Thread: "MagicTile solves"

From: "Eduard" <baumann@mcnet.ch>
Date: Wed, 02 May 2012 13:34:09 -0000
Subject: MagicTile solves



A torus-klein pair:
MT eucl torus {4,4} 8c E 0-1-0 ( 55 twists)
MT eucl klein {4,4} 8c E 0-1-0 (277 twists)

A skew
MT skew {4,4-5} 25c E 0-1-0 (537 twists)

There is always time for these small and funny MT puzzles. Melinda and Roic=
e add your solutions to get a first "rowspan=3D3 or 4" in wiki ;-)




From: "Eduard" <baumann@mcnet.ch>
Date: Wed, 02 May 2012 17:46:06 -0000
Subject: MagicTile solves



2x2x2x2x2
=20
(31!/2)(60^31) =3D
=20
545356551753081970586352633891109632137647267774464000000000000000000000000=
00000000000000
=20
approx. =3D 5.4 x 10^88


Should the 31 be 32?=20

Mine is=20

32!/2*(5!/2)^(32) =3D

104708457936591738352579705707093049370428275412697088000000000000000000000=
000000000000000000

approx. =3D 1.0 x 10^92




From: "Eduard" <baumann@mcnet.ch>
Date: Sat, 05 May 2012 10:15:19 -0000
Subject: MagicTile solves



Three new MagicTile solves more:

MT skew {4,4|5} 25 v 0:0:1 1221 twists
MT skew {4,4|6} 36 f 0:0:1 539 twists
MT skew {4,4|6} 36 e 0:1:0 1091 twists

For the last one here: I was left with one single inversion ... then a big =
AHA !

No macros for these three solves.

Still no "rowspan 3 or 4" im Wiki (=3D 3 or 4 persons which have solved the=
same puzzle).




From: "Eduard" <baumann@mcnet.ch>
Date: Sat, 5 May 2012 14:34:11 -0500
Subject: MagicTile solves



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Wow Ed, you've really been solving a lot of these! I think you should add
a new table to the wiki to track the number of puzzles that you and others
have solved. It would be interesting to know your grand total at this
point :)

Best,
Roice

On Sat, May 5, 2012 at 5:15 AM, Eduard wrote:

> Three new MagicTile solves more:
>
> MT skew {4,4|5} 25 v 0:0:1 1221 twists
> MT skew {4,4|6} 36 f 0:0:1 539 twists
> MT skew {4,4|6} 36 e 0:1:0 1091 twists
>
> For the last one here: I was left with one single inversion ... then a big
> AHA !
>
> No macros for these three solves.
>
> Still no "rowspan 3 or 4" im Wiki (= 3 or 4 persons which have solved the
> same puzzle).
>
>

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Wow Ed, you've really been solving a lot of these!=A0 I think you =
should add a new table to the wiki to track the number of puzzles that you =
and others have solved.=A0 It would be interesting to know your grand total=
at this point :)

=A0
Best,
Roice

uote">On Sat, May 5, 2012 at 5:15 AM, Eduard <=3D"mailto:baumann@mcnet.ch" target=3D"_blank">baumann@mcnet.ch>an> wrote:

color:rgb(204,204,204);border-left-width:1px;border-left-style:solid" class=
=3D"gmail_quote">Three new MagicTile solves more:



MT skew {4,4|5} 25 v 0:0:1 =A0 =A0 =A01221 twists

MT skew {4,4|6} 36 f 0:0:1 =A0 =A0 =A0539 twists

MT skew {4,4|6} 36 e 0:1:0 =A0 =A0 =A01091 twists



For the last one here: I was left with one single inversion ... then a big =
AHA !



No macros for these three solves.



Still no "rowspan 3 or 4" im Wiki (=3D 3 or 4 persons which have =
solved the same puzzle).





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From: "Eduard" <baumann@mcnet.ch>
Date: Sun, 06 May 2012 14:25:46 -0000
Subject: MagicTile solves



To create the first "rowspan=3D3" (three persons for one puzzle) I have sol=
ved the following two MT

MT ell sph [3,5} 8c f 0:0:.8 + v .8:0:0 with 1494 twists
MT proj hemidodeka 6c v 1.5:0:0 with 205 twists

For me solving the puzzle is 95% of the pleasure, optimizing the number of =
twists is only 5% of the pleasure.

The actual list of MT vs2 performers is
1. Nan Ma 83 solves
2. Eduard Baumann 38 solves [only inversed digits ;-) ]=20
3. Roice Nelson 2 solves
4. Andrey Astrelin 1 solve




From: "Eduard" <baumann@mcnet.ch>
Date: Mon, 07 May 2012 12:42:59 -0000
Subject: MagicTile solves



New solve for MT skew {4,4|7} 49c v 0:0:1 with 3455 twists.




