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

Definitions and theorems

Throughout, let S be a scale word in steps x, y, z (and assume all three of these letters are used).

PMOS

Definition: S is pairwise MOS (PMOS) if the result of equating any two of the step sizes is a MOS.

AG

Definition: S satisfies the alternating generator (AG) property if it satisfies the following equivalent properties:

1. S is generated by two chains of generators, either both of size m or one with size m and one of size m-1.
2. S can be built by stacking alternating generators, resulting in a circle of the form either g1 g2 ... g1 g2 g1 g3 or g1 g2 ... g1 g2 g3.

MV3 Theorem 1

The following are equivalent for a scale word S with steps x, y, z:

1. S is MV3.
2. S is PMOS, or S is of the form x'y'z'y'x' or its repetitions.
3. S is AG, or S is of the form x'y'z'y'x' or its repetitions, or x'y'x'z'x'y'x' or its repetitions.

Below, assume that the scale word S is not multiperiod.

Lemma 1: S is pairwise MOS (PMOS) except in the case "xyzyx"

TODO: account for case xyzyx.

To eliminate xyzyx we manually check all words up to length 5... (todo)

Now assume len(S) >= 6.

WOLOG consider chunks of x. Use q for any occurrence of either 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.

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[chunk]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...

PMOS implies AG (except in the case xyxzxyx)

We now prove that except in the case xyxzxyx, if the scale is pairwise MOS, then it is AG.

To eliminate xyxzxyx we manually check all words up to length 7... (todo)

Now assume len(S) >= 8.

PMOS -> Consider mos temperings

• in x, ξ (ξ = y or z), with gen g1 -> g1g1...g1g1' (g1' = imperfect gen)
• in y, η (η = x or z), with gen g2
• in z, ζ (ζ = x or y), with gen g3.

Denote their detemperings as G11, G12, G13, G21, G22, G23, G31, G32, G33.

To be continued...

AG implies "ax by bz"

Assuming the alternating generator property, we have two chains of generator g0 (going right). The two cases are:

O-O-...-O (m notes)
O-O-...-O (m notes)

and

O-O-O-...-O (m notes)
O-O-...-O (m-1 notes).

Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively.

In case 1, let g1 = (2,1)-(1,m) and g2 = (1,1)-(2,m). The circle of stacked g0 generators is (starting from top left): (m-1 g0's) g1 (m-1 g0's) g2. A scale step is always a same number k (which must be odd) of such generators gi. Assume (after taking octave complement) that a single step takes less than half of the generators. So a word corresponding to the scale step is formed by one of:

• k g0
• (k-1) g0 + g1
• (k-1) g0 + g2.

It is clear that the last two sizes must occur the same number of times.

In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. Let g3 be the leftover generator after stacking alternating g1 and g2. Then the generator circle looks like g1 g2 g1 g2 ... g1 g2 g3. Then the generators corresponding to a step are:

• k g1 + (k-1) g2
• (k-1) g1 + k g2
• (k-1) g1 + (k-1) g2 + g3

if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times. QED.

3-DE implies MV3

We prove that 3-DE + not abcba implies PMOS, which is known to imply MV3.

MV3 Theorem 2

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).