# Message #2110

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

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

Date: Sun, 06 May 2012 20:14:50 -0500

I did not download Melinda’s message until after I posted my

own preceding. Thus mine was not written as a response to

Melinda. Nevertheless, there is a sense in which it does

respond:

I would argue that the way one should identify colors is by

the direction in which the corresponding stickers face in

the initial state. Actual color, as in red vs. yellow, is

really just a rendering issue and does not bear on the

nature of the puzzle. So, in this sense, changing the color

used for a given direction is not changing anything at all.

Now you could take it that reassigning the colors

corresponds to redefining the standard orientation of the

pile; and I am guessing that Melinda may have been thinking

of it in this sense. However, you would have to be careful

about that, as the pair of colors facing in opposite

directions on a given axis would have to remain the same on

any axis you might use for the pair in initial state.

Otherwise, it would be difficult to give ‘physical’ meaning

to a rearrangement of sticker colors. With that

restriction, we come back to the n!*2^(n-1) expression,

where n is the dimension. The 2n colors have to retain

their pairings, so that the 2^n factor corresponds to

choosing which of each pair faces in positive direction on

their axis. The n! factor comes from ways of associating

the color pairs with axes. (The factor of 1/2 is to count

only orientations and to exclude ways that create mirror

images.)

In this case, where the issue is to count all possible

distinct arrangements; I think the approach of counting only

arrangements when the pile has a standard orientation and

identifying the colors by the directions they face in

initial state is all that’s needed.

For really gory details, check out this article from way

back in ‘94:

http://www.math.rwth-aachen.de/~Martin.Schoenert/Cube-Lovers/Dan_Hoey__The_real_size_of_cube_space.html

or http://preview.tinyurl.com/ys3pf2

which is referenced, taking a simplified point of view, from

here: http://www.cube20.org/.

Though the above is just for the regular 3D puzzle, the

basic idea remains the same. In their approach, they

admitted not just orientations of the pile (of which there

are 24 in the 3D case) but also more general symmetry

transformations - i.e., including transformations which

create mirror images - for a total of 48 (the full 3!*2^3).

From a computational point of view that makes perfectly good

sense. From a point of view of manipulating the puzzles, it

does not make sense unless one’s implementation allows

reflections of the pile.

Regards,

David V.

—– Original Message —–

From: Melinda Green

To: 4D_Cubing@yahoogroups.com

Sent: Sunday, May 06, 2012 6:39 PM

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

2by2by2by2by2 (2^5)

I may have answered my own question which is that to account

for color symmetry we can simply divide by (n - 1)! where n

is the number of colors. For a 2-colored puzzle there is

only one color permutation whereas for 3 color puzzles there

are two because we can fix one color and either swap or not

swap the other two. (n - 1)! gives the number of unique

color swapping patterns. Is that right? I usually don’t

expect to fully understand the equations here.

-Melinda

On 5/6/2012 4:18 PM, Melinda Green wrote:

```
I feel that it's not just tricky but it is wrong in most<br>
conceptualizations of the idea of puzzle state<br>
spaces. Taking this natural idea one step further, I<br>
would argue that states that have identical patterns of<br>
stickers should be thought of as the same state. For<br>
example, if you scramble any twisty puzzle and then swap<br>
all red and green stickers, then I feel that you still<br>
have the same state in terms of permutations since<br>
anything you can say about one version also applies to<br>
the other. For example, twist one face of a Rubik's<br>
cube. For our purposes, it doesn't matter which face was<br>
twisted. When talking about that state with each other<br>
we will never think to ask about the particular colors.
Would anyone like to attempt to find the formula for the 3D<br>
and 4D cubes with this extra "color symmetry" constraint?
-Melinda
On 5/6/2012 2:35 PM, Andrew Gould wrote:
The choice between 31 and 32 comes down to how you<br>
define the locations of pieces. If you define all<br>
their locations relative to one of the pieces it's<br>
31, but if you define what moves and what doesn't<br>
for each twist you can make it 32. I note that for<br>
32, it would be tricky to say that rotating the<br>
entire puzzle doesn't change the state.
```