Message #2107

From: David Vanderschel <DvdS@Austin.RR.com>
Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)
Date: Sun, 06 May 2012 18:43:37 -0500

I think I can provide a more clear-cut answer to Charlie’s
question: "Should the 31 be 32?"

My answer:

Only if you wish to regard as being distinct arrangements
which differ only by a reorientation of the whole pile.

Personally, I do not think one should wish that. Indeed, the
way the arrangements are counted in the odd order cases takes
care of this by not counting permutations of the facet-center
cubies - i.e., those cubies that are in center slices on all
axes but one. In the odd order cases, those provide a
convenient reference for the orientation of the pile. So we
are only counting arrangements for which the pile is in a
standard orientation as defined by the stickers on those
cubies. The factor left out by using 31 is 5!*2^4 which is
the number of ways in which you can arrange the facet-centers
(or orient a 5-cube).

When the order is even, there is not such a convenient
reference for a standard orientation. However, one way to do
it is to pick one of the corner cubies and say that the pile
is in standard orientation when that one cubie has all its
stickers facing in the same direction as in the initial
state. Again, there 5!*2^4 ways of positioning that one
corner. Now saying that the pile is in standard orientation
is equivalent to saying that there is no choice in the
placement of that reference corner; so there are only 31
corners which we are free to place arbitrarily.

In the odd order situation, you really only need one of the
facet-center stickers on each axis for a reference. Thus
there are 5 (or the dimension of the puzzle) stickers for an
orientation reference in either case. To be consistent I
tend to favor the stickers facing in negative direction on the
facet-center cubies (for the odd order cases) and the
stickers on the corner cubie whose position coordinates are
all negative (for even order). It would actually be even
more consistent to base standard orientation on a corner
cubie for all orders, in which case you would count the
arrangements of facet-center cubies when they are present.
You get the same count either way.

It may be noted that, in Smith’s formulae for odd order
cases, he does use 32 and not 31 for the corners.

Regards,
David V.

—– Original Message —–
From: Andrew Gould
To: 4D_Cubing@yahoogroups.com
Sent: Sunday, May 06, 2012 4:35 PM
Subject: RE: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)

The choice between 31 and 32 comes down to how you define the
locations of pieces. If you define all their locations
relative to one of the pieces it’s 31, but if you define what
moves and what doesn’t for each twist you can make it 32. I
note that for 32, it would be tricky to say that rotating the
entire puzzle doesn’t change the state.

–

Andy

—– Original Message —–
From: 4D_Cubing@yahoogroups.com
[mailto:4D_Cubing@yahoogroups.com] On Behalf Of
charliemckiz@rocketmail.com
Sent: Wednesday, May 02, 2012 12:46
To: 4D_Cubing@yahoogroups.com
Subject: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)

2x2x2x2x2

(31!/2)(60^31) =

545356551753081970586352633891109632137647267774464
00000000000000000000000000000000000000

approx. = 5.4 x 10^88

Should the 31 be 32?

Mine is

32!/2*(5!/2)^(32) =

104708457936591738352579705707093049370428275412697088
000000000000000000000000000000000000000

approx. = 1.0 x 10^92