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Necklaces

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

1/23/2006 10:13:44 AM

Almost done reading Michael Keith's "From Polychords to Polya". It's
a really neat book. I sent him an email and he sent a reply. The book
filled me in on aspects of Polya's theory that I was missing. My goal
is to combine what I learned from this book, and the
complementability problem (when you can swap colors in a necklace)
and some of my own ideas.

Although I may be mixing apples and oranges, I would like to develop
cycle indexes (and then, a Polya Polynomial) that could mix the Sn
groups for complementability with other things, like D4 X S3 symmetry.
For example, in 12-tET for hexachords this would be 26, which are the
26 letters of my hexachord system. It gets trickier with 3 colors,
(S3 complementability) because you are dealing with sets besides just
N/2 that you collapse for (that is, for example, hexachords in 12-et).

Why deal with 3 colors or more? Well, Michael Keith uses these
necklaces formulas to create "meta-beads" and "meta-necklaces" that
can be used to calculate maximum intervals, adjacencies, span, and
different sets in an interval class (for example, the diatonic
interval class of 1122222). Reducing for complementability could help
create other meta-necklaces, but I don't have an application for that
yet. (Okay, I am working backwards).

In the works is developing a good technique to pinpoint, for example
6-bead necklaces, with 2 beads of each of three different colors.
With Polya's technique you obtain 16 necklaces. Reducing for
complementability you obtain 5 possible necklaces. Right now
I am finding the pattern for which kinds of color-swaps are used
by different parts of the formula (that I learned in the Gilbert
and Riordan paper of 1961). For example, for 3 color necklaces
you have the S3 symmetry group. This is 6 possible swaps (1)(2)(3),
(1)(23), (3)(12), (123), (132). Figuring out which ones are used
for particular identity mappings under complementability will
lead me to a technique to pinpoint specific necklaces.

The technique for complementability took me a while to learn,
so that is why I would like to find some good applications. It's
a neat technique, and I mean that literally. For example, take
a neckace with 5 beads and three colors. You can divide each row
by 5 (transpositions) and each column by 6 (S3 complementability) and
the whole thing by 30. Here's the table below:

3^1 3^1 3^1 3^1 3^5
1^1 1^1 1^1 1^1 1^1 (x3)
___________________
6 6 6 6 246

Divide each column by 6:

1 1 1 1 41

Add this up, you get 45, divide by 5 (tranpositions) and Voila! You
get 9 necklaces reducing for transposition and complementability.
Of course it is worse for, say 12 beads. You can also throw in
mirror-inversion, and it gets more difficult. Expanding 2^n into
binomials of 3^n into multinomials give counts for specific subsets
of necklaces (Like 4 white and 8 black beads, or 4 white, 4 red and 4
blue beads) which is the method I learned on this newsgroup.

Why all this work on necklaces? And what does that have to do with Z-
Z-relations, Steiner Systems etc. (That's a work in progress).
I see that Z-relation is defined in Wikipedia now. And it's not the
Z-relation itself that is important. It is that fact that reducing
for Z-relations gives sets that are based on interval-vectors, which
I believe act like a kind of "harmonic-spectra" of a set. I have
found a certain clustering pattern of Z-relations, which holds up
completely for 22-tET, and almost for 31-tET. I am working on
analyzing 31-tET, and see if the clustering pattern holds up better
if you do NOT reduce for mirror-inversion. (One obtains Z-relations
in all subsets and classes (pairs, triplets, etc) which are multiples
of 5, for practically all Z-relations.)

Okay, Steiner Systems. Of note are S(5,6,12) and S(3,6,22). In S
(5,6,12) It's kind of interesting. 64 of 80 hexachord types are used.
also 28/35 of sets (reduced for Z-relation, based on interval
vectors) are used. I think 35/44 partitions are used, but I don't
have my notes with me. So I've talked for enough now. Thoughts anyone?

Thanks

Paul Hj