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| == MV3 Examples ==
| | #redirect [[Rank-3 scale theorems#MV3 proofs]] |
| === Diasem ===
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| "Diasem[5]" is: LMLMs
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| For example (one rotation of) [[diasem]] is (right is 4/3, down right is 7/6)
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| x-x-x-x-x
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| x-x-x-x
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| This is LM LS LM LS L
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| here you're stacking 1/1-7/6-4/3
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| but if you extend to 14 notes you have to do it like this
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| x-x-x-x-x-x-x
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| x-x-x-x-x-x-x
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| <!--
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| XY X YX Z XY X YX Z XY
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| X = 28/27; Y = 9/8 minus 28/27; Z = 64/63
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| L = XY, M = X, S = Z
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| The tokens show what the old steps break down into
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| here you have two parallel chains of 4/3's, a 28/27 apart
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| Could i have switched the roles of X and Z here
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| Y' = 9/8 minus 64/63
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| Y'Z X ZY' Z Y'Z X Z Y'Z
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| L = Y'Z, M = X, S = Z
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| === "Diacot" ===
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| 5-limit MV3 detempering of dicot[10]: 81/80 10/9 16/15 10/9 81/80 10/9 16/15 10/9 81/80 (found by ks26), SLMLSLMLS
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| S LM LS LM LS
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| SL ML SL ML S-->
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| == MV3 Theorem 1==
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| ''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 two exception to this rule are "XYZYX", and "XYXZXYX", along with their repetitions "XYXZXYXXYXZXYX", 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).''
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| === Lemma 1: The word made by any two of the step sizes is a MOS (except in the case "XYZYX") ===
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| Todo: Account for XYZYX
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| Assume the scale word S is not multiperiod. To eliminate XYZYX we manually check all words up to length 5... (todo)
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| Now assume len(S) >= 6.
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| Suppose that some class C in the word of Ys and Zs has three sizes, T1, T2, T3.
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| 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.
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| So it's pretty obvious that you have to have MV2 within a contiguous string, but what about the whole string minus the X's?
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| This part needs to be more careful, taking into account where X can appear:
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| 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...)
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| === Lemma 2: Sizes of chunks of any fixed letter form a MOS (except in the case "XYZYX") ===
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| TODO: account for case XYZYX.
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| Assume the scale word S is not multiperiod. To eliminate XYZYX we manually check all words up to length 5... (todo)
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| Now assume len(S) >= 6.
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| WOLOG consider chunks of X. Use Q for both Y and Z.
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| say you have some intvl class (k steps) with 3 variants in X's and Q's:
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| * S1 = a1X + b1Q, represented by the word s1 in the MV3 scale
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| * S2 = a2X + b2Q, word s2
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| * S3 + a3X + b3Q, word s3
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| (si to be chosen later)
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| Say that the mos formed by the Ys and Zs is rY sZ, wolog r > s.
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| The # of chunks in rY sZ is s. The min chunk size is floor(r/s).
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| The idea is to keep extending si to the right until the chunk size in the MOS guarantees an extra variety.
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| By manually eliminating finitely many cases, we can assume that the scale is at least 5 notes and we have at least three Qs.
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| First we prove that chunk sizes can't differ by 2 or more.
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| have some length (say that of biggest chunk of X's) word with no q's and >=2 q's.
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| now y[biggest]z => contradiction bc two kinds of "one q"
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| so some non biggest chunk has to have y[chunk]z (or z[chunk]y)
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| 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)
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| u get [xyxxx...x]z, x[yxxxx...xz]x, y[x...xz], and all x's from the max chunk
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| If we have more q's than x's then we can't have "xx", so we're done.
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| This proves the claim about chunk sizes.
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| ''To be continued...''
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| === Finishing the proof ===
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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.
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| == MV3 Theorem 2 ==
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| ''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).''
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