# Message #2115

From: David Vanderschel <DvdS@Austin.RR.com>

Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)

Date: Mon, 07 May 2012 00:57:46 -0500

Brandon wrote:

> You can think of re-orienting a puzzle as permuting the

> colors in some way. For example, on a cube if you rotate

> the whole cube about the U face you’ve done the equivalent

> of a 4-cycle of the face colors in the equator between the

> U and D faces. In that sense there are 24 ways to

> "permute" the colors on a Rubik’s cube. The way the

> calculation is done though avoids counting this extra

> factor of 24. On the Rubik’s cube it is easy to avoid

> this extra factor of 24 because it has the centers.

Thank you, Brandon, for echoing exactly what I was

originally trying to explain to Melinda. However, you have

overlooked the same issue that I did on first pass. What

she is driving at is that there is a sense in which certain

arrangements, which appear to be distinct based on the

particular arrangement of sticker colors (even when

correcting for pile orientation), may be very closely

related if you are allowed to reassign the colors. The

relevant sense of similarity that Roice and I have tried to

point her to is conjugation by a symmetry. This is not the

same thing as reorientation of the pile, as it does produce

equivalent arrangements which cannot be made to match up

(sticker-color-wise) by straight reorientation. Please

refer to my last post, which attempts to explain this

distinction in greater detail.

Regards,

David V.

—– Original Message —–

From: "Brandon Enright" <bmenrigh@ucsd.edu>

To: <4D_Cubing@yahoogroups.com>

Cc: <melinda@superliminal.com>; <bmenrigh@ucsd.edu>

Sent: Sunday, May 06, 2012 9:45 PM

Subject: Re: [MC4D] Calculating the number of permutation of

2by2by2by2by2 (2^5)

On Sun, 06 May 2012 19:13:13 -0700

Melinda Green <melinda@superliminal.com> wrote:

> I think that your interpretation of my post is correct

> though I wonder about the quantities. For example, I would

> factor out (6-1)! = 120 from the 3D cube due to color

> symmetry, and (8-1)! = 5,040 for the 4D cube but then I’ve

> not yet read your citations.

Hi Melinda,

Regarding the factoring out the color permutations, it would

only be correct to do so in cases where it is possible to

actually "swap" two colors.

For example, on the 3x3x3 the centers of each face are fixed

relative to each other. You can never solve the puzzle such

that the white face is adjacent to the yellow face. Even if

you took away the stickers on the centers (void cube) you

still can’t because the 2-color edges and 3-color corners

define how the colors are related to each other.

You can think of re-orienting a puzzle as permuting the

colors in some way. For example, on a cube if you rotate

the whole cube about the U face you’ve done the equivalent

of a 4-cycle of the face colors in the equator between the U

and D faces. In that sense there are 24 ways to "permute"

the colors on a Rubik’s cube. The way the calculation is

done though avoids counting this extra factor of 24. On the

Rubik’s cube it is easy to avoid this extra factor of 24

because it has the centers.

For the 2x2x2 cube, the color scheme is still fully defined

by the corners but it doesn’t have centers. There are two

basic ways to calculate the number of distinct states for

puzzles like the 2x2x2. The first is to over count and then

divide by the total orientations, and the second is to fix

one of the pieces in space, call it solved, and then count

how all the rest of the puzzle can be permuted about this

piece.

There are some puzzles that can be solved into their mirror

color scheme (such as the Dino Cube) and some puzzles that

can be solved into any permutation of their colors such as

the Big Chop (and Little Chop). In the case where a puzzle

can be solved into the mirror state you’d divide by 2 and in

the case where all permutations of colors are available it

would be #colors!. I’m sure there are puzzles that can only

be solved into an even permutation of the colors so those

would be #colors! / 2.

Brandon