Recursive structure of MOS scales: Difference between revisions

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=== Example ===

LLsLsLsLLsLs becomes LssLs, becomes Ls.

~~=== Proof ===~~

Suppose we had three chunks L...s with n, n+1 and n+2 'L's. Then we have a length n+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.

[to be continued: proving MOS. idea: Each step of the reduced scale can be associated with, not necessarily the whole chunk, but the interval that spans the length of the shortest chunk. These chunks and multiples thereof are max variety 2 in the original scale, and we can maybe lift that to the reduced scale directly]

== Finding the MOS pattern from nLms ==

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If you do that to 5L 7s you get 5L 2s: LLLsLLs. You know that the generator of 5L 2s is LLLs, so using the above procedure for finding the pattern, the generator of 5L 7s is LsLsLss. If you don't know anything you just reduce all the way to 1L 1s and plug everything back in.

~~=== Proof ===~~

The reason why this works is that the reduction both ''preserves'' and ''reflects'' the generator; see the ''proofs'' section.

The reason why this works is that the reduction both ''preserves'' and ''reflects'' the generator.

~~The generator is the generic interval which is always the same specific interval except in one place~~; ~~we call it ''generator'' because this implies it can be chained together to generate~~ the ~~whole pattern.~~

~~==== Preservation ====~~

~~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.

This interval is, in fact, the generator of the reduced scale. Indeed, you can build this generator on all of the chunk boundaries of the non-reduced scale. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as we showed before, so this is an interval that's on all but one tone of the reduced scale. This is a generator.

~~==== Reflection ====~~

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==~~

~~===Proof===~~

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, either L or s. However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator~~.

==Tree of MOS patterns==

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(0\1, 1\3) (1\3, 1\2)

(0\1, 1\4) (1\4, 1\3) (1\3, 2\5) (2\5, 1\2)

== Proofs ==

=== Preservation of the MOS property ===

Suppose we had three chunks L...s with n, n+1 and n+2 'L's. Then we have a length n+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.

[to be continued: proving MOS. idea: Each step of the reduced scale can be associated with, not necessarily the whole chunk, but the interval that spans the length of the shortest chunk. These chunks and multiples thereof are max variety 2 in the original scale, and we can maybe lift that to the reduced scale directly]

=== 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.

This interval is, in fact, the generator of the reduced scale. Indeed, you can build this generator on all of the chunk boundaries of the non-reduced scale. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as we showed before, so this is an interval that's on all but one tone of the reduced scale. This is a generator.

=== 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, either L or s. However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator.

@@ Line 11: / Line 11: @@
 === Example ===
 LLsLsLsLLsLs becomes LssLs, becomes Ls.
-=== Proof ===
-Suppose we had three chunks L...s with n, n+1 and n+2 'L's. Then we have a length n+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.
-[to be continued: proving MOS. idea: Each step of the reduced scale can be associated with, not necessarily the whole chunk, but the interval that spans the length of the shortest chunk. These chunks and multiples thereof are max variety 2 in the original scale, and we can maybe lift that to the reduced scale directly]
 == Finding the MOS pattern from nLms ==
@@ Line 35: / Line 30: @@
 If you do that to 5L 7s you get 5L 2s: LLLsLLs. You know that the generator of 5L 2s is LLLs, so using the above procedure for finding the pattern, the generator of 5L 7s is LsLsLss. If you don't know anything you just reduce all the way to 1L 1s and plug everything back in.
-=== Proof ===
+The reason why this works is that the reduction both ''preserves'' and ''reflects'' the generator; see the ''proofs'' section.
-The reason why this works is that the reduction both ''preserves'' and ''reflects'' the generator.
-The generator is the generic interval which is always the same specific interval except in one place; we call it ''generator'' because this implies it can be chained together to generate the whole pattern.
-==== Preservation ====
-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.
-This interval is, in fact, the generator of the reduced scale. Indeed, you can build this generator on all of the chunk boundaries of the non-reduced scale. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as we showed before, so this is an interval that's on all but one tone of the reduced scale. This is a generator.
-==== Reflection ====
-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==
-===Proof===
-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, either L or s. However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator.
 ==Tree of MOS patterns==
@@ Line 66: / Line 44: @@
   (0\1, 1\3)            (1\3, 1\2)
   (0\1, 1\4) (1\4, 1\3) (1\3, 2\5) (2\5, 1\2)
+== Proofs ==
+=== Preservation of the MOS property ===
+Suppose we had three chunks L...s with n, n+1 and n+2 'L's. Then we have a length n+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.
+[to be continued: proving MOS. idea: Each step of the reduced scale can be associated with, not necessarily the whole chunk, but the interval that spans the length of the shortest chunk. These chunks and multiples thereof are max variety 2 in the original scale, and we can maybe lift that to the reduced scale directly]
+=== 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.
+This interval is, in fact, the generator of the reduced scale. Indeed, you can build this generator on all of the chunk boundaries of the non-reduced scale. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as we showed before, so this is an interval that's on all but one tone of the reduced scale. This is a generator.
+=== 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, either L or s. However, the previous proof showed that reduction reflects generators, and so by induction all MOS scales have a single generator.