From: "Eduard" <baumann@mcnet.ch>
Date: Mon, 7 May 2012 13:19:31 -0500
Subject: MagicTile solves



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Hi David,

Thanks for helping me realize the state space reduction in the 3D case is *not
quite* 48, and similarly for other dimensions. I missed this
previously, but should have considered it when quoting the cube lovers post
about there being 12 equivalent 1 quarter turn moves (rather than 48). It
feels like a somewhat subtle point. It also makes calculating the
*exact*number of states which are "the same" in this discussion more
difficult!

seeya,
Roice

On Sun, May 6, 2012 at 9:44 PM, David Vanderschel wrote:

> Still out of phase. Significant redundancy between my last
> post and Roice's. :-(
>
> It dawns on me that the problem the cube20.org folks were
> addressing is deeper than the one David Smith was addressing
> (and which I was defending) and more like what Melinda was
> probably driving at. I.e., given two arrangements that are
> visually different when the pile is viewed in standard
> orientation and based on the sticker IDs inferred from
> initial state as in Smith's calculations, when can those two
> still be regarded as the 'same' based on a symmetry of the
> pile? This is where conjugation by a symmetry enters the
> picture, and that can be viewed in a sense which remaps axis
> IDs (along with the corresponding sticker 'color' pairs).
> As may be seen from the '94 Cube Lovers article, the precise
> issue gets very messy and David Smith was not trying to
> address this more complex version. What Smith was
> calculating is much more straightforward, is still
> meaningful, and is more easily understood. The deeper view
> reduces the "real size" by a factor less than n!*2^n (but
> not quite n!*2^n), which is actually rather minor compared
> to the sizes of the numbers Smith was computing. In my
> view, the main point of his efforts is not the precise
> values (which are relatively useless) but to get some feel
> for just how immense these numbers are. They are
> incomprehensibly large with or without the factor of n!*2^n.
> The little ol' order 3 3D problem was close enough to being
> tractible that a factor of nearly 48 actually did make a big
> difference in the ultimate analysis there.
>
> One thing that should be pointed out is that, if you admit
> mirroring transformations, then it does create problems for
> the purpose of counting distinct arrangements. The
> problem is that there exist (a small minority of)
> arrangements which possess mirror symmetry, so that the very
> same arrangement would occur for two different symmetry
> transformations of the pile. Accounting for the number of
> distinct positions possessing such symmetry is not at all
> easy; so the basic counting problem is much easier if you do
> not admit reflecting transformations for achieving standard
> orientation. But the cube20.org folks were not interested
> so much in _counting_ arrangements as in analyzing relations
> between explicit instances of state, so the extra complexity
> that gained another factor of 2 was well worth it for them.
>
> Regards,
> David V.
>

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Hi David,
=A0
Thanks for helping me realize the st=
ate space reduction in the 3D case is=A0not quite 48, and =
similarly for other dimensions.=A0 I missed this previously,=A0but should h=
ave considered it when quoting the cube lovers post about there being 12 eq=
uivalent 1 quarter turn moves (rather than 48).=A0 It feels like a somewhat=
subtle point.=A0 It also makes calculating the exact numb=
er of states which are "the same" in this discussion more difficu=
lt!

=A0
seeya,
Roice

quote">On Sun, May 6, 2012 at 9:44 PM, David Vanderschel =
<DvdS@austin.rr.=
com
>
wrote:

color:rgb(204,204,204);border-left-width:1px;border-left-style:solid" class=
=3D"gmail_quote">Still out of phase. =A0Significant redundancy between my l=
ast


post and Roice's. =A0:-(



It dawns on me that the problem the "_blank">cube20.org folks were

addressing is deeper than the one David Smith was addressing

(and which I was defending) and more like what Melinda was

probably driving at. =A0I.e., given two arrangements that are

visually different when the pile is viewed in standard

orientation and based on the sticker IDs inferred from

initial state as in Smith's calculations, when can those two

still be regarded as the 'same' based on a symmetry of the

pile? =A0This is where conjugation by a symmetry enters the

picture, and that can be viewed in a sense which remaps axis

IDs (along with the corresponding sticker 'color' pairs).

