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

—–BEGIN PGP SIGNED MESSAGE—–
Hash: SHA1

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

—–BEGIN PGP SIGNATURE—–
Version: GnuPG v2.0.17 (GNU/Linux)

iEYEARECAAYFAk+sascACgkQqaGPzAsl94J7xgCcDc0t0/CKzDdax5mqb4QCyjQp
TiYAn1/pk4Sld43+nte4Dn4580EjfT8P
=niwt
—–END PGP SIGNATURE—–