whole-puzzle reorientations that move an arbitrary axis into the "outer"=
2
>> faces. No MC4D analog.
>>
>>
>> Inverses and performing a move more than once
>>
>> To mark that a move should be performed n times let's put ^n after it. F=
or
>> convenience when writing and speaking let ' (prime) represent ^-1 (the
>> inverse) and n represent ^n. The inverse P' of some permutation P is the
>> permutation that satisfies P P' =3D P' P =3D I (the identity). For examp=
le (Rx)'
>> means: do Rx backwards, which corresponds to rotating the right 2x2x2 bl=
ock
>> in the mathematical negative direction (clockwise) around the x-axis and
>> (Rx)2 means: perform Rx twice. However, we can also define powers of jus=
t
>> the lowercase letters, for example, Rx2 =3D Rx^2 =3D Rxx =3D Rx Rx. So x=
2=3Dx^2
>> means: do whatever the capital letter specifies two times with respect t=
o
>> the x-axis. We can see that the capital letter naturally is distributed =
over
>> the two lowercase letters. Rx' =3D Rx^-1 means: do whatever the capital =
letter
>> specifies but in the other direction than you would have if the prime
>> wouldn't have been there (note that thus, x'=3Dx^-1 can be seen as the
>> negative x-axis). Note that (Rx)^2 =3D Rx^2 and (Rx)' =3D Rx' which is t=
rue for
>> all twists and rotations but that doesn't have to be the case for other
>> types of moves (restacks and folds).
>>
>> Restacking moves
>>
>> A restacking move is an S followed by either x, y or z. Here the lowerca=
se
>> letter specifies in what direction to restack. For example, Sy (from the
>> standard rotation) means: take the front 8 pieces and put them at the ba=
ck,
>> whereas Sx means: take the top 8 pieces and put them at the bottom. Note
>> that Sx is equivalent to taking the bottom 8 pieces and putting them at =
the
>> top. However, if we want to be able to make half-rotations we sometimes =
need
>> to restack through a plane that doesn't go through the origin. In the
>> standard rotation, let Sz be the normal restack (taking the 8 right piec=
es,
>> the right 2x2x2 block, and putting them on the left), Sz+ be the restack
>> where you split the puzzle in the plane further in the positive z-direct=
ion
>> (taking the right 2x2x1 cap of 4 pieces and putting it at the left end o=
f
>> the puzzle) and Sz- the restack where you split the puzzle in the plane
>> further in the negative z-direction (taking the left 2x2x1 cap of 4 piec=
es
>> and putting it at the right end of the puzzle). If the longer side of th=
e
>> puzzle is parallel to the x-axis instead, Sx+ would take the top 1x2x2 c=
ap
>> and put it on the bottom. Note that in the standard rotation Sz mod(rot)=
=3D
>> I. For restacks we see that (Sx)' =3D Sx' =3D Sx (true for y and z too o=
f
>> course), that (Sz+)' =3D Sz- and that (Sz+)2 =3D (Sz-)2 =3D Sz. We can d=
efine Sz'+
>> to have meaning by thinking of z=E2=80=99 as the negative z-axis and wit=
h that in
>> mind it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus, (Sz)=E2=80=
=99 =3D Sz=E2=80=99 and (Sz+)=E2=80=99 =3D Sz=E2=80=99+.
>>
>> Fold moves
>>
>> A fold might be a little bit harder to describe in an intuitive way. Fir=
st,
>> let's think about what folds are interesting moves. The folds that canno=
t be
>> expressed as rotations and restacks are unfolding the puzzle to a 4x4 an=
d
>> then folding it back along another axis. If we start with the standard
>> rotation and unfold the puzzle into a 1x4x4 (making it look like a 4x4 f=
rom
>> above) the only folds that will achieve something you can't do with a
>> restack mod(rot) is folding it to a 2x4x2 so that the longer side is
>> parallel with the y-axis after the fold. Thus, there are 8 interesting f=
old
>> moves for any given rotation of the puzzle since there are 4 ways to unf=
old
>> it to a 4x4 and then 2 ways of folding it back that make the move differ=
ent
>> from a restack move mod(rot).
>>
>>
>> Assuming you complete a folding move in the same representation (<><> or
>>><><), then there are only two interesting choices. That's because it
>> doesn't matter which end of a chosen cutting plane you open it from, the=
end
>> result will be the same. That also means that any two consecutive clamsh=
ell
>> moves along the same cutting plane will undo each other. It further sugg=
ests
>> that any interesting sequence of clamshell moves must alternate between =
the
>> two possible long cut directions, meaning there is no choice involved. 1=
2
>> clamshell moves will cycle back to the initial state.
>>
>> There is one other weird folding move where you open it in one direction=
and
>> then fold the two halves back-to-back in a different direction. If you
>> simply kept folding along the initial hinge, you'd simply have a restack=
ing.
>> When completed the other way, it's equivalent to a restacking plus a
>> clamshell, so I don't think it's useful though it is somewhat interestin=
g.
>>
>>
>> Let's call these 8 folds interesting fold moves. Note that an interestin=
g
>> fold move always changes which axis the longer side of the puzzle is
>> parallel with. Further note that both during unfold and fold all pieces =
are
>> moved; it would be possible to have 8 of the pieces fixed during an unfo=
ld
>> and folding the other half 180 degrees but I think that it=E2=80=99s mor=
e intuitive
>> that these moves fold both halves 90 degrees and performing them with
>> 180-degree folds might therefore lead to errors since the puzzle might g=
et
>> rotated differently. To illustrate a correct unfold without a puzzle: Pu=
t
>> your palms together such that your thumbs point upward and your fingers
>> forward. Now turn your right hand 90 degrees clockwise and your left han=
d 90
>> degrees counterclockwise such that the normal to your palms point up, yo=
ur
>> fingers point forward, your right thumb to the right and your left thumb=
to
>> the left. That was what will later be called a Vx unfold and the folds a=
re
>> simply reversed unfolds. (I might have used the word =E2=80=9Cfold=E2=80=
=9D in two different
>> ways but will try to use the term =E2=80=9Cfold move=E2=80=9D when refer=
ring to the move
>> composed by an unfold and a fold rather than simply calling these moves
>> =E2=80=9Cfolds=E2=80=9D.)
>>
>> To specify the unfold let's use V followed by one of x, y, z, x', y' and=
z'.
>> The lowercase letter describes in which direction to unfold. Vx means un=
fold
>> in the direction of the positive x-axis and Vx' in the direction of the
>> negative x-axis, if that makes any sense. I will try to explain more
>> precisely what I mean with the example Vx from the standard rotation (it
>> might also help to read the last sentences in the previous paragraph aga=
in).
>> So, the puzzle is in the standard rotation and thus have the form 2x2x4 =
(x-,
>> y- and z-thickness respectively). The first part (the unfolding) of the =
move
>> specified with Vx is to unfold the puzzle in the x-direction, making it =
a
>> 1x4x4 (note that the thickness in the x-direction is 1 after the Vx unfo=
ld,
>> which is no coincidence). There are two ways to do that; either the side=
s of
>> the pieces that are initially touching another piece (inside of the puzz=
le
>> in the x,z-plane and your palms in the hand example) are facing up or do=
wn
>> after the unfold. Let Vx be the unfold where these sides point in the
>> direction of the positive x-axis (up) and Vx' the other one where these
>> sides point in the direction of the negative x-axis (down) after the unf=
old.
>> Note that if the longer side of the puzzle is parallel to the z-axis onl=
y
>> Vx, Vx', Vy and Vy' are possible. Now we need to specify how to fold the
>> puzzle back to complete the folding move. Given an unfold, say Vx, there=
are
>> only two ways to fold that are interesting (not turning the fold move in=
to a
>> restack mod(rot)) and you have to fold it perpendicular to the unfold to
>> create an interesting fold move. So, if you start with the standard rota=
tion
>> and do Vx you have a 1x4x4 that you have to fold into a 2x4x2. To
>> distinguish the two possibilities, use + or - after the Vx. Let Vx+ be t=
he
>> unfold Vx followed by the interesting fold that makes the sides that are
>> initially touching another piece (before the unfold) touch another piece
>> after the fold move is completed and let Vx- be the other interesting fo=
ld
>> move that starts with the unfold Vx. (Thus, continuing with the hand
>> example, if you want to do a Vx+ first do the Vx unfold described in the=
end
>> of the previous paragraph and then fold your hands such that your finger=
s
>> point up, the normal to your palms point forward, the right palm is touc=
hing
>> the right-hand fingers, the left palm is touching the left-hand fingers,=
the
>> right thumb is pointing to the right and the left thumb is pointing to t=
he
>> left). Note that the two halves of the puzzle always should be folded 90
>> degrees each and you should never make a fold or unfold where you fold j=
ust
>> one half 180-degrees (if you want to use my notation, that is). Further =
note
>> that Vx+ Sx mod(3rot) =3D Vx- and that Vx+ Vx+ =3D I which is equivalent=
to
>> (Vx+)=E2=80=99 =3D Vx+ and this is true for all fold moves (note that af=
ter a Vx+
>> another Vx+ is always possible).
>>
>> The 2x2x2x2 in the MC4D software
>>
>> The notation above can also be applied to the 2x2x2x2 in the MC4D progra=
m.
>> There, you are not allowed to do S or V moves but instead, you are allow=
ed
>> to do the [crtl]+[left-click] moves. This can easily be represented with
>> notation similar to the above. Let=E2=80=99s use C (as in Centering) and=
one of x, y
>> and z. For example, Cx would be to rotate the face in the positive
>> x-direction aka the U face to the center. Thus, Cz' is simply
>> [ctrl]+[left-click] on the L face and similarly for the other C moves. T=
he
>> O, U, D, F, B, R, L, K and A moves are performed in the same way as abov=
e
>> so, for example Rx would be a [left-click] on the top-side of the right
>> face. In this representation of the puzzle almost all moves are allowed;=
all
>> U, D, F, B, R, L, K, O and C moves are possible regardless of rotation a=
nd
>> only A moves (and of course the rightfully forbidden S and V moves) are
>> impossible regardless of rotation. Note that R and L moves in the softwa=
re
>> correspond to the same moves of the physical puzzle but this is not
>> generally true (I will come back to this later).
>>
>> Possible moves (so far) in the standard rotation
>>
>> In the standard rotation, the possible/allowed moves with the definition=
s
>> above are:
>>
>> O moves, all of these are always possible in any state and rotation of t=
he
>> puzzle since they are simply 3D-rotations.
>>
>> R, L moves, all of these as well since the puzzle has a right and left 2=
x2x2
>> block in the standard rotation.
>>
>> U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x4 blocks w=
ith
>> less symmetry than a 2x2x2 block.
>>
>> F, B moves, just Fy2 and By2 since the front and back are 2x1x4 blocks w=
ith
>> less symmetry than a 2x2x2 block.
>>
>> K moves, all possible since this is a rotation of the center 2x2x2 block=
.
>>
>>
>> I think the only legal 90 degree twists of the K face are those about th=
e
>> long axis. I believe this is what Christopher Locke was saying in this
>> message. To see why there is no straightforward way to perform other 90
>> degree twists, you only need to perform a 90 degree twist on an outer (L=
/R)
>> face and then reorient the whole puzzle along a different outer axis. If=
the
>> original twist was not about the new long axis, then there is clearly no
>> straightforward way to undo that twist.
>>
>>
>> A moves, only Az moves since this is two 2x2x1 blocks that have to be
>> rotated together.
>>
>> S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined in the
>> standard rotation.
>>
>> V moves, not Vz+ or Vz- since the definition doesn't give these meaning =
when
>> the long side of the puzzle is parallel with the z-axis.
>>
>> Note that for example Fz2 is not allowed since this won't take you to a
>> state of the puzzle. To allow more moves we need to extend the definitio=
ns
>> (after the extension in the next paragraph all rotations and twists (O, =
R,
>> L, U, D, F, B, K, A) are possible in any rotation and only which S and V
>> moves are possible depend on the state and rotation of the puzzle).
>>
>>
>> Extension of some definitions
>>
>> It's possible to make an extension that allows all O, R, L, U, D, F, B, =
K
>> and A moves in any state. I will explain how this can be done in the
>> standard rotation but it applies analogously to any other rotation where=
the
>> K face is an octahedron. First let's focus on U, D, F and B and because =
of
>> the symmetry of the puzzle all of these are analogous so I will only exp=
lain
>> one. The extension that makes all U moves possible (note that the U face=
is
>> in the positive x-direction) is as follows: when making an U move first
>> detach the 8 top pieces which gives you a 1x2x4 block, fold this block i=
nto
>> a 2x2x2 block in the positive x-direction (similar to the later part of =
a
>> Vx+ move from the standard rotation) such that the U face form an
>> octahedron, rotate this 2x2x2 block around the specified axis (for examp=
le
>> around the z-axis if you are doing an Uz move), reverse the fold you jus=
t
>> did creating a 1x2x4 block again and reattach the block.
>>
>>
>> I noticed something like this the other day but realized that it only se=
ems
>> to work for rotations along the long dimension (z in your example). Thes=
e
>> are already easily accomplished by a simple rotation to put the face in
>> question on the end caps, followed by a double end-cap twist.
>>
>> This is as far as I'm going to comment for the moment because the
>> information gets very dense and I've been mulling and picking over your
>> message for several days already. In short, I really like your attempt t=
o
>> provide a complete system of notation for discussing this puzzle and wil=
l be
>> curious to hear your thoughts on my comments so far. I hope others will
>> chime in too.
>>
>> One final thought is that a real "acid test" of any notation system for =
this
>> puzzle will be attempt to translate some algorithms from MC4D. I would m=
ost
>> like to see a sequence that flips a single piece, like the second 4-colo=
r
>> series on this page of Roice's solution, or his pair of twirled corners =
at
>> the end of this page. One trick will be to minimize the number of
>> whole-puzzle reorientations needed, but really any sequence that works w=
ill
>> be great evidence that the puzzles are equivalent. I suspect that this s=
ort
>> of exercise will never be practical because it will require too many
>> reorientations, and that entirely new methods will be needed to actually
>> solve this puzzle.
>>
>> Best,
>> -Melinda
>>
>> The A moves can be done very similarly but after you have detached the t=
wo
>> 2x2x1 blocks you don't fold them but instead you stack them similar to a=
Sz
>> move, creating a 2x2x2 block with the A face as an octahedron in the mid=
dle
>> and then reverse the process after you have rotated the block as specifi=
ed
>> (for example around the negative y-axis if you are doing an Ay' move). N=
ote
>> that these extended moves are closely related to the normal moves and fo=
r
>> example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99 in the standard rotation=
and note that (Ry
>> Ly=E2=80=99) mod(rot) =3D (Ly Ry=E2=80=99) mod(rot) =3D I (this applies =
to the other extended
>> moves as well).
>>
>> If the cube is in the half-rotated state, where both the R and L faces a=
re
>> octahedra, you can extend the definitions very similarly. The only thing=
you
>> have to change is how you fold the 2x4 blocks when performing a U, D, F =
or B
>> move. Instead of folding the two 2x2 halves of the 2x4 into a 2x2 you ha=
ve
>> to fold the end 1x2 block 180 degrees such that the face forms an
>> octahedron.
>>
>> These moves might be a little bit harder to perform, to me especially th=
e A
>> moves seems a bit awkward, so I don't know if it's good to use them or n=
ot.
>> However, the A moves are not necessary if you allow Sz in the standard
>> rotation (which you really should since Sz mod(rot) =3D I in the standar=
d
>> rotation) and thus it might not be too bad to use this extended version.=
The
>> notation supports both variants so if you don=E2=80=99t want to use thes=
e extended
>> moves that shouldn=E2=80=99t be a problem. Note that, however, for examp=
le Ux
>> (physical puzzle) !=3D Ux (virtual puzzle) where !=3D means =E2=80=9Cnot=
equal to=E2=80=9D (more
>> about legal/illegal moves later).
>>
>> Generalisation of the notation
>>
>> Let=E2=80=99s generalise the notation to make it easier to use and to ma=
ke it work
>> for any n^4 cube in the MC4D software. Previously, we saw that (Rx)2 =3D=
Rx Rx
>> and if we allow ourselves to rewrite this as (Rx)2 =3D Rx2 =3D Rxx =3D R=
x Rx we
>> see that the capital letter naturally can be distributed over the lowerc=
ase
>> letters. We can make this more general and say that any capital letter
>> followed by several lowercase letters means the same thing as the capita=
l
>> letter distributed over the lowercase letters. Like Rxyz =3D Rx Ry Rz an=
d here
>> R can be exchanged with any capital letter and xyz can be exchanged with=
any
>> sequence of lowercase letters. We can also allow several capital letters=
and
>> one lowercase letter, for example RLx and let=E2=80=99s define this as R=
Lx =3D Rx Lx
>> so that the lowercase letter can be distributed over the capital letters=
. We
>> can also define a capital letter followed by =E2=80=98 (prime) like R=E2=
=80=99x =3D Rx=E2=80=99 and
>> R=E2=80=99xy =3D Rx=E2=80=99y=E2=80=99 so the =E2=80=98 (prime) is distr=
ibuted over the lowercase letters.
>> Note that we don=E2=80=99t define a capital to any other power than -1 l=
ike this
>> since for example R2x =3D RRx might seem like a good idea at first but i=
t
>> isn=E2=80=99t very useful since R2 and RR are the same lengths (and powe=
rs greater
>> than two are seldom used) and we will see that we can define R2 in anoth=
er
>> way that generalises the notation to all n^4 cubes.
>>
>> Okay, let=E2=80=99s define R2 and similar moves now and have in mind wha=
t moves we
>> want to be possible for a n^4 cube. The moves that we cannot achieve wit=
h
>> the notation this far is twisting deeper slices. To match the notation w=
ith
>> the controls of the MC4D software let R2x be the move similar to Rx but
>> twisting the 2nd layer instead of the top one and similarly for other
>> capital letters, numbers (up to n) and lowercase letters. Thus, R2x is
>> performed as Rx but holding down the number 2 key. Just as in the progra=
m,
>> when no number is specified 1 is assumed and you can combine several num=
bers
>> like R12x to twist both the first and second layer. This notation does n=
ot
>> apply to rotations (O) folding moves (V) and restacking moves (S) (I sup=
pose
>> you could redefine the S move using this deeper-slice-notation and use S=
1z
>> as Sz+, S2z as Sz and S3z as Sz- but since these moves are only allowed =
for
>> the physical 2x2x2x2 I think that the notation with + and =E2=80=93 is b=
etter since
>> S followed by a lowercase letter without +/- always means splitting the =
cube
>> in a coordinate plane that way, not sure though so input would be great)=
.
>> The direction of the twist R3x should be the same as Rx meaning that if =
Rx
>> takes stickers belonging to K and move them to F, so should R3x, in
>> accordance with the controls of the MC4D software. Note that for a 3x3x3=
x3
>> it=E2=80=99s true that R3z =3D Lz whereas R3x =3D Lx=E2=80=99 (note that=
R and L are the faces
>> in the z-directions so because of the symmetry of the cube it will also =
be
>> true that for example U3x =3D Bx whereas U3y=3DBy=E2=80=99).
>>
>> What about the case with several capital letters and several lowercase
>> letters, for instance, RLxy? I see two natural definitions of this. Eith=
er,
>> we could have RLxy =3D Rxy Lxy or we could define it as RLxy =3D RLx RLy=
. These
>> are generally not the same (if you exchange R and L with any allowed cap=
ital
>> letter and similarly for x and y). I don=E2=80=99t know what is best, wh=
at do you
>> think? The situations I find this most useful in are RL=E2=80=99xy to do=
a rotation
>> and RLxy as a twist. However, since R and L are opposite faces their
>> operations commute which imply RL=E2=80=99xy (1st definition) =3D Rxy L=
=E2=80=99xy =3D Rx Ry
>> Lx=E2=80=99 Ly=E2=80=99 =3D Rx Lx=E2=80=99 Ry Ly=E2=80=99 =3D RL=E2=80=
=99x RL=E2=80=99y =3D RL=E2=80=99xy (2nd definition) and similarly
>> for the other case with RLxy. Hopefully, we can find another useful sequ=
ence
>> of moves where this notation can be used with only one of the definition=
s
>> and can thereby decide which definition to use. Personally, I feel like =
RLxy
>> =3D RLx RLy is the more intuitive definition but I don=E2=80=99t have an=
y good
>> argument for this so I=E2=80=99ll leave the question open.
>>
>> For convenience, it might be good to be able to separate moves like Rxy =
and
>> RLx from the basic moves Rx, Oy etc when speaking and writing. Let=E2=80=
=99s call
>> the basic moves that only contain one capital letter and one lowercase
>> letter (possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or a number) =
simple moves (like
>> Rx, L'y, Ux2 and D3y=E2=80=99) and the moves that contain more than one =
capital
>> letter or more than one lowercase letter composed moves.
>>
>> More about inverses
>>
>> This list can obviously be made longer but here are some identities that=
are
>> good to know and understand. Note that R, L, U, x, y and z below just ar=
e
>> examples, the following is true in general for non-folds (however, S mov=
es
>> are fine).
>>
>> (P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99 P1=E2=80=
=99 (Pi is an arbitrary permutation for
>> i=3D1,2,=E2=80=A6n)
>> (Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx
>> (RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x
>> (Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+ (just as an example with =
restacking moves, note that
>> the inverse doesn=E2=80=99t change the + or -)
>> RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=80=99xyz =
(true for both definitions)
>> (RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=80=99yx =
(true for both definitions)
>>
>> For V moves we have that: (Vx+)=E2=80=99 =3D Vx+ !=3D Vx=E2=80=99+ (=
!=3D means =E2=80=9Cnot equal to=E2=80=9D)
>>
>> Some important notes on legal/illegal moves
>>
>> Although there are a lot of moves possible with this notation we might n=
ot
>> want to use them all. If we really want a 2x2x2x2 and not something else=
I
>> think that we should try to stick to moves that are legal 2x2x2x2 moves =
as
>> far as possible (note that I said legal moves and not permutations (a le=
gal
>> permutation can be made up of one or more legal moves)). Clarification:
>> cycling three of the edge-pieces of a Rubik=E2=80=99s cube is a legal pe=
rmutation
>> but not a legal move, a legal move is a rotation of the cube or a twist =
of
>> one of the layers. In this section I will only address simple moves and
>> simply refer to them as moves (legal composed moves are moves composed b=
y
>> legal simple moves).
>>
>> I do believe that all moves allowed by my notation are legal permutation=
s
>> based on their periodicity (they have a period of 2 or 4 and are all eve=
n
>> permutations of the pieces). So, which of them correspond to legal 2x2x2=
x2
>> moves? The O moves are obviously legal moves since they are equal to the
>> identity mod(rot). The same goes for restacking (S) (with or without +/-=
) in
>> the direction of the longest side of the puzzle (Sz, Sz+ and Sz- in the
>> standard rotation) since these are rotations and half-rotations that don=
=E2=80=99t
>> change the state of the puzzle. Restacking in the other directions and f=
old
>> moves (V) are however not legal moves since they are made of 8 2-cycles =
and
>> change the state of the puzzle (note that they, however, are legal
>> permutations). The rest of the moves (R, L, F, B, U, D, K and A) can be
>> divided into two sets: (1) the moves where you rotate a 2x2x2 block with=
an
>> octahedron inside and (2) the moves where you rotate a 2x2x2 block witho=
ut
>> an octahedron inside. A move belonging to (2) is always legal. We can se=
e
>> this by observing what a Rx does with the pieces in the standard rotatio=
n
>> with just K forming an octahedron. The stickers move in 6 4-cycles and i=
f
>> the puzzle is solved the U and D faces still looks solved after the move=
. A
>> move belonging to set (1) is legal either if it=E2=80=99s an 180-degree =
twist or if
>> it=E2=80=99s a rotation around the axis parallel with the longest side o=
f the puzzle
>> (the z-axis in the standard rotation). Quite interestingly these are exa=
ctly
>> the moves that don=E2=80=99t mix up the R and L stickers with the rest i=
n the
>> standard rotation. I think I know a way to prove that no legal 2x2x2x2 m=
ove
>> can mix up these stickers with the rest and this has to do with the fact
>> that these stickers form an inverted octahedron (with the corners pointi=
ng
>> outward) instead of a normal octahedron (let=E2=80=99s call this hypothe=
sis * for
>> now). Note that all legal twists (R, L, F, B, U, D, K and A) of the phys=
ical
>> puzzle correspond to the same twist in the MC4D software.
>>
>> So, what moves should we add to the set of legal moves be able to get to
>> every state of the 2x2x2x2? I think that we should add the restacking mo=
ves
>> and folding moves since Melinda has already found a pretty short sequenc=
e of
>> those moves to make a rotation that changes which colours are on the R a=
nd L
>> faces. That sequence, starting from the standard rotation, is: Oy Sx+z V=
y+
>> Ozy=E2=80=99 Vy+ Ozy=E2=80=99 Vy+ =3D Oy Sx+z (Vy+ Ozy=E2=80=99)3 mod(3=
rot) =3D I mod(rot) (hopefully I
>> got that right). What I have found (which I mentioned previously) is tha=
t
>> (from the standard rotation): RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz- an=
d since this
>> is equivalent with Sy =3D Ly2 RLx=E2=80=99 Sz- RLx Ly2 By2 the restackin=
g moves that
>> are not legal moves are not very complicated permutations and therefore =
I
>> think that we can accept them since they help us mix up the R and L stic=
kers
>> with other faces. In a sense, the folding moves are =E2=80=9Cmore illega=
l=E2=80=9D since
>> they cannot be composed by the legal moves (according to hypothesis *). =
This
>> is also true for the illegal moves belonging to set (1) discussed above.
>> However, since the folding moves is probably easier to perform and is en=
ough
>> to reach every state of the 2x2x2x2 I think that we should use them and =
not
>> the illegal moves belonging to (1). Note, once again, that all moves
>> described by the notation are legal permutations (even the ones that I j=
ust
>> a few words ago referred to as illegal moves) so if you wish you can use=
all
>> of them and still only reach legal 2x2x2x2 states. However (in a strict
>> sense) one could argue that you are not solving the 2x2x2x2 if you use
>> illegal moves. If you only use illegal moves to compose rotations (that =
is,
>> create a permutation including illegal moves that are equal to I mod(rot=
))
>> and not actually using the illegal moves as twists I would classify that=
as
>> solving a 2x2x2x2. What do you think about this?
>>
>> What moves to use?
>>
>> Here=E2=80=99s a short list of the simple moves that I think should be u=
sed for the
>> physical 2x2x2x2. Note that this is just my thoughts and you may use the
>> notation to describe any move that it can describe if you wish to. The
>> following list assumes that the puzzle is in the standard rotation but i=
s
>> analogous for other representations where the K face is an octahedron.
>>
>> O, all since they are I mod(rot),
>> R, L, all since they are legal (note Rx (physical puzzle) =3D Rx (virtua=
l
>> 2x2x2x2)),
>> U, D, only x2 since these are the only legal easy-to-perform moves,
>> F, B, only y2 since these are the only legal easy-to-perform moves,
>> K, A, only z, z=E2=80=99 and z2 since these are the only legal easy-to-p=
erform
>> moves,
>> S, at least z, z+ and z- since these are equal to I mod(rot),
>> S, possibly x and y since these help us perform rotations and is easy to
>> compose (not necessary to reach all states and not legal though),
>> V, all 8 allowed by the rotation of the puzzle (at least one is necessar=
y to
>> reach all states and if you allow one the others are easy to achieve
>> anyway).
>>
>> If you start with the standard rotation and then perform Sz+ the followi=
ng
>> applies instead (this applies analogously to any other rotation where th=
e R
>> and L faces are octahedra).
>>
>> O, all since they are I mod(rot),
>> R, L, only z, z=E2=80=99 and z2 since these are the only legal easy-to-p=
erform
>> moves,
>> U, D, only x2 since these are the only legal easy-to-perform moves,
>> F, B, only y2 since these are the only legal easy-to-perform moves,
>> K, A, at least z, z=E2=80=99 and z2, possibly all (since they are legal)=
although
>> some might be hard to perform.
>> S, V, same as above.
>>
>> Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D I mod(rot) =
(!=3D
>> for not equal) which implies that the Ux2 move when R and L are octahedr=
a is
>> different from the Ux2 move when K is an octahedron. (Actually, the sequ=
ence
>> above is equal to Uy2).
>>
>> Regarding virtual n^4 cubes I think that all O, C, R, L, U, D, F, B, K a=
nd A
>> moves should be used since they are all legal and really the only thing =
you
>> need (left-clicking on an edge or corner piece in the computer program c=
an
>> be described quite easily with the notation, for example, Kzy2 is
>> left-clicking on the top-front edge piece on the K face).
>>
>> I hope this was possible to follow and understand. Feel free to ask
>> questions about the notation if you find anything ambiguous.
>>
>> Best regards,
>> Joel Karlsson
>>
>> Den 4 maj 2017 12:01 fm skrev "Melinda Green melinda@superliminal.com
>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>
>>
>>
>> Thanks for the correction. A couple of things: First, when assembling on=
e
>> piece at a time, I'd say there is only 1 way to place the first piece, n=
ot
>> 24. Otherwise you'd have to say that the 1x1x1 puzzle has 24 states. I
>> understand that this may be conventional, but to me, that just sounds si=
lly.
>>
>> Second, I have the feeling that the difference between the "two
>> representations" you describe is simply one of those half-rotations I sh=
owed
>> in the video. In the normal solved state there is only one complete
>> octahedron in the very center, and in the half-rotated state there is on=
e in
>> the middle of each half of the "inverted" form. I consider them to be th=
e
>> same solved state.
>>
>> -Melinda
>>
>>
>> On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
>> wrote:
>>
>> Horrible typo... It seems like I made some typos in my email regarding t=
he
>> state count. It should of course be 16!12^16/(6*192) and NOT
>> 12!16^12/(6*192). However, I did calculate the correct number when compa=
ring
>> with previous results so the actual derivation was correct.
>>
>> Something of interest is that the physical pieces can be assembled in
>> 16!24*12^15 ways since there are 16 pieces, the first one can be oriente=
d in
>> 24 ways and the remaining can be oriented in 12 ways (since a corner wit=
h 3
>> colours never touch a corner with just one colour). Dividing with 6 to g=
et a
>> single orbit still gives a factor 2*192 higher than the actual count rat=
her
>> than 192. This shows that every state in the MC4D representation has 2
>> representations in the physical puzzle. These two representations must b=
e
>> the previously discussed, that the two halves either have the same color=
on
>> the outermost corners or the innermost (forming an octahedron) when the
>> puzzle is solved and thus both are complete representations of the 2x2x2=
x2.
>>
>> Best regards,
>> Joel Karlsson
>>
>> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" m>:
>>
>> I am no expert on group theory, so to better understand what twists are
>> legal I read through the part of Kamack and Keane's The Rubik Tesseract
>> about orienting the corners. Since all even permutations are allowed the
>> easiest way to check if a twist is legal might be to:
>> 1. Check that the twist is an even permutation, that is: the same twist =
can
>> be done by performing an even number of piece swaps (2-cycles).
>> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning performin=
g the
>> twist k times and I (the identity) representing the permutation of doing
>> nothing) and k is not divisible by 3 the twist A definitely doesn't viol=
ate
>> the restriction of the orientations since kx mod 3 =3D 0 and k mod 3 !=
=3D 0
>> implies x mod 3 =3D 0 meaning that the change of the total orientation x=
for
>> the twist A mod 3 is 0 (which precisely is the restriction of legal twis=
ts;
>> that they must preserve the orientation mod 3).
>>
>> For instance, this implies that the restacking moves are legal 2x2x2x2 m=
oves
>> since both are composed of 8 2-cycles and both can be performed twice (n=
ote
>> that 2 is not divisible by 3) to obtain the identity.
>>
>> Note that 1 and 2 are sufficient to check if a twist is legal but only 1=
is
>> necessary; there can indeed exist a twist violating 2 that still is lega=
l
>> and in that case, I believe that we might have to study the orientation
>> changes for that specific twist in more detail. However, if a twist can =
be
>> composed by other legal twists it is, of course, legal as well.
>>
>> Best regards,
>> Joel
>>
>> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com [4D_Cub=
ing]
>> <4D_Cubing@yahoogroups.com>:
>>>
>>>
>>>
>>> First off, thanks everyone for the helpful and encouraging feedback!
>>> Thanks Joel for showing us that there are 6 orbits in the 2^4 and for y=
our
>>> rederivation of the state count. And thanks Matt and Roice for pointing=
out
>>> the importance of the inverted views. It looks so strange in that
>>> configuration that I always want to get back to a normal view as quickl=
y as
>>> possible, but it does seem equally valid, and as you've shown, it can b=
e
>>> helpful for more than just finding short sequences.
>>>
>>> I don't understand Matt's "pinwheel" configuration, but I will point ou=
t
>>> that all that is needed to create your twin interior octahedra is a sin=
gle
>>> half-rotation like I showed in the video at 5:29. The two main halves d=
o end
>>> up being mirror images of each other on the visible outside like he
>>> described. Whether it's the pinwheel or the half-rotated version that's
>>> correct, I'm not sure that it's a bummer that the solved state is not a=
t all
>>> obvious, so long as we can operate it in my original configuration and
>>> ignore the fact that the outer faces touch. That would just mean that t=
he
>>> "correct" view is evidence that that the more understandable view is
>>> legitimate.
>>>
>>> I'm going to try to make a snapable V3 which should allow the pieces to=
be
>>> more easily taken apart and reassembled into other forms. Shapeways doe=
s
>>> offer a single, clear translucent plastic that they call "Frosted Detai=
l",
>>> and another called "Transparent Acrylic", but I don't think that any so=
rt of
>>> transparent stickers will help us, especially since this thing is chock=
full
>>> of magnets. The easiest way to let you see into the two hemispheres wou=
ld be
>>> to simply truncate the pointy tips of the stickers. That already happen=
s a
>>> little bit due to the way I've rounded the edges. Here is a close-up of=
a
>>> half-rotation in which you can see that the inner yellow and white face=
s are
>>> solved. Your suggestion of little mapping dots on the corners also work=
s,
>>> but just opening the existing window further would work more directly.
>>>
>>> -Melinda
>>>
>>>
>>> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>>>
>>> I agree with Don's arguments about adjacent sticker colors needing to b=
e
>>> next to each other. I think this can be turned into an accurate 2^4 wi=
th
>>> coloring changes, so I agree with Joel too :)
>>>
>>> To help me think about it, I started adding a new projection option for
>>> spherical puzzles to MagicTile, which takes the two hemispheres of a pu=
zzle
>>> and maps them to two disks with identified boundaries connected at a po=
int,
>>> just like a physical "global chess" game I have. Melinda's puzzle is a=
lot
>>> like this up a dimension, so think about two disjoint balls, each
>>> representing a hemisphere of the 2^4, each a "subcube" of Melinda's puz=
zle.
>>> The two boundaries of the balls are identified with each other and as y=
ou
>>> roll one around, the other half rolls around so that identified points
>>> connect up. We need to have the same restriction on Melinda's puzzle.
>>>
>>> In the pristine state then, I think it'd be nice to have an internal
>>> (hidden), solid colored octahedron on each half. The other 6 faces sho=
uld
>>> all have equal colors split between each hemisphere, 4 stickers on each
>>> half. You should be able to reorient the two subcubes to make a half
>>> octahedron of any color on each subcube. I just saw Matt's email and
>>> picture, and it looks like we were going down the same thought path. I
>>> think with recoloring (mirroring some of the current piece colorings)
>>> though, the windmill's can be avoided (?)
>>>
>>> [...] After staring/thinking a bit more, the coloring Matt came up with=
is
>>> right-on if you want to put a solid color at the center of each hemisph=
ere.
>>> His comment about the "mirrored" pieces on each side helped me understa=
nd
>>> better. 3 of the stickers are mirrored and the 4th is the hidden color
>>> (different on each side for a given pair of "mirrored" pieces). All fa=
ces
>>> behave identically as well, as they should. It's a little bit of a bum=
mer
>>> that it doesn't look very pristine in the pristine state, but it does l=
ook
>>> like it should work as a 2^4.
>>>
>>> I wonder if there might be some adjustments to be made when shapeways
>>> allows printing translucent as a color :)
>>>
>>> [...] Sorry for all the streaming, but I wanted to share one more thoug=
ht.
>>> I now completely agree with Joel/Matt about it behaving as a 2^4, even =
with
>>> the original coloring. You just need to consider the corner colors of =
the
>>> two subcubes (pink/purple near the end of the video) as being a window =
into
>>> the interior of the piece. The other colors match up as desired. (Sor=
ry if
>>> folks already understood this after their emails and I'm just catching =
up!)
>>>
>>> In fact, you could alter the coloring of the pieces slightly so that th=
e
>>> behavior was similar with the inverted coloring. At the corners where =
3
>>> colors meet on each piece, you could put a little circle of color of th=
e
>>> opposite 4th color. In Matt's windmill coloring then, you'd be able to=
see
>>> all four colors of a piece, like you can with some of the pieces on
>>> Melinda's original coloring. And again you'd consider the color circle=
s a
>>> window to the interior that did not require the same matching constrain=
ts
>>> between the subcubes.
>>>
>>> I'm looking forward to having one of these :)
>>>
>>> Happy Friday everyone,
>>> Roice
>>>
>>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.com
>>> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>>>>
>>>>
>>>>
>>>> Seems like there was a slight misunderstanding. I meant that you need =
to
>>>> be able to twist one of the faces and in MC4D the most natural choice=
is
>>>> the center face. In your physical puzzle you can achieve this type of =
twist
>>>> by twisting the two subcubes although this is indeed a twist of the su=
bcubes
>>>> themselves and not the center face, however, this is still the same ty=
pe of
>>>> twist just around another face.
>>>>
>>>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup o=
f
>>>> this puzzle. Hopefully the restrictions will be quite natural and only=
some
>>>> "strange" moves would be illegal. Regarding the "families of states" (=
aka
>>>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed =
twists
>>>> preserves the parity of the pieces, meaning that only half of the
>>>> permutations you can achieve by disassembling and reassembling can be
>>>> reached through legal moves. Because of some geometrical properties of=
the
>>>> 2x2x2x2 and its twists, which would take some time to discuss in detai=
l
>>>> here, the orientation of the stickers mod 3 are preserved, meaning tha=
t the
>>>> last corner only can be oriented in one third of the number of orienta=
tions
>>>> for the other corners. This gives a total number of orbits of 2x3=3D6.=
To
>>>> check this result let's use this information to calculate all the poss=
ible
>>>> states of the 2x2x2x2; if there were no restrictions we would have 16!=
for
>>>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>>>> orientations for each corner). If we now take into account that there =
are 6
>>>> equally sized orbits this gets us to 12!16^12/6. However, we should al=
so
>>>> note that the orientation of the puzzle as a hole is not set by some k=
ind of
>>>> centerpieces and thus we need to devide with the number of orientation=
s of a
>>>> 4D cube if we want all our states to be separated with twists and not =
only
>>>> rotations of the hole thing. The number of ways to orient a 4D cube in=
space
>>>> (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a to=
tal of
>>>> 12!16^12/(6*192) states which is indeed the same number that for examp=
le
>>>> David Smith arrived at during his calculations. Therefore, when deter=
mining
>>>> whether or not a twist on your puzzle is legal or not it is sufficient=
and
>>>> necessary to confirm that the twist is an even permutation of the piec=
es and
>>>> preserves the orientation of stickers mod 3.
>>>>
>>>> Best regards,
>>>> Joel
>>>>
>>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.com
>>>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>>>
>>>>
>>>>
>>>> The new arrangement of magnets allows every valid orientation of piece=
s.
>>>> The only invalid ones are those where the diagonal lines cutting each =
cube's
>>>> face cross each other rather than coincide. In other words, you can as=
semble
>>>> the puzzle in all ways that preserve the overall diamond/harlequin pat=
tern.
>>>> Just about every move you can think of on the whole puzzle is valid th=
ough
>>>> there are definitely invalid moves that the magnets allow. The most ob=
vious
>>>> invalid move is twisting of a single end cap.
>>>>
>>>> I think your description of the center face is not correct though. Twi=
sts
>>>> of the outer faces cause twists "through" the center face, not "of" th=
at
>>>> face. Twists of the outer faces are twists of those faces themselves b=
ecause
>>>> they are the ones not changing, just like the center and outer faces o=
f MC4D
>>>> when you twist the center face. The only direct twist of the center fa=
ce
>>>> that this puzzle allows is a 90 degree twist about the outer axis. Tha=
t
>>>> happens when you simultaneously twist both end caps in the same direct=
ion.
>>>>
>>>> Yes, it's quite straightforward reorienting the whole puzzle to put an=
y
>>>> of the four axes on the outside. This is a very nice improvement over =
the
>>>> first version and should make it much easier to solve. You may be righ=
t that
>>>> we just need to find the right way to think about the outside faces. I=
'll
>>>> leave it to the math geniuses on the list to figure that out.
>>>>
>>>> -Melinda
>>>>
>>>>
>>>>
>>>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubi=
ng]
>>>> wrote:
>>>>
>>>>
>>>> Hi Melinda,
>>>>
>>>> I do not agree with the criticism regarding the white and yellow stick=
ers
>>>> touching each other, this could simply be an effect of the different
>>>> representations of the puzzle. To really figure out if this indeed is =
a
>>>> representation of a 2x2x2x2 we need to look at the possible moves (twi=
sts
>>>> and rotations) and figure out the equivalent moves in the MC4D softwar=
e.
>>>> From the MC4D software, it's easy to understand that the only moves re=
quired
>>>> are free twists of one of the faces (that is, only twisting the center=
face
>>>> in the standard perspective projection in MC4D) and 4D rotations swapp=
ing
>>>> which face is in the center (ctrl-clicking in MC4D). The first is poss=
ible
>>>> in your physical puzzle by rotating the white and yellow subcubes (fro=
m here
>>>> on I use subcube to refer to the two halves of the puzzle and the colo=
urs of
>>>> the subcubes to refer to the "outer colours"). The second is possible =
if
>>>> it's possible to reach a solved state with any two colours on the subc=
ubes
>>>> that still allow you to perform the previously mentioned twists. This =
seems
>>>> to be the case from your demonstration and is indeed true if the magne=
ts
>>>> allow the simple twists regardless of the colours of the subcubes. Thu=
s, it
>>>> is possible to let your puzzle be a representation of a 2x2x2x2, howev=
er, it
>>>> might require that some moves that the magnets allow aren't used.
>>>>
>>>> Best regards,
>>>> Joel
>>>>
>>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>>>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>>>>
>>>>>
>>>>>
>>>>> Dear Cubists,
>>>>>
>>>>> I've finished version 2 of my physical puzzle and uploaded a video of=
it
>>>>> here:
>>>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>>>> Again, please don't share these videos outside this group as their
>>>>> purpose is just to get your feedback. I'll eventually replace them wi=
th a
>>>>> public video.
>>>>>
>>>>> Here is an extra math puzzle that I bet you folks can answer: How man=
y
>>>>> families of states does this puzzle have? In other words, if disassem=
bled
>>>>> and reassembled in any random configuration the magnets allow, what a=
re the
>>>>> odds that it can be solved? This has practical implications if all su=
ch
>>>>> configurations are solvable because it would provide a very easy way =
to
>>>>> fully scramble the puzzle.
>>>>>
>>>>> And finally, a bit of fun: A relatively new friend of mine and new li=
st
>>>>> member, Marc Ringuette, got excited enough to make his own version. H=
e built
>>>>> it from EPP foam and colored tape, and used honey instead of magnets =
to hold
>>>>> it together. Check it out here:
>>>>> http://superliminal.com/cube/dessert_cube.jpg I don't know how practi=
cal a
>>>>> solution this is but it sure looks delicious! Welcome Marc!
>>>>>
>>>>> -Melinda
>>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>
>>
>>
>>=20




