## 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.

## Partial Permutations

Submitted by Jerry Bryan on Mon, 12/26/2005 - 21:41.
A **permutation** is generally defined as a bijection on a
non-empty set.

Given a permutation on a set W, a **partial permutation** is a bijection
from one subset W_{X} of W to another subset W_{Y} of W.
W_{X} and W_{Y} are the domain and range, respectively,
of the partial permutation. Because a partial permutation
is a bijection, W_{X} and W_{Y} must contain the
same number of elements (or must be of the same cardinality, if they
are infinite).
Note that a partial permutation is
defined only in the context of a specific and previously defined
permutation.
Generally speaking, a partial permutation
is not a permutation, and indeed a partial permutation is a permutation
only if its domain and range are the same set.

## Solving Rubik's cube in 28 face turns

Submitted by silviu on Thu, 12/22/2005 - 20:53.The above method can also be formulated in following way:

Given an arbitrary element g in the cube group we multiply it by an element B^-1 from the right such that gB^-1 is in the group H and the length of B^-1 is at most 12 face turns. Then we multiply gB^-1 by an element A^-1 from the right such that gB^-1A^-1=id --> g=AB. And the length of A is at most 18 face turns.

## Optimal solutions for two important subgroups

Submitted by silviu on Thu, 12/22/2005 - 20:17.Group that fixes cubies Distance Nr of pos Unique wrt M Unique wrt M + inv 0q 1 1 1 2q 0 0 0 4q 0 0 0 6q 0 0 0 8q 0 0 0 10q 0 0 0 12q 441 11 8 14q 3944 87 52 16q 32110 708 396

## Square One God's Algorithm Computed

Submitted by masonjones on Tue, 12/06/2005 - 09:04.The main stumbling block to reporting results at the moment is that I discovered a unexpected position-counting problem, where the shape degeneracy interferes with the 16-fold starting position degeneracy, and which requires me to do a recount on the tables. I will post the results back here in a few days when the recount is finished.

## Rubik's cube can be solved in 38 quarter turns max

Submitted by Bruce Norskog on Sat, 11/26/2005 - 14:30.Silviu recently posted a result where he provides move sequences to solve all positions of the corners with an even corner permutation in 22 quarter turns or less without any net effect on the edges. This, combined with a result by Tomas Rokicki that the edges can be solved in 18 quarter turns or less, allowed Silviu to conclude that Rubik's cube can always be solved in 18+22 = 40 quarter turns or less. I have used a C++ program to verify Silviu's sequences do not exceed 22 quarter turns; and combined with applying "M-symmetry" and inverses to those sequences, correctly solve all even permutation and orientation possibilities for the corners with no net effect on the edges.

Looking at Rokicki's analysis further, one can see that there are relatively few positions in the edges-only case that require more than 16 quarter turns. For the cases of 16 or less quarter turns, the edges can be solved, and then the corners can be solved with Silviu's sequences, in no more than 38 quarter turns. Showing that the other cases in the edges-only analysis of more than 16 quarter turns can be solved in 38 quarter turns or less (including corners) would therefore mean that all positions of Rubik's cube can be solved in 38 quarter turns or less.

## Solving Rubik's cube in 40 quarter turns

Submitted by silviu on Mon, 11/14/2005 - 06:56.found at http://www.cubeman.org .This means that given an arbitrary cube in the whole cube group one can solve the edges in at most 18 quarter turns. Once the edges are solved there are 44089920 possibilities for the corners. We prove that each of this configurations can be solved in at most 22 quarter turns. A list of all this configurations expressed in the generators can be found at: http://www.efd.lth.se/~f01sr/ under the link "Data file" (txt document). This list contains only representatives up to M-symmetry+inverse.

## Upgraded to php 4.4.1 last (all) night

Submitted by cubex on Wed, 11/09/2005 - 08:04.Let me know if anything is broken.

## An analysis of the corner and edge permutations of the 3x3x3 cube

Submitted by Bruce Norskog on Sat, 10/01/2005 - 11:15.I have generated a distance table using the quarter-turn metric for positions of the Rubik's cube (the standard 3x3x3 cube) where the orientation of the cubies is ignored. That is, 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. To the best of my knowledge, this is (now) the largest "subset" of the 3x3x3 cube that this has been done for!

I used symmetry in the corner permutations, so that my programs "only" needed to store values for 984*12!/2 or 235,668,787,200 positions. The "real" size (considering symmetry in both the corner permutations and edge permutations) I found to be 201,181,803,792 positions. So about 17% of the stored positions are redundant with respect to the set of symmetrically distinct positions.