Recursive structure of MOS scales: Difference between revisions

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=== Preservation of the MOS property ===

~~Suppose we had three chunks L...s with n, n+1 and n+2 'L's. Then we have a length n+2 subword that's only 'L's, one~~ that ~~has one s at~~ the ~~end and one that~~ has ~~two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes that differ by 1~~.

(Assume that the mos has length n.)

[to be ~~continued: proving MOS~~. ~~idea: Each~~ step of the reduced ~~scale can be associated~~ with, ~~not necessarily~~ the ~~whole chunk~~, ~~but~~ the ~~interval that spans~~ the ~~length of~~ the ~~shortest chunk~~. ~~These chunks~~ and ~~multiples thereof are max variety~~ 2 in the original ~~scale~~, and ~~we can maybe lift~~ that to the ~~reduced scale directly~~]

Suppose we had three chunks L...s with r, r+1 and r+2 'L's. Then we have a length r+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes.

Proving that the chunk sizes form a mos is equivalent to proving that the places where the s's are in the mos form a mos in n-edo (which must be [[maximally even]] by above).

For now assume r ≥ 1. Let W' be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that the s's does not form a mos. Then for some k, W' must have k-steps of the following sizes:

# p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in the reduced word, corresponding to an interval in the mos with (p₁(r + 1) + q₁r) L's and k s's

# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in the reduced word, corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking the word to the right if necessary we assume it begins in λ.

where pᵢ + qᵢ = k and p₂ - p₁ ≥ 2

Let K = p₁(r + 1) + q₁r + k. Then we have at least 3 different sizes for (K+1)-steps:

# w₁ = the word sW₁(Lr+1s, Lrs) [W₁ interpreted as a subword of the original mos], with (k + 1) s's

# w₂ = the first K+1 letters of W₂(Lr+1s, Lrs)

# w₃ = the last K+1 letters of W₂(Lr+1s, Lrs)

If w₂ contains fewer complete chunks than w₃, we are done.

If w₂ and w₃ (which have the same length) contain the same number of complete chunks, one case is (X denotes a chunk boundary, < > are chunk boundaries that are also the boundary of the word, [] are non chunk boundary word boundaries.

w2: <L ... sXL ... sXL ... sXL ... sXL ...]

w3: <L ... sXL ... sXL ... sXL ... s>

This implies that the last chunk is bigger than the first one, a contradiction because w₂ begins in λ.

In the case

w1: sL ... [k s's]

w2: <L ... sXL ... sXL ... sXL ... s>

w3: <L ... sXL ... sXL ... sXL ... s>

Truncate the strings as follows:

w1': L ... [k s's]

w2': <L ... sXL ... sXL ... sXL ... ] (lop off one s, so one fewer s than w3)

w3': <L ... sXL ... sXL ... sXL ... s> (at most k-1 s's)

to get the desired counterexample to mos.

So we must have

w2: <L ... sXL ... sXL ... sXL ... sXL ...]

w3: [... sXL ... sXL ... sXL ... sXL ... s>

or

w2: <L ... sXL ... sXL ... sXL ... sX]

w3: [... sXL ... sXL ... sXL ... sXL ... L s>

(⇒ w₃ has one more s).

If w₂ has more complete chunks than w₃: the only case to consider is when w₂ has one more chunk.

w2: <L ... sXL ... sXL ... sXL ... sXL ... sX(L ...)

w3: <L ... sXL ... sXL ... sXL ... s>

⇒ w₂ has more s's.

The following two cases remain:

w2: <L ... sXL ... sXL ... sXL ... sXL ... sXL ...]

w3: [ ... sXL ... sXL ... sXL ... sXL ... s>

w2: <L ... sXL ... sXL ... sXL ... sXL ... sXL ...]

w3: <L ... sXL ... sXL ... sXL ... s>

Both are contradictions since the first chunk has to be λ.

=== Preservation of generators ===

@@ Line 340: / Line 340: @@
 === Preservation of the MOS property ===
-Suppose we had three chunks L...s with n, n+1 and n+2 'L's. Then we have a length n+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes that differ by 1.
+(Assume that the mos has length n.)
-[to be continued: proving MOS. idea: Each step of the reduced scale can be associated with, not necessarily the whole chunk, but the interval that spans the length of the shortest chunk. These chunks and multiples thereof are max variety 2 in the original scale, and we can maybe lift that to the reduced scale directly]
+Suppose we had three chunks L...s with r, r+1 and r+2 'L's. Then we have a length r+2 subword that's only 'L's, one that has one s at the end and one that has two 's's on either side, which means that the original scale was not MOS. Therefore the reduced word has two step sizes.
+Proving that the chunk sizes form a mos is equivalent to proving that the places where the s's are in the mos form a mos in n-edo (which must be [[maximally even]] by above).
+For now assume r ≥ 1. Let W' be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that the s's does not form a mos. Then for some k, W' must have k-steps of the following sizes:
+# p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in the reduced word, corresponding to an interval in the mos with (p₁(r + 1) + q₁r) L's and k s's
+# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in the reduced word, corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking the word to the right if necessary we assume it begins in λ.
+where pᵢ + qᵢ = k and p₂ - p₁ ≥ 2
+Let K = p₁(r + 1) + q₁r + k. Then we have at least 3 different sizes for (K+1)-steps:
+# w₁ = the word sW₁(L<sup>r+1</sup>s, L<sup>r</sup>s) [W₁ interpreted as a subword of the original mos], with (k + 1) s's
+# w₂ = the first K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s)
+# w₃ = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s)
+If w₂ contains fewer complete chunks than w₃, we are done.
+If w₂ and w₃ (which have the same length) contain the same number of complete chunks, one case is (X denotes a chunk boundary, < > are chunk boundaries that are also the boundary of the word, [] are non chunk boundary word boundaries.
+ w2: <L ... sXL ... sXL ... sXL ... sXL ...]
+ w3:         <L ... sXL ... sXL ... sXL ... s>
+This implies that the last chunk is bigger than the first one, a contradiction because w₂ begins in λ.
+In the case
+ w1:  sL ... [k s's]
+ w2: <L ... sXL ... sXL ... sXL ... s>
+ w3:         <L ... sXL ... sXL ... sXL ... s>
+Truncate the strings as follows:
+ w1':  L ... [k s's]
+ w2': <L ... sXL ... sXL ... sXL ... ] (lop off one s, so one fewer s than w3)
+ w3':         <L ... sXL ... sXL ... sXL ... s> (at most k-1 s's)
+to get the desired counterexample to mos.
+So we must have
+ w2: <L ... sXL ... sXL ... sXL ... sXL ...]
+ w3:   [... sXL ... sXL ... sXL ... sXL ... s>
+or
+ w2: <L ... sXL ... sXL ... sXL ... sX]
+ w3:   [... sXL ... sXL ... sXL ... sXL ... L s>
+(⇒ w₃ has one more s).
+If w₂ has more complete chunks than w₃: the only case to consider is when w₂ has one more chunk.
+ w2: <L ... sXL ... sXL ... sXL ... sXL ... sX(L ...)
+ w3:                 <L ... sXL ... sXL ... sXL ... s>
+⇒ w₂ has more s's.
+The following two cases remain:
+ w2: <L ... sXL ... sXL ... sXL ... sXL ... sXL ...]
+ w3:          [ ... sXL ... sXL ... sXL ... sXL ... s>
+ w2: <L ... sXL ... sXL ... sXL ... sXL ... sXL ...]
+ w3:                 <L ... sXL ... sXL ... sXL ... s>
+Both are contradictions since the first chunk has to be λ.
 === Preservation of generators ===