From: Melinda Green <melinda@superliminal.com>
Date: Mon, 29 May 2017 18:57:41 -0700
Subject: Re: [MC4D] Physical 4D puzzle V2



Thanks for the clarification, Joel. Just to be complete, are you sure that =
that the count of MC4D states are being counted in the same way?

One other thought I had regarding terminology: You called the two forms of =
the physical puzzle "representations", but I wonder whether a somewhat bett=
er term might be "projection". This puzzle is definitely not any sort of ge=
ometric projection into 3-space, but it seems to share a number of analogou=
s properties with them. I often think of it as viewing the 4D object throug=
h a 2x2x4 "viewport".

The half rotations are sort of like translating that viewport along or arou=
nd the surface of that object. Since you point out that it is a 90 degree r=
otation, perhaps "half rotation" isn't the best term for that move. Whateve=
r we call these rotations, the two forms feel to me like the difference bet=
ween cell-first and face-first projections.

-Melinda

On 5/29/2017 2:57 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing] wr=
ote:
> Hello,
>
> Just a quick correction regarding a previous statement. From the
> calculation of the states of the puzzle, we can see that if we choose
> a state and rotation there are still two representations of that state
> and rotation in the physical puzzle. Previously, we also said that
> these states are separated by a half-rotation such as Rx+ from an RL
> rep or AKx rep. This is not the case. When I calculated the states of
> the puzzle I assumed that the longer side of the puzzle should be
> parallel with the x-axis. If we don't make that assumption we find
> that the number of states (not distinct states, some of these are just
> separated by a rotation) are 3*16!*24*12^15 since there are three
> possible choices for which axis should be parallel with the longer
> side. This means that, in fact, each distinct state has 6
> representations in the physical puzzle. This is actually what we
> should expect since it should be possible to represent every state in
> an RL rep, an UD rep, an FB rep, an AKx rep, an AKy rep and an AKz
> rep. In the solved state, the rotation of the puzzle is determined by
> which colour belongs to each face and given such a rotation there are
> indeed six representations: first, choose an axis that the longer side
> should be parallel with (3 alternatives) and then choose either the AK
> rep or the non-AK rep (2 alternatives). However, the so-called
> "half-rotation" changes which colour belongs to which face so this is
> not a move that gets you from one representation of a state to
> another. From an UD rep with:
> P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x ,
> P3 Ox'z' takes you to an FB rep of the same state
> Oy P3 Oz' takes you to an RL rep of the same state
> UD'x P3 UD'z' Sy- Oy takes you to an AKy rep of the same state
> UD'x P3 Oyx' Sz+ takes you to an AKz rep of the same state and
> UD'x P3 UD'x' Oz' Sx- takes you to an AKx rep of the same state
> >From this, we can see that the AK states are closely related and it's
> possible to change between them without illegal moves. For example
> going from an AKy rep to an AKz rep of the same state could be done
> with:
> (UD'x P3 UD'z' Sy- Oy)' UD'x P3 Oyx' Sz+ =3D Oy' Sy+ UD'z
> P3' UD'x' UD'x P3 Oyx' Sz+ =3D Oy' Sy+ UD'z Oyx' Sz+
>
> Note that these sequences don't change the state nor the rotation of
> the puzzle but they do change the representation. Let's call these
> types of sequences J moves. Note that what I have previously called a
> type three rotation is actually a J move and a rotation. Further note
> that J =3D I mod(rep) (modulo representation). The J moves changes which
> pair of faces are symmetry breaking and this is the type of moves that
> needs to be added to the set of legal moves to make the physical
> puzzle an actual 2x2x2x2. A J move is a move that is equal to I
> mod(rep, rot) but that isn't equal to I mod(rot) (which means that it
> has to preserve the state, is allowed to change the rotation and has
> to change the representation/what faces (not in the left/right sense
> but rather in the sense of colours) are symmetry-breaking ).
>
> Best regards,
> Joel Karlsson
>
> 2017-05-22 22:32 GMT+02:00 Joel Karlsson :
>> Hi Melinda,
>>
>> Thank you for the feedback. Regarding the coordinate system, it=E2=80=99=
s just
>> a matter of preference. I thought it would be nice to have a
>> xy-symmetry but understand that it might be more practical to follow
>> conventions. So, adapting your suggestion, let's redefine the axes as
>> x pointing right, y up and z towards you. Note that this means that
>> the longer side of the puzzle is parallel with the x-axis in the
>> standard rotation. From here on, I will use this new coordinate
>> system.
>>
>> Regarding the name of the faces. Since which faces are the "outer
>> ones" changes with how you rotate the puzzle and what state the puzzle
>> is in, I think that the labelling of the faces should be independent
>> of which faces are the outer faces (forming what will be referred to
>> as inverted octahedra). Since the puzzle is a representation of a
>> 4d-cube in 3d-space it will (with our coordinate system) always have
>> two faces "belonging" to every axis and two faces that don't belong to
>> any axis. Therefore, it makes sense to label the faces in such a way
>> that the face in for example the positive x direction is R, always.
>> So, the K face (which is one of the faces not belonging to a
>> particular axis) is always the face only belonging to the center 2x2x2
>> block (and this can be either an octahedron or an inverted octahedron
>> (the outer corners) depending on the representation of the puzzle).
>> The distinction between left and right never disappears; the R face is
>> always the face only belonging to the right half of the puzzle and can
>> be an octahedron, an inverted octahedron, two half octahedra (<><>) or
>> one half and two quarter octahedra (><><). Note that the
>> half-rotations are 90-degree rotations of the puzzle; for example, Sx+
>> (physical puzzle) =3D Cx' (virtual puzzle) is the rotation that does L
>> -> K -> R -> A -> L so the face that previously was the L face is now
>> the K face. From the standard rotation (where R and L are the inverted
>> octahedra before the rotation), this would mean that the K and A faces
>> are inverted octahedra after a Sx+ rotation. What faces are the
>> "mysterious" (or more precisely symmetry breaking, forming inverted
>> octahedra instead of regular octahedra) depend on the rotation of the
>> puzzle and can be any two opposite faces (R and L, U and D, F and B or
>> A and K). It might be useful to be able to describe what faces are
>> inverted octahedra since this determines what moves are legal so let's
>> say that the puzzle is in an RL representation if R and L are inverted
>> octahedra and similarly for other states. Thus, Sx+ can take you from
>> an RL representation to an AK representation (what a long word let's
>> use rep for short). Note that the AK rep doesn't specify which axis
>> the longer side should be parallel with so let's just add a lowercase
>> letter to indicate this (an AKx rep is thus a state where the A and K
>> faces are inverted octahedra and the longer side is parallel with the
>> x-axis). In conclusion, I would like to keep the names of the faces as
>> I first defined them and hope that it's clearer what I mean with them
>> and that the names of the faces are not related to what faces form
>> inverted octahedra.
>>
>> You also wrote:
>> "There are several, distinct types of rotations, none of which change
>> the state of the puzzle, and I think we need a way to be unambiguous
>> about them. The types I see are
>>
>> 1. Simple reorientation of the physical puzzle in the hand, no magnets
>> involved. IE your 'O' moves and maybe analogous to mouse-dragging in
>> MC4D?
>> 2. Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-=
click?
>> 3. Half-rotations. Maybe analogous to a "face first" view? (ctrl-click
>> on a 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by
>> Cubie")
>> 4. Whole-puzzle reorientations that move an arbitrary axis into the
>> "outer" 2 faces. No MC4D analog."
>>
>> The rotations (as far as I know) are the O moves (type one rotation),
>> which are indeed analogous to mouse-dragging in MC4D, rolling the
>> 2x2x2 halves (type two rotation) (these are easily described as for
>> example RL'y in an RL rep) which (in a non-AK rep) is a ctrl-click on
>> a non-inverted face (that is ctrl-click on a face that is currently
>> represented with an octahedron), half-rotations (also type two
>> rotations) which is a ctrl-click on an inverted face and sequences
>> that for example from an RL rep can take the R face to K without
>> turning the puzzle into an AK rep (type three rotation). Type one and
>> two rotations are legal moves but type three contain illegal 2x2x2x2
>> moves according to hypothesis * in my previous email (however, they
>> are very important and needed if we wish to be able to reach all
>> states).
>>
>> Regarding fold moves, from the standard rotation (RL rep) are you
>> saying that Vy+ =3D Vy'+ or something like Vy+ =3D Vy'+ FB'x2 =3D Vy'+
>> mod(rot)? The former seems not to be correct but I believe the latter
>> is, please correct me if I'm wrong. I don't assume that you fold the
>> puzzle back to the same representation. This is what + and -
>> indicates, + preserve the representation mod(rot) (i.e AK rep both
>> before and after or neither before nor after) and - changes it (going
>> from AK rep to non-AK rep or vice versa).
>>
>> "I think the only legal 90 degree twists of the K face are those about
>> the long axis. I believe this is what Christopher Locke was saying in
>> this message. To see why there is no straightforward way to perform
>> other 90 degree twists, you only need to perform a 90 degree twist on
>> an outer (L/R) face and then reorient the whole puzzle along a
>> different outer axis. If the original twist was not about the new long
>> axis, then there is clearly no straightforward way to undo that
>> twist."
>>
>> Yes, as pointed out further down in my previous email. The notation
>> allows all 90-degree rotations after the section "extensions of some
>> definitions" although only 180-degree rotations are legal for
>> octahedral faces around axes not parallel with the longer side of the
>> puzzle (see the section =E2=80=9Csome important notes on legal/illegal
>> moves=E2=80=9D).
>>
>> "I noticed something like this the other day but realized that it only
>> seems to work for rotations along the long dimension (z in your
>> example). These are already easily accomplished by a simple rotation
>> to put the face in question on the end caps, followed by a double
>> end-cap twist."
>>
>> It works for the other rotations as well although these are not legal
>> moves. They could possibly be used instead of the illegal S moves
>> (along axes not parallel with the longer side) and the illegal V moves
>> (V- (minus) moves which are closely related to the illegal S moves,
>> example from RL rep: Vy- =3D Vy+ Sx Oz2) but as I mentioned in my
>> previous email I do believe that it's better to use the S and V moves
>> since you have found a relatively short way to perform a type three
>> rotation with those.
>>
>> Let:
>> P1 =3D Sy+ Uyz2 Sy-
>> P1=E2=80=99 =3D P1 (P1 is its own inverse)
>> P2 =3D (P1 UD=E2=80=99z P1 UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z=E2=
=80=99 P1 UD=E2=80=99z)2 (P2=3DP_2 not P^2
>> whereas the two at the end means: perform twice)
>> P2=E2=80=99 =3D (UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z P1 UD=E2=80=
=99z P1 UD=E2=80=99z=E2=80=99 P1)2
>> P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x =3D I mo=
d(rot) (type three rotation)
>> P3=E2=80=99 =3D Oy=E2=80=99 P3 Oy=E2=80=99
>> P4 =3D P2 P3 P2 P3=E2=80=99 P2=E2=80=99 P3 P2=E2=80=99 P3=E2=80=
=99
>>
>> P4 from UD rep is a 164 move sequence rotating only one corner in its
>> place. The sequence is inspired by Roice =E2=80=9Csecond four-color seri=
es=E2=80=9D
>> but I have changed the =E2=80=9CTop 9=E2=80=9D moves to pure rotations s=
ince it=E2=80=99s all
>> that=E2=80=99s necessary and a bit shorter to perform. P3 is your type t=
hree
>> rotation (with a rotation added at the end) and P2 is Roice =E2=80=9Cthi=
rd
>> three-color series=E2=80=9D. Written with my notation but for the virtua=
l
>> puzzle, the sequences (Roice original since it=E2=80=99s easier to perfo=
rm K
>> twists than O rotations in MC4D) are:
>> Q1 =3D (Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=99 Rz2y=E2=80=99)2 (=
analogue to P2)
>> Q1=E2=80=99 =3D (Rz2y=E2=80=99 Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=
=99)2
>> Q2 =3D Q1 Kxy=E2=80=99 Q1 Kyx=E2=80=99 Q1=E2=80=99 Kxy=E2=80=99 Q1=
=E2=80=99 Kyx=E2=80=99 (analogue to P4 but
>> only 36 moves)
>>
>> How to read faster (the example applies to a 3x3x3x3): moves like
>> Kz2y=E2=80=99 are clicking on 3C-pieces in MC4D. If the move is written =
in
>> this way, [uppercase letter] [lowercase letter]2 [lowercase letter
>> possibly with =E2=80=98 (prime)], there=E2=80=99s a quite quick way to r=
ealize which
>> piece this is. The uppercase letter specifies which face the piece to
>> press is on and the first lowercase letter (followed by 2) specifies
>> one of the sides of that face that the piece belong to. There are then
>> 4 possible pieces. Sadly, the piece do not lie in the direction of the
>> last letter from the center of the side of the face but you have to
>> move one edge clockwise from this. So, Kz2y=E2=80=99 is a click on the e=
dge on
>> the front side of the K face one step clockwise from the negative
>> y-axis (thus, the front left 3C-piece on the K face). It=E2=80=99s a bit
>> unfortunate that this =E2=80=9Crule=E2=80=9D isn=E2=80=99t even simpler =
but it's at least true
>> for all of these moves (as far as I know). Moves like Kxy=E2=80=99 are
>> left/right-clicking on a corner piece but currently I don=E2=80=99t know=
a
>> fast way to determine which corner. Any ideas?
>>
>> Best regards,
>> Joel Karlsson
>>
>> 2017-05-19 3:33 GMT+02:00 Melinda Green melinda@superliminal.com
>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>>
>>> Hello Joel,
>>>
>>> Thanks for drilling into this puzzle. Finding good ways to discuss and =
think
>>> about moves and representations will be key. I'll comment on some detai=
ls
>>> in-line.
>>>
>>> On 5/14/2017 6:16 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing=
]
>>> wrote:
>>>
>>>
>>> Yes, that is correct and in fact, you should divide not only with 24 fo=
r the
>>> orientation but also with 16 for the placement if you want to calculate
>>> unique states (since the 2x2x2x2 doesn't have fixed centerpieces). The
>>> point, however, was that if you don=E2=80=99t take that into account yo=
u get a
>>> factor of 24*16=3D384 (meaning that the puzzle has 384 representations =
of
>>> every unique state) instead of the factor of 192 which you get when
>>> calculating the states from the virtual puzzle and hence every state of=
the
>>> virtual puzzle has two representations in the physical puzzle. Yes exac=
tly,
>>> they are indeed the same solved (or other) state and you are correct th=
at
>>> the half rotation (taking off a 2x2 layer and placing it at the other e=
nd of
>>> the puzzle) takes you from one representation to the same state with th=
e
>>> other representation. This means that the restacking move (taking off t=
he
>>> front 2x4 layer and placing it behind the other 2x4 layer) can be expre=
ssed
>>> with half-rotations and ordinary twists and rotations (which you might =
have
>>> pointed out already).
>>>
>>>
>>> Yes, I made that claim in the video but didn't show it because I have y=
et to
>>> record such a sequence. I've only stumbled through it a few times. I ta=
lked
>>> about it at 5:53 though I mistakenly called it a twist, when I should h=
ave
>>> called it a sequence.
>>>
>>>
>>> I think I've found six moves including ordinary twists and a restacking=
move
>>> that is identical to a half-rotation and thus it's easy to compose a
>>> restacking move with one half-rotation and five ordinary twists. There =
might
>>> be an error since I've only played with the puzzle in my mind so it wou=
ld be
>>> great if you, Melinda, could confirm this (the sequence is described la=
ter
>>> in this email).
>>>
>>>
>>> You mean "RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-"? Yes, that works. Th=
ere do seem to
>>> be easier ways to do that beginning with an ordinary rolling rotation. =
I
>>> don't see those in your notation, but the equivalent using a pair of tw=
ists
>>> would be Rx Lx' Sx Vy if I got that right.
>>>
>>>
>>> To be able to communicate move sequences properly we need notation for
>>> representing twists, rotations, half-rotations, restacking moves and fo=
lds.
>>> Feel free to come with other suggestion but you can find mine below. Pl=
ease
>>> read the following thoroughly (maybe twice) to make sure that you under=
stand
>>> everything since misinterpreted notation could potentially become a
>>> nightmare and feel free to ask questions if there is something that nee=
ds
>>> clarification.
>>>
>>> Coordinate system and labelling:
>>>
>>> Let's introduce a global coordinate system. In whatever state the puzzl=
e is
>>> let the positive x-axis point upwards, the positive y-axis towards you =
and
>>> the positive z-axis to the right (note that this is a right-hand system=
).
>>>
>>>
>>> I see the utility of a global coordinate system, but this one seems rat=
her
>>> non-standard. I suggest that X be to the right, and Y up since these ar=
e
>>> near-universal standards. Z can be in or out. I have no opinion. If the=
re is
>>> any convention in the twisty puzzle community, I'd go with that.
>>>
>>> Note also that the wiki may be a good place to document and iterate on
>>> terminology, descriptions and diagrams. Ray added a "notation" section =
to
>>> the 3^4 page here, and I know that one other member was thinking of
>>> collecting a set of moves on another wiki page.
>>>
>>>
>>> Now let's name the 8 faces of the puzzle. The right face is denoted wit=
h R,
>>> the left with L, the top U (up), the bottom D (down), the front F, the =
back
>>> B, the center K (kata) and the last one A (ana). The R and L faces are
>>> either the outer corners of the right and left halves respectively or t=
he
>>> inner corners of these halves (forming octahedra) depending on the
>>> representation of the puzzle. The U, D, F and B faces are either two di=
amond
>>> shapes (looking something like this: <><>) on the corresponding side of=
the
>>> puzzle or one whole and two half diamond shapes (><><). The K face is e=
ither
>>> an octahedron in the center of the puzzle or the outer corners of the c=
enter
>>> 2x2x2 block. Lastly, the A face is either two diamond shapes, one on th=
e
>>> right and one on the left side of the puzzle, or another shape that=E2=
=80=99s a
>>> little bit hard to describe with just a few words (the white stickers a=
t
>>> 5:10 in the latest video, after the half-rotation but before the restac=
king
>>> move).
>>>
>>>
>>> I think it's more correct to say that the K face is either an octahedro=
n at
>>> the origin (A<>K<>A) or in the center of one of the main halves, with t=
he A
>>> face inside the other half (>A<>K<). This was what I was getting at in =
my
>>> previous message. You do later talk about octahedral faces being in eit=
her
>>> the center or the two main halves, so this is just terminology. But abo=
ut
>>> "the outer corners of the center 2x2x2 block", this cannot be the A or =
K
>>> face as you've labeled them. You've been calling these the L/R faces, b=
ut
>>> the left-right distinction disappears in the half-rotated state, so may=
be
>>> "left" and "right" aren't the best names. To me, they are always the
>>> "outside" faces, regardless. You can distinguish them as the left and r=
ight
>>> outside faces in one representation, or as the center and end outside f=
aces
>>> in the other. (Or perhaps "end" versus "ind" if we want to be cute.)
>>>
>>> I'm also a little torn about naming the interior faces ana and kata, no=
t
>>> because of the names themselves which I like, but because the mysteriou=
s
>>> faces to me are the outermost ones you're calling R and L. It only requ=
ires
>>> a simple rotation to move faces in and out of the interior (octahedral)
>>> positions, but it's much more difficult to move another axis into the
>>> L/R/outer direction.
>>>
>>> So maybe the directions can be
>>>
>>> Up-Down
>>> Front-Back
>>> Ind-End
>>> Ana-Kata
>>>
>>> I'm not in love with it and will be happy with anything that works. Tho=
ughts
>>> anyone?
>>>
>>>
>>> We also need a name for normal 3D-rotations, restacking moves and folds
>>> (note that a half-rotation is a kind of restacking move). Let O be the =
name
>>> for a rOtation (note that the origin O doesn't move during a rotation, =
by
>>> the way, these are 3D rotations of the physical puzzle), let S represen=
t a
>>> reStacking move and V a folding/clamshell move (you can remember this b=
y
>>> thinking of V as a folded line).
>>>
>>>
>>> I think it's fine to call the clamshell move a fold or denote it as V. =
I
>>> just wouldn't consider it to be a basic move since it's a simple compos=
ite
>>> of 3 basic twists as shown here. In general, I think there are so many
>>> useful composite moves that we need to be able to easily make them up a=
d hoc
>>> with substitutions like Let =E2=86=93 =3D Rx Lx'. These are really macr=
o moves which
>>> can be nested. That said, it's a particularly useful move so it's proba=
bly
>>> worth describing in some formal way like you do in detail below.
>>>
>>>
>>>
>>> Further, let I (capital i) be the identity, preserving the state and
>>> rotation of the puzzle. We cannot use I to indicate what moves should b=
e
>>> performed but it's still useful as we will see later. Since we also wan=
t to
>>> be able to express if a sequence of moves is a rotation, preserving the
>>> state of the puzzle but possibly representing it in a different way, we=
can
>>> introduce mod(rot) (modulo rotation). So, if a move sequence P satisfie=
s P
>>> mod(rot) =3D I, that means that the state of the puzzle is the same bef=
ore and
>>> after P is performed although the rotation and representation of the pu=
zzle
>>> are allowed to change. I do also want to introduce mod(3rot) (modulo
>>> 3D-rotation) and P mod(3rot) =3D I means that if the right 3D-rotation =
(a
>>> combination of O moves as we will see later) is applied to the puzzle a=
fter
>>> P you get the identity I. Moreover, let the standard rotation of the pu=
zzle
>>> be any rotation such that the longer side is parallel with the z-axis, =
that
>>> is the puzzle forms a 2x2x4 (2 pieces thick in the x-direction, 2 in th=
e
>>> y-direction and 4 in the z-direction), and the K face is an octahedron.
>>>
>>> Rotations and twists:
>>>
>>> Now we can move on to name actual moves. The notation of a move is a
>>> combination of a capital letter and a lowercase letter. O followed by x=
, y
>>> or z is a rotation of the whole puzzle around the corresponding axis in=
the
>>> mathematical positive direction (counterclockwise/the way your right-ha=
nd
>>> fingers curl if you point in the direction of the axis with your thumb)=
. For
>>> example, Ox is a rotation around the x-axis that turns a 2x2x4 into a 2=
x4x2.
>>> A name of a face (U, D, F, B, R, L, K or A) followed by x, y or z means=
:
>>> detach the 8 pieces that have a sticker belonging to the face and then =
turn
>>> those pieces around the global axis. For example, if the longer side of=
the
>>> puzzle is parallel to the z-axis (the standard rotation), Rx means: tak=
e the
>>> right 2x2x2 block and turn it around the global x-axis in the mathemati=
cal
>>> positive direction. Note that what moves are physically possible and al=
lowed
>>> is determined by the rotation of the puzzle (I will come back to this
>>> later). Further note that Ox mod(rot) =3D Oy mod(rot) =3D Oz mod(rot) =
=3D I,
>>> meaning that the 3D-rotations of the physical puzzle corresponds to a
>>> 4D-rotation of the represented 2x2x2x2.
>>>
>>>
>>> There are several, distinct types of rotations, none of which change th=
e
>>> state of the puzzle, and I think we need a way to be unambiguous about =
them.
>>> The types I see are
>>>
>>> Simple reorientation of the physical puzzle in the hand, no magnets
>>> involved. IE your 'O' moves and maybe analogous to mouse-dragging in MC=
4D?
>>> Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-cl=
ick?
>>> Half-rotations. Maybe analogous to a "face first" view? (ctrl-click on =
a
>>> 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by Cubie=
")
>>> Whole-puzzle reorientations that move an arbitrary axis into the "outer=
" 2
>>> faces. No MC4D analog.
>>>
>>>
>>> Inverses and performing a move more than once
>>>
>>> To mark that a move should be performed n times let's put ^n after it. =
For
>>> convenience when writing and speaking let ' (prime) represent ^-1 (the
>>> inverse) and n represent ^n. The inverse P' of some permutation P is th=
e
>>> permutation that satisfies P P' =3D P' P =3D I (the identity). For exam=
ple (Rx)'
>>> means: do Rx backwards, which corresponds to rotating the right 2x2x2 b=
lock
>>> in the mathematical negative direction (clockwise) around the x-axis an=
d
>>> (Rx)2 means: perform Rx twice. However, we can also define powers of ju=
st
>>> the lowercase letters, for example, Rx2 =3D Rx^2 =3D Rxx =3D Rx Rx. So =
x2=3Dx^2
>>> means: do whatever the capital letter specifies two times with respect =
to
>>> the x-axis. We can see that the capital letter naturally is distributed=
over
>>> the two lowercase letters. Rx' =3D Rx^-1 means: do whatever the capital=
letter
>>> specifies but in the other direction than you would have if the prime
>>> wouldn't have been there (note that thus, x'=3Dx^-1 can be seen as the
>>> negative x-axis). Note that (Rx)^2 =3D Rx^2 and (Rx)' =3D Rx' which is =
true for
>>> all twists and rotations but that doesn't have to be the case for other
>>> types of moves (restacks and folds).
>>>
>>> Restacking moves
>>>
>>> A restacking move is an S followed by either x, y or z. Here the lowerc=
ase
>>> letter specifies in what direction to restack. For example, Sy (from th=
e
>>> standard rotation) means: take the front 8 pieces and put them at the b=
ack,
>>> whereas Sx means: take the top 8 pieces and put them at the bottom. Not=
e
>>> that Sx is equivalent to taking the bottom 8 pieces and putting them at=
the
>>> top. However, if we want to be able to make half-rotations we sometimes=
need
>>> to restack through a plane that doesn't go through the origin. In the
>>> standard rotation, let Sz be the normal restack (taking the 8 right pie=
ces,
>>> the right 2x2x2 block, and putting them on the left), Sz+ be the restac=
k
>>> where you split the puzzle in the plane further in the positive z-direc=
tion
>>> (taking the right 2x2x1 cap of 4 pieces and putting it at the left end =
of
>>> the puzzle) and Sz- the restack where you split the puzzle in the plane
>>> further in the negative z-direction (taking the left 2x2x1 cap of 4 pie=
ces
>>> and putting it at the right end of the puzzle). If the longer side of t=
he
>>> puzzle is parallel to the x-axis instead, Sx+ would take the top 1x2x2 =
cap
>>> and put it on the bottom. Note that in the standard rotation Sz mod(rot=
) =3D
>>> I. For restacks we see that (Sx)' =3D Sx' =3D Sx (true for y and z too =
of
>>> course), that (Sz+)' =3D Sz- and that (Sz+)2 =3D (Sz-)2 =3D Sz. We can =
define Sz'+
>>> to have meaning by thinking of z=E2=80=99 as the negative z-axis and wi=
th that in
>>> mind it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus, (Sz)=E2=
=80=99 =3D Sz=E2=80=99 and (Sz+)=E2=80=99 =3D Sz=E2=80=99+.
>>>
>>> Fold moves
>>>
>>> A fold might be a little bit harder to describe in an intuitive way. Fi=
rst,
>>> let's think about what folds are interesting moves. The folds that cann=
ot be
>>> expressed as rotations and restacks are unfolding the puzzle to a 4x4 a=
nd
>>> then folding it back along another axis. If we start with the standard
>>> rotation and unfold the puzzle into a 1x4x4 (making it look like a 4x4 =
from
>>> above) the only folds that will achieve something you can't do with a
>>> restack mod(rot) is folding it to a 2x4x2 so that the longer side is
>>> parallel with the y-axis after the fold. Thus, there are 8 interesting =
fold
>>> moves for any given rotation of the puzzle since there are 4 ways to un=
fold
>>> it to a 4x4 and then 2 ways of folding it back that make the move diffe=
rent
>>> from a restack move mod(rot).
>>>
>>>
>>> Assuming you complete a folding move in the same representation (<><> o=
r
>>>> <><), then there are only two interesting choices. That's because it
>>> doesn't matter which end of a chosen cutting plane you open it from, th=
e end
>>> result will be the same. That also means that any two consecutive clams=
hell
>>> moves along the same cutting plane will undo each other. It further sug=
gests
>>> that any interesting sequence of clamshell moves must alternate between=
the
>>> two possible long cut directions, meaning there is no choice involved. =
12
>>> clamshell moves will cycle back to the initial state.
>>>
>>> There is one other weird folding move where you open it in one directio=
n and
>>> then fold the two halves back-to-back in a different direction. If you
>>> simply kept folding along the initial hinge, you'd simply have a restac=
king.
>>> When completed the other way, it's equivalent to a restacking plus a
>>> clamshell, so I don't think it's useful though it is somewhat interesti=
ng.
>>>
>>>
>>> Let's call these 8 folds interesting fold moves. Note that an interesti=
ng
>>> fold move always changes which axis the longer side of the puzzle is
>>> parallel with. Further note that both during unfold and fold all pieces=
are
>>> moved; it would be possible to have 8 of the pieces fixed during an unf=
old
>>> and folding the other half 180 degrees but I think that it=E2=80=99s mo=
re intuitive
>>> that these moves fold both halves 90 degrees and performing them with
>>> 180-degree folds might therefore lead to errors since the puzzle might =
get
>>> rotated differently. To illustrate a correct unfold without a puzzle: P=
ut
>>> your palms together such that your thumbs point upward and your fingers
>>> forward. Now turn your right hand 90 degrees clockwise and your left ha=
nd 90
>>> degrees counterclockwise such that the normal to your palms point up, y=
our
>>> fingers point forward, your right thumb to the right and your left thum=
b to
>>> the left. That was what will later be called a Vx unfold and the folds =
are
>>> simply reversed unfolds. (I might have used the word =E2=80=9Cfold=E2=
=80=9D in two different
>>> ways but will try to use the term =E2=80=9Cfold move=E2=80=9D when refe=
rring to the move
>>> composed by an unfold and a fold rather than simply calling these moves
>>> =E2=80=9Cfolds=E2=80=9D.)
>>>
>>> To specify the unfold let's use V followed by one of x, y, z, x', y' an=
d z'.
>>> The lowercase letter describes in which direction to unfold. Vx means u=
nfold
>>> in the direction of the positive x-axis and Vx' in the direction of the
>>> negative x-axis, if that makes any sense. I will try to explain more
>>> precisely what I mean with the example Vx from the standard rotation (i=
t
>>> might also help to read the last sentences in the previous paragraph ag=
ain).
>>> So, the puzzle is in the standard rotation and thus have the form 2x2x4=
(x-,
>>> y- and z-thickness respectively). The first part (the unfolding) of the=
move
>>> specified with Vx is to unfold the puzzle in the x-direction, making it=
a
>>> 1x4x4 (note that the thickness in the x-direction is 1 after the Vx unf=
old,
>>> which is no coincidence). There are two ways to do that; either the sid=
es of
>>> the pieces that are initially touching another piece (inside of the puz=
zle
>>> in the x,z-plane and your palms in the hand example) are facing up or d=
own
>>> after the unfold. Let Vx be the unfold where these sides point in the
>>> direction of the positive x-axis (up) and Vx' the other one where these
>>> sides point in the direction of the negative x-axis (down) after the un=
fold.
>>> Note that if the longer side of the puzzle is parallel to the z-axis on=
ly
>>> Vx, Vx', Vy and Vy' are possible. Now we need to specify how to fold th=
e
>>> puzzle back to complete the folding move. Given an unfold, say Vx, ther=
e are
>>> only two ways to fold that are interesting (not turning the fold move i=
nto a
>>> restack mod(rot)) and you have to fold it perpendicular to the unfold t=
o
>>> create an interesting fold move. So, if you start with the standard rot=
ation
>>> and do Vx you have a 1x4x4 that you have to fold into a 2x4x2. To
>>> distinguish the two possibilities, use + or - after the Vx. Let Vx+ be =
the
>>> unfold Vx followed by the interesting fold that makes the sides that ar=
e
>>> initially touching another piece (before the unfold) touch another piec=
e
>>> after the fold move is completed and let Vx- be the other interesting f=
old
>>> move that starts with the unfold Vx. (Thus, continuing with the hand
>>> example, if you want to do a Vx+ first do the Vx unfold described in th=
e end
>>> of the previous paragraph and then fold your hands such that your finge=
rs
>>> point up, the normal to your palms point forward, the right palm is tou=
ching
>>> the right-hand fingers, the left palm is touching the left-hand fingers=
, the
>>> right thumb is pointing to the right and the left thumb is pointing to =
the
>>> left). Note that the two halves of the puzzle always should be folded 9=
0
>>> degrees each and you should never make a fold or unfold where you fold =
just
>>> one half 180-degrees (if you want to use my notation, that is). Further=
note
>>> that Vx+ Sx mod(3rot) =3D Vx- and that Vx+ Vx+ =3D I which is equivalen=
t to
>>> (Vx+)=E2=80=99 =3D Vx+ and this is true for all fold moves (note that a=
fter a Vx+
>>> another Vx+ is always possible).
>>>
>>> The 2x2x2x2 in the MC4D software
>>>
>>> The notation above can also be applied to the 2x2x2x2 in the MC4D progr=
am.
>>> There, you are not allowed to do S or V moves but instead, you are allo=
wed
>>> to do the [crtl]+[left-click] moves. This can easily be represented wit=
h
>>> notation similar to the above. Let=E2=80=99s use C (as in Centering) an=
d one of x, y
>>> and z. For example, Cx would be to rotate the face in the positive
>>> x-direction aka the U face to the center. Thus, Cz' is simply
>>> [ctrl]+[left-click] on the L face and similarly for the other C moves. =
The
>>> O, U, D, F, B, R, L, K and A moves are performed in the same way as abo=
ve
>>> so, for example Rx would be a [left-click] on the top-side of the right
>>> face. In this representation of the puzzle almost all moves are allowed=
; all
>>> U, D, F, B, R, L, K, O and C moves are possible regardless of rotation =
and
>>> only A moves (and of course the rightfully forbidden S and V moves) are
>>> impossible regardless of rotation. Note that R and L moves in the softw=
are
>>> correspond to the same moves of the physical puzzle but this is not
>>> generally true (I will come back to this later).
>>>
>>> Possible moves (so far) in the standard rotation
>>>
>>> In the standard rotation, the possible/allowed moves with the definitio=
ns
>>> above are:
>>>
>>> O moves, all of these are always possible in any state and rotation of =
the
>>> puzzle since they are simply 3D-rotations.
>>>
>>> R, L moves, all of these as well since the puzzle has a right and left =
2x2x2
>>> block in the standard rotation.
>>>
>>> U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x4 blocks =
with
>>> less symmetry than a 2x2x2 block.
>>>
>>> F, B moves, just Fy2 and By2 since the front and back are 2x1x4 blocks =
with
>>> less symmetry than a 2x2x2 block.
>>>
>>> K moves, all possible since this is a rotation of the center 2x2x2 bloc=
k.
>>>
>>>
>>> I think the only legal 90 degree twists of the K face are those about t=
he
>>> long axis. I believe this is what Christopher Locke was saying in this
>>> message. To see why there is no straightforward way to perform other 90
>>> degree twists, you only need to perform a 90 degree twist on an outer (=
L/R)
>>> face and then reorient the whole puzzle along a different outer axis. I=
f the
>>> original twist was not about the new long axis, then there is clearly n=
o
>>> straightforward way to undo that twist.
>>>
>>>
>>> A moves, only Az moves since this is two 2x2x1 blocks that have to be
>>> rotated together.
>>>
>>> S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined in the
>>> standard rotation.
>>>
>>> V moves, not Vz+ or Vz- since the definition doesn't give these meaning=
when
>>> the long side of the puzzle is parallel with the z-axis.
>>>
>>> Note that for example Fz2 is not allowed since this won't take you to a
>>> state of the puzzle. To allow more moves we need to extend the definiti=
ons
>>> (after the extension in the next paragraph all rotations and twists (O,=
R,
>>> L, U, D, F, B, K, A) are possible in any rotation and only which S and =
V
>>> moves are possible depend on the state and rotation of the puzzle).
>>>
>>>
>>> Extension of some definitions
>>>
>>> It's possible to make an extension that allows all O, R, L, U, D, F, B,=
K
>>> and A moves in any state. I will explain how this can be done in the
>>> standard rotation but it applies analogously to any other rotation wher=
e the
>>> K face is an octahedron. First let's focus on U, D, F and B and because=
of
>>> the symmetry of the puzzle all of these are analogous so I will only ex=
plain
>>> one. The extension that makes all U moves possible (note that the U fac=
e is
>>> in the positive x-direction) is as follows: when making an U move first
>>> detach the 8 top pieces which gives you a 1x2x4 block, fold this block =
into
>>> a 2x2x2 block in the positive x-direction (similar to the later part of=
a
>>> Vx+ move from the standard rotation) such that the U face form an
>>> octahedron, rotate this 2x2x2 block around the specified axis (for exam=
ple
>>> around the z-axis if you are doing an Uz move), reverse the fold you ju=
st
>>> did creating a 1x2x4 block again and reattach the block.
>>>
>>>
>>> I noticed something like this the other day but realized that it only s=
eems
>>> to work for rotations along the long dimension (z in your example). The=
se
>>> are already easily accomplished by a simple rotation to put the face in
>>> question on the end caps, followed by a double end-cap twist.
>>>
>>> This is as far as I'm going to comment for the moment because the
>>> information gets very dense and I've been mulling and picking over your
>>> message for several days already. In short, I really like your attempt =
to
>>> provide a complete system of notation for discussing this puzzle and wi=
ll be
>>> curious to hear your thoughts on my comments so far. I hope others will
>>> chime in too.
>>>
>>> One final thought is that a real "acid test" of any notation system for=
this
>>> puzzle will be attempt to translate some algorithms from MC4D. I would =
most
>>> like to see a sequence that flips a single piece, like the second 4-col=
or
>>> series on this page of Roice's solution, or his pair of twirled corners=
at
>>> the end of this page. One trick will be to minimize the number of
>>> whole-puzzle reorientations needed, but really any sequence that works =
will
>>> be great evidence that the puzzles are equivalent. I suspect that this =
sort
>>> of exercise will never be practical because it will require too many
>>> reorientations, and that entirely new methods will be needed to actuall=
y
>>> solve this puzzle.
>>>
>>> Best,
>>> -Melinda
>>>
>>> The A moves can be done very similarly but after you have detached the =
two
>>> 2x2x1 blocks you don't fold them but instead you stack them similar to =
a Sz
>>> move, creating a 2x2x2 block with the A face as an octahedron in the mi=
ddle
>>> and then reverse the process after you have rotated the block as specif=
ied
>>> (for example around the negative y-axis if you are doing an Ay' move). =
Note
>>> that these extended moves are closely related to the normal moves and f=
or
>>> example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99 in the standard rotatio=
n and note that (Ry
>>> Ly=E2=80=99) mod(rot) =3D (Ly Ry=E2=80=99) mod(rot) =3D I (this applies=
to the other extended
>>> moves as well).
>>>
>>> If the cube is in the half-rotated state, where both the R and L faces =
are
>>> octahedra, you can extend the definitions very similarly. The only thin=
g you
>>> have to change is how you fold the 2x4 blocks when performing a U, D, F=
or B
>>> move. Instead of folding the two 2x2 halves of the 2x4 into a 2x2 you h=
ave
>>> to fold the end 1x2 block 180 degrees such that the face forms an
>>> octahedron.
>>>
>>> These moves might be a little bit harder to perform, to me especially t=
he A
>>> moves seems a bit awkward, so I don't know if it's good to use them or =
not.
>>> However, the A moves are not necessary if you allow Sz in the standard
>>> rotation (which you really should since Sz mod(rot) =3D I in the standa=
rd
>>> rotation) and thus it might not be too bad to use this extended version=
. The
>>> notation supports both variants so if you don=E2=80=99t want to use the=
se extended
>>> moves that shouldn=E2=80=99t be a problem. Note that, however, for exam=
ple Ux
>>> (physical puzzle) !=3D Ux (virtual puzzle) where !=3D means =E2=80=9Cno=
t equal to=E2=80=9D (more
>>> about legal/illegal moves later).
>>>
>>> Generalisation of the notation
>>>
>>> Let=E2=80=99s generalise the notation to make it easier to use and to m=
ake it work
>>> for any n^4 cube in the MC4D software. Previously, we saw that (Rx)2 =
=3D Rx Rx
>>> and if we allow ourselves to rewrite this as (Rx)2 =3D Rx2 =3D Rxx =3D =
Rx Rx we
>>> see that the capital letter naturally can be distributed over the lower=
case
>>> letters. We can make this more general and say that any capital letter
>>> followed by several lowercase letters means the same thing as the capit=
al
>>> letter distributed over the lowercase letters. Like Rxyz =3D Rx Ry Rz a=
nd here
>>> R can be exchanged with any capital letter and xyz can be exchanged wit=
h any
>>> sequence of lowercase letters. We can also allow several capital letter=
s and
>>> one lowercase letter, for example RLx and let=E2=80=99s define this as =
RLx =3D Rx Lx
>>> so that the lowercase letter can be distributed over the capital letter=
s. We
>>> can also define a capital letter followed by =E2=80=98 (prime) like R=
=E2=80=99x =3D Rx=E2=80=99 and
>>> R=E2=80=99xy =3D Rx=E2=80=99y=E2=80=99 so the =E2=80=98 (prime) is dist=
ributed over the lowercase letters.
>>> Note that we don=E2=80=99t define a capital to any other power than -1 =
like this
>>> since for example R2x =3D RRx might seem like a good idea at first but =
it
>>> isn=E2=80=99t very useful since R2 and RR are the same lengths (and pow=
ers greater
>>> than two are seldom used) and we will see that we can define R2 in anot=
her
>>> way that generalises the notation to all n^4 cubes.
>>>
>>> Okay, let=E2=80=99s define R2 and similar moves now and have in mind wh=
at moves we
>>> want to be possible for a n^4 cube. The moves that we cannot achieve wi=
th
>>> the notation this far is twisting deeper slices. To match the notation =
with
>>> the controls of the MC4D software let R2x be the move similar to Rx but
>>> twisting the 2nd layer instead of the top one and similarly for other
>>> capital letters, numbers (up to n) and lowercase letters. Thus, R2x is
>>> performed as Rx but holding down the number 2 key. Just as in the progr=
am,
>>> when no number is specified 1 is assumed and you can combine several nu=
mbers
>>> like R12x to twist both the first and second layer. This notation does =
not
>>> apply to rotations (O) folding moves (V) and restacking moves (S) (I su=
ppose
>>> you could redefine the S move using this deeper-slice-notation and use =
S1z
>>> as Sz+, S2z as Sz and S3z as Sz- but since these moves are only allowed=
for
>>> the physical 2x2x2x2 I think that the notation with + and =E2=80=93 is =
better since
>>> S followed by a lowercase letter without +/- always means splitting the=
cube
>>> in a coordinate plane that way, not sure though so input would be great=
).
>>> The direction of the twist R3x should be the same as Rx meaning that if=
Rx
>>> takes stickers belonging to K and move them to F, so should R3x, in
>>> accordance with the controls of the MC4D software. Note that for a 3x3x=
3x3
>>> it=E2=80=99s true that R3z =3D Lz whereas R3x =3D Lx=E2=80=99 (note tha=
t R and L are the faces
>>> in the z-directions so because of the symmetry of the cube it will also=
be
>>> true that for example U3x =3D Bx whereas U3y=3DBy=E2=80=99).
>>>
>>> What about the case with several capital letters and several lowercase
>>> letters, for instance, RLxy? I see two natural definitions of this. Eit=
her,
>>> we could have RLxy =3D Rxy Lxy or we could define it as RLxy =3D RLx RL=
y. These
>>> are generally not the same (if you exchange R and L with any allowed ca=
pital
>>> letter and similarly for x and y). I don=E2=80=99t know what is best, w=
hat do you
>>> think? The situations I find this most useful in are RL=E2=80=99xy to d=
o a rotation
>>> and RLxy as a twist. However, since R and L are opposite faces their
>>> operations commute which imply RL=E2=80=99xy (1st definition) =3D Rxy =
L=E2=80=99xy =3D Rx Ry
>>> Lx=E2=80=99 Ly=E2=80=99 =3D Rx Lx=E2=80=99 Ry Ly=E2=80=99 =3D RL=E2=80=
=99x RL=E2=80=99y =3D RL=E2=80=99xy (2nd definition) and similarly
>>> for the other case with RLxy. Hopefully, we can find another useful seq=
uence
>>> of moves where this notation can be used with only one of the definitio=
ns
>>> and can thereby decide which definition to use. Personally, I feel like=
RLxy
>>> =3D RLx RLy is the more intuitive definition but I don=E2=80=99t have a=
ny good
>>> argument for this so I=E2=80=99ll leave the question open.
>>>
>>> For convenience, it might be good to be able to separate moves like Rxy=
and
>>> RLx from the basic moves Rx, Oy etc when speaking and writing. Let=E2=
=80=99s call
>>> the basic moves that only contain one capital letter and one lowercase
>>> letter (possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or a number)=
simple moves (like
>>> Rx, L'y, Ux2 and D3y=E2=80=99) and the moves that contain more than one=
capital
>>> letter or more than one lowercase letter composed moves.
>>>
>>> More about inverses
>>>
>>> This list can obviously be made longer but here are some identities tha=
t are
>>> good to know and understand. Note that R, L, U, x, y and z below just a=
re
>>> examples, the following is true in general for non-folds (however, S mo=
ves
>>> are fine).
>>>
>>> (P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99 P1=E2=80=
=99 (Pi is an arbitrary permutation for
>>> i=3D1,2,=E2=80=A6n)
>>> (Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx
>>> (RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x
>>> (Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+ (just as an example with=
restacking moves, note that
>>> the inverse doesn=E2=80=99t change the + or -)
>>> RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=80=99xyz =
(true for both definitions)
>>> (RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=80=99yx =
(true for both definitions)
>>>
>>> For V moves we have that: (Vx+)=E2=80=99 =3D Vx+ !=3D Vx=E2=80=99+ =
(!=3D means =E2=80=9Cnot equal to=E2=80=9D)
>>>
>>> Some important notes on legal/illegal moves
>>>
>>> Although there are a lot of moves possible with this notation we might =
not
>>> want to use them all. If we really want a 2x2x2x2 and not something els=
e I
>>> think that we should try to stick to moves that are legal 2x2x2x2 moves=
as
>>> far as possible (note that I said legal moves and not permutations (a l=
egal
>>> permutation can be made up of one or more legal moves)). Clarification:
>>> cycling three of the edge-pieces of a Rubik=E2=80=99s cube is a legal p=
ermutation
>>> but not a legal move, a legal move is a rotation of the cube or a twist=
of
>>> one of the layers. In this section I will only address simple moves and
>>> simply refer to them as moves (legal composed moves are moves composed =
by
>>> legal simple moves).
>>>
>>> I do believe that all moves allowed by my notation are legal permutatio=
ns
>>> based on their periodicity (they have a period of 2 or 4 and are all ev=
en
>>> permutations of the pieces). So, which of them correspond to legal 2x2x=
2x2
>>> moves? The O moves are obviously legal moves since they are equal to th=
e
>>> identity mod(rot). The same goes for restacking (S) (with or without +/=
-) in
>>> the direction of the longest side of the puzzle (Sz, Sz+ and Sz- in the
>>> standard rotation) since these are rotations and half-rotations that do=
n=E2=80=99t
>>> change the state of the puzzle. Restacking in the other directions and =
fold
>>> moves (V) are however not legal moves since they are made of 8 2-cycles=
and
>>> change the state of the puzzle (note that they, however, are legal
>>> permutations). The rest of the moves (R, L, F, B, U, D, K and A) can be
>>> divided into two sets: (1) the moves where you rotate a 2x2x2 block wit=
h an
>>> octahedron inside and (2) the moves where you rotate a 2x2x2 block with=
out
>>> an octahedron inside. A move belonging to (2) is always legal. We can s=
ee
>>> this by observing what a Rx does with the pieces in the standard rotati=
on
>>> with just K forming an octahedron. The stickers move in 6 4-cycles and =
if
>>> the puzzle is solved the U and D faces still looks solved after the mov=
e. A
>>> move belonging to set (1) is legal either if it=E2=80=99s an 180-degree=
twist or if
>>> it=E2=80=99s a rotation around the axis parallel with the longest side =
of the puzzle
>>> (the z-axis in the standard rotation). Quite interestingly these are ex=
actly
>>> the moves that don=E2=80=99t mix up the R and L stickers with the rest =
in the
>>> standard rotation. I think I know a way to prove that no legal 2x2x2x2 =
move
>>> can mix up these stickers with the rest and this has to do with the fac=
t
>>> that these stickers form an inverted octahedron (with the corners point=
ing
>>> outward) instead of a normal octahedron (let=E2=80=99s call this hypoth=
esis * for
>>> now). Note that all legal twists (R, L, F, B, U, D, K and A) of the phy=
sical
>>> puzzle correspond to the same twist in the MC4D software.
>>>
>>> So, what moves should we add to the set of legal moves be able to get t=
o
>>> every state of the 2x2x2x2? I think that we should add the restacking m=
oves
>>> and folding moves since Melinda has already found a pretty short sequen=
ce of
>>> those moves to make a rotation that changes which colours are on the R =
and L
>>> faces. That sequence, starting from the standard rotation, is: Oy Sx+z =
Vy+
>>> Ozy=E2=80=99 Vy+ Ozy=E2=80=99 Vy+ =3D Oy Sx+z (Vy+ Ozy=E2=80=99)3 mod(=
3rot) =3D I mod(rot) (hopefully I
>>> got that right). What I have found (which I mentioned previously) is th=
at
>>> (from the standard rotation): RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz- a=
nd since this
>>> is equivalent with Sy =3D Ly2 RLx=E2=80=99 Sz- RLx Ly2 By2 the restacki=
ng moves that
>>> are not legal moves are not very complicated permutations and therefore=
I
>>> think that we can accept them since they help us mix up the R and L sti=
ckers
>>> with other faces. In a sense, the folding moves are =E2=80=9Cmore illeg=
al=E2=80=9D since
>>> they cannot be composed by the legal moves (according to hypothesis *).=
This
>>> is also true for the illegal moves belonging to set (1) discussed above=
.
>>> However, since the folding moves is probably easier to perform and is e=
nough
>>> to reach every state of the 2x2x2x2 I think that we should use them and=
not
>>> the illegal moves belonging to (1). Note, once again, that all moves
>>> described by the notation are legal permutations (even the ones that I =
just
>>> a few words ago referred to as illegal moves) so if you wish you can us=
e all
>>> of them and still only reach legal 2x2x2x2 states. However (in a strict
>>> sense) one could argue that you are not solving the 2x2x2x2 if you use
>>> illegal moves. If you only use illegal moves to compose rotations (that=
is,
>>> create a permutation including illegal moves that are equal to I mod(ro=
t))
>>> and not actually using the illegal moves as twists I would classify tha=
t as
>>> solving a 2x2x2x2. What do you think about this?
>>>
>>> What moves to use?
>>>
>>> Here=E2=80=99s a short list of the simple moves that I think should be =
used for the
>>> physical 2x2x2x2. Note that this is just my thoughts and you may use th=
e
>>> notation to describe any move that it can describe if you wish to. The
>>> following list assumes that the puzzle is in the standard rotation but =
is
>>> analogous for other representations where the K face is an octahedron.
>>>
>>> O, all since they are I mod(rot),
>>> R, L, all since they are legal (note Rx (physical puzzle) =3D Rx (virtu=
al
>>> 2x2x2x2)),
>>> U, D, only x2 since these are the only legal easy-to-perform moves,
>>> F, B, only y2 since these are the only legal easy-to-perform moves,
>>> K, A, only z, z=E2=80=99 and z2 since these are the only legal easy-to-=
perform
>>> moves,
>>> S, at least z, z+ and z- since these are equal to I mod(rot),
>>> S, possibly x and y since these help us perform rotations and is easy t=
o
>>> compose (not necessary to reach all states and not legal though),
>>> V, all 8 allowed by the rotation of the puzzle (at least one is necessa=
ry to
>>> reach all states and if you allow one the others are easy to achieve
>>> anyway).
>>>
>>> If you start with the standard rotation and then perform Sz+ the follow=
ing
>>> applies instead (this applies analogously to any other rotation where t=
he R
>>> and L faces are octahedra).
>>>
>>> O, all since they are I mod(rot),
>>> R, L, only z, z=E2=80=99 and z2 since these are the only legal easy-to-=
perform
>>> moves,
>>> U, D, only x2 since these are the only legal easy-to-perform moves,
>>> F, B, only y2 since these are the only legal easy-to-perform moves,
>>> K, A, at least z, z=E2=80=99 and z2, possibly all (since they are legal=
) although
>>> some might be hard to perform.
>>> S, V, same as above.
>>>
>>> Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D I mod(rot)=
(!=3D
>>> for not equal) which implies that the Ux2 move when R and L are octahed=
ra is
>>> different from the Ux2 move when K is an octahedron. (Actually, the seq=
uence
>>> above is equal to Uy2).
>>>
>>> Regarding virtual n^4 cubes I think that all O, C, R, L, U, D, F, B, K =
and A
>>> moves should be used since they are all legal and really the only thing=
you
>>> need (left-clicking on an edge or corner piece in the computer program =
can
>>> be described quite easily with the notation, for example, Kzy2 is
>>> left-clicking on the top-front edge piece on the K face).
>>>
>>> I hope this was possible to follow and understand. Feel free to ask
>>> questions about the notation if you find anything ambiguous.
>>>
>>> Best regards,
>>> Joel Karlsson
>>>
>>> Den 4 maj 2017 12:01 fm skrev "Melinda Green melinda@superliminal.com
>>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>>
>>>
>>>
>>> Thanks for the correction. A couple of things: First, when assembling o=
ne
>>> piece at a time, I'd say there is only 1 way to place the first piece, =
not
>>> 24. Otherwise you'd have to say that the 1x1x1 puzzle has 24 states. I
>>> understand that this may be conventional, but to me, that just sounds s=
illy.
>>>
>>> Second, I have the feeling that the difference between the "two
>>> representations" you describe is simply one of those half-rotations I s=
howed
>>> in the video. In the normal solved state there is only one complete
>>> octahedron in the very center, and in the half-rotated state there is o=
ne in
>>> the middle of each half of the "inverted" form. I consider them to be t=
he
>>> same solved state.
>>>
>>> -Melinda
>>>
>>>
>>> On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
>>> wrote:
>>>
>>> Horrible typo... It seems like I made some typos in my email regarding =
the
>>> state count. It should of course be 16!12^16/(6*192) and NOT
>>> 12!16^12/(6*192). However, I did calculate the correct number when comp=
aring
>>> with previous results so the actual derivation was correct.
>>>
>>> Something of interest is that the physical pieces can be assembled in
>>> 16!24*12^15 ways since there are 16 pieces, the first one can be orient=
ed in
>>> 24 ways and the remaining can be oriented in 12 ways (since a corner wi=
th 3
>>> colours never touch a corner with just one colour). Dividing with 6 to =
get a
>>> single orbit still gives a factor 2*192 higher than the actual count ra=
ther
>>> than 192. This shows that every state in the MC4D representation has 2
>>> representations in the physical puzzle. These two representations must =
be
>>> the previously discussed, that the two halves either have the same colo=
r on
>>> the outermost corners or the innermost (forming an octahedron) when the
>>> puzzle is solved and thus both are complete representations of the 2x2x=
2x2.
>>>
>>> Best regards,
>>> Joel Karlsson
>>>
>>> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" om>:
>>>
>>> I am no expert on group theory, so to better understand what twists are
>>> legal I read through the part of Kamack and Keane's The Rubik Tesseract
>>> about orienting the corners. Since all even permutations are allowed th=
e
>>> easiest way to check if a twist is legal might be to:
>>> 1. Check that the twist is an even permutation, that is: the same twist=
can
>>> be done by performing an even number of piece swaps (2-cycles).
>>> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning performi=
ng the
>>> twist k times and I (the identity) representing the permutation of doin=
g
>>> nothing) and k is not divisible by 3 the twist A definitely doesn't vio=
late
>>> the restriction of the orientations since kx mod 3 =3D 0 and k mod 3 !=
=3D 0
>>> implies x mod 3 =3D 0 meaning that the change of the total orientation =
x for
>>> the twist A mod 3 is 0 (which precisely is the restriction of legal twi=
sts;
>>> that they must preserve the orientation mod 3).
>>>
>>> For instance, this implies that the restacking moves are legal 2x2x2x2 =
moves
>>> since both are composed of 8 2-cycles and both can be performed twice (=
note
>>> that 2 is not divisible by 3) to obtain the identity.
>>>
>>> Note that 1 and 2 are sufficient to check if a twist is legal but only =
1 is
>>> necessary; there can indeed exist a twist violating 2 that still is leg=
al
>>> and in that case, I believe that we might have to study the orientation
>>> changes for that specific twist in more detail. However, if a twist can=
be
>>> composed by other legal twists it is, of course, legal as well.
>>>
>>> Best regards,
>>> Joel
>>>
>>> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com [4D_Cu=
bing]
>>> <4D_Cubing@yahoogroups.com>:
>>>>
>>>>
>>>> First off, thanks everyone for the helpful and encouraging feedback!
>>>> Thanks Joel for showing us that there are 6 orbits in the 2^4 and for =
your
>>>> rederivation of the state count. And thanks Matt and Roice for pointin=
g out
>>>> the importance of the inverted views. It looks so strange in that
>>>> configuration that I always want to get back to a normal view as quick=
ly as
>>>> possible, but it does seem equally valid, and as you've shown, it can =
be
>>>> helpful for more than just finding short sequences.
>>>>
>>>> I don't understand Matt's "pinwheel" configuration, but I will point o=
ut
>>>> that all that is needed to create your twin interior octahedra is a si=
ngle
>>>> half-rotation like I showed in the video at 5:29. The two main halves =
do end
>>>> up being mirror images of each other on the visible outside like he
>>>> described. Whether it's the pinwheel or the half-rotated version that'=
s
>>>> correct, I'm not sure that it's a bummer that the solved state is not =
at all
>>>> obvious, so long as we can operate it in my original configuration and
>>>> ignore the fact that the outer faces touch. That would just mean that =
the
>>>> "correct" view is evidence that that the more understandable view is
>>>> legitimate.
>>>>
>>>> I'm going to try to make a snapable V3 which should allow the pieces t=
o be
>>>> more easily taken apart and reassembled into other forms. Shapeways do=
es
>>>> offer a single, clear translucent plastic that they call "Frosted Deta=
il",
>>>> and another called "Transparent Acrylic", but I don't think that any s=
ort of
>>>> transparent stickers will help us, especially since this thing is choc=
k full
>>>> of magnets. The easiest way to let you see into the two hemispheres wo=
uld be
>>>> to simply truncate the pointy tips of the stickers. That already happe=
ns a
>>>> little bit due to the way I've rounded the edges. Here is a close-up o=
f a
>>>> half-rotation in which you can see that the inner yellow and white fac=
es are
>>>> solved. Your suggestion of little mapping dots on the corners also wor=
ks,
>>>> but just opening the existing window further would work more directly.
>>>>
>>>> -Melinda
>>>>
>>>>
>>>> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing] wrote:
>>>>
>>>> I agree with Don's arguments about adjacent sticker colors needing to =
be
>>>> next to each other. I think this can be turned into an accurate 2^4 w=
ith
>>>> coloring changes, so I agree with Joel too :)
>>>>
>>>> To help me think about it, I started adding a new projection option fo=
r
>>>> spherical puzzles to MagicTile, which takes the two hemispheres of a p=
uzzle
>>>> and maps them to two disks with identified boundaries connected at a p=
oint,
>>>> just like a physical "global chess" game I have. Melinda's puzzle is =
a lot
>>>> like this up a dimension, so think about two disjoint balls, each
>>>> representing a hemisphere of the 2^4, each a "subcube" of Melinda's pu=
zzle.
>>>> The two boundaries of the balls are identified with each other and as =
you
>>>> roll one around, the other half rolls around so that identified points
>>>> connect up. We need to have the same restriction on Melinda's puzzle.
>>>>
>>>> In the pristine state then, I think it'd be nice to have an internal
>>>> (hidden), solid colored octahedron on each half. The other 6 faces sh=
ould
>>>> all have equal colors split between each hemisphere, 4 stickers on eac=
h
>>>> half. You should be able to reorient the two subcubes to make a half
>>>> octahedron of any color on each subcube. I just saw Matt's email and
>>>> picture, and it looks like we were going down the same thought path. =
I
>>>> think with recoloring (mirroring some of the current piece colorings)
>>>> though, the windmill's can be avoided (?)
>>>>
>>>> [...] After staring/thinking a bit more, the coloring Matt came up wit=
h is
>>>> right-on if you want to put a solid color at the center of each hemisp=
here.
>>>> His comment about the "mirrored" pieces on each side helped me underst=
and
>>>> better. 3 of the stickers are mirrored and the 4th is the hidden colo=
r
>>>> (different on each side for a given pair of "mirrored" pieces). All f=
aces
>>>> behave identically as well, as they should. It's a little bit of a bu=
mmer
>>>> that it doesn't look very pristine in the pristine state, but it does =
look
>>>> like it should work as a 2^4.
>>>>
>>>> I wonder if there might be some adjustments to be made when shapeways
>>>> allows printing translucent as a color :)
>>>>
>>>> [...] Sorry for all the streaming, but I wanted to share one more thou=
ght.
>>>> I now completely agree with Joel/Matt about it behaving as a 2^4, even=
with
>>>> the original coloring. You just need to consider the corner colors of=
the
>>>> two subcubes (pink/purple near the end of the video) as being a window=
into
>>>> the interior of the piece. The other colors match up as desired. (So=
rry if
>>>> folks already understood this after their emails and I'm just catching=
up!)
>>>>
>>>> In fact, you could alter the coloring of the pieces slightly so that t=
he
>>>> behavior was similar with the inverted coloring. At the corners where=
3
>>>> colors meet on each piece, you could put a little circle of color of t=
he
>>>> opposite 4th color. In Matt's windmill coloring then, you'd be able t=
o see
>>>> all four colors of a piece, like you can with some of the pieces on
>>>> Melinda's original coloring. And again you'd consider the color circl=
es a
>>>> window to the interior that did not require the same matching constrai=
nts
>>>> between the subcubes.
>>>>
>>>> I'm looking forward to having one of these :)
>>>>
>>>> Happy Friday everyone,
>>>> Roice
>>>>
>>>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson joelkarlsson97@gmail.co=
m
>>>> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
>>>>>
>>>>>
>>>>> Seems like there was a slight misunderstanding. I meant that you need=
to
>>>>> be able to twist one of the faces and in MC4D the most natural choic=
e is
>>>>> the center face. In your physical puzzle you can achieve this type of=
twist
>>>>> by twisting the two subcubes although this is indeed a twist of the s=
ubcubes
>>>>> themselves and not the center face, however, this is still the same t=
ype of
>>>>> twist just around another face.
>>>>>
>>>>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup =
of
>>>>> this puzzle. Hopefully the restrictions will be quite natural and onl=
y some
>>>>> "strange" moves would be illegal. Regarding the "families of states" =
(aka
>>>>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed=
twists
>>>>> preserves the parity of the pieces, meaning that only half of the
>>>>> permutations you can achieve by disassembling and reassembling can be
>>>>> reached through legal moves. Because of some geometrical properties o=
f the
>>>>> 2x2x2x2 and its twists, which would take some time to discuss in deta=
il
>>>>> here, the orientation of the stickers mod 3 are preserved, meaning th=
at the
>>>>> last corner only can be oriented in one third of the number of orient=
ations
>>>>> for the other corners. This gives a total number of orbits of 2x3=3D6=
. To
>>>>> check this result let's use this information to calculate all the pos=
sible
>>>>> states of the 2x2x2x2; if there were no restrictions we would have 16=
! for
>>>>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>>>>> orientations for each corner). If we now take into account that there=
are 6
>>>>> equally sized orbits this gets us to 12!16^12/6. However, we should a=
lso
>>>>> note that the orientation of the puzzle as a hole is not set by some =
kind of
>>>>> centerpieces and thus we need to devide with the number of orientatio=
ns of a
>>>>> 4D cube if we want all our states to be separated with twists and not=
only
>>>>> rotations of the hole thing. The number of ways to orient a 4D cube i=
n space
>>>>> (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a t=
otal of
>>>>> 12!16^12/(6*192) states which is indeed the same number that for exam=
ple
>>>>> David Smith arrived at during his calculations. Therefore, when dete=
rmining
>>>>> whether or not a twist on your puzzle is legal or not it is sufficien=
t and
>>>>> necessary to confirm that the twist is an even permutation of the pie=
ces and
>>>>> preserves the orientation of stickers mod 3.
>>>>>
>>>>> Best regards,
>>>>> Joel
>>>>>
>>>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green melinda@superliminal.co=
m
>>>>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
>>>>>
>>>>>
>>>>>
>>>>> The new arrangement of magnets allows every valid orientation of piec=
es.
>>>>> The only invalid ones are those where the diagonal lines cutting each=
cube's
>>>>> face cross each other rather than coincide. In other words, you can a=
ssemble
>>>>> the puzzle in all ways that preserve the overall diamond/harlequin pa=
ttern.
>>>>> Just about every move you can think of on the whole puzzle is valid t=
hough
>>>>> there are definitely invalid moves that the magnets allow. The most o=
bvious
>>>>> invalid move is twisting of a single end cap.
>>>>>
>>>>> I think your description of the center face is not correct though. Tw=
ists
>>>>> of the outer faces cause twists "through" the center face, not "of" t=
hat
>>>>> face. Twists of the outer faces are twists of those faces themselves =
because
>>>>> they are the ones not changing, just like the center and outer faces =
of MC4D
>>>>> when you twist the center face. The only direct twist of the center f=
ace
>>>>> that this puzzle allows is a 90 degree twist about the outer axis. Th=
at
>>>>> happens when you simultaneously twist both end caps in the same direc=
tion.
>>>>>
>>>>> Yes, it's quite straightforward reorienting the whole puzzle to put a=
ny
>>>>> of the four axes on the outside. This is a very nice improvement over=
the
>>>>> first version and should make it much easier to solve. You may be rig=
ht that
>>>>> we just need to find the right way to think about the outside faces. =
I'll
>>>>> leave it to the math geniuses on the list to figure that out.
>>>>>
>>>>> -Melinda
>>>>>
>>>>>
>>>>>
>>>>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cub=
ing]
>>>>> wrote:
>>>>>
>>>>>
>>>>> Hi Melinda,
>>>>>
>>>>> I do not agree with the criticism regarding the white and yellow stic=
kers
>>>>> touching each other, this could simply be an effect of the different
>>>>> representations of the puzzle. To really figure out if this indeed is=
a
>>>>> representation of a 2x2x2x2 we need to look at the possible moves (tw=
ists
>>>>> and rotations) and figure out the equivalent moves in the MC4D softwa=
re.
>>>>> From the MC4D software, it's easy to understand that the only moves =
required
>>>>> are free twists of one of the faces (that is, only twisting the cente=
r face
>>>>> in the standard perspective projection in MC4D) and 4D rotations swap=
ping
>>>>> which face is in the center (ctrl-clicking in MC4D). The first is pos=
sible
>>>>> in your physical puzzle by rotating the white and yellow subcubes (fr=
om here
>>>>> on I use subcube to refer to the two halves of the puzzle and the col=
ours of
>>>>> the subcubes to refer to the "outer colours"). The second is possible=
if
>>>>> it's possible to reach a solved state with any two colours on the sub=
cubes
>>>>> that still allow you to perform the previously mentioned twists. This=
seems
>>>>> to be the case from your demonstration and is indeed true if the magn=
ets
>>>>> allow the simple twists regardless of the colours of the subcubes. Th=
us, it
>>>>> is possible to let your puzzle be a representation of a 2x2x2x2, howe=
ver, it
>>>>> might require that some moves that the magnets allow aren't used.
>>>>>
>>>>> Best regards,
>>>>> Joel
>>>>>
>>>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
>>>>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
>>>>>>
>>>>>>
>>>>>> Dear Cubists,
>>>>>>
>>>>>> I've finished version 2 of my physical puzzle and uploaded a video o=
f it
>>>>>> here:
>>>>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
>>>>>> Again, please don't share these videos outside this group as their
>>>>>> purpose is just to get your feedback. I'll eventually replace them w=
ith a
>>>>>> public video.
>>>>>>
>>>>>> Here is an extra math puzzle that I bet you folks can answer: How ma=
ny
>>>>>> families of states does this puzzle have? In other words, if disasse=
mbled
>>>>>> and reassembled in any random configuration the magnets allow, what =
are the
>>>>>> odds that it can be solved? This has practical implications if all s=
uch
>>>>>> configurations are solvable because it would provide a very easy way=
to
>>>>>> fully scramble the puzzle.
>>>>>>
>>>>>> And finally, a bit of fun: A relatively new friend of mine and new l=
ist
>>>>>> member, Marc Ringuette, got excited enough to make his own version. =
He built
>>>>>> it from EPP foam and colored tape, and used honey instead of magnets=
to hold
>>>>>> it together. Check it out here:
>>>>>> http://superliminal.com/cube/dessert_cube.jpg I don't know how pract=
ical a
>>>>>> solution this is but it sure looks delicious! Welcome Marc!
>>>>>>
>>>>>> -Melinda
>>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>
>>>
>>>
>>>
>
> ------------------------------------
> Posted by: Joel Karlsson
> ------------------------------------
>
>
> ------------------------------------
>
> Yahoo Groups Links
>
>
>
>




