# 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:

D B2 D' L2 U L' U F L F U' R' B2 D B U B F' D R' (20f*) //C4
L2 F2 U F2 U2 R F U' L2 D' L' B2 R2 D2 F' L' B' R' U R' (20f*)
B2 F2 U L2 B2 L R B L2 U' B L2 D' U' B' F' D2 U' B R2 (20f*)
B R' F' R' D' L R' B D2 F D R D2 R D B R' D' R' U' (20f*)
U2 L2 F2 D F L D2 B2 R D' R' U' F2 R2 B L' F R2 D' R' (20f*)
F2 D2 F2 D' U' R B' U L' D B2 L' B F' D2 F2 R' U' F2 R' (20f*)
D2 U' B2 D2 L2 R U2 B' R2 D2 L' B' D U2 B2 U' B D' F' R (20f*)
D' L2 R2 D L' F D R2 B L2 R' U2 L2 D2 B' D U' R2 F R2 (20f*)
R2 F2 L2 U' R' F R2 D' B U' L' D' F' U2 L2 F2 L F U' R' (20f*)
F L2 R' B2 D' U L B' L R' B D' F D F2 D' R2 D' R' U2 (20f*)
D B R U F2 L F2 U2 L2 B' U' R' B' R2 U2 R U L R2 U2 (20f*)
D' L2 D' B2 U' F R2 F2 U F2 U F D2 L' D' F2 U L2 B' R (20f*)
D L2 D2 L2 U2 R B' U L' D B2 L' B F' D2 F2 R' U' F2 R' (20f*)
D' B2 D2 L2 B2 R D' B' R' U' L' F L2 D B R F2 U2 B' R2 (20f*)
L2 U L2 R2 B2 R' B2 U' L B D R2 U B U2 R B' U2 F R' (20f*)
D' F' L' R2 B L2 F' D L B2 U L' F' U R2 D U2 F' R' U' (20f*)
F2 R2 B2 D R' D' F L F2 L2 D2 F' U' L' D' B U' R2 F R' (20f*)
D L2 D' B2 U' F R2 F2 U F2 U F D2 L' D' F2 U L2 B' R (20f*)
D' L2 D2 L2 U2 R B' U L' D B2 L' B F' D2 F2 R' U' F2 R' (20f*)
F2 U L2 U2 B2 L' U' R' B2 R' U2 B F D U F R D L2 R2 (20f*)
F2 D B' D' F2 U2 B U2 L D' R' B2 D' R2 B' F' U2 L R2 U' (20f*)
F2 U2 B2 D' R' F' R2 D L R2 B' D R F' U L2 R2 B U R2 (20f*)
D' L2 U L2 R F U R B L' U B U R' U' B' F' D' U' F' (20f*)
L2 D U L2 F' D' L' B R2 U' B' D' B L' F R2 F U2 F R' (20f*)
R2 B F2 U B2 D2 R' B F' D' L2 B2 D2 F' U' B' D U' R' U' (20f*)
L2 U' F L R' B' F2 U2 R2 U B R D F R U F2 U' R' U' (20f*)
U R2 D2 R2 D2 R B' U L' D B2 L' B F' D2 F2 R' U' F2 R' (20f*)
F2 D2 L2 D' U' B2 R B' D' U2 F D2 R U' F2 U2 L' R2 F R2 (20f*)
F2 U F2 D' R U2 L2 F D' U' R2 B D' B R2 D L B F R (20f*)
L B2 R2 U2 B' U' B F2 D2 F' D2 L' R F' L2 D B' U2 R' U' (20f*)
D' R2 F2 D' U' F' R B D' L2 R2 B F2 D' B2 F2 D2 L B' R' (20f*)
U R2 F2 D2 U' F D2 B' D' L U2 B2 F2 R2 D' L' B' L2 F R' (20f*)
B2 L2 D' L' U2 R F' D' B U' F R' F2 D' F2 L D2 B R U' (20f*)
U' R2 F2 D' U' F' R B D' L2 R2 B F2 D' B2 F2 D2 L B' R' (20f*)
R2 U' B2 F2 U' B' D2 R' U' L D R' F2 R2 B D R B F' R' (20f*)
B' U B L B F L2 R U' B2 F' D F D' F' R2 D2 U2 R U' (20f*)
B2 R' B F U2 B' D' L' D' R' B' D' F2 L2 R U B' D' R' U' (20f*)
L' D2 F' U R2 B' L R' U2 B' U2 B2 F D' F' D2 L2 F2 R U' (20f*)
L' F U' R' B D R2 F' D U2 B' D B L R B' D2 L2 R' U' (20f*)

B2 U R2 B D U2 R2 B R' D R2 D2 F' D' L' R' B' F' R' U' (20f*) //D2 (face)
D' F2 L2 U B2 F' R D L D' B D' F2 D' B F2 R2 D' B R' (20f*)
D U F2 U' F2 L' D2 F' R D2 R F L' D' R B' R' F D U' (20f*)
L2 B R2 D' B2 L2 R' F2 D B F' D L' B U L U F2 R' U' (20f*)

To complete the analysis of all symmetric subgroups of order 4, there are 6 groups left. The largest has 290880 elements, so this is at the edge what I can compute with one PC within a reasonable time. I strongly believe that all will be solvable in 20 moves too.

## Comment viewing options

### Excellent work! I wonder how

Excellent work! I wonder how much more effort it would be to get complete distance distributions for the symmetry classes, since you've already optimally solved the hard ones? Also, do you have a summary table for all the symmetry classes you've explored, listing the name of the class, the count of elements (maybe broken into antisymmetric and not antisymmetric), and the number that were at distance 20?

Personally, I think that while most 20's we've found are symmetric, that most positions at length 20 will not be (since there are *so* many more non-symmetric positions than symmetric positions, this will overcome the greater tendency of the symmetric positions to be at a large distance). On the other hand, most tails of the subgroups, cosets, and other classes we have explored have been symmetric positions, so maybe I'm wrong.

### I always run the two-phase-so

I always run the two-phase-solver over the complete set telling it to stop when it finds a 19-move solution. For maybe 10% of the cubes the search does not stop within a few days, so for analyzing 30000 cubes there are about 3000 cubes which I have to solve optimally. This takes another few days. If I want to solve all cubes optimally, with an average of maybe 1-2 minute/cube (I actually did not verify this, may also be more) this would take 30000 minutes which would be about three weeks or more. So I content myself with just finding the antipodes in the moment.

I did not make any statistics yet for all the classes. I will try to finish the computations for the 4-symmetries first.

I also have your opinion that there are many more unsymmetric 20's than symmetric. I estimate more than 1,000,000,000 20f* cubes. That the tails of all distributions we know have significantly many cubes with a higher degree of symmetry is no contradiction to this. But I think it is highly improbable that *all* the antipodes will have less than 4 symmetries. So if we do not find any 21f* cube with 4 symmetries I think we never will find any 21f* cube at all.