Message #1628

From: djs314djs314 <>
Subject: Order-2 Klein’s Quartic permutation count
Date: Sat, 30 Apr 2011 15:35:30 -0000

Hello everyone,

I’ve calculated the number of permutations of the order-2 Klein’s Quartic from MagicTile and got the following:

209!/(2*((7!)^29)*(6!)) =


I have done this calculation in half an hour, so there is a possibility of error. I’m leaving for some appointments soon, but still wanted to get this post in before I leave. :) I apologize if this value needs to be corrected.

I’m planning to calculate all of the MagicTile puzzles’ permutations, and all of the MagicCube4D puzzles’ permutations. Perhaps I’ll even try the ‘invent your own’ option which would cover all of the puzzles and an infinitude of others, if I grow two more brains by then. ;) Also, I’m going to calculate the specific values for each puzzle (e.g. the order-2 Klein’s Quartic, the order-3 Klein’s quartic, etc.) Then, I will generalize them into a general formula for each type of puzzle for arbitrary order.

It’s up to Melinda, Don and Roice if they want to use my results at all. There is obviously no pressure to make use of them in any way. I’m just doing it in my spare time for enjoyment, and to try to contribute to this mailing list in any small way I can. :) There are going to be many results, and who needs all of these formulas anyway? :)

I should mention that the above count is the number of visually distinguishable permutations of Klein’s Quartic. Note that this number is different from the number of visually distinguishable permutations we see on the screen of MagicTile, because in MagicTile the center is fixed in both position and rotation. (The fact that there are a finite number of pieces would actually not come into play from this point of view.) The goal of the permutation count is to count the number of actual Klein’s Quartic positions that can occur in the hyperbolic plane, not what we can see on the screen. (‘Visually distinguishable’ is from the point of view of being inside the hyperbolic plane, as we always count visually distinguishable permutations from the point of view of the space it resides in, just as I did for the n^4 and n^d cubes. Otherwise, from a Euclidean point of view we would see the hyperbolic plane as a unit disc, and some pieces would appear larger than others.)

I would like to once again thank Melinda, Don, Jay, Roice, Audrey and anyone else I have missed for creating these excellent puzzles! :) They have provided much enjoyment for myself and so many others.

(It looks like my other message is going to be duplicated. My apologies.)

All the best,