As may be seen from the '94 Cube Lovers article, the precise

issue gets very messy and David Smith was not trying to

address this more complex version. =A0What Smith was

calculating is much more straightforward, is still

meaningful, and is more easily understood. =A0The deeper view

reduces the "real size" by a factor less than n!*2^n (but

not quite n!*2^n), which is actually rather minor compared

to the sizes of the numbers Smith was computing. =A0In my

view, the main point of his efforts is not the precise

values (which are relatively useless) but to get some feel

for just how immense these numbers are. =A0They are

incomprehensibly large with or without the factor of n!*2^n.

The little ol' order 3 3D problem was close enough to being

tractible that a factor of nearly 48 actually did make a big

difference in the ultimate analysis there.



One thing that should be pointed out is that, if you admit

mirroring transformations, then it does create problems for

the purpose of counting distinct arrangements. =A0The

problem is that there exist (a small minority of)

arrangements which possess mirror symmetry, so that the very

same arrangement would occur for two different symmetry

transformations of the pile. =A0Accounting for the number of

distinct positions possessing such symmetry is not at all

easy; so the basic counting problem is much easier if you do

not admit reflecting transformations for achieving standard

orientation. =A0But the cub=
e20.org
folks were not interested

so much in _counting_ arrangements as in analyzing relations

between explicit instances of state, so the extra complexity

that gained another factor of 2 was well worth it for them.



Regards,

=A0David V.



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From: "Eduard" <baumann@mcnet.ch>
Date: Wed, 09 May 2012 00:40:26 -0000
Subject: MagicTile solves



My MT solve no 40 and 41.

MT IRP {4,6}c 18 E 0:1:0 (187 twists)
MT IRP {4,6}c 18 V 0:0:1 (697 twists)
Avoid commutators with more than 1 common element!




From: "Eduard" <baumann@mcnet.ch>
Date: Wed, 9 May 2012 10:56:52 +0200
Subject: MagicTile solves



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Congratulations: Very interesting and absolutely original!

=20

I have also made a model of the compound of 5 tetrahedras with cardboard. I=
t's a "hollow" version with small pentagonal windows. I have the other hand=
edness than your puzzle.

=20

Dkwan wrote (2009):

"With regards to solving this puzzle, I find it visually very confusing, bu=
t I have worked out the following solve order with commutators that may not=
be optimal:
Corners (20 3-sticker pieces): [1,1]
Centers (60 pieces): [5,1] (I feel like this is pretty bad, but I didn't fi=
nd a shorter one)
Corner-Edges (60 pieces): [3,1]"

=20

I also have detected the three pieces: corners (3 stickers), centres and co=
rner-edges (both 2 stickers).

And the three commutators with three different distances.

=20

2009, hence three years ago.

Kind regards
Ed
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>




style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman">Congratulations: Very interesting and absolutely=20
original!"urn:schemas-microsoft-com:office:office" />
>


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman"> 


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman">I have also made a model of the compound of 5 tetr=
ahedras=20
with cardboard. It=92s a =93hollow=94 version with small pentagonal windows=
. I have=20
the other handedness than your puzzle.


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman"> 


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman">Dkwan wrote (2009):>


style=3D"FONT-FAMILY: Verdana; COLOR: #666666; FONT-SIZE: 9pt; mso-ansi-lan=
guage: EN-GB"=20
lang=3DEN-GB>"With regards to solving this puzzle, I find it visually very=
=20
confusing, but I have worked out the following solve order with commutators=
that=20
may not be optimal:
Corners (20 3-sticker pieces): [1,1]
Centers (60=
=20
pieces): [5,1] (I feel like this is pretty bad, but I didn't find a shorter=
=20
one)
Corner-Edges (60 pieces): [3,1]"


style=3D"FONT-FAMILY: Verdana; COLOR: #666666; FONT-SIZE: 9pt; mso-ansi-lan=
guage: EN-GB"=20
lang=3DEN-GB> 


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman">I also have detected the three pieces: corners (3=
=20
stickers), centres and corner-edges (both 2=20
stickers).


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman">And the three commutators with three different=20
distances.


style=3D"mso-ansi-language: EN-GB" lang=3DEN-GB>face=3D"Times New Roman"> 


style=3D"FONT-FAMILY: 'Times New Roman'; FONT-SIZE: 12pt; mso-ansi-language=
: EN-GB; mso-fareast-font-family: 'Times New Roman'; mso-fareast-language: =
DE; mso-bidi-language: AR-SA"=20
lang=3DEN-GB>2009, hence three years ago.

