[next] [prev] [up] Date: Mon, 12 Nov 90 18:38:49 -0500 (EST)
[next] [prev] [up] From: Dan Hoey <hoey@aic.nrl.navy.mil >
~~~ [prev] [up] Subject: Re: Rubik's Cube reassembly problem and solution

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

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

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

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.


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