Message #2113

From: Brandon Enright <>
Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)
Date: Mon, 07 May 2012 02:45:27 +0000

Hash: SHA1

On Sun, 06 May 2012 19:13:13 -0700
Melinda Green <> 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

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

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.


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