Discussions on the mathematics of the cube

Rubik can be solved in 27f

In this paper we give a proof that Rubiks cube can be solved in 27f.
The idea is to eliminate the 476 cosets at distance 12 in the group H=< U,D,L2,F2,R2,B2 >.
In this way we never have to consider in the 2 phase algorithm that a coset is at distance 12.
So we only solve cosets at distance 11. Together with my earlier result of 28 this gives a proof of 27.
The same idea was used by Bruce Norskog in his 38q proof.




However we do not really need to compute all 476 cosets. In fact we only need to compute 7 cosets of the group
T = Intersection ( < U,D,L2,F2,R2,B2 > , < F,B,L2,U2,R2,D2 > , < L,R,F2,B2,U2,D2 > )
The group H is not invariant under all symmetries. But the group T is invariant under all 48.

Two more classes with exacly 4 symmetries done - most 20f* are antisymmetric

I finished the analysis of two more classes with 4 symmetries now. The computation took more than two weeks. All can be solved within 20 moves.
The definition of the classes D2(edge) and C2v(b) are explained on this page. Here you also can get some more information about these and other classes.

What is interesting, that from the 12 20f*-cubes of the class D2 (edge), 10 also have antisymmetry and from the 94 20f*-cubes of the class C2v(b) 92 also are antisymmetric.

Results of two more cosets of the H group, this time face turn metric.

After seeing Silviu have such success with H group (U, D, F2, B2, R2, L2)
cosets, I decided to give it a shot in the face turn metric. So far
I've completed the identity coset and the flip8 (upper and lower edges)
cosets; the superflip coset and flip4 (middle edges) are still running.
The identity, flip4, flip8, and superflip are the four centers of the
H group. I've also shown that of the approximately 234,101,145,600
positions represented by these four cosets, none have a depth greater
than 21. This exploration covers more than 5/1,000,000,000 of the total
cube space.

The identity coset has the following depths. For comparison on the right

Rubik can be solved in 35q

Let H be the group < U,D,L2,F2,B2,R2 > and let N be the subgroup of H that contains all even elements in H. I have run an exhaustive search on the coset space G/N and got the following table:
      0q             1
      1q             9
      2q            68
      3q           624
      4q          5544
      5q         49992
      6q        451898
      7q       4034156
      8q      35109780
      9q     278265460
     10q    1516294722
     11q    2364757036
     12q     235188806
     13q         28144
The group N contains no elements of odd length and the maximum length is 24.

New optimal solutions for an important group

I have computed optimal solutions for every element in the group <U,D,L^2,R^2,F^2,B^2>. For this task I made some minor modifications to Reid's solver. I would like to thank him for sharing it.

   0q                1 
   1q                4 
   2q               10 
   3q               36 
   4q              123 
   5q              368 
   6q            1,320 
   7q            4,800 
   8q           15,495 
   9q           54,016 
  10q          194,334 
  11q          656,752 
  12q        2,222,295 
  13q        7,814,000 
  14q       26,402,962 
  15q       89,183,776

Analysis of another two symmetry subgroups of order 4

The symmetry class C4 defines a 1/4-rotational symmetry around a face (I chose the UD-axis). It took about 8 days to show that all 36160 cubes, which exactly have this symmetry (M-reduced) are solvable in at most 20 moves. There are 39 20f*-cubes. 35 of them also have antisymmetry, 4 only have symmetry, so reduced wrt M+inv there are 37 cubes.

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

Hi...
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

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

After the analysis of cubes with more than 4 symmetries I now try to analyze cubes with 4 symetries. The smallest class has 15552 cubes wrt to M-symmetry. All cubes of this class can be solved in 20f moves. All positions which could not be reduced with the two phase solver to less than 20f moves within a day were solved with my optimal solver within about 4 days. I did not check if the positions are local maxima, so they still are candidates for a 21f maneuver when appending a move.

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

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.