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 wolog we're looking at 3rds. Suppose I have 3 thirds (within a contiguous string of Y's and Z's). Then you get a 4th kind of third (within the whole scale) by using a string including an X, say XY or XZ.

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 T1 has an X inserted in the middle, then we can scoot it left or right until there are no X's in the middle then we (can assume we) have one of T1+X, T2+X, or T3+X. Scoot this 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.

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

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

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

By elimination we can assume that we have at least three Qs.

have some length (say that of max chunk) word with no q's,one q and 2 q's.

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

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

then by using size of [y[non-max 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...