Message #2114

From: David Vanderschel <>
Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)
Date: Mon, 07 May 2012 00:23:16 -0500

Melinda wrote:
>David: Regarding your comments, I didn’t completely follow
>your description but I do not think we are talking about the
>same thing but I could easily misunderstand. Here is a way
>that I think might clarify what I am saying: Imagine that
>you toss me a scrambled Ribik’s cube. I then pick two
>sticker colors, peal them off the cube and swap them. I may
>repeat that process several times and then toss it back to
>you. I claim that I’ve not changed your puzzle state at

I agree with that. That was the point I was making about
the specific colors just being a rendering issue. The
assignment of colors to directions in initial state (or a
subsequent state resulting from some specific twist
sequence) does not change the nature of the puzzle one whit.

>You said:

>"you could take it that reassigning the colors corresponds
>to redefining the standard orientation of the pile."

>This is not what I am saying.

Sorry. I did not claim that you were. The sentence is in
subjunctive mood. I was searching for a meaningful
something that you might have meant. Nevertheless, whether
you want to agree yet or not, I do still believe that, for
what you are trying to do to work right, it _does_ need to
correspond to a reorientation of the pile.

>If for example I take a pristine Rubik’s cube and swap the
>colors of adjacent faces I will get a coloration that you
>cannot duplicate by reorienting the puzzle.

I don’t see the relevance of this sort of recoloring, as it
does not change the puzzle at all. Not only do there exist
such recolorings that cannot be achieved by reorientation
(or even twisting), there is no motivation for wishing to be
able to do so.

The relevant issue would be to consider two puzzles which
started with the same color scheme and consider different
arrangements of the these two puzzles. Now you may ask, "Is
there a way to remap the sticker colors on one of them
(still maintaining, relative to initial state, the same
opposing pairs for any given axis), so that the resulting
pattern of sticker colors is identical?" If so, we can
regard the two patterns, which initially appeared to be
different based on the particular sticker colors as
presented, as being the same for all practical purposes. I
think this is the sort of equivalence you seek.

It is possible to explain the above approach in terms of
conjugation, without having to worry about specific colors
(which I claim are irrelevant). The way to do it is to
regard a state as being that which results from a particular
permutation of the stickers relative to their positions in
the space (no matter what colors). Now a symmetry
transformation of the cube, when applied to the whole pile,
can also be regarded as inducing a permutation of the
stickers. So consider two states A and B resulting from
permutations Pa and Pb. If there is a permutation Ps which
results from a symmetry transformation of the n-cube such
that Pa = Ps*Pb*Ps’ (‘ for inverse, * for composition), then
we say that the states A and B are equivalent. Indeed, they
are equivalent under conjugation by a symmetry.

One way to understand the above is as follows. Suppose I
have a twist sequence that generates state B starting from
initial state, and I represent that sequence in a fixed
system of space coordinates. Now suppose, starting from
initial state, that I reorient the whole puzzle using the
symmetry transform that produces permutation Ps, and then
apply my twist sequence with the (otherwise initialized)
pile in that different orientation. Then I put the pile
back in its original orientation by applying the inverse of
the symmetry transformation to the whole pile. The
resulting permutation of stickers is now Pa, resulting in
state A. There is a strong sense in which the very same
twist sequence produces both states A and B.

Coming back to Melinda’s sticker color point of view, one
can perceive the reorientation of the puzzle before applying
the twist sequence as a reassignment of the colors; as this
is indeed the effect it has on the resulting state in the
sense in which I think she wishes to compare them. (But do
note that you cannot violate the opposing color pair matchup
by simply reorienting the pile.) In dimensions higher than
3, the possibilities for reorienting the pile become much
more complex.

>Your permutations are therefore only a subset of it’s color
>symmetry. Agreement or not, all three of us seem to come
>up with different numbers. Can someone set us straight?

Melinda, I don’t know about "different numbers", but I think
I have clarified the issue of what it means for arrangements
to be equivalent under conjugation by a symmetry transform.
Furthermore, I do think that it is the same basic idea that
you are trying to get at with your concept of rearranging
sticker colors. Not only is the conjugation approach more
amenable to mathematical treatment, but, as far as I am
concerned, the conjugation approach is much clearer because
there is no need to talk about specific colors. They can be
whatever they are as determined by the direction they face
in initial state. Your reassignment corresponds to what
results from the reorientation by the symmetry transform.
It is better to view that as primarily a permutation of
coordinate axes with an associated direction sign on each
axis, so as to prevent violating the opposing sticker color
pair matchup. I do believe that the process is meaningless
if you do violate the opposing pair matchups, as there is no
corresponding ‘physical’ process achievable with the
puzzles. (E.g., as you observed, there is no linear
transformation of n-space that would swap the stickers on
just two adjacent facets of a puzzle.)

I do not think there is any disparity between what I am
saying and what Roice was saying. I am hoping that you can
understand it so that all three of us can be in agreement.
If you really believe that you are driving at some sort of
equivalence that goes beyond conjugation by a symmetry, then
I believe the burden is on you to spell it out in much
greater detail, identify its utility, and make it meaningful
in terms of the ‘mechanics’ of the puzzles.

David V.