Recursive structure of MOS scales: Difference between revisions

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=== Reflection of generators ===

Suppose there is a generator after chunking. Assume also that there are more L's than s's, and that the imperfect generator is larger than the perfect generator. This means that this interval, built on the chunk boundaries after expanding, will still be "a generator" on the chunk boundaries: it will be the same size on all but one of the chunk boundaries. Take one of the "perfect" (smaller) intervals. We showed previously that, before expansion, the step to its right was an L, just like its first step; therefore we can scoot it over for the entirety of the first chunk and keep it perfect. We now come to another chunk. If a "perfect" interval can be built on it then we can repeat. If the interval is imperfect, then we know that the step to its right is an s, which will make it smaller and thus perfect, and we can keep doing this for the remainder of the chunk. Therefore, the interval is a generator after expansion as well, since it is only imperfect in one position.

~~=== Uniqueness and existence of the generator===~~

Every MOS can be reduced to nL 1s (or 1L ns). Each step of the reduction decreases either the number of L's or the number of s's (or both), so one of them must reach 1 at some point. ''(note that reducing further gets us to 1L 0s, which has a period, but no generator per se)''

It is clear that the MOS nL 1s has a unique generator, L (or its inversion). However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator.

=== Reduced words from chunking are binary and preserve the gcd of the step signature ===

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 === Reflection of generators ===
 Suppose there is a generator after chunking. Assume also that there are more L's than s's, and that the imperfect generator is larger than the perfect generator. This means that this interval, built on the chunk boundaries after expanding, will still be "a generator" on the chunk boundaries: it will be the same size on all but one of the chunk boundaries. Take one of the "perfect" (smaller) intervals. We showed previously that, before expansion, the step to its right was an L, just like its first step; therefore we can scoot it over for the entirety of the first chunk and keep it perfect. We now come to another chunk. If a "perfect" interval can be built on it then we can repeat. If the interval is imperfect, then we know that the step to its right is an s, which will make it smaller and thus perfect, and we can keep doing this for the remainder of the chunk. Therefore, the interval is a generator after expansion as well, since it is only imperfect in one position.
-=== Uniqueness and existence of the generator===
-Every MOS can be reduced to nL 1s (or 1L ns). Each step of the reduction decreases either the number of L's or the number of s's (or both), so one of them must reach 1 at some point. ''(note that reducing further gets us to 1L 0s, which has a period, but no generator per se)''
-It is clear that the MOS nL 1s has a unique generator, L (or its inversion). However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator.
 === Reduced words from chunking are binary and preserve the gcd of the step signature ===