style=3D"FONT-FAMILY: 'Times New Roman'; FONT-SIZE: 12pt; mso-ansi-language=
: EN-GB; mso-fareast-font-family: 'Times New Roman'; mso-fareast-language: =
DE; mso-bidi-language: AR-SA"=20
lang=3DEN-GB> 

style=3D"FONT-FAMILY: 'Times New Roman'; FONT-SIZE: 12pt; mso-ansi-language=
: EN-GB; mso-fareast-font-family: 'Times New Roman'; mso-fareast-language: =
DE; mso-bidi-language: AR-SA"=20
lang=3DEN-GB>Kind regards

style=3D"FONT-FAMILY: 'Times New Roman'; FONT-SIZE: 12pt; mso-ansi-language=
: EN-GB; mso-fareast-font-family: 'Times New Roman'; mso-fareast-language: =
DE; mso-bidi-language: AR-SA"=20
lang=3DEN-GB>Ed
Y>

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From: "Eduard" <baumann@mcnet.ch>
Date: Wed, 09 May 2012 22:40:17 -0000
Subject: MagicTile solves



Oufff!
I needed 20919 twists for MT irp {4,6} 12 F 0:0:1.
Na Ma made it in fewer than 400 twists.




From: "Eduard" <baumann@mcnet.ch>
Date: Wed, 09 May 2012 15:59:56 -0700
Subject: MagicTile solves



Well this one was worth the effort because it is perhaps the closest IRP
analog to Rubik's cube. Or maybe we need one somewhere between F0:0:1
and F0:2:0 to be the true analog? Congratulations either way!

-Melinda

On 5/9/2012 3:40 PM, Eduard wrote:
> Oufff!
> I needed 20919 twists for MT irp {4,6} 12 F 0:0:1.
> Na Ma made it in fewer than 400 twists.




From: Melinda Green <melinda@superliminal.com>
Date: Wed, 9 May 2012 23:11:14 -0500
Subject: Re: [MC4D] MagicTile solves



Melinda wrote:
>On 5/7/2012 6:34 PM, David Vanderschel wrote:
>>Melinda wrote:
>>>It sounds like we are getting closer to closure.

>>>>Does this mean that you now understand why you must
>>>>maintain the pairs of opposing colors when reassigning
>>>>sticker colors? With all my effort at trying to explain
>>>>it for you, I was hoping for an "I see now!" I was also
>>>>hoping that you would realize that it is easier to talk
>>>>about conjugation by a symmetry than to get involved in
>>>>talking about remapping sticker colors, which are
>>>>actually somewhat tangential to the real problem.

>I'm not sure that I feel much closer to that "ah ha" moment
>but I do feel as if we are converging on a story that will
>feel right for all of us.

Unfortunately, I can tell from your later questions that you
have yet to get it.

>If conjugation is the operation that allows us to compose
>all the symmetries into one state space, then it does feel
>like a useful way to describe what we are doing.

I don't think there is any "if" to it. However, I am not
sure that I can interpret your "compose all the symmetries
into one state space" meaningfully. In particular, we are
not composing symmetries. Without reducing to the
equivalence classes of states related by conjugation by a
symmetry, we have a perfectly meaningful state space; and I
don't really understand why you are fighting it. Why do you
suppose Dan Hoey titled his article "The _real_ size of cube
space"? It is because the approach of regarding states as
distinct just by sticker color arrangement was already well
established even back in '94. He was "injecting the
complication" of grouping those states which were related by
conjugation by a symmetry, and he realized that that was not
the accepted approach. His observation became relevant to
those who were actually trying to analyze relationships
between _particular_ states for the conventional 3D puzzle
because the reduction actually brought the problem into a
range that was computationally feasible. When we are just
interested in counting distinct states, that is not our
issue. You get just as good a sense of how many states
there are without the much more complex additional reduction
by a factor somewhat less than n!*2^n.

I think the important thing about the distinction is that,
when we reduce to equivalence classes of the (first pass)
distinct states with the conjugation approach, we are
focusing more on the nature of the _process_ (aka permutation)
which gives rise to a state rather than to the state itself.
The result of that process depends on the initial
orientation of the pile.

>>> The main MC4D page claims to give the exact counts for
>>> the 3D and 4D puzzles. Is that not true? If not, then I
>>> need to change it.

>>>>It is true. Also the counts that David Smith came up
>>>>with for 5D are intended to be exact. (I am not ruling
>>>>out the possibility of an error.) As I said, the
>>>>problem that Eric and David were solving is much more
>>>>straightforward and admits a precise answer. The
>>>>problem only arises when you start trying to get the
>>>>number of equivalence classes of the distinct states
>>>>they counted that are related through conjugation by a
>>>>symmetry, which is the complication that it would seem
>>>>you were trying to inject. That would result in a
>>>>reduction of Eric's and David's numbers by a factor
>>>>close to (but less than) n!*2^n. But it is not
>>>>practical to do that in an exact sense.

