## Representation of edge permutations and move table

Submitted by Herbert Kociemba on Mon, 06/18/2007 - 15:02.This MoveTable would have 12!*18 4 Byte entries when we take the coordinate from 0..12!-1 and of course is far too big. Of course we could reduce this by 48 symmetries, but then we still would have a very large table.

## Two Face Group

Submitted by B MacKenzie on Wed, 06/06/2007 - 09:47.Has the Rubik cube subgroup generated by the turns of two orthogonal faces been exhaustively expanded? My computer runs out of physical memory and bogs down after 18 q turns:

Shell Classes Elements 0 1 1 1 1 4 2 3 10 3 6 24 4 15 58 5 35 140 6 85 338 7 204 816 8 493 1970 9 1189 4756 10 2863 11448 11 6862 27448 12 16324 65260 13 38550 154192 14 90192 360692 15 206898 827540 16 462893 1851345 17 992268 3968840 18 1973209 7891990 Totals 3792091 15166872

## 26f now claimed proven sufficient

Submitted by Bruce Norskog on Thu, 05/31/2007 - 20:52.http://www.physorg.com/news99843195.html

## More on Branching Factors

Submitted by Jerry Bryan on Mon, 05/21/2007 - 23:32.I was pretty sure that I posted an article to this site about Starts-With and Ends-With, but if so I can't find it. In any case, for a position x we define Starts-With(x) to be the set of moves with which a minimal process for x can start, and Ends-With(x) to be the set of moves with which a minimal process for x can end. If Ends-With(x)=Q (the set of quarter turns), then x is a local maximum. A similar formulation of the same idea is that if |Ends-With(x)|=12, then x is a local maximum.

## Number of maneuvers for Rubik's Cube

Submitted by Herbert Kociemba on Sat, 05/19/2007 - 10:35.Let r = Sqrt(6) and k the maneuver length, then we have

N(k) = [(3+r)(6+3r)^n + (3-r)(6-3r)^n]/4

which gives 1, 18, 243, 3240, 43254, ...

Round[(3+r)(6+3r)^n] is a good approximation even for small n. and we see that 6+3r = 13.348... is the asymtotic branching factor.

## Disjoint Cycles and Twist/Flip Parity Rules

Submitted by B MacKenzie on Wed, 05/16/2007 - 00:13.(-1,-1,-1) being the coordinate for the (left,down,back) cublet through to (1,1,1) being the coordinate for the (right,up,front) cublet. The transformed position and orientation of each cublet is then specified as an element of the O symmetry point group, there being a one to one correspondence between the 24 elements of the O point group and the 24 states a cublet may assume via Rubik cube face turns: 12 edge positions with two flip states each or 8 corner positions with 3 twist states each.

## Irreducible Loops

Submitted by Peter on Wed, 04/18/2007 - 09:20.Hi, I am new to the Cube problem, so probably the ideas are not new, or too naive, but I could not find them anywhere. This is possibly due to my lack of knowledge of terminology. (My background is theoretical solid states physics.)

Thus I would like to share these ideas with you, which you hopefully find useful, or can tell me that these ideas are not new or possibly that they are useless. I kindly ask you to comment. I just go ahead...

How could one calculate the diameter of the Group?

Let A be a random permutation. I start by choosing a (not optimal) path from id to A, say in quarter turn metric, with A=prod_i(ai) (i=1,...,N) where ai in {U,U',D,...}.

## Welcome to my Blog

Submitted by Peter on Wed, 04/18/2007 - 07:21.Only a couple of days ago I got very much interested in the Cube. At wikipedia I found a note that the diameter of the Cube group is not yet known, and a link to this site.

Great work!

Sniffing into the problem, it seems to be quite complex. But some ideas that came to me these days I could not find. That's the purpose of this visit: To ask whether attempts have been made along these lines of thought, and if so, what is the outcome. And if not, I would like to contribute some analysis.

## Antisymmetry and the 2x2x2 Cube

Submitted by Bruce Norskog on Sat, 03/17/2007 - 22:48.Someone on the Yahoo forum asked about how to do a 2x2x2 God's algorithm calculation and mentioned the "1152-fold" symmetry for the 2x2x2. I got to looking at some of the messages in the Cube-Lovers archives that Jerry Bryan had made about B-conjugation and the 1152-fold symmetry of the 2x2x2. He found that there were 77802 equivalence classes for the 2x2x2.

I have used antisymmetry to further reduce the number of equivalence classes for the 2x2x2 to 40296. The following table shows the class sizes of these equivalence classes.

class size class size/24 count ---------- ------------- ----- 24 1 1 48 2 1 72 3 3 96 4 1 144 6 14 192 8 11 288 12 49 384 16 22 576 24 337 768 32 6 1152 48 3353 2304 96 36498 ----- total 40296

I then performed God's algorithm calculations (HTM and QTM) to find the number of equivalence classes at each distance from the solved 2x2x2 cube. The results are given below. Because the 2x2x2 has no centers that provide a reference for the positions of the other cubies, the number of positions of the corners for the 2x2x2 (the only cubies it has) can be considered to be 1/24 the number of positions of the corners of the 3x3x3 (3674160 instead of 88179840). So in the tables below, I use the factor-of-24 reduced numbers for simplicity. The tables further break down the positions with respect to different class sizes.

## Odd Permutations of the Cube Shape of Square-1

Submitted by Mike G on Fri, 03/02/2007 - 11:41.
The method I used was suggested by Tom and Silviu's coset searches for
the Rubik's Cube: Starting from a cube-shaped, *odd*-parity
position of Square-1, an iterated depth-first search was made for
all *even*-parity cube-shaped positions, with the search being
pruned on [shape]x[parity].