# 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