User:Inthar/MV3

MV3 Examples

Diasem

"Diasem[5]" is: LMLMs

For example (one rotation of) diasem is (right is 4/3, down right is 7/6)

x-x-x-x-x
 x-x-x-x

This is LM LS LM LS L here you're stacking 1/1-7/6-4/3 but if you extend to 14 notes you have to do it like this

x-x-x-x-x-x-x
     x-x-x-x-x-x-x

MV3 proof

Prove the following: "It is a mathematical fact that, with only one exception, at least two of the three steps must occur the same number of times. For example, it is possible to have a max-variety-3 scale with 3 small steps, 5 medium steps, and 3 large steps, because there are the same number of small steps as large steps. But a max-variety-3 scale with 3 small steps, 5 medium steps, and 4 large steps is impossible. (The one exception to this rule is "aabacab", along with its repetitions "aabacabaabacab", etc.) Because of this, there always exists some "generator" interval for any max-variety-3 scale (other than the one exception) such that the scale can be expressed as two parallel chains of this generator which are almost equal in length (the lengths are either equal, or differ by 1). Once you have chosen a rank-3 temperament and a specific generator interval, there is a mechanical procedure to generate all max-variety-3 scales of a certain size (of which there are, however, infinitely many)."

Claim: A MV3 scale always has some subset (made by combining some of the two-step intervals into one) that is also MV3.

Lemma 1: The word made by any two of the step sizes is a MOS

Suppose we have three sizes of a class, T1, T2, T3 of Y's and Z's.

Assume T1, T2, T3 occur within a contiguous string of Y's and Z's. Then you get a 4th variant of this class (within the whole scale) by using a string of the same length including an X.

So it's pretty obvious that you havve to have MV2 within a contiguous string, but what about the whole string minus the X's?

If the occurrence of any class has an X inserted in the middle, then we can scoot it left or right until we have one of T1(possibly with inserted X's)+X, T2(possibly with inserted X's)+X, or T3 (possibly with inserted X's)+X. Scoot the string with the least X's to the left and you lose the X on the right, and gain another non-X letter on the left so you get a fourth variant of this interval class that contains T1 + X, a contradiction. (Check this again...)

Lemma 2: Sizes of chunks of any fixed letter form a MOS

WOLOG consider chunks of X. Use Q for both Y and Z.

say you have some intvl class (k steps) with 3 variants in X's and Q's:

S1 = a1X + b1Q, represented by the word s1 in the MV3 scale
S2 = a2X + b2Q, word s2
S3 + a3X + b3Q, word s3

(si to be chosen later) Say that the mos formed by the Ys and Zs is rY sZ, wolog r > s. The # of chunks in rY sZ is s. The min chunk size is floor(r/s). The idea is to keep extending si to the right until the chunk size in the MOS guarantees an extra variety.

By manually eliminating finitely many cases, we can assume that the scale is at least 5 notes and we have at least three Qs.

First we prove that chunk sizes can't differ by 2 or more.

have some length (say that of biggest chunk of X's) word with no q's and >=2 q's.

now y[biggest]z => contradiction bc two kinds of "one q"

so some non biggest chunk has to have y[chunk]z (or z...y)

then by using size of [y[non-biggest chunk]z] you get a contradiction bc you can scoot to get an x (since consecutive q's cant happen if there are consecutive x's)

u get [xyxxx...x]z, x[yxxxx...xz]x, y[x...xz], and all x's from the max chunk

If we have more q's than x's then we can't have "xx", so we're done.

This proves the claim about chunk sizes.

To be continued...