An unsolved problem: how many solutions are there to the 8-Color Cube?

Let us call a position clean if all faces are of the form

A A B          B A A
D   B    or    B   D
D C C          C C D

and pretty if all faces are of the form

A B C
H   D
G F E

There are 4 different clean positions (apply superflip and superinversion to get the others). The existence of a clean position ensures that we use the right set of edge cubies, with proper orientation. Namely, if we take out an edge cubie, and reinsert it flipped, the cube gets flipped out, and a clean position cannot be reached any more with maneuvers. This is the only way to make maneuvers insufficient, though. We can get a pretty position from a clean position if we apply half turns on all slices.

Suppose that the 8-color cube has N pretty positions. If we omit the SO3 symmetry (special orthogonal) reduction, then we get a variant of the 8-color cube with 24 times as many positions, and 24 * N pretty positions. Next, if we renumber to restore the corner cubies after each move on the variant of the 8-color cube, a process which we call tag reduction, we get 48 times as less positions, and 24 * N / 48 = N / 2 pretty positions. Let us call the last cube with N / 2 pretty positions the tag reduced cube.

To find pretty positions on the tag reduced cube, one can simply take the corners fixed, and rearrange the edge cubies. There is however one problem: the tag reduction affects the set of edge cubies being used. This problem is tackled as follows. First, we make four extra edge cubies, corresponding to the interior diagonals of the clean cube. After that, our set of edge cubies corresponds to the edges of a complete bipartite graph (between the numbers on the corners {ULB, URF, DLF, DRB} and the numbers on the corners {ULF, URB, DLB, DRF} of the clean cube). There are 35 ways to split {1,2,3,4,5,6,7,8} in two equal parts, and for each split, we get a set of 16 edge cubies. Next, one can find all pretty placements of 12 of the 16 edge cubies. One can show that in a pretty placement, the set of 12 edge cubies which are used always correspond to the edges of a clean cube, and the 4 edges which are not used correspond to its interior diagonals. For the 8-color cube as I saw it in pictures, the split is {{1,3,5,7},{2,4,6,8}} and the 4 unused edges are {1,6},{2,5},{3,8},{4,7}.

I wrote a program which computes all tag reduced cubes within a minute. Here are the results.

Tag reduced:        290976 normal,   145568 flipped out,   436544 total
Not reduced:      13966848 normal,  6987264 flipped out, 20954112 total
SO3 reduced:        581952 normal,   291136 flipped out,   873088 total
GO3 reduced:        290976 normal,   145568 flipped out,   436544 total
Tag+SO3 reduced:     12398 normal,     6126 flipped out,    18524 total
Tag+SO3 reduced:     12398 normal,     6126 flipped out,    18524 total
Tag+GO3 reduced:      6213 normal,     3063 flipped out,     9276 total
Tag+GO3 reduced:      6213 normal,     3063 flipped out,     9276 total
Tag+SO3 red sym:        28 normal,        0 flipped out,       28 total
Tag+GO3 red sym:        28 normal,        0 flipped out,       28 total

The first row was computed in the above-described way. The second, third and fourth row were obtained by scaling the first row, as indicated above for the second and third row. For people who do not read: the answer is in the third row. It appears that the numbers for flipped out positions are about half of those numbers for normal positions. Perhaps, there is a heuristic argument which indicates this.

In the fifth and sixth row, the original SO3 symmetry reduction is combined with tag reduction. To obtain these numbers, all 24 SO3 symmetries are applied to the found solutions. The numbers are obtained in two ways for verification. This is also the case for the seventh and eighth row, in which GO3 symmetry (general orthogonal) reduction is combined with tag reduction. To obtain these numbers, all 48 GO3 symmetries are applied to the found solutions (except the 24 symmetries that were already applied). In the ninth and tenth row, positions are only counted if they are symmetric with respect to superinversion. One can show theoretically that the numbers in both rows correspond to each other, and that they can be obtained by adding rows 7 and 8 and subtracting row 6.