From:

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This problem of counting the number of solvable restickerings seems to

be a lot easier than I had thought, but a lot trickier than you might

think.

Haym Hirsh sent in a buggy analysis, then corrected himself, but not

quite enough. The fix was to account for cases where the stickers

corresponded to a cube recoloring, but he just multiplied by 30 (cube

recolorings up to rotational symmetry) rather than by 720 (total cube

recolorings). We are dividing by 54!, which includes positions

differing only by a rotation, so when figuring how many are solvable

you have to count such positions also.

Another way of figuring this is 6! ways of coloring the face centers,

then (8! 3^8 12! 2^12)/12 ways of coloring the rest of the cube, then

9!^6 ways of arranging stickers among identically-colored faces, out

of 54! ways of arranging stickers randomly.

So the probability that a random restickering will be solvable is

71107747385251871439716429038520049557443706880000000000 ------------------------------------------------------------------------ 230843697339241380472092742683027581083278564571807941132288000000000000

40122452017152 = ------------------------------ ~ 3.0803 X 10^-16. 130253249618151492335575683325

It seems odd to me that this is not the reciprocal of an integer, but

I guess that's because we are dealing with color cosets rather than

some cube group.

Haym Hirsch also asked me how to figure out the minimum number of

stickers to fix to make an unsolvable stickering solvable. Sounds

hard to me. His question arises in the same way that I recall the

original problem arising: trying to clean up after someone who tried

to solve the cube by restickering. Since the adhesive isn't designed

for moving the stickers around, this leads rapidly to Dik Winter's

problem: dealing cubes that have lost some of their stickers.

Dan

Hoey@AIC.NRL.Navy.Mil