Recursive structure of MOS scales: Difference between revisions

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=== Preservation of the MOS property ===

(Assume that the mos has length ''n''; the notation ''w''(X, Y) for a word ''w''(L, s) in L, s means ''w'' with step sizes X substituted for L and Y substituted for s.)

(Assume that the mos has length n; the notation w(X, Y) for a word w(L, s) in L, s means w with step sizes X substituted for L and Y substituted for s.)

Suppose w(L, s) 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.

Suppose w(L, s) 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.

~~We claim~~ that ''T''<~~sub~~>s</~~sub~~> = the ~~set~~ of ~~indices~~ where ~~s steps occur~~ is a ~~[[Maximal evenness|maximally even]]~~ subset of ~~size |T_s| in '''Z'''~~/~~n'''Z''~~'.

[Proof is incomplete]<!--

Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes λ (corresponding to the chunk of L's of size r+1) and σ (corresponding the chunk of size r), and assume that W' is not 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 W'.

# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W'. We can assume W₂ begins in λ; otherwise we can slink W₂ to the right until it begins in λ, which is guaranteed to never decrease the number of λ's.

Here, pᵢ + qᵢ = k and we assume p₂ - p₁ ≥ 2.

Let K = p₁(r + 1) + q₁r + k. Consider the following sizes for (K+1)-steps in w:

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

# w₂(L, s) = the first K+1 letters of W₂(Lr+1s, Lrs) (which must be longer than w₁(L, s), since W₂ had more λ's than W₁)

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

Note that the latter two words have at most k s's, and that W₂(Lr+1s, Lrs) has k chunks in total.

It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise we would contradict either the length of W₂(λ, σ) or the mos property of w (Todo: bring back the cases and diagrams to prove this rigorously). If the intersection had exactly k-1 chunks, this implies that one substring is a proper subset of the other, a contradiction. Thus the intersection has to have exactly k chunks, implying w₂ = w₃, and that W₂(Lr+1s, Lrs) is exactly K+1 letters long, only one more than w₁(L, s). This contradicts the fact that W₂(λ, σ) has at least two more λ's than W₁(λ, σ).-->

=== Preservation of generators ===

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 === Preservation of the MOS property ===
-(Assume that the mos has length ''n''; the notation ''w''(X, Y) for a word ''w''(L, s) in L, s means ''w'' with step sizes X substituted for L and Y substituted for s.)
+(Assume that the mos has length n; the notation w(X, Y) for a word w(L, s) in L, s means w with step sizes X substituted for L and Y substituted for s.)
-Suppose w(L, s) 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.
+Suppose w(L, s) 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.
-We claim that ''T''<sub>s</sub> = the set of indices where s steps occur is a [[Maximal evenness|maximally even]] subset of size |T_s| in '''Z'''/n'''Z'''.
+[Proof is incomplete]<!--
+Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes λ (corresponding to the chunk of L's of size r+1) and σ (corresponding the chunk of size r), and assume that W' is not 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 W'.
+# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W'. We can assume W₂ begins in λ; otherwise we can slink W₂ to the right until it begins in λ, which is guaranteed to never decrease the number of λ's.
+Here, pᵢ + qᵢ = k and we assume p₂ - p₁ ≥ 2.
+Let K = p₁(r + 1) + q₁r + k. Consider the following sizes for (K+1)-steps in w:
+# w₁(L, s) = 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. w₁ is K+1 letters long.
+# w₂(L, s) = the first K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) (which must be longer than w₁(L, s), since W₂ had more λ's than W₁)
+# w₃(L, s) = the last K+1 letters of W₂(L<sup>r+1</sup>s, L<sup>r</sup>s)
+Note that the latter two words have at most k s's, and that W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) has k chunks in total.
+It suffices to consider the case where the intersection w₂ ∩ w₃ contains at least k-1 complete chunks, since otherwise we would contradict either the length of W₂(λ, σ) or the mos property of w (Todo: bring back the cases and diagrams to prove this rigorously). If the intersection had exactly k-1 chunks, this implies that one substring is a proper subset of the other, a contradiction. Thus the intersection has to have exactly k chunks, implying w₂ = w₃, and that W₂(L<sup>r+1</sup>s, L<sup>r</sup>s) is exactly K+1 letters long, only one more than w₁(L, s). This contradicts the fact that W₂(λ, σ) has at least two more λ's than W₁(λ, σ).-->
 === Preservation of generators ===