Thread: "3^4 one 4C left to orient"

------=_NextPart_000_000B_01CB92EF.BC28CA60
Content-Type: text/plain;
charset="us-ascii"
Content-Transfer-Encoding: 7bit

I'm having trouble proving something for the 4D cubes. In either case, ALL
but one 4C piece is perfectly solved. I can see that there are at least 4
possible states for this remaining 4C piece (the solved state and three
states where 2 independent pairs of stickers are permuted).and although I
haven't seen it, I can't prove that there are more possible states. Anyone
have an argument as to why you can't have all stickers solved except a
3-cycle of stickers on the remaining 4C piece?



I see in 5 dimensions (and higher) you can have a single 3-cycle on the
remaining 5C piece (and higher).



I also had similar trouble with the original 3D cubes: trying to prove why
you can't have everything solved except one 3C piece (where its stickers
would be in an unsolved 3-cycle). My sketchy argument there had to define
what it means to 'permute 3C pieces without disorienting them'.I defined it
as yellow/white corner stickers must be facing the yellow/white face (yellow
is opposite white on my cube). I then argued that a simple twist always
orients the SUM of all 3C pieces by a multiple of 360 degrees. I don't know
if a similar argument will work in 4D.



--

Andy


------=_NextPart_000_000B_01CB92EF.BC28CA60
Content-Type: text/html;
charset="us-ascii"
Content-Transfer-Encoding: quoted-printable

osoft-com:office:office" xmlns:w=3D"urn:schemas-microsoft-com:office:word" =
xmlns:m=3D"http://schemas.microsoft.com/office/2004/12/omml" xmlns=3D"http:=
//www.w3.org/TR/REC-html40">=3D"text/html; charset=3Dus-ascii">oft Word 12 (filtered medium)">nk=3Dpurple>

Andy,
what we need there is to map orientations of 4C piece to the group Z_3=3D=
{-1,0,1} in such way that: (1) changing of the code of some piece orientati=
on during a twist depends only on the piece position (not on its orientatio=
n before twist!) and is additive (i.e. m'=3Dm+f(t,p) mod 3, where m is code=
of orientation "before", m' is code of orientation of the same piece "afte=
r", t is twist description, p is piece position) and (2) that sum(f(t,p)) f=
or all positions p is zero for every twist.
I think that the following construction will work:
Let faces of the cube be {a,b,c,d,A,B,C,D} ("A" is opposite to "a" and so=
on). Divide this faces to 2 sets (tubes): one set is {a,b,A,B} and another=
is {c,d,C,D}. Call a piece well-oriented if its stickers that initially be=
long to one tube (say, "a" and "B") are laying in faces that also belong to=
one tube (e.g. sticker "a" is in the face "C" and sticker "B" is in the fa=
ce "D"). This orientation has code=3D0. If sticker in not well-oriented do =
the following. Enumerate faces of the cube that contains this sticker now s=
uch that first is "a" or "A", second is "b" or "B" and two last faces give =
positive (right-handed,...) orientation of the vertex (like abcd, aBDC, Abc=
D etc.) Take the sticker S of the piece that lays in the first face and loo=
k for sticker S' that initially was in the same tube with S. Now it can lay=
in second, third or fourth faces wrt our enumeration. The code of the orie=
ntation will be 0,1 and -1 respectively.=20
The rest is to prove properties (1) and (2) for this mapping. I didn't do=
it yet but hope that they both are true.
Even if this construction works, it's not easy to generalize it to other =
figures - shallow-cut simplex and 120-cell. But somehow I think that the in=
variant shoud be true for them too.

And I think that couple of years ago there was long post about the same p=
roblem for 120-cell. Don't remember when was it and how was the author.

