# 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.