Recursive structure of MOS scales: Difference between revisions

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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 ===

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

It now remains to show that reduction and expansion preserve 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 are MOS ===

@@ Line 436: / Line 436: @@
 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 ===
+=== Reduced words from chunking are binary and preserve the gcd of the step signature ===
 It now remains to show that reduction and expansion preserve 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 are MOS ===