>I am not trying to inject a complication, rather I am trying
>to account for what appears to be an error in our claim. I
>certainly don't want to count two states as unique when they
>share all the same properties other than handedness. I
>understand that attempting to account for this one rare
>symmetry won't change the state count by much and that maybe
>it is currently impractical to attempt to do that.

Melinda, it is not an error! The configurations are truly
different in the sense that no amount of reorientation will
cause a matchup of sticker colors all around. There is no
unavoidable reason to want to regard the assignment of
sticker colors as being mutable.

>This may be covered in previous messages, but would you
>please explain how you derive the n!*2^n and why it is not
>(n-1)! as I had thought?

Melinda, I am losing patience on this one. Brandon tried to
explain it in his recent post. Nan mentioned it in that old
one I cited. It comes down to the fact that many of the
remappings of sticker colors you are proposing are just not
physical - the puzzles can't do that! In particular, the
two sticker colors that are on the same axis in initial
state must always be on the same axis no matter how you
otherwise remap the stickers. Thus what is really possible
for remapping colors comes down to a permutation of the axes
and possible swapping of the two colors on an axis. E.g.,
on a standard Rubik's cube, yellow/white and blue/green are
so paired. Thus you could not swap yellow with blue unless
you also swapped white and green. You could also swap just
blue and green. The n in the formulae Roice and I were
tossing about is the dimension of the puzzle. So, with
that understanding, I thought you were saying (2n-1)!, where
2n is the number of sticker colors. (Not sure why you took
out one factor of 2n. I guess you were thinking of one
color as being a reference with respect to which the others
were being permuted.)

I will make one more brief attempt at the end; but you
should really try to reread earlier postings in this thread
more carefully. It has been explained.

>>You have already worked out a simple example giving
>>representative members of the equivalence classes for the
>>case of the 2D puzzle with mirroring twists allowed. I
>>would hope you would remember the previous occasion on which
>>some folks on the list tried to explain the relationship of
>>your equivalence classes to conjugation by a symmetry:
>>http://games.groups.yahoo.com/group/4D_Cubing/message/1843
>>Perhaps that previous discussion will make better sense to
>>you now.

>A little bit. Nan's link to
>http://kociemba.org/math/symmetries.htm on that message is
>definitely what I am talking about. I think that the fact
>that MC2D appears to require reflections to operate at all
>probably confused the issue more than clarifying, because
>the real issue applies to all of our puzzles, not just to
>that one.

I don't think that allowing mirroring twists makes any
difference at all with respect to the problems of
enumerating the distinct states or the equivalence classes
of these distinct states under conjugation by a symmetry.

>I definitely do not get the bit about how using axes and
>color pairs is better than thinking about color
>equivalencies but it is good enough if the result is the
>same.

It is important, because, if you fail to maintain the opposing
color pairs, then you do _not_ get the same result.

More generally, I regard the whole issue of remapping
colors as a red herring. If you understand conjugation
by a symmetry, then the recoloring is just an obvious side
effect which is not worth getting worked up about. It is
true; but it is not what is important. The actual colors
themselves were always unimportant. What you should
actually regard as identifying a sticker 'color' is the
direction it faces in initial state - not some particular
hue that is being used to render it for visualization
purposes.

>I am still a little troubled by the "exact" values that you
>guys seem to prefer. MC2D is indeed a great example of the
>problem and I just can't see how we can really say that
>there are 24 states in that puzzle that I think David S
>claimed in Wikipedia. Looking at my state graph on the MC2D
>page there are clearly only 8 that make any real sense to
>me.

It is really quite simple. The state of this particular
puzzle is characterized solely by the permutation of the 4
corner cubies, and any permutation of the 4 cubies is
possible. Thus there are 4! different arrangements. This
is a perfectly legitimate way to count distinct states. You
have observed that lots of them are not essentially
different. E.g., you would regard as being essentially the
same any two states for which just two adjacent corners have
been transposed without regard to which pair that is. That
is an additional complication by which you are grouping the
states into equivalence classes; and, yes, the relation is
conjugation by a symmetry.

The relevant conjugator for the example is rotation by a
multiple of 90 degrees in the plane of the puzzle. So start
with "swap the top 2". Then, if I rotate a quarter turn
clockwise, swap the top 2, and rotate back, I will have
achieved "swap the left 2". Those two _different_
permutations of the cubies are related by conjugation by a
90 degree rotation. It really is a different pair of cubies
that get swapped in the two cases. It is perfectly
legitimate to regard the two states as being different
without worrying about their relationship by conjugation.
Melinda, whether you wish to recognize it or not, you really
are trying to unnecessarily complicate the problem of
counting distinct puzzle states. And, as may be seen from
the "real size" article, the complication is by no means
trivial.

