## Analysis of another two symmetry subgroups of order 4

Submitted by Herbert Kociemba on Wed, 03/08/2006 - 12:46.The class D2 (face) consists of all cubes which have a 1/2-rotational symmetry around all faces. Up to M-symmetry there are 23356 cubes, which exactly have this symmetry. It took about 4 days to show, that all cubes of this symmetry class can be solved in 20 moves. There are only 4 cubes which are 20f*, all of them also are antisymmetric. Here are the results:

## Regarding God's Algorithm Calculations and lower bounds on face metric

Submitted by xintax on Mon, 03/06/2006 - 10:35.Im kinda new here. Just a few intros: Im sheelah from the philippines and im doing an overnight research on the length of God's Algorithm... due tomorrow.

I was kinda ready but then I encountered this page that says that the real size of the cube universe is not 43 quintillion but rather 901 quadrillion.

For one, the repercussions of this on my project are:

1.) Theoretically less runtime and

2.) A different criteria for 'similarity'.

I was trying to read the calculations but I got lost between the numbers. Can someone please clarify this for me?

Secondly, I want to ask about the current lower-bound of the length of God's Algorithm in face turn metric. I keep seeing 20f* but I was wondering if there is a higher lowerbound...

## Permutations of the corners and edges: FTM

Submitted by Bruce Norskog on Tue, 02/21/2006 - 17:00.I have finally completed the face-turn metric version of the analysis of the permutations of the corners and edges of the 3x3x3 Rubik's Cube. That is, I have generated a table of the Cayley graph distances for the positions where the permutation of the corner cubies and the permutation of the edge cubies are considered, but not the orientation of either the edge or corner cubies. This is a set of 8!*12!/2 or 9,656,672,256,000 positions. As with the quarter-turn metric analysis, I used symmetry in the corner permutations to reduce the number of stored distances to 984*12!/2 or 235,668,787,200.

## 39 cubes with 20f moves in a class with 4 symmetries

Submitted by Herbert Kociemba on Tue, 02/21/2006 - 14:58.D L2 U' B2 L2 B2 D B2 U L2 U2 R2 B F L' D U F2 L' R' (20f*) //C2v (a1)

## Another four interesting cosets

Submitted by rokicki on Fri, 02/03/2006 - 22:38.## Symmetric Cube Positions with more than 4 symmetries

Submitted by Herbert Kociemba on Fri, 02/03/2006 - 15:31.For the labeling of cube symmetries there does not seem to exist a really consistent procedure. Michael Reid uses a different labeling than Jaap Scherhuis which also differs from my notation. My notation uses the Schoenflies symbols, I used for example this site for a deeper understanding.

## 44 million cubes

Submitted by rokicki on Wed, 02/01/2006 - 01:03.## Some more interesting groups

Submitted by rokicki on Mon, 01/30/2006 - 18:10.## Calculating Symmetry using Representative Elements

Submitted by Jerry Bryan on Wed, 01/25/2006 - 14:19.I have long been curious how other people who write cube programs and who incorporate symmetry into their programs actually do the symmetry calculations. The general way I do it has been outlined in Cube-Lovers.

For a given position x, I calculate a representative element of its m-conjugates, where M is the standard Cube-Lovers terminology for the 48 symmetries of the cube. I then store and manipulate only the representatives.

We denote this calculation as
y = Rep(x) = min{x^{M}} = min{x^{m} | m in M}.
The minimum element of x^{M} is taken to be the one that
is first in lexicographic order. And as usual,
x^{m} means m^{-1}xm, and x^{M} means
{x^{m} | m in M}.

## Solving Rubik's cube in 36 quarter turns

Submitted by silviu on Sat, 01/14/2006 - 12:44.We call the group of corner edge permutations of the cube for CEP and cube group for C. Let N be the normal subgroup that fixes cubies. Then we have a homomorphism hom:C->C/N=CEP such that hom(g)="permtutation of cubies done by g". Let

**a**be the unique antipode in CEP. Every element in CEP can be written as x*

**a**. Where x is an element requiring at most 18 quarter turns according to Norskog's analysis. There are 220 elements at distance 17 and 1 element at distance 18. So we could say that all elements of CEP except 220+1 elements can be written as x'*

**a**where x' is an element requiring at most 16 quarter turns.