Message #2118

From: Roice Nelson <>
Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)
Date: Mon, 07 May 2012 13:19:31 -0500

Hi David,

Thanks for helping me realize the state space reduction in the 3D case is *not
quite* 48, and similarly for other dimensions. I missed this
previously, but should have considered it when quoting the cube lovers post
about there being 12 equivalent 1 quarter turn moves (rather than 48). It
feels like a somewhat subtle point. It also makes calculating the
*exact*number of states which are "the same" in this discussion more


On Sun, May 6, 2012 at 9:44 PM, David Vanderschel <>wrote:

> Still out of phase. Significant redundancy between my last
> post and Roice’s. :-(
> It dawns on me that the problem the 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 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.