Recursive structure of MOS scales: Difference between revisions

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== Proofs ==

=== The equave splits into gcd(#L, #s) periods in a MOS ===

=== Preservation of generators ===

Assume there are more L's and s's, and that there is more than one s, and thus more than one chunk. Assume there is a generator. Assume the imperfect generator is bigger than the perfect generator. (If this isn't true, just use the inverted generator.) Because there are at least two chunk boundaries, and only one imperfect generator, there must be a chunk boundary with a perfect generator on top (the chunk boundary is on the left of the generator, e.g. <code>...|Ls|LLs|Ls...</code>). Because there's a chunk boundary to the left, there's an s just to the left of the left endpoint of the generator. If the rightmost step of this generator were an L, scooting the generator one step to the left would make it smaller, which contradicts the assumption that the imperfect generator is larger than perfect. Thus, the rightmost step of this generator is an s. That means the right endpoint of this generator falls on a chunk boundary. We already know the left endpoint also falls on a chunk boundary, so this perfect generator is still present as an interval in the reduced MOS.

@@ Line 423: / Line 423: @@
 == Proofs ==
+=== The equave splits into gcd(#L, #s) periods in a MOS ===
 === Preservation of generators ===
 Assume there are more L's and s's, and that there is more than one s, and thus more than one chunk. Assume there is a generator. Assume the imperfect generator is bigger than the perfect generator. (If this isn't true, just use the inverted generator.) Because there are at least two chunk boundaries, and only one imperfect generator, there must be a chunk boundary with a perfect generator on top (the chunk boundary is on the left of the generator, e.g. <code>...|Ls|LLs|Ls...</code>). Because there's a chunk boundary to the left, there's an s just to the left of the left endpoint of the generator. If the rightmost step of this generator were an L, scooting the generator one step to the left would make it smaller, which contradicts the assumption that the imperfect generator is larger than perfect. Thus, the rightmost step of this generator is an s. That means the right endpoint of this generator falls on a chunk boundary. We already know the left endpoint also falls on a chunk boundary, so this perfect generator is still present as an interval in the reduced MOS.