From hirsh@cs.rutgers.edu Sun Nov 11 15:34:28 1990
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Sender: Haym Hirsh
Date: Sun, 11 Nov 90 15:34:23 EST
From: Haym Hirsh
Reply-To: Haym Hirsh
To: Cube-Lovers@life.ai.mit.edu
Subject: Re: Rubik's Cube reassembly problem and solution
In-Reply-To: Your message of Sat, 10 Nov 90 18:49:48 EST
Cc: Haym Hirsh
Message-Id:
I just caught a bug in my reasoning. The restickering need not yield
something equivalent to the original undestickered cube, but rather
just one that can be solved to obtain solid colors on each face.
Since there are 5*3*2 different distinguishable cubes (i.e., 30
different ways to label a die with the numbers 1-6) (6! labelings, but
rotational symmetry removes 24 -- six faces can be brought to the top,
and for each it can be rotated around the axis perpendicular to that
face in one of 4 ways), the numerator should be multiplied by 30, and
hence the probability is actually
2962822807718827976654851209938335398226821120000000000
------------------------------------------------------------------------
230843697339241380472092742683027581083278564571807941132288000000000000
= 3.0*10^54/2.3*10^71 = 1.3*10^-17
Haym