Let me go back once more to a more complex puzzle and some
non-trivial twist sequence. Do that sequence starting with
the pile in initial state and you get one particular
arrangement. Now get another identical puzzle in initial
state, but, without twisting any slices, first reorient the
whole pile so that it is no longer in initial state. Now
perform the same twist sequence again. You get a new
arrangement which is not the same as the arrangement on the
first puzzle. You cannot (in general) make them match up
sticker-for-sticker just by reorienting one. However, your
sticker color remapping approach will make them match up
with appropriate relative orientations. But there is no
mystery about the required sticker remapping!: It is
precisely that which was induced by the initial
reorientation of the second puzzle before performing the
twist sequence. So there is no need to talk about it.
Furthermore, it is clear that a reorientation of the pile
cannot change which pairs of sticker colors occur facing in
opposite directions on the same axis. Other than that
restriction, lots of sticker color mappings are possible via
the orientation step preceding the twist sequence. Indeed,
that number is n!*2^n.

Mirroring transforms for the conjugator are perfectly
legitimate even if they are not allowed for the puzzle,
because the conjugate will not induce any mirroring if none
of the twists do. (Indeed, you can actually practice
conjugation by a reflection on a 3D cube by watching what
you are doing in a mirror. Left/right and up/down do not
change, but front/back switches as well as the directional
sense of all twists (always looking towards the twisted
slice from its side of the cube).)

Regards,
David V.




From: "Eduard" <baumann@mcnet.ch>
Date: Sat, 12 May 2012 22:36:48 -0000
Subject: MagicTile solves



I have solved MT skew {4,4|6} 36 v 0:0:1 with 643 twists.

Level of the pictures in the solver list:

Is the level now okay?
Nan did chose 2 levels lower. I have raised his picture two levels.

One level lower would also be possible (differentiating the turning axes). =
Or do
we want only one axis representing all axes?




From: "Eduard" <baumann@mcnet.ch>
Date: Sun, 13 May 2012 10:16:59 -0000
Subject: MagicTile solves



I have solved

(1) MT proj hemi octa e 0:1:0 with 17 twists, EASY
(2) MT proj hemi icosa e 0:1:0 with 67 twists, EASY
(3) MT proj hemi cube e 0:1:0 with 39 twists, not so easy




From: "Eduard" <baumann@mcnet.ch>
Date: Fri, 11 May 2012 21:48:49 -0500
Subject: MagicTile solves



Andrew Gould wrote (about me):
>He claimed that an even length cubie permutation cycle (of
>3C pieces) implies 3 even length permutation cycles of
>stickers. Counterexample: a 2-cycle of the 3C pieces, but
>a single 6-cycle of their stickers.

Andy is absolutely correct. I had figured out long ago,
using the proof he provides below, based on the effect of
quarter turn twists, the fact that the sticker permutation
parity was equal to the cubie permutation parity. For some
reason, when I thought to provide a proof for the fact in my
post, I did not remember how I had done it before and I came
up with the incomplete proof based on the permutation cycles
in the whole pile's actual state. I did not think about it
enough because I knew I was getting the right result. :-(

(The proof using the state cycles can be resuscitated by
observing that the length of any sticker cycle corresponding
to a cubie cycle must be the cubie cycle length (for 3 of
them) or 3 times the cubie cycle length (for just 1), but it
is not worth pursuing this approach because the proof based
incrementally on the twists that generate the state is much
easier. (The tricky part is proving that you cannot get a
sticker cycle of length twice that of the cubie cycle.))

>I believe you would still be able to arrive at the last
>sentence of that paragraph ("Thus the parity of the sticker
>permutation is equal to the parity of the cubie
>permutation."), but I believe this approach may be easier
>(it seems to be what David Smith was getting at in Roice's
>last link):

I did not even see where David Smith attempted to prove
equality of the permutation parities for the cubies and
their stickers. He just seemed to take it as given.

>Note that an element of the generating set (of twists)
>performs an odd permutation on the 3C pieces iff (if and
>only if) it performs an odd permutation on the 3C
>stickers. Thus, any composition of generators performs
>an odd permutation on the 3C pieces iff it performs an
>odd permutation on the 3C stickers.

And, yes, it is pretty clear based on the twisting argument.