Andrey


--- In 4D_Cubing@yahoogroups.com, "Andrew Gould" wrote:
>
> I'm having trouble proving something for the 4D cubes. In either case, A=
LL
> but one 4C piece is perfectly solved. I can see that there are at least =
4
> possible states for this remaining 4C piece (the solved state and three
> states where 2 independent pairs of stickers are permuted).and although I
> haven't seen it, I can't prove that there are more possible states. Anyo=
ne
> have an argument as to why you can't have all stickers solved except a
> 3-cycle of stickers on the remaining 4C piece?
>=20
>=20=20
>=20
> I see in 5 dimensions (and higher) you can have a single 3-cycle on the
> remaining 5C piece (and higher).
>=20
>=20=20
>=20
> I also had similar trouble with the original 3D cubes: trying to prove w=
hy
> you can't have everything solved except one 3C piece (where its stickers
> would be in an unsolved 3-cycle). My sketchy argument there had to defin=
e
> what it means to 'permute 3C pieces without disorienting them'.I defined =
it
> as yellow/white corner stickers must be facing the yellow/white face (yel=
low
> is opposite white on my cube). I then argued that a simple twist always
> orients the SUM of all 3C pieces by a multiple of 360 degrees. I don't k=
now
> if a similar argument will work in 4D.
>=20
>=20=20
>=20
> --
>=20
> Andy
>

Here is the message about 120cell: http://games.groups.yahoo.com/group/4D_C=
ubing/message/658
Less than 2 years ago, but already almost in the middle of the archive!


--- In 4D_Cubing@yahoogroups.com, "Andrey" wrote:
>
> Andy,
> what we need there is to map orientations of 4C piece to the group Z_3=
=3D{-1,0,1} in such way that: (1) changing of the code of some piece orient=
ation during a twist depends only on the piece position (not on its orienta=
tion before twist!) and is additive (i.e. m'=3Dm+f(t,p) mod 3, where m is c=
ode of orientation "before", m' is code of orientation of the same piece "a=
fter", t is twist description, p is piece position) and (2) that sum(f(t,p)=
) for all positions p is zero for every twist.
> I think that the following construction will work:
> Let faces of the cube be {a,b,c,d,A,B,C,D} ("A" is opposite to "a" and =
so on). Divide this faces to 2 sets (tubes): one set is {a,b,A,B} and anoth=
er is {c,d,C,D}. Call a piece well-oriented if its stickers that initially =
belong to one tube (say, "a" and "B") are laying in faces that also belong =
to one tube (e.g. sticker "a" is in the face "C" and sticker "B" is in the =
face "D"). This orientation has code=3D0. If sticker in not well-oriented d=
o the following. Enumerate faces of the cube that contains this sticker now=
such that first is "a" or "A", second is "b" or "B" and two last faces giv=
e positive (right-handed,...) orientation of the vertex (like abcd, aBDC, A=
bcD etc.) Take the sticker S of the piece that lays in the first face and l=
ook for sticker S' that initially was in the same tube with S. Now it can l=
ay in second, third or fourth faces wrt our enumeration. The code of the or=
ientation will be 0,1 and -1 respectively.=20
> The rest is to prove properties (1) and (2) for this mapping. I didn't =
do it yet but hope that they both are true.
> Even if this construction works, it's not easy to generalize it to othe=
r figures - shallow-cut simplex and 120-cell. But somehow I think that the =
invariant shoud be true for them too.
>=20
> And I think that couple of years ago there was long post about the same=
problem for 120-cell. Don't remember when was it and how was the author.
>=20
> Andrey
>=20
>=20
> --- In 4D_Cubing@yahoogroups.com, "Andrew Gould" wrote:
> >
> > I'm having trouble proving something for the 4D cubes. In either case,=
ALL
> > but one 4C piece is perfectly solved. I can see that there are at leas=
t 4
> > possible states for this remaining 4C piece (the solved state and three
> > states where 2 independent pairs of stickers are permuted).and although=
I
> > haven't seen it, I can't prove that there are more possible states. An=
yone
> > have an argument as to why you can't have all stickers solved except a
> > 3-cycle of stickers on the remaining 4C piece?
> >=20
> >=20=20
> >=20
> > I see in 5 dimensions (and higher) you can have a single 3-cycle on the
> > remaining 5C piece (and higher).
> >=20
> >=20=20
> >=20
> > I also had similar trouble with the original 3D cubes: trying to prove=
why
> > you can't have everything solved except one 3C piece (where its sticker=
s
> > would be in an unsolved 3-cycle). My sketchy argument there had to def=
ine
> > what it means to 'permute 3C pieces without disorienting them'.I define=
d it
> > as yellow/white corner stickers must be facing the yellow/white face (y=
ellow
> > is opposite white on my cube). I then argued that a simple twist alway=
s
> > orients the SUM of all 3C pieces by a multiple of 360 degrees. I don't=
know
> > if a similar argument will work in 4D.
> >=20
> >=20=20
> >=20
> > --
> >=20
> > Andy
> >
>

