Message #2110

From: David Vanderschel <>
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

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

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

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.

David V.

—– Original Message —–
From: Melinda Green
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.


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 &quot;color symmetry&quot; constraint?


On 5/6/2012 2&#58;35 PM, Andrew Gould wrote&#58;

    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.