User:Inthar/MV3: Difference between revisions

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== MV3 Examples ==
#redirect [[Rank-3 scale theorems#MV3 proofs]]
=== 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
<!--
XY X YX Z XY X YX Z XY
 
X = 28/27; Y = 9/8 minus 28/27; Z = 64/63
 
L = XY, M = X, S = Z
 
The tokens show what the old steps break down into
here you have two parallel chains of 4/3's, a 28/27 apart
 
Could i have switched the roles of X and Z here
Y' = 9/8 minus 64/63
 
Y'Z X ZY' Z Y'Z X Z Y'Z
 
L = Y'Z, M = X, S = Z
=== "Diacot" ===
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
 
S LM LS LM LS
SL ML SL ML S-->
 
== 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).''
 
=== Lemma 1: Sizes of chunks of any fixed letter form a MOS (except in the case "xyzyx") ===
TODO: account for case xyzyx.
 
Assume the scale word S is not multiperiod. 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.
 
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[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...''
 
=== Lemma 2: The word made by any two of the step sizes is a MOS (except in the case "xyzyx") ===
Use Lemma 1?
 
=== Proof of the alternating-gens property (except in the case xyxzxyx) ===
We now prove that except in the case xyxzxyx, if the scale is pairwise well-formed, the scale is generated by alternating generators.
 
Assume the scale word S is not multiperiod. To eliminate xyxzxyx we manually check all words up to length 7... (todo)
 
Now assume len(S) >= 8
 
=== Proof of "aX bY bZ" ===
 
== 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).''

Latest revision as of 21:48, 23 June 2021

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