From: Joel Karlsson <joelkarlsson97@gmail.com>
Date: Tue, 30 May 2017 09:06:01 +0200
Subject: Re: [MC4D] Physical 4D puzzle V2



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Well, usually we want to calculate distinct states and for the 2x2x2x2
(virtual) that become 16!12^16/(6*192), 16! for permutation of pieces,
12^16 for orienting the pieces, 1/6 to only get legal states (the puzzle
has 6 orbits) and 1/192 to fix one corner since the puzzle doesn't have
fixed pieces like the 3x3x3x3. However, when I checked how many
representations there are of every state I didn't count distinct states but
all states, even illegal ones. The count for the virtual 2x2x2x2 thus
becomes 16!*12^16 and with this count, every distinct state has 192
representations and only a sixth of them are legal (you can think of it as:
how many legal and illegal states are there if I'm not allowed to rotate
the puzzle? or: in how many ways can I put the pieces together?). The same
type of count for the physical puzzle would be 3*16!*24*12^15. The ratio of
these are 1:6 which means that even if you only count legal states and fix
the rotation of the puzzle the count for the physical puzzle, which would
be 3*16!*24*12^15/(6*192), is still too high with a factor 6 and thus, if
we claim that the physical puzzle is a representation of a 2x2x2x2, there
must be six representations of every distinct state that are not separated
with only rotations. Luckily, this is the case and these come from the
different projections (I agree that projection might be a better word than
representation although it might not be completely accurate).

Best regards,
Joel Karlsson

2017-05-30 3:57 GMT+02:00 Melinda Green melinda@superliminal.com
[4D_Cubing] <4D_Cubing@yahoogroups.com>:

