# Message #512

From: David Smith <djs314djs314@yahoo.com>

Subject: Re: [MC4D] On positions of the non-full puzzles

Date: Mon, 12 May 2008 15:16:39 -0700

Hello All,

Roice, your idea of trying to calculate how many possible puzzles could

exist is great! Once again, I have changed my plans and am now attempting

to solve this problem. My plan is to find a formula for the number of possible

puzzles with exactly k colors. Then, the total for all puzzles could easily be

found. In my mind, for a puzzle to be identical to another, it should function

entirely the same and have all of the same permutations. Therefore, the

colors themselves chosen for the faces of the puzzle are irrelevant; the relationship

between the colors is what counts. (Would you say an entirely blue 1-colored

Hyperminx is different from a red-colored one?)

Also, I would count 4D reflections as different puzzles because they would technically

be different (can’t turn one into the other in 4-space), although they would have the same

number of permutations and all of their permutations would be mirror images of each other.

So far, I have had to do a lot of research to try and find a solution. A lemma called

Burnside’s Lemma (also called the lemma that is not Burnside’s, apparently it was

well-known when he published it but somehow it got attributed to him) is critical

in my strategy. It is not to hard to see that only the colors and relative positions of

the centers matter; the problem is equivalent to finding out how many ways one

can distinctly color the faces of a 120-cell.

First, I am going to solve this problem for the Megaminx, then the Hyperminx.

I also had an idea for what seems would be an incomprehensibly hard problem:

Finding the total number of permutations of all possible puzzles! If I even

attempt to solve this, I would try the Megaminx first.

As for your question about each face of the Hyperminx having all different colors,

I can’t yet think of a mathematical way to approach this problem, although I

believe it almost definitely is possible, with all those permutations!

Thanks, Roice for giving me more puzzles to solve! I also appreciate the advice

you gave in your email to me.

All the Best,

David

Roice Nelson <roice3@gmail.com> wrote:

I changed the name of the email thread for this because my gmail conversation view was getting out of control with "Magic120Cell Realized" replies ;)

Anyway, along these lines, it is also interesting to think about the number of possible puzzles having a given number of colors. There is only one puzzle with 120 colors and one puzzle with a single color, but how many different puzzles with 9 colors are there? An upper bound is 120P9 = 3.79E18, but that has multiple counts of visually identical ones (equivalent after 4D rotations like you described). Since understanding the 4D view transforms will be key for what you are looking at too, maybe some of your investigations will help be able to give the final answer here. It does sound quite difficult.

Naturally once the calc for that is done, we’ll have to wonder what the total number of possible puzzles is (given the freedom to set any number and pattern of repeat colors desired), but that will just the be the sum of the answers for the 1…120 cases. Btw, I felt justified in having these extra puzzles because there is a common version of Megaminx that has 6 colors instead of 12 (opposite colors are the same). Hmmm, that just made me realize I guess I didn’t include the most relevant variant. It’d be nice to add at least one more puzzle then, a 60 colored puzzle where antipodal cells are the same color.

Also, somewhat related and worthy of note is that the full version of Hyperminx is unique because it has more colors (120) than stickers-per-cell (63). Contrast this with all the other puzzles (I’ve watched many frustratingly try to scramble my Rubik’s cube so well that no colors are repeated on a face, which is of course impossible and enjoyable to see someone discover). I wonder if it is possible to scramble the Hyperminx so that every sticker on any given cell is a different color? I’m not sure.

Roice

On Thu, May 8, 2008 at 8:29 PM, David Smith <djs314djs314@yahoo.com> wrote:

Roice, that was a great article! Some of those numbers make

the number I found look like nothing! Thank you again for

putting my result on your website.

spel_werdz_rite, thank you for verifying this result! I had

no idea anyone else had calculated this number.

I recently had another idea for Magic120Cell before I go

back to the n^4 cube. It seems like it will be very difficult,

but I am going to try to find the number of visually different

positions of each of the other variations of puzzles (the

2-colored, both 9-colored, and the 12-colored versions) of

Magic120Cell. This will involve accounting for the similarly

colored pieces (4-colored pieces with the same colors may not be

visually identical due to their orientation, and counting the pieces

will require the use of the Magic120Cell program), and the similarly

colored centers (accounting for apparently different positions

acctually being visually identical due to rotations of the entire

puzzle in 4-space; the corner orientation logic would also apply

to the centers for counting how many ways the they can be visually

identical when rotated. This would be made eaiser by imagining the

0-colored piece that Roice mentioned.) These are just a few quick

observations, there may be more complications I am not yet aware of.

All the best,

David

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