Recursive structure of MOS scales: Difference between revisions
formatting; added Saga's proof of generator invariance |
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To find the generator you reduce to a simple pattern you know the generator of, then plug everything back in. | 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 | 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. | ||
=== Proof === | === Proof === | ||
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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. | 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. | ||
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. | 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. | ||
== Tree of MOS patterns == | == Tree of MOS patterns == | ||