>
>
> Thanks for the clarification, Joel. Just to be complete, are you sure tha=
t
> that the count of MC4D states are being counted in the same way?
>
> One other thought I had regarding terminology: You called the two forms o=
f
> the physical puzzle "representations", but I wonder whether a somewhat
> better term might be "projection". This puzzle is definitely not any sort
> of geometric projection into 3-space, but it seems to share a number of
> analogous properties with them. I often think of it as viewing the 4D
> object through a 2x2x4 "viewport".
>
> The half rotations are sort of like translating that viewport along or
> around the surface of that object. Since you point out that it is a 90
> degree rotation, perhaps "half rotation" isn't the best term for that mov=
e.
> Whatever we call these rotations, the two forms feel to me like the
> difference between cell-first and face-first projections.
>
> -Melinda
>
>
> On 5/29/2017 2:57 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing]
> wrote:
> > Hello,
> >
> > Just a quick correction regarding a previous statement. From the
> > calculation of the states of the puzzle, we can see that if we choose
> > a state and rotation there are still two representations of that state
> > and rotation in the physical puzzle. Previously, we also said that
> > these states are separated by a half-rotation such as Rx+ from an RL
> > rep or AKx rep. This is not the case. When I calculated the states of
> > the puzzle I assumed that the longer side of the puzzle should be
> > parallel with the x-axis. If we don't make that assumption we find
> > that the number of states (not distinct states, some of these are just
> > separated by a rotation) are 3*16!*24*12^15 since there are three
> > possible choices for which axis should be parallel with the longer
> > side. This means that, in fact, each distinct state has 6
> > representations in the physical puzzle. This is actually what we
> > should expect since it should be possible to represent every state in
> > an RL rep, an UD rep, an FB rep, an AKx rep, an AKy rep and an AKz
> > rep. In the solved state, the rotation of the puzzle is determined by
> > which colour belongs to each face and given such a rotation there are
> > indeed six representations: first, choose an axis that the longer side
> > should be parallel with (3 alternatives) and then choose either the AK
> > rep or the non-AK rep (2 alternatives). However, the so-called
> > "half-rotation" changes which colour belongs to which face so this is
> > not a move that gets you from one representation of a state to
> > another. From an UD rep with:
> > P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x ,
> > P3 Ox'z' takes you to an FB rep of the same state
> > Oy P3 Oz' takes you to an RL rep of the same state
> > UD'x P3 UD'z' Sy- Oy takes you to an AKy rep of the same state
> > UD'x P3 Oyx' Sz+ takes you to an AKz rep of the same state and
> > UD'x P3 UD'x' Oz' Sx- takes you to an AKx rep of the same state
> > >From this, we can see that the AK states are closely related and it's
> > possible to change between them without illegal moves. For example
> > going from an AKy rep to an AKz rep of the same state could be done
> > with:
> > (UD'x P3 UD'z' Sy- Oy)' UD'x P3 Oyx' Sz+ =3D Oy' Sy+ UD'z
> > P3' UD'x' UD'x P3 Oyx' Sz+ =3D Oy' Sy+ UD'z Oyx' Sz+
> >
> > Note that these sequences don't change the state nor the rotation of
> > the puzzle but they do change the representation. Let's call these
> > types of sequences J moves. Note that what I have previously called a
> > type three rotation is actually a J move and a rotation. Further note
> > that J =3D I mod(rep) (modulo representation). The J moves changes whic=
h
> > pair of faces are symmetry breaking and this is the type of moves that
> > needs to be added to the set of legal moves to make the physical
> > puzzle an actual 2x2x2x2. A J move is a move that is equal to I
> > mod(rep, rot) but that isn't equal to I mod(rot) (which means that it
> > has to preserve the state, is allowed to change the rotation and has
> > to change the representation/what faces (not in the left/right sense
> > but rather in the sense of colours) are symmetry-breaking ).
> >
> > Best regards,
> > Joel Karlsson
> >
> > 2017-05-22 22:32 GMT+02:00 Joel Karlsson :
> >> Hi Melinda,
> >>
> >> Thank you for the feedback. Regarding the coordinate system, it=E2=80=
=99s just
> >> a matter of preference. I thought it would be nice to have a
> >> xy-symmetry but understand that it might be more practical to follow
> >> conventions. So, adapting your suggestion, let's redefine the axes as
> >> x pointing right, y up and z towards you. Note that this means that
> >> the longer side of the puzzle is parallel with the x-axis in the
> >> standard rotation. From here on, I will use this new coordinate
> >> system.
> >>
> >> Regarding the name of the faces. Since which faces are the "outer
> >> ones" changes with how you rotate the puzzle and what state the puzzle
> >> is in, I think that the labelling of the faces should be independent
> >> of which faces are the outer faces (forming what will be referred to
> >> as inverted octahedra). Since the puzzle is a representation of a
> >> 4d-cube in 3d-space it will (with our coordinate system) always have
> >> two faces "belonging" to every axis and two faces that don't belong to
> >> any axis. Therefore, it makes sense to label the faces in such a way
> >> that the face in for example the positive x direction is R, always.
> >> So, the K face (which is one of the faces not belonging to a
> >> particular axis) is always the face only belonging to the center 2x2x2
> >> block (and this can be either an octahedron or an inverted octahedron
> >> (the outer corners) depending on the representation of the puzzle).
> >> The distinction between left and right never disappears; the R face is
> >> always the face only belonging to the right half of the puzzle and can
> >> be an octahedron, an inverted octahedron, two half octahedra (<><>) or
> >> one half and two quarter octahedra (><><). Note that the
> >> half-rotations are 90-degree rotations of the puzzle; for example, Sx+
> >> (physical puzzle) =3D Cx' (virtual puzzle) is the rotation that does L
> >> -> K -> R -> A -> L so the face that previously was the L face is now
> >> the K face. From the standard rotation (where R and L are the inverted
> >> octahedra before the rotation), this would mean that the K and A faces
> >> are inverted octahedra after a Sx+ rotation. What faces are the
> >> "mysterious" (or more precisely symmetry breaking, forming inverted
> >> octahedra instead of regular octahedra) depend on the rotation of the
> >> puzzle and can be any two opposite faces (R and L, U and D, F and B or
> >> A and K). It might be useful to be able to describe what faces are
> >> inverted octahedra since this determines what moves are legal so let's
> >> say that the puzzle is in an RL representation if R and L are inverted
> >> octahedra and similarly for other states. Thus, Sx+ can take you from
> >> an RL representation to an AK representation (what a long word let's
> >> use rep for short). Note that the AK rep doesn't specify which axis
> >> the longer side should be parallel with so let's just add a lowercase
> >> letter to indicate this (an AKx rep is thus a state where the A and K
> >> faces are inverted octahedra and the longer side is parallel with the
> >> x-axis). In conclusion, I would like to keep the names of the faces as
> >> I first defined them and hope that it's clearer what I mean with them
> >> and that the names of the faces are not related to what faces form
> >> inverted octahedra.
> >>
> >> You also wrote:
> >> "There are several, distinct types of rotations, none of which change
> >> the state of the puzzle, and I think we need a way to be unambiguous
> >> about them. The types I see are
> >>
> >> 1. Simple reorientation of the physical puzzle in the hand, no magnets
> >> involved. IE your 'O' moves and maybe analogous to mouse-dragging in
> >> MC4D?
> >> 2. Rolling of one 2x2x2 half against the other. Maybe analogous to
> ctrl-click?
> >> 3. Half-rotations. Maybe analogous to a "face first" view? (ctrl-click
> >> on a 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by
> >> Cubie")
> >> 4. Whole-puzzle reorientations that move an arbitrary axis into the
> >> "outer" 2 faces. No MC4D analog."
> >>
> >> The rotations (as far as I know) are the O moves (type one rotation),
> >> which are indeed analogous to mouse-dragging in MC4D, rolling the
> >> 2x2x2 halves (type two rotation) (these are easily described as for
> >> example RL'y in an RL rep) which (in a non-AK rep) is a ctrl-click on
> >> a non-inverted face (that is ctrl-click on a face that is currently
> >> represented with an octahedron), half-rotations (also type two
> >> rotations) which is a ctrl-click on an inverted face and sequences
> >> that for example from an RL rep can take the R face to K without
> >> turning the puzzle into an AK rep (type three rotation). Type one and
> >> two rotations are legal moves but type three contain illegal 2x2x2x2
> >> moves according to hypothesis * in my previous email (however, they
> >> are very important and needed if we wish to be able to reach all
> >> states).
> >>
> >> Regarding fold moves, from the standard rotation (RL rep) are you
> >> saying that Vy+ =3D Vy'+ or something like Vy+ =3D Vy'+ FB'x2 =3D Vy'+
> >> mod(rot)? The former seems not to be correct but I believe the latter
> >> is, please correct me if I'm wrong. I don't assume that you fold the
> >> puzzle back to the same representation. This is what + and -
> >> indicates, + preserve the representation mod(rot) (i.e AK rep both
> >> before and after or neither before nor after) and - changes it (going
> >> from AK rep to non-AK rep or vice versa).
> >>
> >> "I think the only legal 90 degree twists of the K face are those about
> >> the long axis. I believe this is what Christopher Locke was saying in
> >> this message. To see why there is no straightforward way to perform
> >> other 90 degree twists, you only need to perform a 90 degree twist on
> >> an outer (L/R) face and then reorient the whole puzzle along a
> >> different outer axis. If the original twist was not about the new long
> >> axis, then there is clearly no straightforward way to undo that
> >> twist."
> >>
> >> Yes, as pointed out further down in my previous email. The notation
> >> allows all 90-degree rotations after the section "extensions of some
> >> definitions" although only 180-degree rotations are legal for
> >> octahedral faces around axes not parallel with the longer side of the
> >> puzzle (see the section =E2=80=9Csome important notes on legal/illegal
> >> moves=E2=80=9D).
> >>
> >> "I noticed something like this the other day but realized that it only
> >> seems to work for rotations along the long dimension (z in your
> >> example). These are already easily accomplished by a simple rotation
> >> to put the face in question on the end caps, followed by a double
> >> end-cap twist."
> >>
> >> It works for the other rotations as well although these are not legal
> >> moves. They could possibly be used instead of the illegal S moves
> >> (along axes not parallel with the longer side) and the illegal V moves
> >> (V- (minus) moves which are closely related to the illegal S moves,
> >> example from RL rep: Vy- =3D Vy+ Sx Oz2) but as I mentioned in my
> >> previous email I do believe that it's better to use the S and V moves
> >> since you have found a relatively short way to perform a type three
> >> rotation with those.
> >>
> >> Let:
> >> P1 =3D Sy+ Uyz2 Sy-
> >> P1=E2=80=99 =3D P1 (P1 is its own inverse)
> >> P2 =3D (P1 UD=E2=80=99z P1 UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z=E2=80=
=99 P1 UD=E2=80=99z)2 (P2=3DP_2 not P^2
> >> whereas the two at the end means: perform twice)
> >> P2=E2=80=99 =3D (UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z P1 UD=E2=80=99z=
P1 UD=E2=80=99z=E2=80=99 P1)2
> >> P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x =3D I mod(ro=
t) (type three
> rotation)
> >> P3=E2=80=99 =3D Oy=E2=80=99 P3 Oy=E2=80=99
> >> P4 =3D P2 P3 P2 P3=E2=80=99 P2=E2=80=99 P3 P2=E2=80=99 P3=E2=80=99
> >>
> >> P4 from UD rep is a 164 move sequence rotating only one corner in its
> >> place. The sequence is inspired by Roice =E2=80=9Csecond four-color se=
ries=E2=80=9D
> >> but I have changed the =E2=80=9CTop 9=E2=80=9D moves to pure rotations=
since it=E2=80=99s all
> >> that=E2=80=99s necessary and a bit shorter to perform. P3 is your type=
three
> >> rotation (with a rotation added at the end) and P2 is Roice =E2=80=9Ct=
hird
> >> three-color series=E2=80=9D. Written with my notation but for the virt=
ual
> >> puzzle, the sequences (Roice original since it=E2=80=99s easier to per=
form K
> >> twists than O rotations in MC4D) are:
> >> Q1 =3D (Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=99 Rz2y=E2=80=99)2 (ana=
logue to P2)
> >> Q1=E2=80=99 =3D (Rz2y=E2=80=99 Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=
=99)2
> >> Q2 =3D Q1 Kxy=E2=80=99 Q1 Kyx=E2=80=99 Q1=E2=80=99 Kxy=E2=80=99 Q1=E2=
=80=99 Kyx=E2=80=99 (analogue to P4 but
> >> only 36 moves)
> >>
> >> How to read faster (the example applies to a 3x3x3x3): moves like
> >> Kz2y=E2=80=99 are clicking on 3C-pieces in MC4D. If the move is writte=
n in
> >> this way, [uppercase letter] [lowercase letter]2 [lowercase letter
> >> possibly with =E2=80=98 (prime)], there=E2=80=99s a quite quick way to=
realize which
> >> piece this is. The uppercase letter specifies which face the piece to
> >> press is on and the first lowercase letter (followed by 2) specifies
> >> one of the sides of that face that the piece belong to. There are then
> >> 4 possible pieces. Sadly, the piece do not lie in the direction of the
> >> last letter from the center of the side of the face but you have to
> >> move one edge clockwise from this. So, Kz2y=E2=80=99 is a click on the=
edge on
> >> the front side of the K face one step clockwise from the negative
> >> y-axis (thus, the front left 3C-piece on the K face). It=E2=80=99s a b=
it
> >> unfortunate that this =E2=80=9Crule=E2=80=9D isn=E2=80=99t even simple=
r but it's at least true
> >> for all of these moves (as far as I know). Moves like Kxy=E2=80=99 are
> >> left/right-clicking on a corner piece but currently I don=E2=80=99t kn=
ow a
> >> fast way to determine which corner. Any ideas?
> >>
> >> Best regards,
> >> Joel Karlsson
> >>
> >> 2017-05-19 3:33 GMT+02:00 Melinda Green melinda@superliminal.com
> >> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
> >>>
> >>> Hello Joel,
> >>>
> >>> Thanks for drilling into this puzzle. Finding good ways to discuss an=
d
> think
> >>> about moves and representations will be key. I'll comment on some
> details
> >>> in-line.
> >>>
> >>> On 5/14/2017 6:16 AM, Joel Karlsson joelkarlsson97@gmail.com
> [4D_Cubing]
> >>> wrote:
> >>>
> >>>
> >>> Yes, that is correct and in fact, you should divide not only with 24
> for the
> >>> orientation but also with 16 for the placement if you want to calcula=
te
> >>> unique states (since the 2x2x2x2 doesn't have fixed centerpieces). Th=
e
> >>> point, however, was that if you don=E2=80=99t take that into account =
you get a
> >>> factor of 24*16=3D384 (meaning that the puzzle has 384 representation=
s of
> >>> every unique state) instead of the factor of 192 which you get when
> >>> calculating the states from the virtual puzzle and hence every state
> of the
> >>> virtual puzzle has two representations in the physical puzzle. Yes
> exactly,
> >>> they are indeed the same solved (or other) state and you are correct
> that
> >>> the half rotation (taking off a 2x2 layer and placing it at the other
> end of
> >>> the puzzle) takes you from one representation to the same state with
> the
> >>> other representation. This means that the restacking move (taking off
> the
> >>> front 2x4 layer and placing it behind the other 2x4 layer) can be
> expressed
> >>> with half-rotations and ordinary twists and rotations (which you migh=
t
> have
> >>> pointed out already).
> >>>
> >>>
> >>> Yes, I made that claim in the video but didn't show it because I have
> yet to
> >>> record such a sequence. I've only stumbled through it a few times. I
> talked
> >>> about it at 5:53 though I mistakenly called it a twist, when I should
> have
> >>> called it a sequence.
> >>>
> >>>
> >>> I think I've found six moves including ordinary twists and a
> restacking move
> >>> that is identical to a half-rotation and thus it's easy to compose a
> >>> restacking move with one half-rotation and five ordinary twists. Ther=
e
> might
> >>> be an error since I've only played with the puzzle in my mind so it
> would be
> >>> great if you, Melinda, could confirm this (the sequence is described
> later
> >>> in this email).
> >>>
> >>>
> >>> You mean "RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-"? Yes, that works. =
There do
> seem to
> >>> be easier ways to do that beginning with an ordinary rolling rotation=
.
> I
> >>> don't see those in your notation, but the equivalent using a pair of
> twists
> >>> would be Rx Lx' Sx Vy if I got that right.
> >>>
> >>>
> >>> To be able to communicate move sequences properly we need notation fo=
r
> >>> representing twists, rotations, half-rotations, restacking moves and
> folds.
> >>> Feel free to come with other suggestion but you can find mine below.
> Please
> >>> read the following thoroughly (maybe twice) to make sure that you
> understand
> >>> everything since misinterpreted notation could potentially become a
> >>> nightmare and feel free to ask questions if there is something that
> needs
> >>> clarification.
> >>>
> >>> Coordinate system and labelling:
> >>>
> >>> Let's introduce a global coordinate system. In whatever state the
> puzzle is
> >>> let the positive x-axis point upwards, the positive y-axis towards yo=
u
> and
> >>> the positive z-axis to the right (note that this is a right-hand
> system).
> >>>
> >>>
> >>> I see the utility of a global coordinate system, but this one seems
> rather
> >>> non-standard. I suggest that X be to the right, and Y up since these
> are
> >>> near-universal standards. Z can be in or out. I have no opinion. If
> there is
> >>> any convention in the twisty puzzle community, I'd go with that.
> >>>
> >>> Note also that the wiki may be a good place to document and iterate o=
n
> >>> terminology, descriptions and diagrams. Ray added a "notation" sectio=
n
> to
> >>> the 3^4 page here, and I know that one other member was thinking of
> >>> collecting a set of moves on another wiki page.
> >>>
> >>>
> >>> Now let's name the 8 faces of the puzzle. The right face is denoted
> with R,
> >>> the left with L, the top U (up), the bottom D (down), the front F, th=
e
> back
> >>> B, the center K (kata) and the last one A (ana). The R and L faces ar=
e
> >>> either the outer corners of the right and left halves respectively or
> the
> >>> inner corners of these halves (forming octahedra) depending on the
> >>> representation of the puzzle. The U, D, F and B faces are either two
> diamond
> >>> shapes (looking something like this: <><>) on the corresponding side
> of the
> >>> puzzle or one whole and two half diamond shapes (><><). The K face is
> either
> >>> an octahedron in the center of the puzzle or the outer corners of the
> center
> >>> 2x2x2 block. Lastly, the A face is either two diamond shapes, one on
> the
> >>> right and one on the left side of the puzzle, or another shape that=
=E2=80=99s a
> >>> little bit hard to describe with just a few words (the white stickers
> at
> >>> 5:10 in the latest video, after the half-rotation but before the
> restacking
> >>> move).
> >>>
> >>>
> >>> I think it's more correct to say that the K face is either an
> octahedron at
> >>> the origin (A<>K<>A) or in the center of one of the main halves, with
> the A
> >>> face inside the other half (>A<>K<). This was what I was getting at i=
n
> my
> >>> previous message. You do later talk about octahedral faces being in
> either
> >>> the center or the two main halves, so this is just terminology. But
> about
> >>> "the outer corners of the center 2x2x2 block", this cannot be the A o=
r
> K
> >>> face as you've labeled them. You've been calling these the L/R faces,
> but
> >>> the left-right distinction disappears in the half-rotated state, so
> maybe
> >>> "left" and "right" aren't the best names. To me, they are always the
> >>> "outside" faces, regardless. You can distinguish them as the left and
> right
> >>> outside faces in one representation, or as the center and end outside
> faces
> >>> in the other. (Or perhaps "end" versus "ind" if we want to be cute.)
> >>>
> >>> I'm also a little torn about naming the interior faces ana and kata,
> not
> >>> because of the names themselves which I like, but because the
> mysterious
> >>> faces to me are the outermost ones you're calling R and L. It only
> requires
> >>> a simple rotation to move faces in and out of the interior (octahedra=
l)
> >>> positions, but it's much more difficult to move another axis into the
> >>> L/R/outer direction.
> >>>
> >>> So maybe the directions can be
> >>>
> >>> Up-Down
> >>> Front-Back
> >>> Ind-End
> >>> Ana-Kata
> >>>
> >>> I'm not in love with it and will be happy with anything that works.
> Thoughts
> >>> anyone?
> >>>
> >>>
> >>> We also need a name for normal 3D-rotations, restacking moves and fol=
ds
> >>> (note that a half-rotation is a kind of restacking move). Let O be th=
e
> name
> >>> for a rOtation (note that the origin O doesn't move during a rotation=
,
> by
> >>> the way, these are 3D rotations of the physical puzzle), let S
> represent a
> >>> reStacking move and V a folding/clamshell move (you can remember this
> by
> >>> thinking of V as a folded line).
> >>>
> >>>
> >>> I think it's fine to call the clamshell move a fold or denote it as V=
.
> I
> >>> just wouldn't consider it to be a basic move since it's a simple
> composite
> >>> of 3 basic twists as shown here. In general, I think there are so man=
y
> >>> useful composite moves that we need to be able to easily make them up
> ad hoc
> >>> with substitutions like Let =E2=86=93 =3D Rx Lx'. These are really ma=
cro moves
> which
> >>> can be nested. That said, it's a particularly useful move so it's
> probably
> >>> worth describing in some formal way like you do in detail below.
> >>>
> >>>
> >>>
> >>> Further, let I (capital i) be the identity, preserving the state and
> >>> rotation of the puzzle. We cannot use I to indicate what moves should
> be
> >>> performed but it's still useful as we will see later. Since we also
> want to
> >>> be able to express if a sequence of moves is a rotation, preserving t=
he
> >>> state of the puzzle but possibly representing it in a different way,
> we can
> >>> introduce mod(rot) (modulo rotation). So, if a move sequence P
> satisfies P
> >>> mod(rot) =3D I, that means that the state of the puzzle is the same
> before and
> >>> after P is performed although the rotation and representation of the
> puzzle
> >>> are allowed to change. I do also want to introduce mod(3rot) (modulo
> >>> 3D-rotation) and P mod(3rot) =3D I means that if the right 3D-rotatio=
n (a
> >>> combination of O moves as we will see later) is applied to the puzzle
> after
> >>> P you get the identity I. Moreover, let the standard rotation of the
> puzzle
> >>> be any rotation such that the longer side is parallel with the z-axis=
,
> that
> >>> is the puzzle forms a 2x2x4 (2 pieces thick in the x-direction, 2 in
> the
> >>> y-direction and 4 in the z-direction), and the K face is an octahedro=
n.
> >>>
> >>> Rotations and twists:
> >>>
> >>> Now we can move on to name actual moves. The notation of a move is a
> >>> combination of a capital letter and a lowercase letter. O followed by
> x, y
> >>> or z is a rotation of the whole puzzle around the corresponding axis
> in the
> >>> mathematical positive direction (counterclockwise/the way your
> right-hand
> >>> fingers curl if you point in the direction of the axis with your
> thumb). For
> >>> example, Ox is a rotation around the x-axis that turns a 2x2x4 into a
> 2x4x2.
> >>> A name of a face (U, D, F, B, R, L, K or A) followed by x, y or z
> means:
> >>> detach the 8 pieces that have a sticker belonging to the face and the=
n
> turn
> >>> those pieces around the global axis. For example, if the longer side
> of the
> >>> puzzle is parallel to the z-axis (the standard rotation), Rx means:
> take the
> >>> right 2x2x2 block and turn it around the global x-axis in the
> mathematical
> >>> positive direction. Note that what moves are physically possible and
> allowed
> >>> is determined by the rotation of the puzzle (I will come back to this
> >>> later). Further note that Ox mod(rot) =3D Oy mod(rot) =3D Oz mod(rot)=
=3D I,
> >>> meaning that the 3D-rotations of the physical puzzle corresponds to a
> >>> 4D-rotation of the represented 2x2x2x2.
> >>>
> >>>
> >>> There are several, distinct types of rotations, none of which change
> the
> >>> state of the puzzle, and I think we need a way to be unambiguous abou=
t
> them.
> >>> The types I see are
> >>>
> >>> Simple reorientation of the physical puzzle in the hand, no magnets
> >>> involved. IE your 'O' moves and maybe analogous to mouse-dragging in
> MC4D?
> >>> Rolling of one 2x2x2 half against the other. Maybe analogous to
> ctrl-click?
> >>> Half-rotations. Maybe analogous to a "face first" view? (ctrl-click o=
n
> a
> >>> 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by
> Cubie")
> >>> Whole-puzzle reorientations that move an arbitrary axis into the
> "outer" 2
> >>> faces. No MC4D analog.
> >>>
> >>>
> >>> Inverses and performing a move more than once
> >>>
> >>> To mark that a move should be performed n times let's put ^n after it=
.
> For
> >>> convenience when writing and speaking let ' (prime) represent ^-1 (th=
e
> >>> inverse) and n represent ^n. The inverse P' of some permutation P is
> the
> >>> permutation that satisfies P P' =3D P' P =3D I (the identity). For ex=
ample
> (Rx)'
> >>> means: do Rx backwards, which corresponds to rotating the right 2x2x2
> block
> >>> in the mathematical negative direction (clockwise) around the x-axis
> and
> >>> (Rx)2 means: perform Rx twice. However, we can also define powers of
> just
> >>> the lowercase letters, for example, Rx2 =3D Rx^2 =3D Rxx =3D Rx Rx. S=
o x2=3Dx^2
> >>> means: do whatever the capital letter specifies two times with respec=
t
> to
> >>> the x-axis. We can see that the capital letter naturally is
> distributed over
> >>> the two lowercase letters. Rx' =3D Rx^-1 means: do whatever the capit=
al
> letter
> >>> specifies but in the other direction than you would have if the prime
> >>> wouldn't have been there (note that thus, x'=3Dx^-1 can be seen as th=
e
> >>> negative x-axis). Note that (Rx)^2 =3D Rx^2 and (Rx)' =3D Rx' which i=
s
> true for
> >>> all twists and rotations but that doesn't have to be the case for oth=
er
> >>> types of moves (restacks and folds).
> >>>
> >>> Restacking moves
> >>>
> >>> A restacking move is an S followed by either x, y or z. Here the
> lowercase
> >>> letter specifies in what direction to restack. For example, Sy (from
> the
> >>> standard rotation) means: take the front 8 pieces and put them at the
> back,
> >>> whereas Sx means: take the top 8 pieces and put them at the bottom.
> Note
> >>> that Sx is equivalent to taking the bottom 8 pieces and putting them
> at the
> >>> top. However, if we want to be able to make half-rotations we
> sometimes need
> >>> to restack through a plane that doesn't go through the origin. In the
> >>> standard rotation, let Sz be the normal restack (taking the 8 right
> pieces,
> >>> the right 2x2x2 block, and putting them on the left), Sz+ be the
> restack
> >>> where you split the puzzle in the plane further in the positive
> z-direction
> >>> (taking the right 2x2x1 cap of 4 pieces and putting it at the left en=
d
> of
> >>> the puzzle) and Sz- the restack where you split the puzzle in the pla=
ne
> >>> further in the negative z-direction (taking the left 2x2x1 cap of 4
> pieces
> >>> and putting it at the right end of the puzzle). If the longer side of
> the
> >>> puzzle is parallel to the x-axis instead, Sx+ would take the top 1x2x=
2
> cap
> >>> and put it on the bottom. Note that in the standard rotation Sz
> mod(rot) =3D
> >>> I. For restacks we see that (Sx)' =3D Sx' =3D Sx (true for y and z to=
o of
> >>> course), that (Sz+)' =3D Sz- and that (Sz+)2 =3D (Sz-)2 =3D Sz. We ca=
n
> define Sz'+
> >>> to have meaning by thinking of z=E2=80=99 as the negative z-axis and =
with that
> in
> >>> mind it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus, (Sz)=E2=
=80=99 =3D Sz=E2=80=99 and (Sz+)=E2=80=99 =3D
> Sz=E2=80=99+.
> >>>
> >>> Fold moves
> >>>
> >>> A fold might be a little bit harder to describe in an intuitive way.
> First,
> >>> let's think about what folds are interesting moves. The folds that
> cannot be
> >>> expressed as rotations and restacks are unfolding the puzzle to a 4x4
> and
> >>> then folding it back along another axis. If we start with the standar=
d
> >>> rotation and unfold the puzzle into a 1x4x4 (making it look like a 4x=
4
> from
> >>> above) the only folds that will achieve something you can't do with a
> >>> restack mod(rot) is folding it to a 2x4x2 so that the longer side is
> >>> parallel with the y-axis after the fold. Thus, there are 8 interestin=
g
> fold
> >>> moves for any given rotation of the puzzle since there are 4 ways to
> unfold
> >>> it to a 4x4 and then 2 ways of folding it back that make the move
> different
> >>> from a restack move mod(rot).
> >>>
> >>>
> >>> Assuming you complete a folding move in the same representation (<><>
> or
> >>>> <><), then there are only two interesting choices. That's because it
> >>> doesn't matter which end of a chosen cutting plane you open it from,
> the end
> >>> result will be the same. That also means that any two consecutive
> clamshell
> >>> moves along the same cutting plane will undo each other. It further
> suggests
> >>> that any interesting sequence of clamshell moves must alternate
> between the
> >>> two possible long cut directions, meaning there is no choice involved=
.
> 12
> >>> clamshell moves will cycle back to the initial state.
> >>>
> >>> There is one other weird folding move where you open it in one
> direction and
> >>> then fold the two halves back-to-back in a different direction. If yo=
u
> >>> simply kept folding along the initial hinge, you'd simply have a
> restacking.
> >>> When completed the other way, it's equivalent to a restacking plus a
> >>> clamshell, so I don't think it's useful though it is somewhat
> interesting.
> >>>
> >>>
> >>> Let's call these 8 folds interesting fold moves. Note that an
> interesting
> >>> fold move always changes which axis the longer side of the puzzle is
> >>> parallel with. Further note that both during unfold and fold all
> pieces are
> >>> moved; it would be possible to have 8 of the pieces fixed during an
> unfold
> >>> and folding the other half 180 degrees but I think that it=E2=80=99s =
more
> intuitive
> >>> that these moves fold both halves 90 degrees and performing them with
> >>> 180-degree folds might therefore lead to errors since the puzzle migh=
t
> get
> >>> rotated differently. To illustrate a correct unfold without a puzzle:
> Put
> >>> your palms together such that your thumbs point upward and your finge=
rs
> >>> forward. Now turn your right hand 90 degrees clockwise and your left
> hand 90
> >>> degrees counterclockwise such that the normal to your palms point up,
> your
> >>> fingers point forward, your right thumb to the right and your left
> thumb to
> >>> the left. That was what will later be called a Vx unfold and the fold=
s
> are
> >>> simply reversed unfolds. (I might have used the word =E2=80=9Cfold=E2=
=80=9D in two
> different
> >>> ways but will try to use the term =E2=80=9Cfold move=E2=80=9D when re=
ferring to the
> move
> >>> composed by an unfold and a fold rather than simply calling these mov=
es
> >>> =E2=80=9Cfolds=E2=80=9D.)
> >>>
> >>> To specify the unfold let's use V followed by one of x, y, z, x', y'
> and z'.
> >>> The lowercase letter describes in which direction to unfold. Vx means
> unfold
> >>> in the direction of the positive x-axis and Vx' in the direction of t=
he
> >>> negative x-axis, if that makes any sense. I will try to explain more
> >>> precisely what I mean with the example Vx from the standard rotation
> (it
> >>> might also help to read the last sentences in the previous paragraph
> again).
> >>> So, the puzzle is in the standard rotation and thus have the form
> 2x2x4 (x-,
> >>> y- and z-thickness respectively). The first part (the unfolding) of
> the move
> >>> specified with Vx is to unfold the puzzle in the x-direction, making
> it a
> >>> 1x4x4 (note that the thickness in the x-direction is 1 after the Vx
> unfold,
> >>> which is no coincidence). There are two ways to do that; either the
> sides of
> >>> the pieces that are initially touching another piece (inside of the
> puzzle
> >>> in the x,z-plane and your palms in the hand example) are facing up or
> down
> >>> after the unfold. Let Vx be the unfold where these sides point in the
> >>> direction of the positive x-axis (up) and Vx' the other one where the=
se
> >>> sides point in the direction of the negative x-axis (down) after the
> unfold.
> >>> Note that if the longer side of the puzzle is parallel to the z-axis
> only
> >>> Vx, Vx', Vy and Vy' are possible. Now we need to specify how to fold
> the
> >>> puzzle back to complete the folding move. Given an unfold, say Vx,
> there are
> >>> only two ways to fold that are interesting (not turning the fold move
> into a
> >>> restack mod(rot)) and you have to fold it perpendicular to the unfold
> to
> >>> create an interesting fold move. So, if you start with the standard
> rotation
> >>> and do Vx you have a 1x4x4 that you have to fold into a 2x4x2. To
> >>> distinguish the two possibilities, use + or - after the Vx. Let Vx+ b=
e
> the
> >>> unfold Vx followed by the interesting fold that makes the sides that
> are
> >>> initially touching another piece (before the unfold) touch another
> piece
> >>> after the fold move is completed and let Vx- be the other interesting
> fold
> >>> move that starts with the unfold Vx. (Thus, continuing with the hand
> >>> example, if you want to do a Vx+ first do the Vx unfold described in
> the end
> >>> of the previous paragraph and then fold your hands such that your
> fingers
> >>> point up, the normal to your palms point forward, the right palm is
> touching
> >>> the right-hand fingers, the left palm is touching the left-hand
> fingers, the
> >>> right thumb is pointing to the right and the left thumb is pointing t=
o
> the
> >>> left). Note that the two halves of the puzzle always should be folded
> 90
> >>> degrees each and you should never make a fold or unfold where you fol=
d
> just
> >>> one half 180-degrees (if you want to use my notation, that is).
> Further note
> >>> that Vx+ Sx mod(3rot) =3D Vx- and that Vx+ Vx+ =3D I which is equival=
ent to
> >>> (Vx+)=E2=80=99 =3D Vx+ and this is true for all fold moves (note that=
after a Vx+
> >>> another Vx+ is always possible).
> >>>
> >>> The 2x2x2x2 in the MC4D software
> >>>
> >>> The notation above can also be applied to the 2x2x2x2 in the MC4D
> program.
> >>> There, you are not allowed to do S or V moves but instead, you are
> allowed
> >>> to do the [crtl]+[left-click] moves. This can easily be represented
> with
> >>> notation similar to the above. Let=E2=80=99s use C (as in Centering) =
and one
> of x, y
> >>> and z. For example, Cx would be to rotate the face in the positive
> >>> x-direction aka the U face to the center. Thus, Cz' is simply
> >>> [ctrl]+[left-click] on the L face and similarly for the other C moves=
.
> The
> >>> O, U, D, F, B, R, L, K and A moves are performed in the same way as
> above
> >>> so, for example Rx would be a [left-click] on the top-side of the rig=
ht
> >>> face. In this representation of the puzzle almost all moves are
> allowed; all
> >>> U, D, F, B, R, L, K, O and C moves are possible regardless of rotatio=
n
> and
> >>> only A moves (and of course the rightfully forbidden S and V moves) a=
re
> >>> impossible regardless of rotation. Note that R and L moves in the
> software
> >>> correspond to the same moves of the physical puzzle but this is not
> >>> generally true (I will come back to this later).
> >>>
> >>> Possible moves (so far) in the standard rotation
> >>>
> >>> In the standard rotation, the possible/allowed moves with the
> definitions
> >>> above are:
> >>>
> >>> O moves, all of these are always possible in any state and rotation o=
f
> the
> >>> puzzle since they are simply 3D-rotations.
> >>>
> >>> R, L moves, all of these as well since the puzzle has a right and lef=
t
> 2x2x2
> >>> block in the standard rotation.
> >>>
> >>> U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x4 block=
s
> with
> >>> less symmetry than a 2x2x2 block.
> >>>
> >>> F, B moves, just Fy2 and By2 since the front and back are 2x1x4 block=
s
> with
> >>> less symmetry than a 2x2x2 block.
> >>>
> >>> K moves, all possible since this is a rotation of the center 2x2x2
> block.
> >>>
> >>>
> >>> I think the only legal 90 degree twists of the K face are those about
> the
> >>> long axis. I believe this is what Christopher Locke was saying in thi=
s
> >>> message. To see why there is no straightforward way to perform other =
90
> >>> degree twists, you only need to perform a 90 degree twist on an outer
> (L/R)
> >>> face and then reorient the whole puzzle along a different outer axis.
