Message #2112

From: David Vanderschel <DvdS@Austin.RR.com>
Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)
Date: Sun, 06 May 2012 21:44:50 -0500

Still out of phase. Significant redundancy between my last
post and Roice’s. :-(

It dawns on me that the problem the cube20.org folks were
addressing is deeper than the one David Smith was addressing
(and which I was defending) and more like what Melinda was
probably driving at. I.e., given two arrangements that are
visually different when the pile is viewed in standard
orientation and based on the sticker IDs inferred from
initial state as in Smith’s calculations, when can those two
still be regarded as the ‘same’ based on a symmetry of the
pile? This is where conjugation by a symmetry enters the
picture, and that can be viewed in a sense which remaps axis
IDs (along with the corresponding sticker ‘color’ pairs).
As may be seen from the ‘94 Cube Lovers article, the precise
issue gets very messy and David Smith was not trying to
address this more complex version. What Smith was
calculating is much more straightforward, is still
meaningful, and is more easily understood. The deeper view
reduces the "real size" by a factor less than n!*2^n (but
not quite n!*2^n), which is actually rather minor compared
to the sizes of the numbers Smith was computing. In my
view, the main point of his efforts is not the precise
values (which are relatively useless) but to get some feel
for just how immense these numbers are. They are
incomprehensibly large with or without the factor of n!*2^n.
The little ol’ order 3 3D problem was close enough to being
tractible that a factor of nearly 48 actually did make a big
difference in the ultimate analysis there.

One thing that should be pointed out is that, if you admit
mirroring transformations, then it does create problems for
the purpose of counting distinct arrangements. The
problem is that there exist (a small minority of)
arrangements which possess mirror symmetry, so that the very
same arrangement would occur for two different symmetry
transformations of the pile. Accounting for the number of
distinct positions possessing such symmetry is not at all
easy; so the basic counting problem is much easier if you do
not admit reflecting transformations for achieving standard
orientation. But the cube20.org folks were not interested
so much in _counting_ arrangements as in analyzing relations
between explicit instances of state, so the extra complexity
that gained another factor of 2 was well worth it for them.

Regards,
David V.

—– Original Message —–
From: Roice Nelson
To: 4D_Cubing@yahoogroups.com
Sent: Sunday, May 06, 2012 8:00 PM
Subject: Re: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)

Hi Melinda,

To help solve the "God’s Number" problem, the cube20 guys
used the trick of considering states that only differ by a
symmetry of the cube to be the same. I think this is
precisely what you are describing. Their site links to a
cube lovers post titled "The real size of cube space", which
writes:

…