# Antisymmetry and enumeration of LL algorithms

Submitted by Joe Miller on Thu, 08/04/2005 - 07:38.

I have been trying to enumerate all LL algorithms, (following Bernard Helmstetter's work) and cannot seem to reduce the 62208 permutations to the 1211 he has come up with. I am unaware of whether or not he used antisymmetry for his reductions. Using GAP and U rotations and reflections I have come up with a total of 8020. Initially considering antisymmetry as well (removing order 2 permutations and the identity from consideration. Note: I did not consider antisymmetric positions with dihedral symmetry or reflections) still does not bring it close to 1211. I was using Martin Schoenert's method in GAP and a very slow algorithm I wrote to determine equivalence classes from conjugation with a group (automorphism) separately with the same results. If anyone would like to see my definitions I will post them.

Also: Does anyone know of a method to employ the inverse homomorphism in GAP on the cube without forcing GAP to consider such a large space (the WHOLE cube space?).

Thank you for your time.

-Joe

Also: Does anyone know of a method to employ the inverse homomorphism in GAP on the cube without forcing GAP to consider such a large space (the WHOLE cube space?).

Thank you for your time.

-Joe

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### Thanks for the response.A

Submitted by Joe Miller on Fri, 08/05/2005 - 07:03.

Thanks for the response.

Actually, I should have been more precise than 'U rotations'. I meant conjugation by rotations of the final layer's face. Your original interpretation that I meant permutations that differ only by a face turn was what I was missing!

Thanks again,

-Joe

Actually, I should have been more precise than 'U rotations'. I meant conjugation by rotations of the final layer's face. Your original interpretation that I meant permutations that differ only by a face turn was what I was missing!

Thanks again,

-Joe

## Using GAP and U rotations and

When you say you did U rotations, do you mean you consider two LL perms the same if they differ by a final U turn? Doing only that is not enough. Two algs LL perms may also differ by preceding it by a U turn, and that would reduce your number by a further factor 4 approx.

You can consider two LLs equivalent if they are conjugates by a U turn (or a rotation of the whole cube around the U axis, which amounts to the same thing). Together with inversion, that should get you down to 1211 (or maybe 1212 if you count the identity).

Jaap's Puzzle Page:

http://www.geocities.com/jaapsch/puzzles/