> If the
> >>> original twist was not about the new long axis, then there is clearly
> no
> >>> straightforward way to undo that twist.
> >>>
> >>>
> >>> A moves, only Az moves since this is two 2x2x1 blocks that have to be
> >>> rotated together.
> >>>
> >>> S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined in the
> >>> standard rotation.
> >>>
> >>> V moves, not Vz+ or Vz- since the definition doesn't give these
> meaning when
> >>> the long side of the puzzle is parallel with the z-axis.
> >>>
> >>> Note that for example Fz2 is not allowed since this won't take you to=
a
> >>> state of the puzzle. To allow more moves we need to extend the
> definitions
> >>> (after the extension in the next paragraph all rotations and twists
> (O, R,
> >>> L, U, D, F, B, K, A) are possible in any rotation and only which S an=
d
> V
> >>> moves are possible depend on the state and rotation of the puzzle).
> >>>
> >>>
> >>> Extension of some definitions
> >>>
> >>> It's possible to make an extension that allows all O, R, L, U, D, F,
> B, K
> >>> and A moves in any state. I will explain how this can be done in the
> >>> standard rotation but it applies analogously to any other rotation
> where the
> >>> K face is an octahedron. First let's focus on U, D, F and B and
> because of
> >>> the symmetry of the puzzle all of these are analogous so I will only
> explain
> >>> one. The extension that makes all U moves possible (note that the U
> face is
> >>> in the positive x-direction) is as follows: when making an U move fir=
st
> >>> detach the 8 top pieces which gives you a 1x2x4 block, fold this bloc=
k
> into
> >>> a 2x2x2 block in the positive x-direction (similar to the later part
> of a
> >>> Vx+ move from the standard rotation) such that the U face form an
> >>> octahedron, rotate this 2x2x2 block around the specified axis (for
> example
> >>> around the z-axis if you are doing an Uz move), reverse the fold you
> just
> >>> did creating a 1x2x4 block again and reattach the block.
> >>>
> >>>
> >>> I noticed something like this the other day but realized that it only
> seems
> >>> to work for rotations along the long dimension (z in your example).
> These
> >>> are already easily accomplished by a simple rotation to put the face =
in
> >>> question on the end caps, followed by a double end-cap twist.
> >>>
> >>> This is as far as I'm going to comment for the moment because the
> >>> information gets very dense and I've been mulling and picking over yo=
ur
> >>> message for several days already. In short, I really like your attemp=
t
> to
> >>> provide a complete system of notation for discussing this puzzle and
> will be
> >>> curious to hear your thoughts on my comments so far. I hope others wi=
ll
> >>> chime in too.
> >>>
> >>> One final thought is that a real "acid test" of any notation system
> for this
> >>> puzzle will be attempt to translate some algorithms from MC4D. I woul=
d
> most
> >>> like to see a sequence that flips a single piece, like the second
> 4-color
> >>> series on this page of Roice's solution, or his pair of twirled
> corners at
> >>> the end of this page. One trick will be to minimize the number of
> >>> whole-puzzle reorientations needed, but really any sequence that work=
s
> will
> >>> be great evidence that the puzzles are equivalent. I suspect that thi=
s
> sort
> >>> of exercise will never be practical because it will require too many
> >>> reorientations, and that entirely new methods will be needed to
> actually
> >>> solve this puzzle.
> >>>
> >>> Best,
> >>> -Melinda
> >>>
> >>> The A moves can be done very similarly but after you have detached th=
e
> two
> >>> 2x2x1 blocks you don't fold them but instead you stack them similar t=
o
> a Sz
> >>> move, creating a 2x2x2 block with the A face as an octahedron in the
> middle
> >>> and then reverse the process after you have rotated the block as
> specified
> >>> (for example around the negative y-axis if you are doing an Ay' move)=
.
> Note
> >>> that these extended moves are closely related to the normal moves and
> for
> >>> example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99 in the standard rotat=
ion and note that
> (Ry
> >>> Ly=E2=80=99) mod(rot) =3D (Ly Ry=E2=80=99) mod(rot) =3D I (this appli=
es to the other
> extended
> >>> moves as well).
> >>>
> >>> If the cube is in the half-rotated state, where both the R and L face=
s
> are
> >>> octahedra, you can extend the definitions very similarly. The only
> thing you
> >>> have to change is how you fold the 2x4 blocks when performing a U, D,
> F or B
> >>> move. Instead of folding the two 2x2 halves of the 2x4 into a 2x2 you
> have
> >>> to fold the end 1x2 block 180 degrees such that the face forms an
> >>> octahedron.
> >>>
> >>> These moves might be a little bit harder to perform, to me especially
> the A
> >>> moves seems a bit awkward, so I don't know if it's good to use them o=
r
> not.
> >>> However, the A moves are not necessary if you allow Sz in the standar=
d
> >>> rotation (which you really should since Sz mod(rot) =3D I in the stan=
dard
> >>> rotation) and thus it might not be too bad to use this extended
> version. The
> >>> notation supports both variants so if you don=E2=80=99t want to use t=
hese
> extended
> >>> moves that shouldn=E2=80=99t be a problem. Note that, however, for ex=
ample Ux
> >>> (physical puzzle) !=3D Ux (virtual puzzle) where !=3D means =E2=80=9C=
not equal to=E2=80=9D
> (more
> >>> about legal/illegal moves later).
> >>>
> >>> Generalisation of the notation
> >>>
> >>> Let=E2=80=99s generalise the notation to make it easier to use and to=
make it
> work
> >>> for any n^4 cube in the MC4D software. Previously, we saw that (Rx)2 =
=3D
> Rx Rx
> >>> and if we allow ourselves to rewrite this as (Rx)2 =3D Rx2 =3D Rxx =
=3D Rx Rx
> we
> >>> see that the capital letter naturally can be distributed over the
> lowercase
> >>> letters. We can make this more general and say that any capital lette=
r
> >>> followed by several lowercase letters means the same thing as the
> capital
> >>> letter distributed over the lowercase letters. Like Rxyz =3D Rx Ry Rz
> and here
> >>> R can be exchanged with any capital letter and xyz can be exchanged
> with any
> >>> sequence of lowercase letters. We can also allow several capital
> letters and
> >>> one lowercase letter, for example RLx and let=E2=80=99s define this a=
s RLx =3D
> Rx Lx
> >>> so that the lowercase letter can be distributed over the capital
> letters. We
> >>> can also define a capital letter followed by =E2=80=98 (prime) like R=
=E2=80=99x =3D Rx=E2=80=99
> and
> >>> R=E2=80=99xy =3D Rx=E2=80=99y=E2=80=99 so the =E2=80=98 (prime) is di=
stributed over the lowercase
> letters.
> >>> Note that we don=E2=80=99t define a capital to any other power than -=
1 like
> this
> >>> since for example R2x =3D RRx might seem like a good idea at first bu=
t it
> >>> isn=E2=80=99t very useful since R2 and RR are the same lengths (and p=
owers
> greater
> >>> than two are seldom used) and we will see that we can define R2 in
> another
> >>> way that generalises the notation to all n^4 cubes.
> >>>
> >>> Okay, let=E2=80=99s define R2 and similar moves now and have in mind =
what
> moves we
> >>> want to be possible for a n^4 cube. The moves that we cannot achieve
> with
> >>> the notation this far is twisting deeper slices. To match the notatio=
n
> with
> >>> the controls of the MC4D software let R2x be the move similar to Rx b=
ut
> >>> twisting the 2nd layer instead of the top one and similarly for other
> >>> capital letters, numbers (up to n) and lowercase letters. Thus, R2x i=
s
> >>> performed as Rx but holding down the number 2 key. Just as in the
> program,
> >>> when no number is specified 1 is assumed and you can combine several
> numbers
> >>> like R12x to twist both the first and second layer. This notation doe=
s
> not
> >>> apply to rotations (O) folding moves (V) and restacking moves (S) (I
> suppose
> >>> you could redefine the S move using this deeper-slice-notation and us=
e
> S1z
> >>> as Sz+, S2z as Sz and S3z as Sz- but since these moves are only
> allowed for
> >>> the physical 2x2x2x2 I think that the notation with + and =E2=80=93 i=
s better
> since
> >>> S followed by a lowercase letter without +/- always means splitting
> the cube
> >>> in a coordinate plane that way, not sure though so input would be
> great).
> >>> The direction of the twist R3x should be the same as Rx meaning that
> if Rx
> >>> takes stickers belonging to K and move them to F, so should R3x, in
> >>> accordance with the controls of the MC4D software. Note that for a
> 3x3x3x3
> >>> it=E2=80=99s true that R3z =3D Lz whereas R3x =3D Lx=E2=80=99 (note t=
hat R and L are the
> faces
> >>> in the z-directions so because of the symmetry of the cube it will
> also be
> >>> true that for example U3x =3D Bx whereas U3y=3DBy=E2=80=99).
> >>>
> >>> What about the case with several capital letters and several lowercas=
e
> >>> letters, for instance, RLxy? I see two natural definitions of this.
> Either,
> >>> we could have RLxy =3D Rxy Lxy or we could define it as RLxy =3D RLx =
RLy.
> These
> >>> are generally not the same (if you exchange R and L with any allowed
> capital
> >>> letter and similarly for x and y). I don=E2=80=99t know what is best,=
what do
> you
> >>> think? The situations I find this most useful in are RL=E2=80=99xy to=
do a
> rotation
> >>> and RLxy as a twist. However, since R and L are opposite faces their
> >>> operations commute which imply RL=E2=80=99xy (1st definition) =3D Rxy=
L=E2=80=99xy =3D Rx
> Ry
> >>> Lx=E2=80=99 Ly=E2=80=99 =3D Rx Lx=E2=80=99 Ry Ly=E2=80=99 =3D RL=E2=
=80=99x RL=E2=80=99y =3D RL=E2=80=99xy (2nd definition) and
> similarly
> >>> for the other case with RLxy. Hopefully, we can find another useful
> sequence
> >>> of moves where this notation can be used with only one of the
> definitions
> >>> and can thereby decide which definition to use. Personally, I feel
> like RLxy
> >>> =3D RLx RLy is the more intuitive definition but I don=E2=80=99t have=
any good
> >>> argument for this so I=E2=80=99ll leave the question open.
> >>>
> >>> For convenience, it might be good to be able to separate moves like
> Rxy and
> >>> RLx from the basic moves Rx, Oy etc when speaking and writing. Let=E2=
=80=99s
> call
> >>> the basic moves that only contain one capital letter and one lowercas=
e
> >>> letter (possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or a numbe=
r) simple moves
> (like
> >>> Rx, L'y, Ux2 and D3y=E2=80=99) and the moves that contain more than o=
ne capital
> >>> letter or more than one lowercase letter composed moves.
> >>>
> >>> More about inverses
> >>>
> >>> This list can obviously be made longer but here are some identities
> that are
> >>> good to know and understand. Note that R, L, U, x, y and z below just
> are
> >>> examples, the following is true in general for non-folds (however, S
> moves
> >>> are fine).
> >>>
> >>> (P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99 P1=E2=
=80=99 (Pi is an arbitrary permutation for
> >>> i=3D1,2,=E2=80=A6n)
> >>> (Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx
> >>> (RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x
> >>> (Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+ (just as an example with=
restacking moves, note that
> >>> the inverse doesn=E2=80=99t change the + or -)
> >>> RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=80=99xyz=
(true for both definitions)
> >>> (RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=80=99yx=
(true for both definitions)
> >>>
> >>> For V moves we have that: (Vx+)=E2=80=99 =3D Vx+ !=3D Vx=E2=80=99+ (!=
=3D means =E2=80=9Cnot equal
> to=E2=80=9D)
> >>>
> >>> Some important notes on legal/illegal moves
> >>>
> >>> Although there are a lot of moves possible with this notation we migh=
t
> not
> >>> want to use them all. If we really want a 2x2x2x2 and not something
> else I
> >>> think that we should try to stick to moves that are legal 2x2x2x2
> moves as
> >>> far as possible (note that I said legal moves and not permutations (a
> legal
> >>> permutation can be made up of one or more legal moves)). Clarificatio=
n:
> >>> cycling three of the edge-pieces of a Rubik=E2=80=99s cube is a legal
> permutation
> >>> but not a legal move, a legal move is a rotation of the cube or a
> twist of
> >>> one of the layers. In this section I will only address simple moves a=
nd
> >>> simply refer to them as moves (legal composed moves are moves compose=
d
> by
> >>> legal simple moves).
> >>>
> >>> I do believe that all moves allowed by my notation are legal
> permutations
> >>> based on their periodicity (they have a period of 2 or 4 and are all
> even
> >>> permutations of the pieces). So, which of them correspond to legal
> 2x2x2x2
> >>> moves? The O moves are obviously legal moves since they are equal to
> the
> >>> identity mod(rot). The same goes for restacking (S) (with or without
> +/-) in
> >>> the direction of the longest side of the puzzle (Sz, Sz+ and Sz- in t=
he
> >>> standard rotation) since these are rotations and half-rotations that
> don=E2=80=99t
> >>> change the state of the puzzle. Restacking in the other directions an=
d
> fold
> >>> moves (V) are however not legal moves since they are made of 8
> 2-cycles and
> >>> change the state of the puzzle (note that they, however, are legal
> >>> permutations). The rest of the moves (R, L, F, B, U, D, K and A) can =
be
> >>> divided into two sets: (1) the moves where you rotate a 2x2x2 block
> with an
> >>> octahedron inside and (2) the moves where you rotate a 2x2x2 block
> without
> >>> an octahedron inside. A move belonging to (2) is always legal. We can
> see
> >>> this by observing what a Rx does with the pieces in the standard
> rotation
> >>> with just K forming an octahedron. The stickers move in 6 4-cycles an=
d
> if
> >>> the puzzle is solved the U and D faces still looks solved after the
> move. A
> >>> move belonging to set (1) is legal either if it=E2=80=99s an 180-degr=
ee twist
> or if
> >>> it=E2=80=99s a rotation around the axis parallel with the longest sid=
e of the
> puzzle
> >>> (the z-axis in the standard rotation). Quite interestingly these are
> exactly
> >>> the moves that don=E2=80=99t mix up the R and L stickers with the res=
t in the
> >>> standard rotation. I think I know a way to prove that no legal 2x2x2x=
2
> move
> >>> can mix up these stickers with the rest and this has to do with the
> fact
> >>> that these stickers form an inverted octahedron (with the corners
> pointing
> >>> outward) instead of a normal octahedron (let=E2=80=99s call this hypo=
thesis *
> for
> >>> now). Note that all legal twists (R, L, F, B, U, D, K and A) of the
> physical
> >>> puzzle correspond to the same twist in the MC4D software.
> >>>
> >>> So, what moves should we add to the set of legal moves be able to get
> to
> >>> every state of the 2x2x2x2? I think that we should add the restacking
> moves
> >>> and folding moves since Melinda has already found a pretty short
> sequence of
> >>> those moves to make a rotation that changes which colours are on the =
R
> and L
> >>> faces. That sequence, starting from the standard rotation, is: Oy Sx+=
z
> Vy+
> >>> Ozy=E2=80=99 Vy+ Ozy=E2=80=99 Vy+ =3D Oy Sx+z (Vy+ Ozy=E2=80=99)3 mod=
(3rot) =3D I mod(rot)
> (hopefully I
> >>> got that right). What I have found (which I mentioned previously) is
> that
> >>> (from the standard rotation): RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-=
and since
> this
> >>> is equivalent with Sy =3D Ly2 RLx=E2=80=99 Sz- RLx Ly2 By2 the restac=
king moves
> that
> >>> are not legal moves are not very complicated permutations and
> therefore I
> >>> think that we can accept them since they help us mix up the R and L
> stickers
> >>> with other faces. In a sense, the folding moves are =E2=80=9Cmore ill=
egal=E2=80=9D
> since
> >>> they cannot be composed by the legal moves (according to hypothesis
> *). This
> >>> is also true for the illegal moves belonging to set (1) discussed
> above.
> >>> However, since the folding moves is probably easier to perform and is
> enough
> >>> to reach every state of the 2x2x2x2 I think that we should use them
> and not
> >>> the illegal moves belonging to (1). Note, once again, that all moves
> >>> described by the notation are legal permutations (even the ones that =
I
> just
> >>> a few words ago referred to as illegal moves) so if you wish you can
> use all
> >>> of them and still only reach legal 2x2x2x2 states. However (in a stri=
ct
> >>> sense) one could argue that you are not solving the 2x2x2x2 if you us=
e
> >>> illegal moves. If you only use illegal moves to compose rotations
> (that is,
> >>> create a permutation including illegal moves that are equal to I
> mod(rot))
> >>> and not actually using the illegal moves as twists I would classify
> that as
> >>> solving a 2x2x2x2. What do you think about this?
> >>>
> >>> What moves to use?
> >>>
> >>> Here=E2=80=99s a short list of the simple moves that I think should b=
e used
> for the
> >>> physical 2x2x2x2. Note that this is just my thoughts and you may use
> the
> >>> notation to describe any move that it can describe if you wish to. Th=
e
> >>> following list assumes that the puzzle is in the standard rotation bu=
t
> is
> >>> analogous for other representations where the K face is an octahedron=
.
> >>>
> >>> O, all since they are I mod(rot),
> >>> R, L, all since they are legal (note Rx (physical puzzle) =3D Rx (vir=
tual
> >>> 2x2x2x2)),
> >>> U, D, only x2 since these are the only legal easy-to-perform moves,
> >>> F, B, only y2 since these are the only legal easy-to-perform moves,
> >>> K, A, only z, z=E2=80=99 and z2 since these are the only legal easy-t=
o-perform
> >>> moves,
> >>> S, at least z, z+ and z- since these are equal to I mod(rot),
> >>> S, possibly x and y since these help us perform rotations and is easy
> to
> >>> compose (not necessary to reach all states and not legal though),
> >>> V, all 8 allowed by the rotation of the puzzle (at least one is
> necessary to
> >>> reach all states and if you allow one the others are easy to achieve
> >>> anyway).
> >>>
> >>> If you start with the standard rotation and then perform Sz+ the
> following
> >>> applies instead (this applies analogously to any other rotation where
> the R
> >>> and L faces are octahedra).
> >>>
> >>> O, all since they are I mod(rot),
> >>> R, L, only z, z=E2=80=99 and z2 since these are the only legal easy-t=
o-perform
> >>> moves,
> >>> U, D, only x2 since these are the only legal easy-to-perform moves,
> >>> F, B, only y2 since these are the only legal easy-to-perform moves,
> >>> K, A, at least z, z=E2=80=99 and z2, possibly all (since they are leg=
al)
> although
> >>> some might be hard to perform.
> >>> S, V, same as above.
> >>>
> >>> Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D I mod(ro=
t)
> (!=3D
> >>> for not equal) which implies that the Ux2 move when R and L are
> octahedra is
> >>> different from the Ux2 move when K is an octahedron. (Actually, the
> sequence
> >>> above is equal to Uy2).
> >>>
> >>> Regarding virtual n^4 cubes I think that all O, C, R, L, U, D, F, B, =
K
> and A
> >>> moves should be used since they are all legal and really the only
> thing you
> >>> need (left-clicking on an edge or corner piece in the computer progra=
m
> can
> >>> be described quite easily with the notation, for example, Kzy2 is
> >>> left-clicking on the top-front edge piece on the K face).
> >>>
> >>> I hope this was possible to follow and understand. Feel free to ask
> >>> questions about the notation if you find anything ambiguous.
> >>>
> >>> Best regards,
> >>> Joel Karlsson
> >>>
> >>> Den 4 maj 2017 12:01 fm skrev "Melinda Green melinda@superliminal.com
> >>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
> >>>
> >>>
> >>>
> >>> Thanks for the correction. A couple of things: First, when assembling
> one
> >>> piece at a time, I'd say there is only 1 way to place the first piece=
,
> not
> >>> 24. Otherwise you'd have to say that the 1x1x1 puzzle has 24 states. =
I
> >>> understand that this may be conventional, but to me, that just sounds
> silly.
> >>>
> >>> Second, I have the feeling that the difference between the "two
> >>> representations" you describe is simply one of those half-rotations I
> showed
> >>> in the video. In the normal solved state there is only one complete
> >>> octahedron in the very center, and in the half-rotated state there is
> one in
> >>> the middle of each half of the "inverted" form. I consider them to be
> the
> >>> same solved state.
> >>>
> >>> -Melinda
> >>>
> >>>
> >>> On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com
> [4D_Cubing]
> >>> wrote:
> >>>
> >>> Horrible typo... It seems like I made some typos in my email regardin=
g
> the
> >>> state count. It should of course be 16!12^16/(6*192) and NOT
> >>> 12!16^12/(6*192). However, I did calculate the correct number when
> comparing
> >>> with previous results so the actual derivation was correct.
> >>>
> >>> Something of interest is that the physical pieces can be assembled in
> >>> 16!24*12^15 ways since there are 16 pieces, the first one can be
> oriented in
> >>> 24 ways and the remaining can be oriented in 12 ways (since a corner
> with 3
> >>> colours never touch a corner with just one colour). Dividing with 6 t=
o
> get a
> >>> single orbit still gives a factor 2*192 higher than the actual count
> rather
> >>> than 192. This shows that every state in the MC4D representation has =
2
> >>> representations in the physical puzzle. These two representations mus=
t
> be
> >>> the previously discussed, that the two halves either have the same
> color on
> >>> the outermost corners or the innermost (forming an octahedron) when t=
he
> >>> puzzle is solved and thus both are complete representations of the
> 2x2x2x2.
> >>>
> >>> Best regards,
> >>> Joel Karlsson
> >>>
> >>> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" <
> joelkarlsson97@gmail.com>:
> >>>
> >>> I am no expert on group theory, so to better understand what twists a=
re
> >>> legal I read through the part of Kamack and Keane's The Rubik Tessera=
ct
> >>> about orienting the corners. Since all even permutations are allowed
> the
> >>> easiest way to check if a twist is legal might be to:
> >>> 1. Check that the twist is an even permutation, that is: the same
> twist can
> >>> be done by performing an even number of piece swaps (2-cycles).
> >>> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning
> performing the
> >>> twist k times and I (the identity) representing the permutation of
> doing
> >>> nothing) and k is not divisible by 3 the twist A definitely doesn't
> violate
> >>> the restriction of the orientations since kx mod 3 =3D 0 and k mod 3 =
!=3D 0
> >>> implies x mod 3 =3D 0 meaning that the change of the total orientatio=
n x
> for
> >>> the twist A mod 3 is 0 (which precisely is the restriction of legal
> twists;
> >>> that they must preserve the orientation mod 3).
> >>>
> >>> For instance, this implies that the restacking moves are legal 2x2x2x=
2
> moves
> >>> since both are composed of 8 2-cycles and both can be performed twice
> (note
> >>> that 2 is not divisible by 3) to obtain the identity.
> >>>
> >>> Note that 1 and 2 are sufficient to check if a twist is legal but onl=
y
> 1 is
> >>> necessary; there can indeed exist a twist violating 2 that still is
> legal
> >>> and in that case, I believe that we might have to study the orientati=
on
> >>> changes for that specific twist in more detail. However, if a twist
> can be
> >>> composed by other legal twists it is, of course, legal as well.
> >>>
> >>> Best regards,
> >>> Joel
> >>>
> >>> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com
> [4D_Cubing]
> >>> <4D_Cubing@yahoogroups.com>:
> >>>>
> >>>>
> >>>> First off, thanks everyone for the helpful and encouraging feedback!
> >>>> Thanks Joel for showing us that there are 6 orbits in the 2^4 and fo=
r
> your
> >>>> rederivation of the state count. And thanks Matt and Roice for
> pointing out
> >>>> the importance of the inverted views. It looks so strange in that
> >>>> configuration that I always want to get back to a normal view as
> quickly as
> >>>> possible, but it does seem equally valid, and as you've shown, it ca=
n
> be
> >>>> helpful for more than just finding short sequences.
> >>>>
> >>>> I don't understand Matt's "pinwheel" configuration, but I will point
> out
> >>>> that all that is needed to create your twin interior octahedra is a
> single
> >>>> half-rotation like I showed in the video at 5:29. The two main halve=
s
> do end
> >>>> up being mirror images of each other on the visible outside like he
> >>>> described. Whether it's the pinwheel or the half-rotated version
> that's
> >>>> correct, I'm not sure that it's a bummer that the solved state is no=
t
> at all
> >>>> obvious, so long as we can operate it in my original configuration a=
nd
> >>>> ignore the fact that the outer faces touch. That would just mean tha=
t
> the
> >>>> "correct" view is evidence that that the more understandable view is
> >>>> legitimate.
> >>>>
> >>>> I'm going to try to make a snapable V3 which should allow the pieces
> to be
> >>>> more easily taken apart and reassembled into other forms. Shapeways
> does
> >>>> offer a single, clear translucent plastic that they call "Frosted
> Detail",
> >>>> and another called "Transparent Acrylic", but I don't think that any
> sort of
> >>>> transparent stickers will help us, especially since this thing is
> chock full
> >>>> of magnets. The easiest way to let you see into the two hemispheres
> would be
> >>>> to simply truncate the pointy tips of the stickers. That already
> happens a
> >>>> little bit due to the way I've rounded the edges. Here is a close-up
> of a
> >>>> half-rotation in which you can see that the inner yellow and white
> faces are
> >>>> solved. Your suggestion of little mapping dots on the corners also
> works,
> >>>> but just opening the existing window further would work more directl=
y.
> >>>>
> >>>> -Melinda
> >>>>
> >>>>
> >>>> On 4/28/2017 2:15 PM, Roice Nelson roice3@gmail.com [4D_Cubing]
> wrote:
> >>>>
> >>>> I agree with Don's arguments about adjacent sticker colors needing t=
o
> be
> >>>> next to each other. I think this can be turned into an accurate 2^4
> with
> >>>> coloring changes, so I agree with Joel too :)
> >>>>
> >>>> To help me think about it, I started adding a new projection option
> for
> >>>> spherical puzzles to MagicTile, which takes the two hemispheres of a
> puzzle
> >>>> and maps them to two disks with identified boundaries connected at a
> point,
> >>>> just like a physical "global chess" game I have. Melinda's puzzle is
> a lot
> >>>> like this up a dimension, so think about two disjoint balls, each
> >>>> representing a hemisphere of the 2^4, each a "subcube" of Melinda's
> puzzle.
> >>>> The two boundaries of the balls are identified with each other and a=
s
> you
> >>>> roll one around, the other half rolls around so that identified poin=
ts
> >>>> connect up. We need to have the same restriction on Melinda's puzzle=
.
> >>>>
> >>>> In the pristine state then, I think it'd be nice to have an internal
> >>>> (hidden), solid colored octahedron on each half. The other 6 faces
> should
> >>>> all have equal colors split between each hemisphere, 4 stickers on
> each
> >>>> half. You should be able to reorient the two subcubes to make a half
> >>>> octahedron of any color on each subcube. I just saw Matt's email and
> >>>> picture, and it looks like we were going down the same thought path.=
I
> >>>> think with recoloring (mirroring some of the current piece colorings=
)
> >>>> though, the windmill's can be avoided (?)
> >>>>
> >>>> [...] After staring/thinking a bit more, the coloring Matt came up
> with is
> >>>> right-on if you want to put a solid color at the center of each
> hemisphere.
> >>>> His comment about the "mirrored" pieces on each side helped me
> understand
> >>>> better. 3 of the stickers are mirrored and the 4th is the hidden col=
or
> >>>> (different on each side for a given pair of "mirrored" pieces). All
> faces
> >>>> behave identically as well, as they should. It's a little bit of a
> bummer
> >>>> that it doesn't look very pristine in the pristine state, but it doe=
s
> look
> >>>> like it should work as a 2^4.
> >>>>
> >>>> I wonder if there might be some adjustments to be made when shapeway=
s
> >>>> allows printing translucent as a color :)
> >>>>
> >>>> [...] Sorry for all the streaming, but I wanted to share one more
> thought.
> >>>> I now completely agree with Joel/Matt about it behaving as a 2^4,
> even with
> >>>> the original coloring. You just need to consider the corner colors o=
f
> the
> >>>> two subcubes (pink/purple near the end of the video) as being a
> window into
> >>>> the interior of the piece. The other colors match up as desired.
> (Sorry if
> >>>> folks already understood this after their emails and I'm just
> catching up!)
> >>>>
> >>>> In fact, you could alter the coloring of the pieces slightly so that
> the
> >>>> behavior was similar with the inverted coloring. At the corners wher=
e
> 3
> >>>> colors meet on each piece, you could put a little circle of color of
> the
> >>>> opposite 4th color. In Matt's windmill coloring then, you'd be able
> to see
> >>>> all four colors of a piece, like you can with some of the pieces on
> >>>> Melinda's original coloring. And again you'd consider the color
> circles a
> >>>> window to the interior that did not require the same matching
> constraints
> >>>> between the subcubes.
> >>>>
> >>>> I'm looking forward to having one of these :)
> >>>>
> >>>> Happy Friday everyone,
> >>>> Roice
> >>>>
> >>>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson
> joelkarlsson97@gmail.com
> >>>> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:
> >>>>>
> >>>>>
> >>>>> Seems like there was a slight misunderstanding. I meant that you
> need to
> >>>>> be able to twist one of the faces and in MC4D the most natural
> choice is
> >>>>> the center face. In your physical puzzle you can achieve this type
> of twist
> >>>>> by twisting the two subcubes although this is indeed a twist of the
> subcubes
> >>>>> themselves and not the center face, however, this is still the same
> type of
> >>>>> twist just around another face.
> >>>>>
> >>>>> If the magnets are that allowing the 2x2x2x2 is obviously a subgrou=
p
> of
> >>>>> this puzzle. Hopefully the restrictions will be quite natural and
> only some
> >>>>> "strange" moves would be illegal. Regarding the "families of states=
"
> (aka
> >>>>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all
> allowed twists
> >>>>> preserves the parity of the pieces, meaning that only half of the
> >>>>> permutations you can achieve by disassembling and reassembling can =
be
> >>>>> reached through legal moves. Because of some geometrical properties
> of the
> >>>>> 2x2x2x2 and its twists, which would take some time to discuss in
> detail
> >>>>> here, the orientation of the stickers mod 3 are preserved, meaning
> that the
> >>>>> last corner only can be oriented in one third of the number of
> orientations
> >>>>> for the other corners. This gives a total number of orbits of 2x3=
=3D6.
> To
> >>>>> check this result let's use this information to calculate all the
> possible
> >>>>> states of the 2x2x2x2; if there were no restrictions we would have
> 16! for
> >>>>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
> >>>>> orientations for each corner). If we now take into account that
> there are 6
> >>>>> equally sized orbits this gets us to 12!16^12/6. However, we should
> also
> >>>>> note that the orientation of the puzzle as a hole is not set by som=
e
> kind of
> >>>>> centerpieces and thus we need to devide with the number of
> orientations of a
> >>>>> 4D cube if we want all our states to be separated with twists and
> not only
> >>>>> rotations of the hole thing. The number of ways to orient a 4D cube
> in space
> >>>>> (only allowing rotations and not mirroring) is 8x6x4=3D192 giving a
> total of
> >>>>> 12!16^12/(6*192) states which is indeed the same number that for
> example
> >>>>> David Smith arrived at during his calculations. Therefore, when
> determining
> >>>>> whether or not a twist on your puzzle is legal or not it is
> sufficient and
> >>>>> necessary to confirm that the twist is an even permutation of the
> pieces and
> >>>>> preserves the orientation of stickers mod 3.
> >>>>>
> >>>>> Best regards,
> >>>>> Joel
> >>>>>
> >>>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green
> melinda@superliminal.com
> >>>>> [4D_Cubing]" <4D_Cubing@yahoogroups.com>:
> >>>>>
> >>>>>
> >>>>>
> >>>>> The new arrangement of magnets allows every valid orientation of
> pieces.
> >>>>> The only invalid ones are those where the diagonal lines cutting
> each cube's
> >>>>> face cross each other rather than coincide. In other words, you can
> assemble
> >>>>> the puzzle in all ways that preserve the overall diamond/harlequin
> pattern.
> >>>>> Just about every move you can think of on the whole puzzle is valid
> though
> >>>>> there are definitely invalid moves that the magnets allow. The most
> obvious
> >>>>> invalid move is twisting of a single end cap.
> >>>>>
> >>>>> I think your description of the center face is not correct though.
> Twists
> >>>>> of the outer faces cause twists "through" the center face, not "of"
> that
> >>>>> face. Twists of the outer faces are twists of those faces themselve=
s
> because
> >>>>> they are the ones not changing, just like the center and outer face=
s
> of MC4D
> >>>>> when you twist the center face. The only direct twist of the center
> face
> >>>>> that this puzzle allows is a 90 degree twist about the outer axis.
> That
> >>>>> happens when you simultaneously twist both end caps in the same
> direction.
> >>>>>
> >>>>> Yes, it's quite straightforward reorienting the whole puzzle to put
> any
> >>>>> of the four axes on the outside. This is a very nice improvement
> over the
> >>>>> first version and should make it much easier to solve. You may be
> right that
> >>>>> we just need to find the right way to think about the outside faces=
.
> I'll
> >>>>> leave it to the math geniuses on the list to figure that out.
> >>>>>
> >>>>> -Melinda
> >>>>>
> >>>>>
> >>>>>
> >>>>> On 4/27/2017 10:31 AM, Joel Karlsson joelkarlsson97@gmail.com
> [4D_Cubing]
> >>>>> wrote:
> >>>>>
> >>>>>
> >>>>> Hi Melinda,
> >>>>>
> >>>>> I do not agree with the criticism regarding the white and yellow
> stickers
> >>>>> touching each other, this could simply be an effect of the differen=
t
> >>>>> representations of the puzzle. To really figure out if this indeed
> is a
> >>>>> representation of a 2x2x2x2 we need to look at the possible moves
> (twists
> >>>>> and rotations) and figure out the equivalent moves in the MC4D
> software.
> >>>>> From the MC4D software, it's easy to understand that the only moves
> required
> >>>>> are free twists of one of the faces (that is, only twisting the
> center face
> >>>>> in the standard perspective projection in MC4D) and 4D rotations
> swapping
> >>>>> which face is in the center (ctrl-clicking in MC4D). The first is
> possible
> >>>>> in your physical puzzle by rotating the white and yellow subcubes
> (from here
> >>>>> on I use subcube to refer to the two halves of the puzzle and the
> colours of
> >>>>> the subcubes to refer to the "outer colours"). The second is
> possible if
> >>>>> it's possible to reach a solved state with any two colours on the
> subcubes
> >>>>> that still allow you to perform the previously mentioned twists.
> This seems
> >>>>> to be the case from your demonstration and is indeed true if the
> magnets
> >>>>> allow the simple twists regardless of the colours of the subcubes.
> Thus, it
> >>>>> is possible to let your puzzle be a representation of a 2x2x2x2,
> however, it
> >>>>> might require that some moves that the magnets allow aren't used.
> >>>>>
> >>>>> Best regards,
> >>>>> Joel
> >>>>>
> >>>>> 2017-04-27 3:09 GMT+02:00 Melinda Green melinda@superliminal.com
> >>>>> [4D_Cubing] <4D_Cubing@yahoogroups.com>:
> >>>>>>
> >>>>>>
> >>>>>> Dear Cubists,
> >>>>>>
> >>>>>> I've finished version 2 of my physical puzzle and uploaded a video
> of it
> >>>>>> here:
> >>>>>> https://www.youtube.com/watch?v=3DzqftZ8kJKLo
> >>>>>> Again, please don't share these videos outside this group as their
> >>>>>> purpose is just to get your feedback. I'll eventually replace them
> with a
> >>>>>> public video.
> >>>>>>
> >>>>>> Here is an extra math puzzle that I bet you folks can answer: How
> many
> >>>>>> families of states does this puzzle have? In other words, if
> disassembled
> >>>>>> and reassembled in any random configuration the magnets allow, wha=
t
> are the
> >>>>>> odds that it can be solved? This has practical implications if all
> such
> >>>>>> configurations are solvable because it would provide a very easy
> way to
> >>>>>> fully scramble the puzzle.
> >>>>>>
> >>>>>> And finally, a bit of fun: A relatively new friend of mine and new
> list
> >>>>>> member, Marc Ringuette, got excited enough to make his own version=
.
> He built
> >>>>>> it from EPP foam and colored tape, and used honey instead of
> magnets to hold
> >>>>>> it together. Check it out here:
> >>>>>> http://superliminal.com/cube/dessert_cube.jpg I don't know how
> practical a
> >>>>>> solution this is but it sure looks delicious! Welcome Marc!
> >>>>>>
> >>>>>> -Melinda
> >>>>>>
> >>>>>
> >>>>>
> >>>>>
> >>>>>
> >>>>
> >>>
> >>>
> >>>
> >
> > ------------------------------------
> > Posted by: Joel Karlsson
> > ------------------------------------
> >
> >
> > ------------------------------------
> >
> > Yahoo Groups Links
> >
> >
> >
> >
>
>=20
>

