[next] [prev] [up] Date: Fri, 09 Nov 90 15:02:48 -0500 (EST)
[next] [prev] [up] From: Dan Hoey <hoey@aic.nrl.navy.mil >
[next] ~~~ [up] Subject: Rubik's Cube reassembly problem and solution

In rec.puzzles article <1990Nov8.182534.18625@agate.berkeley.edu>,
greg@math.berkeley.edu (Greg Kuperberg) writes:

Consider a standard Rubik's cube. Disassemble it and put it back
together at random. Find, with proof, the probability that it can be

It depends on how you take it apart. If you just pull out the corner
and edge pieces then put them back in without respect to color, the
probability is one in 12 that you will put it back into the right
orbit. I won't bore you with yet another proof of this; if you spent
the last decade in a box see the archives, Singmaster's NOTES ON
RUBIK'S MAGIC CUBE, J. A. Eidswick's article in the March 1986 Math
Monthly, or even Hofstadter's METAMAGICAL THEMAS.

Now if you take the face centers off and scramble them, then there is
only one chance in 60 of getting it right. Of the 720 permutations of
the six face centers, only 24 can be generated by rigid motions of the
cube. But half of these 24 permutations are odd, and leaving the cube
in an unsolvable orbit. If you put the face centers on in the
``standard'' configuration with opposite faces ``differing by yellow''
(i.e., white opposite yellow, red opposite orange, and blue opposite
green), your chances go up to one in four--half the time you will get
an odd permutation, and half the time you will get a mirror-reversed

But wait, if you took the face centers off you probably noticed that
the corners and edges don't stay on very well. So, say you scrambled
all three kinds of pieces. You will be able to solve the resulting
cube if you could solve the corner/edge permutation and the face-
center permutation. But if the only thing keeping you from solving
the corner/edge permutation and the face-center permutation is that
both permutation parities were odd, then you will be able to solve the
two of them together. Therefore your chances of success are one in
360 (= (1/12)*(1/60)*2), or one in 24 if you preserved opposite pairs
of face centers.

Now suppose you peeled off the 54 colored stickers and stuck them back
on at random (carefully keeping them out of the reach of children, as
there are rumors the paint contains lead, especially on the cheap
Taiwanese knockoffs), what is the probability of getting a solvable
cube? This question was posed years ago (in Singmaster?) but I
believe it is still open.

Dan Hoey

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