On 12/6/2010 12:10 AM, Andrey wrote:
> Here is the message about 120cell: http://games.groups.yahoo.com/group/4D_Cubing/message/658
> Less than 2 years ago, but already almost in the middle of the archive!

That's because you joined us this year and really kicked the anthill
with all of your amazing products and other accomplishments. It also
reminds me that I miss having David on the list so it's great to have
another math wiz to help bridge the gap.

Thanks for the link and the memories,
-Melinda

--0015174c11ca7f67e30496b9f0d0
Content-Type: text/plain; charset=windows-1252
Content-Transfer-Encoding: quoted-printable

Hey guys,

Although I joined this group when I solved the 3^4 awhile ago, I hardly eve=
r
send out an email; I only read them :)
However in this case, I think I can reference a link:
http://www.gravitation3d.com/david/Magic120Cell_Permutations.pdf (might be
what Andrey was referring to)
This is David Smith's paper on the number of permutations to the
Magic120Cell. In it, he has what seems to be a very rigorous proof of the
exact things you mention. Although I haven't taken the time to really think
about what he wrote, a brief glance leads me to believe his proof is indeed
complete. Perhaps this can help :)

-Matt Galla


PS: Yes, that is me credited in the paper, although in hindsight, my
algorithms were so far from optimal that it is embarrassing. Nevertheless,
they complete the proof.


On Mon, Dec 6, 2010 at 2:02 AM, Andrey wrote:

