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.

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

~~We must show that reduction preserves the MOS property.~~

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.

~~The number of chunks is b, and gcd(a%b, (b-a%b)) = gcd(a%b, b) = gcd(a,b) by the Euclidean algorithm.~~

=== Binary generated scales with #L coprime to #s within each period are MOS ===

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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.
-=== Reduced words from chunking are binary and preserve the gcd of the step signature ===
-We must show that reduction preserves the MOS property.
-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.
-The number of chunks is b, and gcd(a%b, (b-a%b)) = gcd(a%b, b) =  gcd(a,b) by the Euclidean algorithm.
 === Binary generated scales with #L coprime to #s within each period are MOS ===