# Message #2132

From: Brandon Enright <bmenrigh@ucsd.edu>

Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)

Date: Fri, 11 May 2012 01:26:21 +0000

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On Thu, 10 May 2012 19:05:51 -0500

"David Vanderschel" <DvdS@Austin.RR.com> wrote:

> I am interested both in the number of distinct states as

> determined by the positions and orientations of the cubies as well as

> the equivalence classes of those states under conjugation by a

> symmetry. Your "interest" in the equivalence classes under

> conjugation (or whatever you have in mind) seems to go so far that

> you would actually exclude the distinct states on which the published

> counts are based - to the extent that you expressed concern that the

> published numbers are in error.

I’ll jump in here with a concrete example. A while back I enumerated

the shortest possible sequence of moves to generate every 3-cycle

the corners on a Pentultimate can be in.

Depending on the relative positions of the corners (the pattern of the

cycle), the pattern can have no symmetry, mirror symmetry, rotational

symmetry, or mirror & rotational symmetry.

Here is an example of a 3-cycle that doesn’t have any symmetry:

http://www.brandonenright.net/twistypuzzles/post_images/pentultimate/01_000.png

This one has mirror symmetry:

http://www.brandonenright.net/twistypuzzles/post_images/pentultimate/02_000.png

This one has rotational symmetry (this pattern is chiral):

http://www.brandonenright.net/twistypuzzles/post_images/pentultimate/09_000.png

This one has both rotational and mirror symmetry:

http://www.brandonenright.net/twistypuzzles/post_images/pentultimate/06_000.png

The reason the symmetry of these patterns matter is that by

re-orientation / mirroring / inverting a sequence you can achieve more

than one distinct "permutation" of the puzzle even though all of these

distinct permutations are equivalent under conjugation.

One of the complications of looking at the permutation of corners on

the Pentultimate is that the corners have orientation in addition to

just their position. This means that there are 9 distinct clockwise

3-cycles depending on the resulting twist of each corner.

For simplicity rather than describing how to re-orient / mirror the

puzzle I have instead inverted / mirrored / re-oriented the sequence of

moves themselves.

By taking the sequence [D’, E’2, D’, E2, C, A’, E2, A, E’2, C’, D2] and

applying these symmetries, you can reach 6 distinct permutations that

are all a clockwise 3-cycle of the same pieces on the Pentultimate:

Twists: 000 / 021 / 102 / 111 / 210 / 222 (11 moves):

000 : [D’, E’2, D’, E2, C, A’, E2, A, E’2, C’, D2]

021 : [H2, C’, I’2, B, I2, B’, C, I2, H’, I’2, H’]

102 : [J2, I’, E’2, F, E2, F’, I, E2, J’, E’2, J’]

111 : [J’, I’2, J’, I2, E, F’, I2, F, I’2, E’, J2]

210 : [D2, E’, C’2, A, C2, A’, E, C2, D’, C’2, D’]

222 : [H’, C’2, H’, C2, I, B’, C2, B, C’2, I’, H2]

The ### designations describe the resulting twist of the corners. 0

being no twist, 1 being clockwise, and 2 being counter-clockwise.

For the sake of counting the number of possible permutations of the

Pentultimate, each of these are counted as distinct positions. I think

they should be. This is the case with every twisty puzzle.

There is also the much more complicated question of how many

"essentially distinct" positions the puzzle has. That count would

consider the above 6 permutations as just one.

Calculating it would be very difficult. For a dodecahedron there are

120 orientations (considering mirroring too) but not all positions have

the symmetries necessary to be equivalent under these re-orientations

and mirroring.

The number of essentially different positions should be less than:

((20! / 2) * (3^19) * (12! / 2)) / 60

But greater than:

(((20! / 2) * (3^19) * (12! / 2)) / 60) / 120

I’d certainly like some insight into how to get at the number but I

don’t have any ideas about how to approach the problem.

Brandon

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