>
>
> Andy,
> what we need there is to map orientations of 4C piece to the group
> Z_3=3D{-1,0,1} in such way that: (1) changing of the code of some piece
> orientation during a twist depends only on the piece position (not on its
> orientation before twist!) and is additive (i.e. m'=3Dm+f(t,p) mod 3, whe=
re m
> is code of orientation "before", m' is code of orientation of the same pi=
ece
> "after", t is twist description, p is piece position) and (2) that
> sum(f(t,p)) for all positions p is zero for every twist.
> I think that the following construction will work:
> Let faces of the cube be {a,b,c,d,A,B,C,D} ("A" is opposite to "a" and so
> on). Divide this faces to 2 sets (tubes): one set is {a,b,A,B} and anothe=
r
> is {c,d,C,D}. Call a piece well-oriented if its stickers that initially
> belong to one tube (say, "a" and "B") are laying in faces that also belon=
g
> to one tube (e.g. sticker "a" is in the face "C" and sticker "B" is in th=
e
> face "D"). This orientation has code=3D0. If sticker in not well-oriented=
do
> the following. Enumerate faces of the cube that contains this sticker now
> such that first is "a" or "A", second is "b" or "B" and two last faces gi=
ve
> positive (right-handed,...) orientation of the vertex (like abcd, aBDC, A=
bcD
> etc.) Take the sticker S of the piece that lays in the first face and loo=
k
> for sticker S' that initially was in the same tube with S. Now it can lay=
in
> second, third or fourth faces wrt our enumeration. The code of the
> orientation will be 0,1 and -1 respectively.
> The rest is to prove properties (1) and (2) for this mapping. I didn't do
> it yet but hope that they both are true.
> Even if this construction works, it's not easy to generalize it to other
> figures - shallow-cut simplex and 120-cell. But somehow I think that the
> invariant shoud be true for them too.
>
> And I think that couple of years ago there was long post about the same
> problem for 120-cell. Don't remember when was it and how was the author.
>
> Andrey
>
> --- In 4D_Cubing@yahoogroups.com <4D_Cubing%40yahoogroups.com>, "Andrew
> Gould" wrote:
> >
> > I'm having trouble proving something for the 4D cubes. In either case,
> ALL
> > but one 4C piece is perfectly solved. I can see that there are at least=
4
> > possible states for this remaining 4C piece (the solved state and three
> > states where 2 independent pairs of stickers are permuted).and although=
I
> > haven't seen it, I can't prove that there are more possible states.
> Anyone
> > have an argument as to why you can't have all stickers solved except a
> > 3-cycle of stickers on the remaining 4C piece?
> >
> >
> >
> > I see in 5 dimensions (and higher) you can have a single 3-cycle on the
> > remaining 5C piece (and higher).
> >
> >
> >
> > I also had similar trouble with the original 3D cubes: trying to prove
> why
> > you can't have everything solved except one 3C piece (where its sticker=
s
> > would be in an unsolved 3-cycle). My sketchy argument there had to defi=
ne
> > what it means to 'permute 3C pieces without disorienting them'.I define=
d
> it
> > as yellow/white corner stickers must be facing the yellow/white face
> (yellow
> > is opposite white on my cube). I then argued that a simple twist always
> > orients the SUM of all 3C pieces by a multiple of 360 degrees. I don't
> know
> > if a similar argument will work in 4D.
> >
> >
> >
> > --
> >
> > Andy
> >
>
>=20
>

--0015174c11ca7f67e30496b9f0d0
Content-Type: text/html; charset=windows-1252
Content-Transfer-Encoding: quoted-printable

=A0

------=_NextPart_000_0013_01CB95AD.230AEAF0
Content-Type: multipart/alternative;
boundary="----=_NextPart_001_0014_01CB95AD.230AEAF0"

------=_NextPart_001_0014_01CB95AD.230AEAF0
Content-Type: text/plain;
charset="us-ascii"
Content-Transfer-Encoding: 7bit

Thanks Andrey,

I never realized that if 'a' and 'b' start in the same tube, then they
stay in the same tube in the klein 4 subgroup:
{{a,b,c,d},{b,a,d,c},{c,d,a,b},{d,c,b,a}}. Interesting and neat.



Your construction seems to be true for (1): either your code stays the
same for all 4C pieces, or else for the 4C pieces that do get twisted: the
code of the 4C pieces in the CUBE positions with C (or c) before D (or d)
does that opposite of the code of the 4C pieces in the CUBE positions with D
(or d) before C (or c). That is, if you travel between 4C pieces via the
1-dimensional edges only, every other 4C piece (that gets twisted) increases
it's code by 1(mod3)--the code of the remaining twisted pieces decreases by
1(mod3). This is true for 90-degree 3D face twists, 120-degree twists, and
180 degree twists.and even my 90-degree 2D face twists. Since the number of
pieces that increase in code = the number of pieces that decrease in code,
sum(f(t,p)) = 0(mod3). Hence it's true for (2).



I'm eager to check the links out when I find time.



--

Andy





From: 4D_Cubing@yahoogroups.com [mailto:4D_Cubing@yahoogroups.com] On Behalf
Of Andrey
Sent: Monday, December 06, 2010 2:10
To: 4D_Cubing@yahoogroups.com
Subject: [MC4D] Re: 3^4 one 4C left to orient





Here is the message about 120cell:
http://games.groups.yahoo.com/group/4D_Cubing/message/658
Less than 2 years ago, but already almost in the middle of the archive!

