User:Inthar/MV3: Difference between revisions
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== MV3 Theorem 1== | == MV3 Theorem 1== | ||
''Suppose we have an MV3 scale word with steps x, y and z. With only one exception ("xyxzxyx"), 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 "xyxzxyx", along with their repetitions "xyxzxyx", etc.) Moreover, there always exists some "generator" interval for any max-variety-3 scale (other than two exceptions, "xyzyx" and "xyxzxyx") 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).'' | ''Suppose we have an MV3 scale word with steps x, y and z. With only one exception ("xyxzxyx"), 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 "xyxzxyx", along with their repetitions "xyxzxyx", etc.) Moreover, there always exists some "generator" interval for any max-variety-3 scale (other than two exceptions, "xyzyx" and "xyxzxyx") 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).'' | ||
Below, assume that the scale word S is not multiperiod. | |||
=== Lemma 1: S is pairwise-well-formed, i.e. sizes of chunks of any fixed letter form a MOS (except in the case "xyzyx") === | === Lemma 1: S is pairwise-well-formed, i.e. sizes of chunks of any fixed letter form a MOS (except in the case "xyzyx") === | ||
TODO: account for 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. | Now assume len(S) >= 6. | ||
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We now prove that except in the case xyxzxyx, if the scale is pairwise well-formed, the scale is generated by alternating generators. | We now prove that except in the case xyxzxyx, if the scale is pairwise well-formed, the scale is generated by alternating generators. | ||
To eliminate xyxzxyx we manually check all words up to length 7... (todo) | |||
Now assume len(S) >= 8 | Now assume len(S) >= 8. | ||
=== Proof of "ax by bz" (except in case xyxzxyx) === | === Proof of "ax by bz" (except in case xyxzxyx) === | ||
Assume len(S) >= 8. Write S in x's and q's, as above. By PWF, we have that the word of x's and q's is a MOS. | |||
*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-2) g0 + g2. | |||
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 (can always ensure this by taking octave complements of all the generators). QED. | |||
== MV3 Theorem 2 == | == 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).'' | ''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).'' | ||