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