--- In 4D_Cubing@yahoogroups.com ,
"Andrey" wrote:
>
> Andy,
> what we need there is to map orientations of 4C piece to the group
Z_3={-1,0,1} in such way that: (1) changing of the code of some piece
orientation during a twist depends only on the piece position (not on its
orientation before twist!) and is additive (i.e. m'=m+f(t,p) mod 3, where m
is code of orientation "before", m' is code of orientation of the same piece
"after", t is twist description, p is piece position) and (2) that
sum(f(t,p)) for all positions p is zero for every twist.
> I think that the following construction will work:
> Let faces of the cube be {a,b,c,d,A,B,C,D} ("A" is opposite to "a" and so
on). Divide this faces to 2 sets (tubes): one set is {a,b,A,B} and another
is {c,d,C,D}. Call a piece well-oriented if its stickers that initially
belong to one tube (say, "a" and "B") are laying in faces that also belong
to one tube (e.g. sticker "a" is in the face "C" and sticker "B" is in the
face "D"). This orientation has code=0. If sticker in not well-oriented do
the following. Enumerate faces of the cube that contains this sticker now
such that first is "a" or "A", second is "b" or "B" and two last faces give
positive (right-handed,...) orientation of the vertex (like abcd, aBDC, AbcD
etc.) Take the sticker S of the piece that lays in the first face and look
for sticker S' that initially was in the same tube with S. Now it can lay in
second, third or fourth faces wrt our enumeration. The code of the
orientation will be 0,1 and -1 respectively.
> The rest is to prove properties (1) and (2) for this mapping. I didn't do
it yet but hope that they both are true.
> Even if this construction works, it's not easy to generalize it to other
figures - shallow-cut simplex and 120-cell. But somehow I think that the
invariant shoud be true for them too.
>
> And I think that couple of years ago there was long post about the same
problem for 120-cell. Don't remember when was it and how was the author.
>
> Andrey
>
>
> --- In 4D_Cubing@yahoogroups.com ,
"Andrew Gould" wrote:
> >
> > I'm having trouble proving something for the 4D cubes. In either case,
ALL
> > but one 4C piece is perfectly solved. I can see that there are at least
4
> > possible states for this remaining 4C piece (the solved state and three
> > states where 2 independent pairs of stickers are permuted).and although
I
> > haven't seen it, I can't prove that there are more possible states.
Anyone
> > have an argument as to why you can't have all stickers solved except a
> > 3-cycle of stickers on the remaining 4C piece?
> >
> >
> >
> > I see in 5 dimensions (and higher) you can have a single 3-cycle on the
> > remaining 5C piece (and higher).
> >
> >
> >
> > I also had similar trouble with the original 3D cubes: trying to prove
why
> > you can't have everything solved except one 3C piece (where its stickers
> > would be in an unsolved 3-cycle). My sketchy argument there had to
define
> > what it means to 'permute 3C pieces without disorienting them'.I defined
it
> > as yellow/white corner stickers must be facing the yellow/white face
(yellow
> > is opposite white on my cube). I then argued that a simple twist always
> > orients the SUM of all 3C pieces by a multiple of 360 degrees. I don't
know
> > if a similar argument will work in 4D.
> >
> >
> >
> > --
> >
> > Andy
> >
>




------=_NextPart_001_0014_01CB95AD.230AEAF0
Content-Type: text/html;
charset="us-ascii"
Content-Transfer-Encoding: quoted-printable

osoft-com:office:office" xmlns:w=3D"urn:schemas-microsoft-com:office:word" =
xmlns:m=3D"http://schemas.microsoft.com/office/2004/12/omml" xmlns=3D"http:=
//www.w3.org/TR/REC-html40">=3D"text/html; charset=3Dus-ascii">oft Word 12 (filtered medium)"> link=3D"#1E66AE" vlink=3D"#1E66AE">

ss=3DMsoNormal> 

  =