Message #2121

From: David Vanderschel <DvdS@Austin.RR.com>
Subject: Re: [MC4D] Calculating the number of permutation of 2by2by2by2by2 (2^5)
Date: Mon, 07 May 2012 20:34:45 -0500

Melinda wrote:
>It sounds like we are getting closer to closure.

Does this mean that you now understand why you must maintain
the pairs of opposing colors when reassigning sticker
colors? With all my effort at trying to explain it for you,
I was hoping for an "I see now!" I was also hoping that you
would realize that it is easier to talk about conjugation by
a symmetry than to get involved in talking about remapping
sticker colors, which are actually somewhat tangential to
the real problem.

>The main MC4D page claims to give the exact counts for the
>3D and 4D puzzles. Is that not true? If not, then I need to
>change it.

It is true. Also the counts that David Smith came up with
for 5D are intended to be exact. (I am not ruling out the
possibility of an error.) As I said, the problem that Eric
and David were solving is much more straightforward and
admits a precise answer. The problem only arises when you
start trying to get the number of equivalence classes of the
distinct states they counted that are related through
conjugation by a symmetry, which is the complication that it
would seem you were trying to inject. That would result in
a reduction of Eric’s and David’s numbers by a factor close
to (but less than) n!*2^n. But it is not practical to do
that in an exact sense.

>Please help me to understand the part that is "not
>quite". Here is what I am hearing: The main thing that makes
>it tricky to be exact seems to involve mirror symmetry. In
>particular, we would need to account for some relatively
>rare cases of states that happen to have mirror symmetry,
>and that counting those special cases appears to be very
>difficult.

That is my understanding; but I must confess that I have not
myself fully absorbed all the gory details of the "real
size" article. There may be other relevant symmetries
besides mirroring.

>If it can’t be done then it can’t be done. If that is the
>case, then what is our best estimate of the truly unique
>positions when accounting for color and mirror symmetries,
>and in particular, what is the term that we need to divide
>by and how is it derived?

I say don’t try. Stay with the distinct arrangement counts
we have. They are meaningful. Since you have an interest
in the equivalence classes of these arrangements under
conjugation by a symmetry, you might want to add the comment
that the number of such equivalence classes will be less by
a factor close to n!*2^n. Regarding the nature of the
issue, you could just link to the "real size" article,
saying that the analogous problem exists for any dimension.

You have already worked out a simple example giving
representative members of the equivalence classes for the
case of the 2D puzzle with mirroring twists allowed. I
would hope you would remember the previous occasion on which
some folks on the list tried to explain the relationship of
your equivalence classes to conjugation by a symmetry:
http://games.groups.yahoo.com/group/4D_Cubing/message/1843
Perhaps that previous discussion will make better sense to
you now.

Regards,
David V.