--001a1146b23cc761080550b86fff
Content-Type: text/html; charset="UTF-8"
Content-Transfer-Encoding: quoted-printable

Well, usually we want to calculate distinct stat=
es and for the 2x2x2x2 (virtual) that become 16!12^16/(6*192), 16! for perm=
utation of pieces, 12^16 for orienting the pieces, 1/6 to only get legal st=
ates (the puzzle has 6 orbits) and 1/192 to fix one corner since the puzzle=
doesn't have fixed pieces like the 3x3x3x3. However, when I checked ho=
w many representations there are of every state I didn't count distinct=
states but all states, even illegal ones. The count for the virtual 2x2x2x=
2 thus becomes 16!*12^16 and with this count, every distinct state has 192 =
representations and only a sixth of them are legal (you can think of it as:=
how many legal and illegal states are there if I'm not allowed to rota=
te the puzzle? or: in how many ways can I put the pieces together?). The sa=
me type of count for the physical puzzle would be 3*16!*24*12^15. The ratio=
of these are 1:6 which means that even if you only count legal states and =
fix the rotation of the puzzle the count for the physical puzzle, which wou=
ld be 3*16!*24*12^15/(6*192), is still too high with a factor 6 and thus, i=
f we claim that the physical puzzle is a representation of a 2x2x2x2, there=
must be six representations of every distinct state that are not separated=
with only rotations. Luckily, this is the case and these come from the dif=
ferent projections (I agree that projection might be a better word than rep=
resentation although it might not be completely accurate).

Bes=
t regards,
Joel Karlsson

<=
div class=3D"gmail_quote">2017-05-30 3:57 GMT+02:00 Melinda Green =3D"mailto:melinda@superliminal.com">melinda@superliminal.com [4D_Cubin=
g] <t=3D"_blank">4D_Cubing@yahoogroups.com>:
=3D"gmail_quote" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padd=
ing-left:1ex">












=20

=C2=A0







=20=20=20=20=20=20
=20=20=20=20=20=20

Thanks for the clarification, Joel. Just to be complete, are you s=
ure that that the count of MC4D states are being counted in the same way?r>


One other thought I had regarding terminology: You called the two forms of =
the physical puzzle "representations", but I wonder whether a som=
ewhat better term might be "projection". This puzzle is definitel=
y not any sort of geometric projection into 3-space, but it seems to share =
a number of analogous properties with them. I often think of it as viewing =
the 4D object through a 2x2x4 "viewport".



The half rotations are sort of like translating that viewport along or arou=
nd the surface of that object. Since you point out that it is a 90 degree r=
otation, perhaps "half rotation" isn't the best term for that=
move. Whatever we call these rotations, the two forms feel to me like the =
difference between cell-first and face-first projections.



-Melinda





On 5/29/2017 2:57 AM, Joel Karlsson com" target=3D"_blank">joelkarlsson97@gmail.com [4D_Cubing] wrote:

> Hello,

>

> Just a quick correction regarding a previous statement. From the

> calculation of the states of the puzzle, we can see that if we choose<=
br>
> a state and rotation there are still two representations of that state=


> and rotation in the physical puzzle. Previously, we also said that

> these states are separated by a half-rotation such as Rx+ from an RLr>
> rep or AKx rep. This is not the case. When I calculated the states of<=
br>
> the puzzle I assumed that the longer side of the puzzle should be

> parallel with the x-axis. If we don't make that assumption we find=


> that the number of states (not distinct states, some of these are just=


> separated by a rotation) are 3*16!*24*12^15 since there are three

> possible choices for which axis should be parallel with the longer

> side. This means that, in fact, each distinct state has 6

> representations in the physical puzzle. This is actually what we

> should expect since it should be possible to represent every state in<=
br>
> an RL rep, an UD rep, an FB rep, an AKx rep, an AKy rep and an AKz

> rep. In the solved state, the rotation of the puzzle is determined by<=
br>
> which colour belongs to each face and given such a rotation there are<=
br>
> indeed six representations: first, choose an axis that the longer side=


> should be parallel with (3 alternatives) and then choose either the AK=


> rep or the non-AK rep (2 alternatives). However, the so-called

> "half-rotation" changes which colour belongs to which face s=
o this is

> not a move that gets you from one representation of a state to

> another. From an UD rep with:

> P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'x ,=


> P3 Ox'z' takes you to an FB rep of the same state

> Oy P3 Oz' takes you to an RL rep of the same state

> UD'x P3 UD'z' Sy- Oy takes you to an AKy rep of the=
same state

> UD'x P3 Oyx' Sz+ takes you to an AKz rep of the same sta=
te and

> UD'x P3 UD'x' Oz' Sx- takes you to an AKx rep o=
f the same state

> >From this, we can see that the AK states are closely related and i=
t's

> possible to change between them without illegal moves. For example

> going from an AKy rep to an AKz rep of the same state could be done>
> with:

> (UD'x P3 UD'z' Sy- Oy)' UD'x P3 Oyx' S=
z+ =3D Oy' Sy+ UD'z

> P3' UD'x' UD'x P3 Oyx' Sz+ =3D Oy' Sy+=
UD'z Oyx' Sz+

>

> Note that these sequences don't change the state nor the rotation =
of

> the puzzle but they do change the representation. Let's call these=


> types of sequences J moves. Note that what I have previously called a<=
br>
> type three rotation is actually a J move and a rotation. Further note<=
br>
> that J =3D I mod(rep) (modulo representation). The J moves changes whi=
ch

> pair of faces are symmetry breaking and this is the type of moves that=


> needs to be added to the set of legal moves to make the physical

> puzzle an actual 2x2x2x2. A J move is a move that is equal to I

> mod(rep, rot) but that isn't equal to I mod(rot) (which means that=
it

> has to preserve the state, is allowed to change the rotation and hasr>
> to change the representation/what faces (not in the left/right senser>
> but rather in the sense of colours) are symmetry-breaking ).

>

> Best regards,

> Joel Karlsson

>

> 2017-05-22 22:32 GMT+02:00 Joel Karlsson <lsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com>:

>> Hi Melinda,

>>

>> Thank you for the feedback. Regarding the coordinate system, it=E2=
=80=99s just

>> a matter of preference. I thought it would be nice to have a

>> xy-symmetry but understand that it might be more practical to foll=
ow

>> conventions. So, adapting your suggestion, let's redefine the =
axes as

>> x pointing right, y up and z towards you. Note that this means tha=
t

>> the longer side of the puzzle is parallel with the x-axis in ther>
>> standard rotation. From here on, I will use this new coordinate>
>> system.

>>

>> Regarding the name of the faces. Since which faces are the "o=
uter

>> ones" changes with how you rotate the puzzle and what state t=
he puzzle

>> is in, I think that the labelling of the faces should be independe=
nt

>> of which faces are the outer faces (forming what will be referred =
to

>> as inverted octahedra). Since the puzzle is a representation of a<=
br>
>> 4d-cube in 3d-space it will (with our coordinate system) always ha=
ve

>> two faces "belonging" to every axis and two faces that d=
on't belong to

>> any axis. Therefore, it makes sense to label the faces in such a w=
ay

>> that the face in for example the positive x direction is R, always=
.

>> So, the K face (which is one of the faces not belonging to a

>> particular axis) is always the face only belonging to the center 2=
x2x2

>> block (and this can be either an octahedron or an inverted octahed=
ron

>> (the outer corners) depending on the representation of the puzzle)=
.

>> The distinction between left and right never disappears; the R fac=
e is

>> always the face only belonging to the right half of the puzzle and=
can

>> be an octahedron, an inverted octahedron, two half octahedra (<=
><>) or

>> one half and two quarter octahedra (><><). Note that t=
he

>> half-rotations are 90-degree rotations of the puzzle; for example,=
Sx+

>> (physical puzzle) =3D Cx' (virtual puzzle) is the rotation tha=
t does L

>> -> K -> R -> A -> L so the face that previously was th=
e L face is now

>> the K face. From the standard rotation (where R and L are the inve=
rted

>> octahedra before the rotation), this would mean that the K and A f=
aces

>> are inverted octahedra after a Sx+ rotation. What faces are the>
>> "mysterious" (or more precisely symmetry breaking, formi=
ng inverted

>> octahedra instead of regular octahedra) depend on the rotation of =
the

>> puzzle and can be any two opposite faces (R and L, U and D, F and =
B or

>> A and K). It might be useful to be able to describe what faces are=


>> inverted octahedra since this determines what moves are legal so l=
et's

>> say that the puzzle is in an RL representation if R and L are inve=
rted

>> octahedra and similarly for other states. Thus, Sx+ can take you f=
rom

>> an RL representation to an AK representation (what a long word let=
's

>> use rep for short). Note that the AK rep doesn't specify which=
axis

>> the longer side should be parallel with so let's just add a lo=
wercase

>> letter to indicate this (an AKx rep is thus a state where the A an=
d K

>> faces are inverted octahedra and the longer side is parallel with =
the

>> x-axis). In conclusion, I would like to keep the names of the face=
s as

>> I first defined them and hope that it's clearer what I mean wi=
th them

>> and that the names of the faces are not related to what faces form=


>> inverted octahedra.

>>

>> You also wrote:

>> "There are several, distinct types of rotations, none of whic=
h change

>> the state of the puzzle, and I think we need a way to be unambiguo=
us

>> about them. The types I see are

>>

>> 1. Simple reorientation of the physical puzzle in the hand, no mag=
nets

>> involved. IE your 'O' moves and maybe analogous to mouse-d=
ragging in

>> MC4D?

>> 2. Rolling of one 2x2x2 half against the other. Maybe analogous to=
ctrl-click?

>> 3. Half-rotations. Maybe analogous to a "face first" vie=
w? (ctrl-click

>> on a 2-color piece of the 3^4 with the setting "Ctrl-Click Ro=
tates: by

>> Cubie")

>> 4. Whole-puzzle reorientations that move an arbitrary axis into th=
e

>> "outer" 2 faces. No MC4D analog."

>>

>> The rotations (as far as I know) are the O moves (type one rotatio=
n),

>> which are indeed analogous to mouse-dragging in MC4D, rolling the<=
br>
>> 2x2x2 halves (type two rotation) (these are easily described as fo=
r

>> example RL'y in an RL rep) which (in a non-AK rep) is a ctrl-c=
lick on

>> a non-inverted face (that is ctrl-click on a face that is currentl=
y

>> represented with an octahedron), half-rotations (also type two

>> rotations) which is a ctrl-click on an inverted face and sequences=


>> that for example from an RL rep can take the R face to K withoutr>
>> turning the puzzle into an AK rep (type three rotation). Type one =
and

>> two rotations are legal moves but type three contain illegal 2x2x2=
x2

>> moves according to hypothesis * in my previous email (however, the=
y

>> are very important and needed if we wish to be able to reach allr>
>> states).

