Recursive structure of MOS scales: Difference between revisions

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To find the generator you reduce to a simple pattern you know the generator of, then plug everything back in.

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 nL 1s and plug everything back in.

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 ''preserves'' the generator.

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.

Assume there is more than one s, and thus more than one chunk. 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, ~~scooching~~ 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 in the reduced ~~mos~~.

==== 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.

~~You~~ can build this generator on all of the chunk boundaries. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as ~~I just~~ showed before~~. So~~ this is an interval that's on all but one tone of the reduced ~~mos~~. ~~That's~~ a generator.

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.

~~To complete~~ the ~~proof~~ that the generator of ~~a mos is unique~~, ~~just reduce all~~ the ~~way down~~ to ~~nL1s~~ for ~~some n. Any generator~~ of the ~~original mos must~~ be ~~a generator of this nL1s~~. ~~The generator of nL1s~~ is ~~obviously unique~~, so the ~~generator~~ of the ~~original mos~~ is ~~unique~~.

==== 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.

== Tree of MOS patterns ==

== Uniqueness an existence of 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==

MOS patterns have parents and children. If 1L 1s is the root of the tree, with any generator between 0\1 (for paucitonic 1L 1s) and 1\2 (for equalized 1L 1s) Every node aL bs has two children, aL (a+b)s and (a+b)L as (these MOS patterns are [[sister]] MOS patterns; they are called such because they have the same parent). The generator range of aL bs splits at the mediant of the endpoints of the parent interval (which is the generator size for [[MOS step ratio spectrum|basic]] aL bs), giving the generator ranges of the two child patterns.

@@ Line 31: / Line 31: @@
 To find the generator you reduce to a simple pattern you know the generator of, then plug everything back in.
-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 nL 1s and plug everything back in.
+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 ''preserves'' the generator.
+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.
-Assume there is more than one s, and thus more than one chunk. 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, scooching 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 in the reduced mos.
+==== 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.
-You can build this generator on all of the chunk boundaries. All but one of these will be perfect. The perfect ones will have their right endpoint be a chunk boundary, as I just showed before. So this is an interval that's on all but one tone of the reduced mos. That's a generator.
+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.
-To complete the proof that the generator of a mos is unique, just reduce all the way down to nL1s for some n. Any generator of the original mos must be a generator of this nL1s. The generator of nL1s is obviously unique, so the generator of the original mos is unique.
+==== 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.
-== Tree of MOS patterns ==
+== Uniqueness an existence of 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==
 MOS patterns have parents and children. If 1L 1s is the root of the tree, with any generator between 0\1 (for paucitonic 1L 1s) and 1\2 (for equalized 1L 1s) Every node aL bs has two children, aL (a+b)s and (a+b)L as (these MOS patterns are [[sister]] MOS patterns; they are called such because they have the same parent). The generator range of aL bs splits at the mediant of the endpoints of the parent interval (which is the generator size for [[MOS step ratio spectrum|basic]] aL bs), giving the generator ranges of the two child patterns.