Thus I offer the following revision to my (hopefully really
now) valid argument:

There are 96 sticker positions in the pile occupied by
stickers from these 32 cubies. When we scramble the
puzzle, we are permuting not only the cubies but the
individual stickers themselves. It may be observed that
the permutation parity of the stickers equals the
permutation parity of the cubies: It suffices to check
the effect of a quarter turn of an external slice, since
any twist can be generated by composing such quarter
turns. For a quarter turn, there are 3 4-cycles of the
cubies and 9 4-cycles of the stickers. Thus permutation
parity always changes for both cubies and stickers. (If
you want to consider twisting a center slice, there are
2 4-cycles of cubies and 6 of stickers for no change in
permutation parity of either.)

Now we can come back to the placement of 31 of the
cubies, each with any of 6 possible orientations. The
permutation parity of the cubies themselves is now
determined and so is that of the stickers. Of the six
possible orientations for the 32nd cubie only half of
them will lead to the one possible overall sticker
permutation parity value that is possible for whatever
the cubie permutation parity is. (Without working out
the actual parity of the cubie permutation, we don't
know what that parity is. Indeed, it can go either way;
but, whichever way that is, only half of the 6 possible
orientations will lead to the correct corresponding
overall sticker permutation parity.)

Thank you, Andy, for the correction.

Regards,
David V.




From: Roice Nelson <roice3@gmail.com>
Date: Sun, 13 May 2012 21:44:50 -0500
Subject: Re: [MC4D] MagicTile solves



--e89a8f22bd89f56c5304bff61170
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Hi Ed,

The level you put the picture at looks good to me :) I like having a
single picture represent all the puzzles of a given tiling, even if the
twisting type shown doesn't apply for all of them. Separate pictures for
each twisting type feels like it would be too cluttered. In any case, the
images are making the solutions page look a lot nicer!

Roice


On Sat, May 12, 2012 at 9:32 AM, Eduard wrote:

> I have solved MT skew {4,4|6} 36 v 0:0:1 with 643 twists.
>
> Level of the pictures in the solver list:
>
> Is the level now okay?
> Nan did chose 2 levels lower. I have rised his picture two levels.
>
> One level lower would also be possible (differentiating the turning axes).
> Or do we want only one axis representing all axes?
>
>
>

--e89a8f22bd89f56c5304bff61170
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

Hi Ed,


The level you put the picture at looks good to me=
:) =A0I like having a single picture represent all the puzzles of a given =
tiling, even if the twisting type shown doesn't apply for all of them. =
=A0Separate pictures for each twisting type feels like it would be too clut=
tered. =A0In any case, the images are making the solutions page look a lot =
nicer!


Roice


O=
n Sat, May 12, 2012 at 9:32 AM, Eduard <lto:baumann@mcnet.ch" target=3D"_blank">baumann@mcnet.ch> wro=
te:


x #ccc solid;padding-left:1ex">I have solved MT skew {4,4|6} 36 v 0:0:1 wit=
h 643 twists.



Level of the pictures in the solver list:



Is the level now okay?

Nan did chose 2 levels lower. I have rised his picture two levels.



One level lower would also be possible (differentiating the turning axes). =
Or do we want only one axis representing all axes?






--e89a8f22bd89f56c5304bff61170--




From: Melinda Green <melinda@superliminal.com>
Date: Sun, 13 May 2012 20:50:31 -0700
Subject: Re: [MC4D] MagicTile solves



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Hello Ed,

I agree that the images really help a lot. I've added a few IRP screen
shots there too. I like them at the top of each section like that and I
agree with Roice about one being enough for each section. One suggestion
however is that I think the placement looks best when they have the same
background color as the page.

-Melinda

On 5/13/2012 7:44 PM, Roice Nelson wrote:
>
>
> Hi Ed,
>
> The level you put the picture at looks good to me :) I like having a
> single picture represent all the puzzles of a given tiling, even if
> the twisting type shown doesn't apply for all of them. Separate
> pictures for each twisting type feels like it would be too cluttered.
> In any case, the images are making the solutions page look a lot nicer!
>
> Roice
>
>
> On Sat, May 12, 2012 at 9:32 AM, Eduard > > wrote:
>
> I have solved MT skew {4,4|6} 36 v 0:0:1 with 643 twists.
>
> Level of the pictures in the solver list:
>
> Is the level now okay?
> Nan did chose 2 levels lower. I have rised his picture two levels.
>
> One level lower would also be possible (differentiating the
> turning axes). Or do we want only one axis representing all axes?
>
>
>
>
>

--------------030905010509050102020102
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit



http-equiv="Content-Type">


Hello Ed,



I agree that the images really help a lot. I've added a few IRP
screen shots there too. I like them at the top of each section like
that and I agree with Roice about one being enough for each section.
One suggestion however is that I think the placement looks best when
they have the same background color as the page.