>>

>> Regarding fold moves, from the standard rotation (RL rep) are you<=
br>
>> saying that Vy+ =3D Vy'+ or something like Vy+ =3D Vy'+ FB=
'x2 =3D Vy'+

>> mod(rot)? The former seems not to be correct but I believe the lat=
ter

>> is, please correct me if I'm wrong. I don't assume that yo=
u fold the

>> puzzle back to the same representation. This is what + and -

>> indicates, + preserve the representation mod(rot) (i.e AK rep both=


>> before and after or neither before nor after) and - changes it (go=
ing

>> from AK rep to non-AK rep or vice versa).

>>

>> "I think the only legal 90 degree twists of the K face are th=
ose about

>> the long axis. I believe this is what Christopher Locke was saying=
in

>> this message. To see why there is no straightforward way to perfor=
m

>> other 90 degree twists, you only need to perform a 90 degree twist=
on

>> an outer (L/R) face and then reorient the whole puzzle along a

>> different outer axis. If the original twist was not about the new =
long

>> axis, then there is clearly no straightforward way to undo that>
>> twist."

>>

>> Yes, as pointed out further down in my previous email. The notatio=
n

>> allows all 90-degree rotations after the section "extensions =
of some

>> definitions" although only 180-degree rotations are legal for=


>> octahedral faces around axes not parallel with the longer side of =
the

>> puzzle (see the section =E2=80=9Csome important notes on legal/ill=
egal

>> moves=E2=80=9D).

>>

>> "I noticed something like this the other day but realized tha=
t it only

>> seems to work for rotations along the long dimension (z in your>
>> example). These are already easily accomplished by a simple rotati=
on

>> to put the face in question on the end caps, followed by a double<=
br>
>> end-cap twist."

>>

>> It works for the other rotations as well although these are not le=
gal

>> moves. They could possibly be used instead of the illegal S moves<=
br>
>> (along axes not parallel with the longer side) and the illegal V m=
oves

>> (V- (minus) moves which are closely related to the illegal S moves=
,

>> example from RL rep: Vy- =3D Vy+ Sx Oz2) but as I mentioned in my<=
br>
>> previous email I do believe that it's better to use the S and =
V moves

>> since you have found a relatively short way to perform a type thre=
e

>> rotation with those.

>>

>> Let:

>> P1 =3D Sy+ Uyz2 Sy-

>> P1=E2=80=99 =3D P1 (P1 is its own inverse)

>> P2 =3D (P1 UD=E2=80=99z P1 UD=E2=80=99z=E2=80=99 P1 UD=E2=80=
=99z=E2=80=99 P1 UD=E2=80=99z)2 (P2=3DP_2 not P^2

>> whereas the two at the end means: perform twice)

>> P2=E2=80=99 =3D (UD=E2=80=99z=E2=80=99 P1 UD=E2=80=99z P1 UD=
=E2=80=99z P1 UD=E2=80=99z=E2=80=99 P1)2

>> P3 =3D Sy+x Vz+ Oxz=E2=80=99 Vz+ Oxz=E2=80=99 Vz+ Oz UD'=
x =3D I mod(rot) (type three rotation)

>> P3=E2=80=99 =3D Oy=E2=80=99 P3 Oy=E2=80=99

>> P4 =3D P2 P3 P2 P3=E2=80=99 P2=E2=80=99 P3 P2=E2=80=99 P3=
=E2=80=99

>>

>> P4 from UD rep is a 164 move sequence rotating only one corner in =
its

>> place. The sequence is inspired by Roice =E2=80=9Csecond four-colo=
r series=E2=80=9D

>> but I have changed the =E2=80=9CTop 9=E2=80=9D moves to pure rotat=
ions since it=E2=80=99s all

>> that=E2=80=99s necessary and a bit shorter to perform. P3 is your =
type three

>> rotation (with a rotation added at the end) and P2 is Roice =E2=80=
=9Cthird

>> three-color series=E2=80=9D. Written with my notation but for the =
virtual

>> puzzle, the sequences (Roice original since it=E2=80=99s easier to=
perform K

>> twists than O rotations in MC4D) are:

>> Q1 =3D (Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=E2=80=99 Rz2y=E2=80=99=
)2 (analogue to P2)

>> Q1=E2=80=99 =3D (Rz2y=E2=80=99 Kz2y=E2=80=99 Lz2y=E2=80=99 Kz2y=
=E2=80=99)2

>> Q2 =3D Q1 Kxy=E2=80=99 Q1 Kyx=E2=80=99 Q1=E2=80=99 Kxy=E2=80=
=99 Q1=E2=80=99 Kyx=E2=80=99 (analogue to P4 but

>> only 36 moves)

>>

>> How to read faster (the example applies to a 3x3x3x3): moves like<=
br>
>> Kz2y=E2=80=99 are clicking on 3C-pieces in MC4D. If the move is wr=
itten in

>> this way, [uppercase letter] [lowercase letter]2 [lowercase letter=


>> possibly with =E2=80=98 (prime)], there=E2=80=99s a quite quick wa=
y to realize which

>> piece this is. The uppercase letter specifies which face the piece=
to

>> press is on and the first lowercase letter (followed by 2) specifi=
es

>> one of the sides of that face that the piece belong to. There are =
then

>> 4 possible pieces. Sadly, the piece do not lie in the direction of=
the

>> last letter from the center of the side of the face but you have t=
o

>> move one edge clockwise from this. So, Kz2y=E2=80=99 is a click on=
the edge on

>> the front side of the K face one step clockwise from the negative<=
br>
>> y-axis (thus, the front left 3C-piece on the K face). It=E2=80=99s=
a bit

>> unfortunate that this =E2=80=9Crule=E2=80=9D isn=E2=80=99t even si=
mpler but it's at least true

>> for all of these moves (as far as I know). Moves like Kxy=E2=80=99=
are

>> left/right-clicking on a corner piece but currently I don=E2=80=99=
t know a

>> fast way to determine which corner. Any ideas?

>>

>> Best regards,

>> Joel Karlsson

>>

>> 2017-05-19 3:33 GMT+02:00 Melinda Green superliminal.com" target=3D"_blank">melinda@superliminal.com

>> [4D_Cubing] <t=3D"_blank">4D_Cubing@yahoogroups.com>:

>>>

>>> Hello Joel,

>>>

>>> Thanks for drilling into this puzzle. Finding good ways to dis=
cuss and think

>>> about moves and representations will be key. I'll comment =
on some details

>>> in-line.

>>>

>>> On 5/14/2017 6:16 AM, Joel Karlsson sson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com [4D_Cubing=
]

>>> wrote:

>>>

>>>

>>> Yes, that is correct and in fact, you should divide not only w=
ith 24 for the

>>> orientation but also with 16 for the placement if you want to =
calculate

>>> unique states (since the 2x2x2x2 doesn't have fixed center=
pieces). The

>>> point, however, was that if you don=E2=80=99t take that into a=
ccount you get a

>>> factor of 24*16=3D384 (meaning that the puzzle has 384 represe=
ntations of

>>> every unique state) instead of the factor of 192 which you get=
when

>>> calculating the states from the virtual puzzle and hence every=
state of the

>>> virtual puzzle has two representations in the physical puzzle.=
Yes exactly,

>>> they are indeed the same solved (or other) state and you are c=
orrect that

>>> the half rotation (taking off a 2x2 layer and placing it at th=
e other end of

>>> the puzzle) takes you from one representation to the same stat=
e with the

>>> other representation. This means that the restacking move (tak=
ing off the

>>> front 2x4 layer and placing it behind the other 2x4 layer) can=
be expressed

>>> with half-rotations and ordinary twists and rotations (which y=
ou might have

>>> pointed out already).

>>>

>>>

>>> Yes, I made that claim in the video but didn't show it bec=
ause I have yet to

>>> record such a sequence. I've only stumbled through it a fe=
w times. I talked

>>> about it at 5:53 though I mistakenly called it a twist, when I=
should have

>>> called it a sequence.

>>>

>>>

>>> I think I've found six moves including ordinary twists and=
a restacking move

>>> that is identical to a half-rotation and thus it's easy to=
compose a

>>> restacking move with one half-rotation and five ordinary twist=
s. There might

>>> be an error since I've only played with the puzzle in my m=
ind so it would be

>>> great if you, Melinda, could confirm this (the sequence is des=
cribed later

>>> in this email).

>>>

>>>

>>> You mean "RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =3D Sz-"? =
Yes, that works. There do seem to

>>> be easier ways to do that beginning with an ordinary rolling r=
otation. I

>>> don't see those in your notation, but the equivalent using=
a pair of twists

>>> would be Rx Lx' Sx Vy if I got that right.

>>>

>>>

>>> To be able to communicate move sequences properly we need nota=
tion for

>>> representing twists, rotations, half-rotations, restacking mov=
es and folds.

>>> Feel free to come with other suggestion but you can find mine =
below. Please

>>> read the following thoroughly (maybe twice) to make sure that =
you understand

>>> everything since misinterpreted notation could potentially bec=
ome a

>>> nightmare and feel free to ask questions if there is something=
that needs

>>> clarification.

>>>

>>> Coordinate system and labelling:

>>>

>>> Let's introduce a global coordinate system. In whatever st=
ate the puzzle is

>>> let the positive x-axis point upwards, the positive y-axis tow=
ards you and

>>> the positive z-axis to the right (note that this is a right-ha=
nd system).

>>>

>>>

>>> I see the utility of a global coordinate system, but this one =
seems rather

>>> non-standard. I suggest that X be to the right, and Y up since=
these are

>>> near-universal standards. Z can be in or out. I have no opinio=
n. If there is

>>> any convention in the twisty puzzle community, I'd go with=
that.

>>>

>>> Note also that the wiki may be a good place to document and it=
erate on

>>> terminology, descriptions and diagrams. Ray added a "nota=
tion" section to

>>> the 3^4 page here, and I know that one other member was thinki=
ng of

>>> collecting a set of moves on another wiki page.

>>>

>>>

>>> Now let's name the 8 faces of the puzzle. The right face i=
s denoted with R,

>>> the left with L, the top U (up), the bottom D (down), the fron=
t F, the back

>>> B, the center K (kata) and the last one A (ana). The R and L f=
aces are

>>> either the outer corners of the right and left halves respecti=
vely or the

>>> inner corners of these halves (forming octahedra) depending on=
the

>>> representation of the puzzle. The U, D, F and B faces are eith=
er two diamond

>>> shapes (looking something like this: <><>) on the =
corresponding side of the

>>> puzzle or one whole and two half diamond shapes (><>&=
lt;). The K face is either

>>> an octahedron in the center of the puzzle or the outer corners=
of the center

>>> 2x2x2 block. Lastly, the A face is either two diamond shapes, =
one on the

>>> right and one on the left side of the puzzle, or another shape=
that=E2=80=99s a

>>> little bit hard to describe with just a few words (the white s=
tickers at

>>> 5:10 in the latest video, after the half-rotation but before t=
he restacking

>>> move).

>>>

>>>

>>> I think it's more correct to say that the K face is either=
an octahedron at

>>> the origin (A<>K<>A) or in the center of one of th=
e main halves, with the A

>>> face inside the other half (>A<>K<). This was what=
I was getting at in my

>>> previous message. You do later talk about octahedral faces bei=
ng in either

>>> the center or the two main halves, so this is just terminology=
. But about

>>> "the outer corners of the center 2x2x2 block", this =
cannot be the A or K

>>> face as you've labeled them. You've been calling these=
the L/R faces, but

>>> the left-right distinction disappears in the half-rotated stat=
e, so maybe

>>> "left" and "right" aren't the best nam=
es. To me, they are always the

>>> "outside" faces, regardless. You can distinguish the=
m as the left and right

>>> outside faces in one representation, or as the center and end =
outside faces

>>> in the other. (Or perhaps "end" versus "ind&quo=
t; if we want to be cute.)

>>>

>>> I'm also a little torn about naming the interior faces ana=
and kata, not

>>> because of the names themselves which I like, but because the =
mysterious

>>> faces to me are the outermost ones you're calling R and L.=
It only requires

>>> a simple rotation to move faces in and out of the interior (oc=
tahedral)

>>> positions, but it's much more difficult to move another ax=
is into the

>>> L/R/outer direction.

>>>

>>> So maybe the directions can be

>>>

>>> Up-Down

>>> Front-Back

>>> Ind-End

>>> Ana-Kata

>>>

>>> I'm not in love with it and will be happy with anything th=
at works. Thoughts

>>> anyone?

>>>

>>>

>>> We also need a name for normal 3D-rotations, restacking moves =
and folds

>>> (note that a half-rotation is a kind of restacking move). Let =
O be the name

>>> for a rOtation (note that the origin O doesn't move during=
a rotation, by

>>> the way, these are 3D rotations of the physical puzzle), let S=
represent a

>>> reStacking move and V a folding/clamshell move (you can rememb=
er this by

>>> thinking of V as a folded line).

>>>

>>>

>>> I think it's fine to call the clamshell move a fold or den=
ote it as V. I

>>> just wouldn't consider it to be a basic move since it'=
s a simple composite

>>> of 3 basic twists as shown here. In general, I think there are=
so many

>>> useful composite moves that we need to be able to easily make =
them up ad hoc

>>> with substitutions like Let =E2=86=93 =3D Rx Lx'. These ar=
e really macro moves which

>>> can be nested. That said, it's a particularly useful move =
so it's probably

>>> worth describing in some formal way like you do in detail belo=
w.

>>>

>>>

>>>

>>> Further, let I (capital i) be the identity, preserving the sta=
te and

>>> rotation of the puzzle. We cannot use I to indicate what moves=
should be

>>> performed but it's still useful as we will see later. Sinc=
e we also want to

>>> be able to express if a sequence of moves is a rotation, prese=
rving the

>>> state of the puzzle but possibly representing it in a differen=
t way, we can

>>> introduce mod(rot) (modulo rotation). So, if a move sequence P=
satisfies P

>>> mod(rot) =3D I, that means that the state of the puzzle is the=
same before and

>>> after P is performed although the rotation and representation =
of the puzzle

>>> are allowed to change. I do also want to introduce mod(3rot) (=
modulo

>>> 3D-rotation) and P mod(3rot) =3D I means that if the right 3D-=
rotation (a

>>> combination of O moves as we will see later) is applied to the=
puzzle after

>>> P you get the identity I. Moreover, let the standard rotation =
of the puzzle

>>> be any rotation such that the longer side is parallel with the=
z-axis, that

>>> is the puzzle forms a 2x2x4 (2 pieces thick in the x-direction=
, 2 in the

>>> y-direction and 4 in the z-direction), and the K face is an oc=
tahedron.

>>>

>>> Rotations and twists:

>>>

>>> Now we can move on to name actual moves. The notation of a mov=
e is a

>>> combination of a capital letter and a lowercase letter. O foll=
owed by x, y

>>> or z is a rotation of the whole puzzle around the correspondin=
g axis in the

>>> mathematical positive direction (counterclockwise/the way your=
right-hand

>>> fingers curl if you point in the direction of the axis with yo=
ur thumb). For

>>> example, Ox is a rotation around the x-axis that turns a 2x2x4=
into a 2x4x2.

>>> A name of a face (U, D, F, B, R, L, K or A) followed by x, y o=
r z means:

>>> detach the 8 pieces that have a sticker belonging to the face =
and then turn

>>> those pieces around the global axis. For example, if the longe=
r side of the

>>> puzzle is parallel to the z-axis (the standard rotation), Rx m=
eans: take the

>>> right 2x2x2 block and turn it around the global x-axis in the =
mathematical

>>> positive direction. Note that what moves are physically possib=
le and allowed

>>> is determined by the rotation of the puzzle (I will come back =
to this

>>> later). Further note that Ox mod(rot) =3D Oy mod(rot) =3D Oz m=
od(rot) =3D I,

>>> meaning that the 3D-rotations of the physical puzzle correspon=
ds to a

>>> 4D-rotation of the represented 2x2x2x2.

>>>

>>>

>>> There are several, distinct types of rotations, none of which =
change the

>>> state of the puzzle, and I think we need a way to be unambiguo=
us about them.

>>> The types I see are

>>>

>>> Simple reorientation of the physical puzzle in the hand, no ma=
gnets

>>> involved. IE your 'O' moves and maybe analogous to mou=
se-dragging in MC4D?

>>> Rolling of one 2x2x2 half against the other. Maybe analogous t=
o ctrl-click?

>>> Half-rotations. Maybe analogous to a "face first" vi=
ew? (ctrl-click on a

>>> 2-color piece of the 3^4 with the setting "Ctrl-Click Rot=
ates: by Cubie")

>>> Whole-puzzle reorientations that move an arbitrary axis into t=
he "outer" 2

>>> faces. No MC4D analog.

>>>

>>>

>>> Inverses and performing a move more than once

>>>

>>> To mark that a move should be performed n times let's put =
^n after it. For

>>> convenience when writing and speaking let ' (prime) repres=
ent ^-1 (the

>>> inverse) and n represent ^n. The inverse P' of some permut=
ation P is the

>>> permutation that satisfies P P' =3D P' P =3D I (the id=
entity). For example (Rx)'

>>> means: do Rx backwards, which corresponds to rotating the righ=
t 2x2x2 block

>>> in the mathematical negative direction (clockwise) around the =
x-axis and

>>> (Rx)2 means: perform Rx twice. However, we can also define pow=
ers of just

>>> the lowercase letters, for example, Rx2 =3D Rx^2 =3D Rxx =3D R=
x Rx. So x2=3Dx^2

>>> means: do whatever the capital letter specifies two times with=
respect to

>>> the x-axis. We can see that the capital letter naturally is di=
stributed over

>>> the two lowercase letters. Rx' =3D Rx^-1 means: do whateve=
r the capital letter

>>> specifies but in the other direction than you would have if th=
e prime

>>> wouldn't have been there (note that thus, x'=3Dx^-1 ca=
n be seen as the

>>> negative x-axis). Note that (Rx)^2 =3D Rx^2 and (Rx)' =3D =
Rx' which is true for

>>> all twists and rotations but that doesn't have to be the c=
ase for other

>>> types of moves (restacks and folds).

>>>

>>> Restacking moves

>>>

>>> A restacking move is an S followed by either x, y or z. Here t=
he lowercase

>>> letter specifies in what direction to restack. For example, Sy=
(from the

>>> standard rotation) means: take the front 8 pieces and put them=
at the back,

>>> whereas Sx means: take the top 8 pieces and put them at the bo=
ttom. Note

>>> that Sx is equivalent to taking the bottom 8 pieces and puttin=
g them at the

>>> top. However, if we want to be able to make half-rotations we =
sometimes need

>>> to restack through a plane that doesn't go through the ori=
gin. In the

>>> standard rotation, let Sz be the normal restack (taking the 8 =
right pieces,

>>> the right 2x2x2 block, and putting them on the left), Sz+ be t=
he restack

>>> where you split the puzzle in the plane further in the positiv=
e z-direction

>>> (taking the right 2x2x1 cap of 4 pieces and putting it at the =
left end of

>>> the puzzle) and Sz- the restack where you split the puzzle in =
the plane

>>> further in the negative z-direction (taking the left 2x2x1 cap=
of 4 pieces

>>> and putting it at the right end of the puzzle). If the longer =
side of the

>>> puzzle is parallel to the x-axis instead, Sx+ would take the t=
op 1x2x2 cap

>>> and put it on the bottom. Note that in the standard rotation S=
z mod(rot) =3D

>>> I. For restacks we see that (Sx)' =3D Sx' =3D Sx (true=
for y and z too of

>>> course), that (Sz+)' =3D Sz- and that (Sz+)2 =3D (Sz-)2 =
=3D Sz. We can define Sz'+

>>> to have meaning by thinking of z=E2=80=99 as the negative z-ax=
is and with that in

>>> mind it=E2=80=99s natural to define Sz=E2=80=99+=3DSz-. Thus, =
(Sz)=E2=80=99 =3D Sz=E2=80=99 and (Sz+)=E2=80=99 =3D Sz=E2=80=99+.

>>>

>>> Fold moves

>>>

>>> A fold might be a little bit harder to describe in an intuitiv=
e way. First,

>>> let's think about what folds are interesting moves. The fo=
lds that cannot be

>>> expressed as rotations and restacks are unfolding the puzzle t=
o a 4x4 and

>>> then folding it back along another axis. If we start with the =
standard

>>> rotation and unfold the puzzle into a 1x4x4 (making it look li=
ke a 4x4 from

>>> above) the only folds that will achieve something you can'=
t do with a

>>> restack mod(rot) is folding it to a 2x4x2 so that the longer s=
ide is

>>> parallel with the y-axis after the fold. Thus, there are 8 int=
eresting fold

>>> moves for any given rotation of the puzzle since there are 4 w=
ays to unfold

>>> it to a 4x4 and then 2 ways of folding it back that make the m=
ove different

>>> from a restack move mod(rot).

>>>

>>>

>>> Assuming you complete a folding move in the same representatio=
n (<><> or

>>>> <><), then there are only two interesting choices=
. That's because it

>>> doesn't matter which end of a chosen cutting plane you ope=
n it from, the end

>>> result will be the same. That also means that any two consecut=
ive clamshell

>>> moves along the same cutting plane will undo each other. It fu=
rther suggests

>>> that any interesting sequence of clamshell moves must alternat=
e between the

>>> two possible long cut directions, meaning there is no choice i=
nvolved. 12

>>> clamshell moves will cycle back to the initial state.

>>>

>>> There is one other weird folding move where you open it in one=
direction and

>>> then fold the two halves back-to-back in a different direction=
. If you

>>> simply kept folding along the initial hinge, you'd simply =
have a restacking.

>>> When completed the other way, it's equivalent to a restack=
ing plus a

>>> clamshell, so I don't think it's useful though it is s=
omewhat interesting.

>>>

>>>

>>> Let's call these 8 folds interesting fold moves. Note that=
an interesting

>>> fold move always changes which axis the longer side of the puz=
zle is

>>> parallel with. Further note that both during unfold and fold a=
ll pieces are

>>> moved; it would be possible to have 8 of the pieces fixed duri=
ng an unfold

>>> and folding the other half 180 degrees but I think that it=E2=
=80=99s more intuitive

>>> that these moves fold both halves 90 degrees and performing th=
em with

>>> 180-degree folds might therefore lead to errors since the puzz=
le might get

>>> rotated differently. To illustrate a correct unfold without a =
puzzle: Put

>>> your palms together such that your thumbs point upward and you=
r fingers

>>> forward. Now turn your right hand 90 degrees clockwise and you=
r left hand 90

>>> degrees counterclockwise such that the normal to your palms po=
int up, your

>>> fingers point forward, your right thumb to the right and your =
left thumb to

>>> the left. That was what will later be called a Vx unfold and t=
he folds are

>>> simply reversed unfolds. (I might have used the word =E2=80=9C=
fold=E2=80=9D in two different

>>> ways but will try to use the term =E2=80=9Cfold move=E2=80=9D =
when referring to the move

>>> composed by an unfold and a fold rather than simply calling th=
ese moves

>>> =E2=80=9Cfolds=E2=80=9D.)

>>>

>>> To specify the unfold let's use V followed by one of x, y,=
z, x', y' and z'.

>>> The lowercase letter describes in which direction to unfold. V=
x means unfold

>>> in the direction of the positive x-axis and Vx' in the dir=
ection of the

>>> negative x-axis, if that makes any sense. I will try to explai=
n more

>>> precisely what I mean with the example Vx from the standard ro=
tation (it

>>> might also help to read the last sentences in the previous par=
agraph again).

>>> So, the puzzle is in the standard rotation and thus have the f=
orm 2x2x4 (x-,

>>> y- and z-thickness respectively). The first part (the unfoldin=
g) of the move

>>> specified with Vx is to unfold the puzzle in the x-direction, =
making it a

>>> 1x4x4 (note that the thickness in the x-direction is 1 after t=
he Vx unfold,

>>> which is no coincidence). There are two ways to do that; eithe=
r the sides of

>>> the pieces that are initially touching another piece (inside o=
f the puzzle

>>> in the x,z-plane and your palms in the hand example) are facin=
g up or down

>>> after the unfold. Let Vx be the unfold where these sides point=
in the

>>> direction of the positive x-axis (up) and Vx' the other on=
e where these

>>> sides point in the direction of the negative x-axis (down) aft=
er the unfold.

>>> Note that if the longer side of the puzzle is parallel to the =
z-axis only

>>> Vx, Vx', Vy and Vy' are possible. Now we need to speci=
fy how to fold the

>>> puzzle back to complete the folding move. Given an unfold, say=
Vx, there are

>>> only two ways to fold that are interesting (not turning the fo=
ld move into a

>>> restack mod(rot)) and you have to fold it perpendicular to the=
unfold to

>>> create an interesting fold move. So, if you start with the sta=
ndard rotation

>>> and do Vx you have a 1x4x4 that you have to fold into a 2x4x2.=
To

>>> distinguish the two possibilities, use + or - after the Vx. Le=
t Vx+ be the

>>> unfold Vx followed by the interesting fold that makes the side=
s that are

>>> initially touching another piece (before the unfold) touch ano=
ther piece

>>> after the fold move is completed and let Vx- be the other inte=
resting fold

>>> move that starts with the unfold Vx. (Thus, continuing with th=
e hand

>>> example, if you want to do a Vx+ first do the Vx unfold descri=
bed in the end

>>> of the previous paragraph and then fold your hands such that y=
our fingers

>>> point up, the normal to your palms point forward, the right pa=
lm is touching

>>> the right-hand fingers, the left palm is touching the left-han=
d fingers, the

>>> right thumb is pointing to the right and the left thumb is poi=
nting to the

>>> left). Note that the two halves of the puzzle always should be=
folded 90

>>> degrees each and you should never make a fold or unfold where =
you fold just

>>> one half 180-degrees (if you want to use my notation, that is)=
. Further note

>>> that Vx+ Sx mod(3rot) =3D Vx- and that Vx+ Vx+ =3D I which is =
equivalent to

>>> (Vx+)=E2=80=99 =3D Vx+ and this is true for all fold moves (no=
te that after a Vx+

>>> another Vx+ is always possible).

>>>

>>> The 2x2x2x2 in the MC4D software

>>>

>>> The notation above can also be applied to the 2x2x2x2 in the M=
C4D program.

>>> There, you are not allowed to do S or V moves but instead, you=
are allowed

>>> to do the [crtl]+[left-click] moves. This can easily be repres=
ented with

>>> notation similar to the above. Let=E2=80=99s use C (as in Cent=
ering) and one of x, y

>>> and z. For example, Cx would be to rotate the face in the posi=
tive

>>> x-direction aka the U face to the center. Thus, Cz' is sim=
ply

>>> [ctrl]+[left-click] on the L face and similarly for the other =
C moves. The

>>> O, U, D, F, B, R, L, K and A moves are performed in the same w=
ay as above

>>> so, for example Rx would be a [left-click] on the top-side of =
the right

>>> face. In this representation of the puzzle almost all moves ar=
e allowed; all

>>> U, D, F, B, R, L, K, O and C moves are possible regardless of =
rotation and

>>> only A moves (and of course the rightfully forbidden S and V m=
oves) are

>>> impossible regardless of rotation. Note that R and L moves in =
the software

>>> correspond to the same moves of the physical puzzle but this i=
s not

>>> generally true (I will come back to this later).

>>>

>>> Possible moves (so far) in the standard rotation

>>>

>>> In the standard rotation, the possible/allowed moves with the =
definitions

>>> above are:

>>>

>>> O moves, all of these are always possible in any state and rot=
ation of the

>>> puzzle since they are simply 3D-rotations.

>>>

>>> R, L moves, all of these as well since the puzzle has a right =
and left 2x2x2

>>> block in the standard rotation.

>>>

>>> U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x=
4 blocks with

>>> less symmetry than a 2x2x2 block.

>>>

>>> F, B moves, just Fy2 and By2 since the front and back are 2x1x=
4 blocks with

>>> less symmetry than a 2x2x2 block.

>>>

>>> K moves, all possible since this is a rotation of the center 2=
x2x2 block.

>>>

>>>

>>> I think the only legal 90 degree twists of the K face are thos=
e about the

>>> long axis. I believe this is what Christopher Locke was saying=
in this

>>> message. To see why there is no straightforward way to perform=
other 90

>>> degree twists, you only need to perform a 90 degree twist on a=
n outer (L/R)

>>> face and then reorient the whole puzzle along a different oute=
r axis. If the

>>> original twist was not about the new long axis, then there is =
clearly no

>>> straightforward way to undo that twist.

>>>

>>>

>>> A moves, only Az moves since this is two 2x2x1 blocks that hav=
e to be

>>> rotated together.

>>>

>>> S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined =
in the

>>> standard rotation.

>>>

>>> V moves, not Vz+ or Vz- since the definition doesn't give =
these meaning when

>>> the long side of the puzzle is parallel with the z-axis.

>>>

>>> Note that for example Fz2 is not allowed since this won't =
take you to a

>>> state of the puzzle. To allow more moves we need to extend the=
definitions

>>> (after the extension in the next paragraph all rotations and t=
wists (O, R,

>>> L, U, D, F, B, K, A) are possible in any rotation and only whi=
ch S and V

>>> moves are possible depend on the state and rotation of the puz=
zle).

>>>

>>>

>>> Extension of some definitions

>>>

>>> It's possible to make an extension that allows all O, R, L=
, U, D, F, B, K

>>> and A moves in any state. I will explain how this can be done =
in the

>>> standard rotation but it applies analogously to any other rota=
tion where the

>>> K face is an octahedron. First let's focus on U, D, F and =
B and because of

>>> the symmetry of the puzzle all of these are analogous so I wil=
l only explain

>>> one. The extension that makes all U moves possible (note that =
the U face is

>>> in the positive x-direction) is as follows: when making an U m=
ove first

>>> detach the 8 top pieces which gives you a 1x2x4 block, fold th=
is block into

>>> a 2x2x2 block in the positive x-direction (similar to the late=
r part of a

>>> Vx+ move from the standard rotation) such that the U face form=
an

>>> octahedron, rotate this 2x2x2 block around the specified axis =
(for example

>>> around the z-axis if you are doing an Uz move), reverse the fo=
ld you just

>>> did creating a 1x2x4 block again and reattach the block.

>>>

>>>

>>> I noticed something like this the other day but realized that =
it only seems

>>> to work for rotations along the long dimension (z in your exam=
ple). These

>>> are already easily accomplished by a simple rotation to put th=
e face in

>>> question on the end caps, followed by a double end-cap twist.<=
br>
>>>

>>> This is as far as I'm going to comment for the moment beca=
use the

>>> information gets very dense and I've been mulling and pick=
ing over your

>>> message for several days already. In short, I really like your=
attempt to

>>> provide a complete system of notation for discussing this puzz=
le and will be

>>> curious to hear your thoughts on my comments so far. I hope ot=
hers will

>>> chime in too.

>>>

>>> One final thought is that a real "acid test" of any =
notation system for this

>>> puzzle will be attempt to translate some algorithms from MC4D.=
I would most

>>> like to see a sequence that flips a single piece, like the sec=
ond 4-color

>>> series on this page of Roice's solution, or his pair of tw=
irled corners at

>>> the end of this page. One trick will be to minimize the number=
of

>>> whole-puzzle reorientations needed, but really any sequence th=
at works will

>>> be great evidence that the puzzles are equivalent. I suspect t=
hat this sort

>>> of exercise will never be practical because it will require to=
o many

>>> reorientations, and that entirely new methods will be needed t=
o actually

>>> solve this puzzle.

>>>

>>> Best,

>>> -Melinda

>>>

>>> The A moves can be done very similarly but after you have deta=
ched the two

>>> 2x2x1 blocks you don't fold them but instead you stack the=
m similar to a Sz

>>> move, creating a 2x2x2 block with the A face as an octahedron =
in the middle

>>> and then reverse the process after you have rotated the block =
as specified

>>> (for example around the negative y-axis if you are doing an Ay=
' move). Note

>>> that these extended moves are closely related to the normal mo=
ves and for

>>> example Ux =3D Ry Ly=E2=80=99 Kx Ly Ry=E2=80=99 in the standar=
d rotation and note that (Ry

>>> Ly=E2=80=99) mod(rot) =3D (Ly Ry=E2=80=99) mod(rot) =3D I (thi=
s applies to the other extended

>>> moves as well).

>>>

>>> If the cube is in the half-rotated state, where both the R and=
L faces are

>>> octahedra, you can extend the definitions very similarly. The =
only thing you

>>> have to change is how you fold the 2x4 blocks when performing =
a U, D, F or B

>>> move. Instead of folding the two 2x2 halves of the 2x4 into a =
2x2 you have

>>> to fold the end 1x2 block 180 degrees such that the face forms=
an

>>> octahedron.

>>>

>>> These moves might be a little bit harder to perform, to me esp=
ecially the A

>>> moves seems a bit awkward, so I don't know if it's goo=
d to use them or not.

