# 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:

…