## Commutator elements of the cube

Submitted by mdlazreg on Sat, 10/31/2009 - 18:35.Commutator elements are of the form ABA'B' where A and B are some sequences.

It seems that the subgroup generated by the commutator elements is half the size of the cube, but is it the case that every element of the commutator subgroup is a commutator element? if not how many are they and how far they are from solved?

## Conjugacy classes of the cube

Submitted by mdlazreg on Sun, 10/18/2009 - 05:03.http://en.wikipedia.org/wiki/Conjugacy_class

and wondered if the number of conjugacy classes of the cube is known?

## Challenge with three faces

Submitted by brac37 on Tue, 10/06/2009 - 11:33.## Linear formula

Submitted by mdlazreg on Mon, 09/28/2009 - 08:13.We have then the following formula:

P(n) = sum(R(k)*P(n-k)) for 1 <= k <= n

Calculating R(k) using rokiki's results we deduce :

1 12 = 12*1 2 114 = 12*12 -30*1 3 1068 = 12*114 -30*12 +60*1 4 10011 = 12*1068 -30*114 +60*12 -105*1

## Algorithm for Counting Identities

Submitted by Jerry Bryan on Sat, 09/26/2009 - 16:45.I've been thinking about writing a program to calculate and count duplicate positions - roughly speaking, those positions that are half way through an identity. What I have in mind will probably be a more time consuming program to write than I would prefer. So I wonder if I could ask Herbert Kociemba and/or mdlazreg to post a little something about the programs they have already written to find identities. It may well be that there is a much simpler approach to calculating duplicate positions than what I have in mind.

What I have in mind is an iterative deepening depth first search beginning at the Start position. If that's all I did, the search would simply count 12^{n} maneuvers for each distance from n, and it would not extract any useful information about how many duplicate positions there are for each n. To solve these problems, I propose to store all the duplicate positions and not to store those positions that are not duplicate. This would be for the quarter turn metric. The program I have in mind would not be able to handle the face turn metric.

## God's algorithm for FTM mod 48, 2. Try

Submitted by kociemba on Sat, 09/26/2009 - 07:32.distance positions mod 48 0 1 1 18 2 3 3 24 4 39 5 12 6 22 7 12 8 40 9 3 10 4 11 20

## Puzzle about the Cube: Coloring the Cayley Graph

Submitted by rokicki on Sat, 09/19/2009 - 14:22.Here's a slightly harder puzzle: What's the chromatic number of the Cayley graph for the half turn metric? If you can't figure it out, can you figure out an upper bound? A lower bound?

This was discussed on speedsolving.com before, but I think it's a good enough puzzle to present here as well.

## God's Algorithm out to 15q*

Submitted by rokicki on Sat, 09/19/2009 - 13:56.In any case, it is finally done; here are the results. First we have positions at exactly that depth:

d mod M + inv mod M positions

## Numerical formula

Submitted by mdlazreg on Tue, 09/15/2009 - 07:55.d positions I4 positions I4&I12 positions ALL -- ------------ ---------------- -------------- 0 1 1 1 1 12 12 12 2 114 114 114

## Drupal database corrupted

Submitted by cubex on Wed, 09/09/2009 - 17:12.I try my best to make sure everything is working but this one slipped through the cracks. Somehow the mysql database ballooned in size to over 2 gigabytes. After that happened the subsequent databases were not backed up correctly.

It would be a good idea for any posts to be buffered in some way before uploading to the forum, especially long ones.