>>> However, the A moves are not necessary if you allow Sz in the =
standard

>>> rotation (which you really should since Sz mod(rot) =3D I in t=
he standard

>>> rotation) and thus it might not be too bad to use this extende=
d version. The

>>> notation supports both variants so if you don=E2=80=99t want t=
o use these extended

>>> moves that shouldn=E2=80=99t be a problem. Note that, however,=
for example Ux

>>> (physical puzzle) !=3D Ux (virtual puzzle) where !=3D means =
=E2=80=9Cnot equal to=E2=80=9D (more

>>> about legal/illegal moves later).

>>>

>>> Generalisation of the notation

>>>

>>> Let=E2=80=99s generalise the notation to make it easier to use=
and to make it work

>>> for any n^4 cube in the MC4D software. Previously, we saw that=
(Rx)2 =3D Rx Rx

>>> and if we allow ourselves to rewrite this as (Rx)2 =3D Rx2 =3D=
Rxx =3D Rx Rx we

>>> see that the capital letter naturally can be distributed over =
the lowercase

>>> letters. We can make this more general and say that any capita=
l letter

>>> followed by several lowercase letters means the same thing as =
the capital

>>> letter distributed over the lowercase letters. Like Rxyz =3D R=
x Ry Rz and here

>>> R can be exchanged with any capital letter and xyz can be exch=
anged with any

>>> sequence of lowercase letters. We can also allow several capit=
al letters and

>>> one lowercase letter, for example RLx and let=E2=80=99s define=
this as RLx =3D Rx Lx

>>> so that the lowercase letter can be distributed over the capit=
al letters. We

>>> can also define a capital letter followed by =E2=80=98 (prime)=
like R=E2=80=99x =3D Rx=E2=80=99 and

>>> R=E2=80=99xy =3D Rx=E2=80=99y=E2=80=99 so the =E2=80=98 (prime=
) is distributed over the lowercase letters.

>>> Note that we don=E2=80=99t define a capital to any other power=
than -1 like this

>>> since for example R2x =3D RRx might seem like a good idea at f=
irst but it

>>> isn=E2=80=99t very useful since R2 and RR are the same lengths=
(and powers greater

>>> than two are seldom used) and we will see that we can define R=
2 in another

>>> way that generalises the notation to all n^4 cubes.

>>>

>>> Okay, let=E2=80=99s define R2 and similar moves now and have i=
n mind what moves we

>>> want to be possible for a n^4 cube. The moves that we cannot a=
chieve with

>>> the notation this far is twisting deeper slices. To match the =
notation with

>>> the controls of the MC4D software let R2x be the move similar =
to Rx but

>>> twisting the 2nd layer instead of the top one and similarly fo=
r other

>>> capital letters, numbers (up to n) and lowercase letters. Thus=
, R2x is

>>> performed as Rx but holding down the number 2 key. Just as in =
the program,

>>> when no number is specified 1 is assumed and you can combine s=
everal numbers

>>> like R12x to twist both the first and second layer. This notat=
ion does not

>>> apply to rotations (O) folding moves (V) and restacking moves =
(S) (I suppose

>>> you could redefine the S move using this deeper-slice-notation=
and use S1z

>>> as Sz+, S2z as Sz and S3z as Sz- but since these moves are onl=
y allowed for

>>> the physical 2x2x2x2 I think that the notation with + and =E2=
=80=93 is better since

>>> S followed by a lowercase letter without +/- always means spli=
tting the cube

>>> in a coordinate plane that way, not sure though so input would=
be great).

>>> The direction of the twist R3x should be the same as Rx meanin=
g that if Rx

>>> takes stickers belonging to K and move them to F, so should R3=
x, in

>>> accordance with the controls of the MC4D software. Note that f=
or a 3x3x3x3

>>> it=E2=80=99s true that R3z =3D Lz whereas R3x =3D Lx=E2=80=99 =
(note that R and L are the faces

>>> in the z-directions so because of the symmetry of the cube it =
will also be

>>> true that for example U3x =3D Bx whereas U3y=3DBy=E2=80=99).r>
>>>

>>> What about the case with several capital letters and several l=
owercase

>>> letters, for instance, RLxy? I see two natural definitions of =
this. Either,

>>> we could have RLxy =3D Rxy Lxy or we could define it as RLxy =
=3D RLx RLy. These

>>> are generally not the same (if you exchange R and L with any a=
llowed capital

>>> letter and similarly for x and y). I don=E2=80=99t know what i=
s best, what do you

>>> think? The situations I find this most useful in are RL=E2=80=
=99xy to do a rotation

>>> and RLxy as a twist. However, since R and L are opposite faces=
their

>>> operations commute which imply RL=E2=80=99xy (1st definition)=
=3D Rxy L=E2=80=99xy =3D Rx Ry

>>> Lx=E2=80=99 Ly=E2=80=99 =3D Rx Lx=E2=80=99 Ry Ly=E2=80=99 =3D =
RL=E2=80=99x RL=E2=80=99y =3D RL=E2=80=99xy (2nd definition) and similarly<=
br>
>>> for the other case with RLxy. Hopefully, we can find another u=
seful sequence

>>> of moves where this notation can be used with only one of the =
definitions

>>> and can thereby decide which definition to use. Personally, I =
feel like RLxy

>>> =3D RLx RLy is the more intuitive definition but I don=E2=80=
=99t have any good

>>> argument for this so I=E2=80=99ll leave the question open.

>>>

>>> For convenience, it might be good to be able to separate moves=
like Rxy and

>>> RLx from the basic moves Rx, Oy etc when speaking and writing.=
Let=E2=80=99s call

>>> the basic moves that only contain one capital letter and one l=
owercase

>>> letter (possibly a + or =E2=80=93, a =E2=80=98 (prime) and/or =
a number) simple moves (like

>>> Rx, L'y, Ux2 and D3y=E2=80=99) and the moves that contain =
more than one capital

>>> letter or more than one lowercase letter composed moves.

>>>

>>> More about inverses

>>>

>>> This list can obviously be made longer but here are some ident=
ities that are

>>> good to know and understand. Note that R, L, U, x, y and z bel=
ow just are

>>> examples, the following is true in general for non-folds (howe=
ver, S moves

>>> are fine).

>>>

>>> (P1 P2 =E2=80=A6 Pn)=E2=80=99 =3D Pn=E2=80=99 ... P2=E2=80=99 =
P1=E2=80=99 (Pi is an arbitrary permutation for

>>> i=3D1,2,=E2=80=A6n)

>>> (Rxy)=E2=80=99 =3D Ry=E2=80=99x=E2=80=99 =3D R=E2=80=99yx

>>> (RLx)=E2=80=99 =3D LRx=E2=80=99 =3D L=E2=80=99R=E2=80=99x

>>> (Sx+z)=E2=80=99 =3D Sz=E2=80=99 Sx=E2=80=99+ (just as an exa=
mple with restacking moves, note that

>>> the inverse doesn=E2=80=99t change the + or -)

>>> RLUx=E2=80=99y=E2=80=99z=E2=80=99=3DR=E2=80=99L=E2=80=99U=E2=
=80=99xyz (true for both definitions)

>>> (RLxy)=E2=80=99 =3D LRy=E2=80=99x=E2=80=99 =3D L=E2=80=99R=E2=
=80=99yx (true for both definitions)

>>>

>>> For V moves we have that: (Vx+)=E2=80=99 =3D Vx+ !=3D Vx=E2=
=80=99+ (!=3D means =E2=80=9Cnot equal to=E2=80=9D)

>>>

>>> Some important notes on legal/illegal moves

>>>

>>> Although there are a lot of moves possible with this notation =
we might not

>>> want to use them all. If we really want a 2x2x2x2 and not some=
thing else I

>>> think that we should try to stick to moves that are legal 2x2x=
2x2 moves as

>>> far as possible (note that I said legal moves and not permutat=
ions (a legal

>>> permutation can be made up of one or more legal moves)). Clari=
fication:

>>> cycling three of the edge-pieces of a Rubik=E2=80=99s cube is =
a legal permutation

>>> but not a legal move, a legal move is a rotation of the cube o=
r a twist of

>>> one of the layers. In this section I will only address simple =
moves and

>>> simply refer to them as moves (legal composed moves are moves =
composed by

>>> legal simple moves).

>>>

>>> I do believe that all moves allowed by my notation are legal p=
ermutations

>>> based on their periodicity (they have a period of 2 or 4 and a=
re all even

>>> permutations of the pieces). So, which of them correspond to l=
egal 2x2x2x2

>>> moves? The O moves are obviously legal moves since they are eq=
ual to the

>>> identity mod(rot). The same goes for restacking (S) (with or w=
ithout +/-) in

>>> the direction of the longest side of the puzzle (Sz, Sz+ and S=
z- in the

>>> standard rotation) since these are rotations and half-rotation=
s that don=E2=80=99t

>>> change the state of the puzzle. Restacking in the other direct=
ions and fold

>>> moves (V) are however not legal moves since they are made of 8=
2-cycles and

>>> change the state of the puzzle (note that they, however, are l=
egal

>>> permutations). The rest of the moves (R, L, F, B, U, D, K and =
A) can be

>>> divided into two sets: (1) the moves where you rotate a 2x2x2 =
block with an

>>> octahedron inside and (2) the moves where you rotate a 2x2x2 b=
lock without

>>> an octahedron inside. A move belonging to (2) is always legal.=
We can see

>>> this by observing what a Rx does with the pieces in the standa=
rd rotation

>>> with just K forming an octahedron. The stickers move in 6 4-cy=
cles and if

>>> the puzzle is solved the U and D faces still looks solved afte=
r the move. A

>>> move belonging to set (1) is legal either if it=E2=80=99s an 1=
80-degree twist or if

>>> it=E2=80=99s a rotation around the axis parallel with the long=
est side of the puzzle

>>> (the z-axis in the standard rotation). Quite interestingly the=
se are exactly

>>> the moves that don=E2=80=99t mix up the R and L stickers with =
the rest in the

>>> standard rotation. I think I know a way to prove that no legal=
2x2x2x2 move

>>> can mix up these stickers with the rest and this has to do wit=
h the fact

>>> that these stickers form an inverted octahedron (with the corn=
ers pointing

>>> outward) instead of a normal octahedron (let=E2=80=99s call th=
is hypothesis * for

>>> now). Note that all legal twists (R, L, F, B, U, D, K and A) o=
f the physical

>>> puzzle correspond to the same twist in the MC4D software.

>>>

>>> So, what moves should we add to the set of legal moves be able=
to get to

>>> every state of the 2x2x2x2? I think that we should add the res=
tacking moves

>>> and folding moves since Melinda has already found a pretty sho=
rt sequence of

>>> those moves to make a rotation that changes which colours are =
on the R and L

>>> faces. That sequence, starting from the standard rotation, is:=
Oy Sx+z Vy+

>>> Ozy=E2=80=99 Vy+ Ozy=E2=80=99 Vy+ =3D Oy Sx+z (Vy+ Ozy=E2=80=
=99)3 mod(3rot) =3D I mod(rot) (hopefully I

>>> got that right). What I have found (which I mentioned previous=
ly) is that

>>> (from the standard rotation): RLx Ly2 Sy By2 Ly2 RLx=E2=80=99 =
=3D Sz- and since this

>>> is equivalent with Sy =3D Ly2 RLx=E2=80=99 Sz- RLx Ly2 By2 the=
restacking moves that

>>> are not legal moves are not very complicated permutations and =
therefore I

>>> think that we can accept them since they help us mix up the R =
and L stickers

>>> with other faces. In a sense, the folding moves are =E2=80=9Cm=
ore illegal=E2=80=9D since

>>> they cannot be composed by the legal moves (according to hypot=
hesis *). This

>>> is also true for the illegal moves belonging to set (1) discus=
sed above.

>>> However, since the folding moves is probably easier to perform=
and is enough

>>> to reach every state of the 2x2x2x2 I think that we should use=
them and not

>>> the illegal moves belonging to (1). Note, once again, that all=
moves

>>> described by the notation are legal permutations (even the one=
s that I just

>>> a few words ago referred to as illegal moves) so if you wish y=
ou can use all

>>> of them and still only reach legal 2x2x2x2 states. However (in=
a strict

>>> sense) one could argue that you are not solving the 2x2x2x2 if=
you use

>>> illegal moves. If you only use illegal moves to compose rotati=
ons (that is,

>>> create a permutation including illegal moves that are equal to=
I mod(rot))

>>> and not actually using the illegal moves as twists I would cla=
ssify that as

>>> solving a 2x2x2x2. What do you think about this?

>>>

>>> What moves to use?

>>>

>>> Here=E2=80=99s a short list of the simple moves that I think s=
hould be used for the

>>> physical 2x2x2x2. Note that this is just my thoughts and you m=
ay use the

>>> notation to describe any move that it can describe if you wish=
to. The

>>> following list assumes that the puzzle is in the standard rota=
tion but is

>>> analogous for other representations where the K face is an oct=
ahedron.

>>>

>>> O, all since they are I mod(rot),

>>> R, L, all since they are legal (note Rx (physical puzzle) =3D =
Rx (virtual

>>> 2x2x2x2)),

>>> U, D, only x2 since these are the only legal easy-to-perform m=
oves,

>>> F, B, only y2 since these are the only legal easy-to-perform m=
oves,

>>> K, A, only z, z=E2=80=99 and z2 since these are the only legal=
easy-to-perform

>>> moves,

>>> S, at least z, z+ and z- since these are equal to I mod(rot),<=
br>
>>> S, possibly x and y since these help us perform rotations and =
is easy to

>>> compose (not necessary to reach all states and not legal thoug=
h),

>>> V, all 8 allowed by the rotation of the puzzle (at least one i=
s necessary to

>>> reach all states and if you allow one the others are easy to a=
chieve

>>> anyway).

>>>

>>> If you start with the standard rotation and then perform Sz+ t=
he following

>>> applies instead (this applies analogously to any other rotatio=
n where the R

>>> and L faces are octahedra).

>>>

>>> O, all since they are I mod(rot),

>>> R, L, only z, z=E2=80=99 and z2 since these are the only legal=
easy-to-perform

>>> moves,

>>> U, D, only x2 since these are the only legal easy-to-perform m=
oves,

>>> F, B, only y2 since these are the only legal easy-to-perform m=
oves,

>>> K, A, at least z, z=E2=80=99 and z2, possibly all (since they =
are legal) although

>>> some might be hard to perform.

>>> S, V, same as above.

>>>

>>> Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- !=3D I=
mod(rot) (!=3D

>>> for not equal) which implies that the Ux2 move when R and L ar=
e octahedra is

>>> different from the Ux2 move when K is an octahedron. (Actually=
, the sequence

>>> above is equal to Uy2).

>>>

>>> Regarding virtual n^4 cubes I think that all O, C, R, L, U, D,=
F, B, K and A

>>> moves should be used since they are all legal and really the o=
nly thing you

>>> need (left-clicking on an edge or corner piece in the computer=
program can

>>> be described quite easily with the notation, for example, Kzy2=
is

>>> left-clicking on the top-front edge piece on the K face).

>>>

>>> I hope this was possible to follow and understand. Feel free t=
o ask

>>> questions about the notation if you find anything ambiguous.r>
>>>

>>> Best regards,

>>> Joel Karlsson

>>>

>>> Den 4 maj 2017 12:01 fm skrev "Melinda Green ailto:melinda@superliminal.com" target=3D"_blank">melinda@superliminal.com<=
/a>

>>> [4D_Cubing]" <
com" target=3D"_blank">4D_Cubing@yahoogroups.com>:

>>>

>>>

>>>

>>> Thanks for the correction. A couple of things: First, when ass=
embling one

>>> piece at a time, I'd say there is only 1 way to place the =
first piece, not

>>> 24. Otherwise you'd have to say that the 1x1x1 puzzle has =
24 states. I

>>> understand that this may be conventional, but to me, that just=
sounds silly.

>>>

>>> Second, I have the feeling that the difference between the &qu=
ot;two

>>> representations" you describe is simply one of those half=
-rotations I showed

>>> in the video. In the normal solved state there is only one com=
plete

>>> octahedron in the very center, and in the half-rotated state t=
here is one in

>>> the middle of each half of the "inverted" form. I co=
nsider them to be the

>>> same solved state.

>>>

>>> -Melinda

>>>

>>>

>>> On 5/3/2017 2:39 PM, Joel Karlsson son97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com [4D_Cubing]=


>>> wrote:

>>>

>>> Horrible typo... It seems like I made some typos in my email r=
egarding the

>>> state count. It should of course be 16!12^16/(6*192) and NOTr>
>>> 12!16^12/(6*192). However, I did calculate the correct number =
when comparing

>>> with previous results so the actual derivation was correct.>
>>>

>>> Something of interest is that the physical pieces can be assem=
bled in

>>> 16!24*12^15 ways since there are 16 pieces, the first one can =
be oriented in

>>> 24 ways and the remaining can be oriented in 12 ways (since a =
corner with 3

>>> colours never touch a corner with just one colour). Dividing w=
ith 6 to get a

>>> single orbit still gives a factor 2*192 higher than the actual=
count rather

>>> than 192. This shows that every state in the MC4D representati=
on has 2

>>> representations in the physical puzzle. These two representati=
ons must be

>>> the previously discussed, that the two halves either have the =
same color on

>>> the outermost corners or the innermost (forming an octahedron)=
when the

>>> puzzle is solved and thus both are complete representations of=
the 2x2x2x2.

>>>

>>> Best regards,

>>> Joel Karlsson

>>>

>>> Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" <=
joelkarlsson9=
7@gmail.com
>:

>>>

>>> I am no expert on group theory, so to better understand what t=
wists are

>>> legal I read through the part of Kamack and Keane's The Ru=
bik Tesseract

>>> about orienting the corners. Since all even permutations are a=
llowed the

>>> easiest way to check if a twist is legal might be to:

>>> 1. Check that the twist is an even permutation, that is: the s=
ame twist can

>>> be done by performing an even number of piece swaps (2-cycles)=
.

>>> 2. Check the periodicity of the twist. If A^k=3DI (A^k meaning=
performing the

>>> twist k times and I (the identity) representing the permutatio=
n of doing

>>> nothing) and k is not divisible by 3 the twist A definitely do=
esn't violate

>>> the restriction of the orientations since kx mod 3 =3D 0 and k=
mod 3 !=3D 0

>>> implies x mod 3 =3D 0 meaning that the change of the total ori=
entation x for

>>> the twist A mod 3 is 0 (which precisely is the restriction of =
legal twists;

>>> that they must preserve the orientation mod 3).

>>>

>>> For instance, this implies that the restacking moves are legal=
2x2x2x2 moves

>>> since both are composed of 8 2-cycles and both can be performe=
d twice (note

>>> that 2 is not divisible by 3) to obtain the identity.

>>>

>>> Note that 1 and 2 are sufficient to check if a twist is legal =
but only 1 is

>>> necessary; there can indeed exist a twist violating 2 that sti=
ll is legal

>>> and in that case, I believe that we might have to study the or=
ientation

>>> changes for that specific twist in more detail. However, if a =
twist can be

>>> composed by other legal twists it is, of course, legal as well=
.

>>>

>>> Best regards,

>>> Joel

>>>

>>> 2017-04-29 1:04 GMT+02:00 Melinda Green nda@superliminal.com" target=3D"_blank">melinda@superliminal.com [4D_Cu=
bing]

>>> <ank">4D_Cubing@yahoogroups.com>:

>>>>

>>>>

>>>> First off, thanks everyone for the helpful and encouraging=
feedback!

>>>> Thanks Joel for showing us that there are 6 orbits in the =
2^4 and for your

>>>> rederivation of the state count. And thanks Matt and Roice=
for pointing out

>>>> the importance of the inverted views. It looks so strange =
in that

>>>> configuration that I always want to get back to a normal v=
iew as quickly as

>>>> possible, but it does seem equally valid, and as you'v=
e shown, it can be

>>>> helpful for more than just finding short sequences.

>>>>

>>>> I don't understand Matt's "pinwheel" con=
figuration, but I will point out

>>>> that all that is needed to create your twin interior octah=
edra is a single

>>>> half-rotation like I showed in the video at 5:29. The two =
main halves do end

>>>> up being mirror images of each other on the visible outsid=
e like he

>>>> described. Whether it's the pinwheel or the half-rotat=
ed version that's

>>>> correct, I'm not sure that it's a bummer that the =
solved state is not at all

>>>> obvious, so long as we can operate it in my original confi=
guration and

>>>> ignore the fact that the outer faces touch. That would jus=
t mean that the

>>>> "correct" view is evidence that that the more un=
derstandable view is

>>>> legitimate.

>>>>

>>>> I'm going to try to make a snapable V3 which should al=
low the pieces to be

>>>> more easily taken apart and reassembled into other forms. =
Shapeways does

>>>> offer a single, clear translucent plastic that they call &=
quot;Frosted Detail",

>>>> and another called "Transparent Acrylic", but I =
don't think that any sort of

>>>> transparent stickers will help us, especially since this t=
hing is chock full

>>>> of magnets. The easiest way to let you see into the two he=
mispheres would be

>>>> to simply truncate the pointy tips of the stickers. That a=
lready happens a

>>>> little bit due to the way I've rounded the edges. Here=
is a close-up of a

>>>> half-rotation in which you can see that the inner yellow a=
nd white faces are

>>>> solved. Your suggestion of little mapping dots on the corn=
ers also works,

>>>> but just opening the existing window further would work mo=
re directly.

>>>>

>>>> -Melinda

>>>>

>>>>

>>>> On 4/28/2017 2:15 PM, Roice Nelson 3@gmail.com" target=3D"_blank">roice3@gmail.com [4D_Cubing] wrote:

>>>>

>>>> I agree with Don's arguments about adjacent sticker co=
lors needing to be

>>>> next to each other. I think this can be turned into an ac=
curate 2^4 with

>>>> coloring changes, so I agree with Joel too :)

>>>>

>>>> To help me think about it, I started adding a new projecti=
on option for

>>>> spherical puzzles to MagicTile, which takes the two hemisp=
heres of a puzzle

>>>> and maps them to two disks with identified boundaries conn=
ected at a point,

>>>> just like a physical "global chess" game I have.=
Melinda's puzzle is a lot

>>>> like this up a dimension, so think about two disjoint ball=
s, each

>>>> representing a hemisphere of the 2^4, each a "subcube=
" of Melinda's puzzle.

>>>> The two boundaries of the balls are identified with each o=
ther and as you

>>>> roll one around, the other half rolls around so that ident=
ified points

>>>> connect up. We need to have the same restriction on Melin=
da's puzzle.

>>>>

>>>> In the pristine state then, I think it'd be nice to ha=
ve an internal

>>>> (hidden), solid colored octahedron on each half. The othe=
r 6 faces should

>>>> all have equal colors split between each hemisphere, 4 sti=
ckers on each

>>>> half. You should be able to reorient the two subcubes to =
make a half

>>>> octahedron of any color on each subcube. I just saw Matt&=
#39;s email and

>>>> picture, and it looks like we were going down the same tho=
ught path. I

>>>> think with recoloring (mirroring some of the current piece=
colorings)

>>>> though, the windmill's can be avoided (?)

>>>>

>>>> [...] After staring/thinking a bit more, the coloring Matt=
came up with is

>>>> right-on if you want to put a solid color at the center of=
each hemisphere.

>>>> His comment about the "mirrored" pieces on each =
side helped me understand

>>>> better. 3 of the stickers are mirrored and the 4th is the=
hidden color

>>>> (different on each side for a given pair of "mirrored=
" pieces). All faces

>>>> behave identically as well, as they should. It's a li=
ttle bit of a bummer

>>>> that it doesn't look very pristine in the pristine sta=
te, but it does look

>>>> like it should work as a 2^4.

>>>>

>>>> I wonder if there might be some adjustments to be made whe=
n shapeways

>>>> allows printing translucent as a color :)

>>>>

>>>> [...] Sorry for all the streaming, but I wanted to share o=
ne more thought.

>>>> I now completely agree with Joel/Matt about it behaving as=
a 2^4, even with

>>>> the original coloring. You just need to consider the corn=
er colors of the

>>>> two subcubes (pink/purple near the end of the video) as be=
ing a window into

>>>> the interior of the piece. The other colors match up as d=
esired. (Sorry if

>>>> folks already understood this after their emails and I'=
;m just catching up!)

>>>>

>>>> In fact, you could alter the coloring of the pieces slight=
ly so that the

>>>> behavior was similar with the inverted coloring. At the c=
orners where 3

>>>> colors meet on each piece, you could put a little circle o=
f color of the

>>>> opposite 4th color. In Matt's windmill coloring then,=
you'd be able to see

>>>> all four colors of a piece, like you can with some of the =
pieces on

>>>> Melinda's original coloring. And again you'd cons=
ider the color circles a

>>>> window to the interior that did not require the same match=
ing constraints

>>>> between the subcubes.

>>>>

>>>> I'm looking forward to having one of these :)

>>>>

>>>> Happy Friday everyone,

>>>> Roice

>>>>

>>>> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson mailto:joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com=


>>>> [4D_Cubing] <m" target=3D"_blank">4D_Cubing@yahoogroups.com> wrote:

>>>>>

>>>>>

>>>>> Seems like there was a slight misunderstanding. I mean=
t that you need to

>>>>> be able to twist one of the faces and in MC4D the mos=
t natural choice is

>>>>> the center face. In your physical puzzle you can achie=
ve this type of twist

>>>>> by twisting the two subcubes although this is indeed a=
twist of the subcubes

>>>>> themselves and not the center face, however, this is s=
till the same type of

>>>>> twist just around another face.

>>>>>

>>>>> If the magnets are that allowing the 2x2x2x2 is obviou=
sly a subgroup of

>>>>> this puzzle. Hopefully the restrictions will be quite =
natural and only some

>>>>> "strange" moves would be illegal. Regarding =
the "families of states" (aka

>>>>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earl=
ier all allowed twists

>>>>> preserves the parity of the pieces, meaning that only =
half of the

>>>>> permutations you can achieve by disassembling and reas=
sembling can be

>>>>> reached through legal moves. Because of some geometric=
al properties of the

>>>>> 2x2x2x2 and its twists, which would take some time to =
discuss in detail

>>>>> here, the orientation of the stickers mod 3 are preser=
ved, meaning that the

>>>>> last corner only can be oriented in one third of the n=
umber of orientations

>>>>> for the other corners. This gives a total number of or=
bits of 2x3=3D6. To

>>>>> check this result let's use this information to ca=
lculate all the possible

>>>>> states of the 2x2x2x2; if there were no restrictions w=
e would have 16! for

>>>>> permuting the pieces (16 pieces) and 12^16 for orient=
ing them (12

>>>>> orientations for each corner). If we now take into acc=
ount that there are 6

>>>>> equally sized orbits this gets us to 12!16^12/6. Howev=
er, we should also

>>>>> note that the orientation of the puzzle as a hole is n=
ot set by some kind of

>>>>> centerpieces and thus we need to devide with the numbe=
r of orientations of a

>>>>> 4D cube if we want all our states to be separated with=
twists and not only

>>>>> rotations of the hole thing. The number of ways to ori=
ent a 4D cube in space

>>>>> (only allowing rotations and not mirroring) is 8x6x4=
=3D192 giving a total of

>>>>> 12!16^12/(6*192) states which is indeed the same numbe=
r that for example

>>>>> David Smith arrived at during his calculations. Theref=
ore, when determining

>>>>> whether or not a twist on your puzzle is legal or not =
it is sufficient and

>>>>> necessary to confirm that the twist is an even permuta=
tion of the pieces and

>>>>> preserves the orientation of stickers mod 3.

>>>>>

>>>>> Best regards,

>>>>> Joel

>>>>>

>>>>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green href=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlim=
inal.com


>>>>> [4D_Cubing]" <ogroups.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:

>>>>>

>>>>>

>>>>>

>>>>> The new arrangement of magnets allows every valid orie=
ntation of pieces.

>>>>> The only invalid ones are those where the diagonal lin=
es cutting each cube's

>>>>> face cross each other rather than coincide. In other w=
ords, you can assemble

>>>>> the puzzle in all ways that preserve the overall diamo=
nd/harlequin pattern.

>>>>> Just about every move you can think of on the whole pu=
zzle is valid though

>>>>> there are definitely invalid moves that the magnets al=
low. The most obvious

>>>>> invalid move is twisting of a single end cap.

>>>>>

>>>>> I think your description of the center face is not cor=
rect though. Twists

>>>>> of the outer faces cause twists "through" th=
e center face, not "of" that

>>>>> face. Twists of the outer faces are twists of those fa=
ces themselves because

>>>>> they are the ones not changing, just like the center a=
nd outer faces of MC4D

>>>>> when you twist the center face. The only direct twist =
of the center face

>>>>> that this puzzle allows is a 90 degree twist about the=
outer axis. That

>>>>> happens when you simultaneously twist both end caps in=
the same direction.

>>>>>

>>>>> Yes, it's quite straightforward reorienting the wh=
ole puzzle to put any

>>>>> of the four axes on the outside. This is a very nice i=
mprovement over the

>>>>> first version and should make it much easier to solve.=
You may be right that

>>>>> we just need to find the right way to think about the =
outside faces. I'll

>>>>> leave it to the math geniuses on the list to figure th=
at out.

>>>>>

>>>>> -Melinda

>>>>>

>>>>>

>>>>>

>>>>> On 4/27/2017 10:31 AM, Joel Karlsson :joelkarlsson97@gmail.com" target=3D"_blank">joelkarlsson97@gmail.com [=
4D_Cubing]

>>>>> wrote:

>>>>>

>>>>>

>>>>> Hi Melinda,

>>>>>

>>>>> I do not agree with the criticism regarding the white =
and yellow stickers

>>>>> touching each other, this could simply be an effect of=
the different

>>>>> representations of the puzzle. To really figure out if=
this indeed is a

>>>>> representation of a 2x2x2x2 we need to look at the pos=
sible moves (twists

>>>>> and rotations) and figure out the equivalent moves in =
the MC4D software.

>>>>> From the MC4D software, it's easy to understand t=
hat the only moves required

>>>>> are free twists of one of the faces (that is, only twi=
sting the center face

>>>>> in the standard perspective projection in MC4D) and 4D=
rotations swapping

>>>>> which face is in the center (ctrl-clicking in MC4D). T=
he first is possible

>>>>> in your physical puzzle by rotating the white and yell=
ow subcubes (from here

>>>>> on I use subcube to refer to the two halves of the puz=
zle and the colours of

>>>>> the subcubes to refer to the "outer colours"=
). The second is possible if

>>>>> it's possible to reach a solved state with any two=
colours on the subcubes

>>>>> that still allow you to perform the previously mention=
ed twists. This seems

>>>>> to be the case from your demonstration and is indeed t=
rue if the magnets

>>>>> allow the simple twists regardless of the colours of t=
he subcubes. Thus, it

>>>>> is possible to let your puzzle be a representation of =
a 2x2x2x2, however, it

>>>>> might require that some moves that the magnets allow a=
ren't used.

>>>>>

>>>>> Best regards,

>>>>> Joel

>>>>>

>>>>> 2017-04-27 3:09 GMT+02:00 Melinda Green lto:melinda@superliminal.com" target=3D"_blank">melinda@superliminal.com>

>>>>> [4D_Cubing] <
s.com" target=3D"_blank">4D_Cubing@yahoogroups.com>:

>>>>>>

>>>>>>

>>>>>> Dear Cubists,

>>>>>>

>>>>>> I've finished version 2 of my physical puzzle =
and uploaded a video of it

>>>>>> here:

>>>>>> 8kJKLo" target=3D"_blank">https://www.youtube.com/watch?v=3DzqftZ8kJKL=
o


>>>>>> Again, please don't share these videos outside=
this group as their

>>>>>> purpose is just to get your feedback. I'll eve=
ntually replace them with a

>>>>>> public video.

>>>>>>

>>>>>> Here is an extra math puzzle that I bet you folks =
can answer: How many

>>>>>> families of states does this puzzle have? In other=
words, if disassembled

>>>>>> and reassembled in any random configuration the ma=
gnets allow, what are the

>>>>>> odds that it can be solved? This has practical imp=
lications if all such

>>>>>> configurations are solvable because it would provi=
de a very easy way to

>>>>>> fully scramble the puzzle.

>>>>>>

>>>>>> And finally, a bit of fun: A relatively new friend=
of mine and new list

>>>>>> member, Marc Ringuette, got excited enough to make=
his own version. He built

>>>>>> it from EPP foam and colored tape, and used honey =
instead of magnets to hold

>>>>>> it together. Check it out here:

>>>>>> be.jpg" target=3D"_blank">http://superliminal.com/cube/dessert_cube.jp=
g
I don't know how practical a

>>>>>> solution this is but it sure looks delicious! Welc=
ome Marc!

>>>>>>

>>>>>> -Melinda

>>>>>>

>>>>>

>>>>>

>>>>>

>>>>>

>>>>

>>>

>>>

>>>

>

> ------------------------------------

> Posted by: Joel Karlsson <m" target=3D"_blank">joelkarlsson97@gmail.com>

> ------------------------------------

>

>

> ------------------------------------

>

> Yahoo Groups Links

>

>

>

>






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