-Melinda



On 5/13/2012 7:44 PM, Roice Nelson wrote:
cite="mid:CAEMuGXqGXGMT+5y6EirUqNQOpP2ii7ZsOOSjZ2zaPaQZm2fRdQ@mail.gmail.com"
type="cite">


Hi Ed,




The level you put the picture at looks good to me :)  I like
having a single picture represent all the puzzles of a given
tiling, even if the twisting type shown doesn't apply for all of
them.  Separate pictures for each twisting type feels like it
would be too cluttered.  In any case, the images are making the
solutions page look a lot nicer!




Roice






On Sat, May 12, 2012 at 9:32 AM,
Eduard < href="mailto:baumann@mcnet.ch" target="_blank">baumann@mcnet.ch>
wrote:

I have
solved MT skew {4,4|6} 36 v 0:0:1 with 643 twists.



Level of the pictures in the solver list:



Is the level now okay?

Nan did chose 2 levels lower. I have rised his picture
two levels.



One level lower would also be possible (differentiating
the turning axes). Or do we want only one axis
representing all axes?
















--------------030905010509050102020102--




From: "Eduard Baumann" <ed.baumann@bluewin.ch>
Date: Thu, 12 Apr 2018 23:14:36 +0200
Subject: MagicTile solves



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MagicTile solves

My new 17 solves

24.6.2017 hyp {6,6} 6 R4.7 f0,7-0-0 f100 eh00 231 moves
25.6.2017 hyp {6,6} 6 R4.8 f0-1,2-0 4 moves
26.6.2017 hyp {6,6} 6 R4.8 f001 14 moves
2.7.2017 hyp {6,6} 6 R4.8 e100 442 moves
2.7.2017 hyp {6,6} 6 R4.8 v010 6 moves
2.7.2017 hyp {6,6} 6 R4.8 f010 v100 119 moves
16.7.2017 hyp {6,6} 6 R4.8 eh00 v100 164 moves
16.7.2017 hyp {6,6} 6 R4.8 f0,7-0-0 f100 eh00 205 moves
18.7.2017 hyp {7,4} 24 e100 206 moves
19.7.2017 hyp {7,4} 24 f001 787 moves
21.7.2017 hyp {7,4} 24 v010 602 moves
11.9.2017 hyp {6,6} 6 R4.7 e100 1958 moves
31.3.2018 hyp {6,6} 6 R4.7 e0.61-0-0 e100 2870 moves
5.4.2018 hyp {6,6} 6 R4.8 e0.61-0-0 e100 1212 moves
5.4.2018 hyp {6,6} 6 R4.8 e0,7-0-0 64 moves
7.4.2018 skew {4,4|4} 16 f0-2.24-0 5274 moves
12.4.2018 hyp {4,5} R5.3 40 f001 2114 moves

Best regards
Ed
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pe>




MagicTile solves

 

My new 17 solves

 

24.6.2017 hyp {6,6} 6 R4.7 f0,7-0-0 f100=20
eh00   231 moves
25.6.2017 hyp {6,6} 6 R4.8 f0-1,2-0 &nbs=
p; 4=20
moves
26.6.2017 hyp {6,6} 6 R4.8 f001   14 moves
2.7.2017 h=
yp=20
{6,6} 6 R4.8 e100   442 moves
2.7.2017 hyp {6,6} 6 R4.8=20
v010   6 moves
2.7.2017 hyp {6,6} 6 R4.8 f010 v100  =
119=20
moves
16.7.2017 hyp {6,6} 6 R4.8 eh00 v100   164 moves
16.7=
.2017=20
hyp {6,6} 6 R4.8 f0,7-0-0 f100 eh00   205 moves
18.7.2017 hyp =
{7,4}=20
24 e100   206 moves
19.7.2017 hyp {7,4} 24 f001   78=
7=20
moves
21.7.2017 hyp {7,4} 24 v010   602 moves
11.9.2017 hyp=
=20
{6,6} 6 R4.7 e100   1958 moves
31.3.2018 hyp {6,6} 6 R4.7 e0.6=
1-0-0=20
e100   2870 moves
5.4.2018 hyp {6,6} 6 R4.8 e0.61-0-0=20
e100   1212 moves
5.4.2018 hyp {6,6} 6 R4.8 e0,7-0-0 &nbs=
p; 64=20
moves
7.4.2018 skew {4,4|4} 16 f0-2.24-0   5274 moves
12.4.=
2018=20
hyp {4,5} R5.3 40 f001   2114 moves

 

Best regards

Ed


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