## 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].

## God's Algorithm to 13q, Summarized by Symmetry Class

Submitted by Jerry Bryan on Mon, 02/26/2007 - 12:25.|x| Symmetry Size Patterns Positions Class of x^{M}0 M 1 1 1 Total 1 1 1 CR 12 1 12 Total 1 12 2 I 48 2 96 Q 6 1 6

## God's Algorithm to 11f, Summarized by Symmetry Class

Submitted by Jerry Bryan on Mon, 02/26/2007 - 12:21.|x| Symmetry Size Patterns Positions Class of x^{M}0 M 1 1 1 Total 1 1 1 CR 12 1 12 Q 6 1 6 Total 2 18 2 I 48 4 192

## Dan Hoey's Taxonomy of Symmetry Groups and Shoenflies Symbols

Submitted by Jerry Bryan on Thu, 02/22/2007 - 23:40.I have received permission to post Dan Hoey's taxonomy of symmetry groups of Rubik's Cube. Also, I will relate Dan's taxonomy to Shoenflies symbols as implemented in Herbert Kociemba's Cube Explorer. (Go to http://kociemba.org/cube.htm and then click on Symmetric Patterns.) To that end, some preparatory comments are in order.

In order to define any terminology for the symmetry groups of Rubik's Cube,
it's necessary first
to define some terminology for the symmetries of the cube.
To the best of my knowledge,
no standard terminology has been adopted by the Rubik's cube
community for the symmetries of the cube.
The terminology I'm going to use is very similar to
some terminology I have seen before, but I can't remember the
reference. It may have been Christoph Bandelow's book,
*Inside Rubik's Cube and Beyond*. In any case, if I can
find the reference I want to give proper credit.

## Solving the 4x4x4 in 68 turns

Submitted by Bruce Norskog on Wed, 02/14/2007 - 01:03.I have completed a five-stage analysis of the 4x4x4 cube showing that
it can always be solved using at most 68 turns.
The analysis used the same five stages that were used in my prior posts where I claimed the 4x4x4 cube
can be solved in 79 single-slice turns, or alternatively in 85 twists.
The difference in this analysis is that it allows any single layer turn or double layer turn
(where the two layers are any two adjacent layers and moved together) to be counted as a "turn."
In some prior posts, I referred to these turns as "block turns."
So the set of turns about the U-D axis that count as one turn are the following:

U,U',U^{2},u,u',u^{2},(Uu),(Uu)',(Uu)^{2},

D,D',D^{2},d,d',d^{2},(Dd),(Dd)',(Dd)^{2},

(ud'),(u'd),(ud')^{2}

## Lower bound using the Edges antipode

Submitted by mdlazreg on Mon, 01/29/2007 - 15:01.1) First solve all the edges [less than 18 moves as proved by Tom Rokicki ].

2) Take it to the cube identity [less than 22 moves as proved by Silviu Radu]. Which gives a maximum of 40 moves.

Is it possible to follow the same logic but use the Edges Antipode instead?. So the two phases become:

1) Solve all the edges by taking them to the Antipode edges instead of the identity edges. [this is guaranteed to be 18 moves].

2) From the edges antipode position take it to the cube identity. [Does anyone know the maximum moves needed for this phase?]. To find this maximum it would take the same computational effort to prove the 22 moves I guess. Let's call this maximum X.

## A Little More on Odd and Even Permutations

Submitted by Jerry Bryan on Thu, 01/11/2007 - 12:50.If you take a cube apart and put it back together, you can put it back together in twelve times as many ways as there are positions in the cube group. Twelve is because 3 * 2 * 2 = 12. Three of the ways are because the twist of the last corner cubie you put back together can be set in one of three different ways, only one of which is legal. Two of the ways are because the flip of the last edge cubie you put back together can be set in one of two different ways, only one of which is legal. The last two of the ways are because the corners and edges can by put back together either with the same odd/even parity (legal) or with the opposite odd/even parity (illegal).