Discussions on the mathematics of the cube

Another four interesting cosets

These are the distributions of the optimal solution lengths for the cosets where the edges start at the given M-symmetric position, using the quarter turn metric. The first run took longer than the other three combined; not sure why. Only a handful of positions at length 24; none at 26 or higher. I'm currently running the coset where the edges begin in the known length-26 position. It's interesting to note that with the edges superflipped and reflected across the center, all solutions took either 18, 20, or 22 moves.

Symmetric Cube Positions with more than 4 symmetries

Symmetric cube positions tend to be deeper positions than positions without symmetry. So in the search for a 21 FTM-positon it seems natural to look for positions with a higher degree of symmetry.

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

Here's another coset distribution. This one is face turn metric, edges in a random initial position (with no symmetry at all), and the distribution of the lengths of the optimal solutions for all corner possibilities:

Some more interesting groups

Here are the results of an exploration of four groups, all of whom leave the edges in some M-symmetric class, for the face turn metric:

Calculating Symmetry using Representative Elements

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{xM} = min{xm | m in M}.  The minimum element of xM is taken to be the one that is first in lexicographic order.  And as usual, xm means m-1xm, and xM means {xm | m in M}.

Solving Rubik's cube in 36 quarter turns

In this paper we present a method of solving Rubik's cube in 36 quarter turns. The proof have many common facts with other methods presented on the forum. In able to prove our claim we have used the corner and edge permutation analysis done for the first time by Bruce Norskog.

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

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 WX of W to another subset WY of W.  WX and WY are the domain and range, respectively, of the partial permutation.  Because a partial permutation is a bijection, WX and WY 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

In this paper we prove that Rubiks cube can be solved in at most 28 face turns. The proof uses exactly the same method as Michael Reid used in 1995. The only difference is that we show how to avoid the positions at distance 29. The proof was based on the fact that it takes 12 face turns max to take an arbitrary element of the Rubiks cube group into the group H=< U,D,L2,F2,B2,R2 >. And at most 18 face turns to take an element of H to the identity.

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

I have computed optimal solutions for following subgroups of the cube:
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

I reported this to the most visible Square One guy I could find on the web (Jaap Scherphius), and he sent me here. Yes, I have spent the last year or so computing God's Algorithm for Square One, in the center-flip-count domain, taking advantage of all top/bottom rotational degeneracies available. The table calculation itself is finished, but I am not finished analyzing it yet. I can report that the maximum number of flips required to get back to START is 13, with many shapes (not just square-square) contributing to that number.

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.