Rank-3 scale theorems: Difference between revisions

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* Assume ''S'' is a 2-step scale. Then ''S'' is ''slope-LQ'' if the slope between any two pair of points (representing a ''k''-mosstep) is one of the two nearest possible slopes (in the set {k/0,...,0/k}) to b/a.

* Say that ''S'' is ''floor-LQ'' if some mode of ''S'' gives the graph of floor(b/a*x).

* Say that a 2-step scale ''S'' is ''floor-LQ'' if some mode of ''S'' gives the graph of floor(b/a*x).

===== MV2 is equivalent to LQ in 2-step scales (WIP) =====

===== MV2 is equivalent to floor-LQ in 2-step scales (WIP) =====

Every mos is slope-LQ. This follows from the fact that every k-step is of the form pL qs or (p+1)L (q-1)s, where p/q <= a/b <= (p+1)/(q-1). This in turn follows from the fact that a mos has a generator, since every k-step is an octave reduction of either k perfect generators or k-1 perfect generators + 1 imperfect generator, and the perfect generator and the imperfect generator differ by changing one L to an s. If both the perfect generator and the imperfect generator had #L/#s > b/a (it must definitely be smaller than ceil(b/a) + 1, since the perfect generator is a collection of L...Ls chunks, and the size of such a chunk is <= ceil(b/a)), this would imply that n-1 perfect generators + 1 imperfect generator would not close at a period multiple. It is obvious that slope-LQ implies MV2.

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 * Assume ''S'' is a 2-step scale. Then ''S'' is ''slope-LQ'' if the slope between any two pair of points (representing a ''k''-mosstep) is one of the two nearest possible slopes (in the set {k/0,...,0/k}) to b/a.
-* Say that ''S'' is ''floor-LQ'' if some mode of ''S'' gives the graph of floor(b/a*x).
+* Say that a 2-step scale ''S'' is ''floor-LQ'' if some mode of ''S'' gives the graph of floor(b/a*x).
-===== MV2 is equivalent to LQ in 2-step scales (WIP) =====
+===== MV2 is equivalent to floor-LQ in 2-step scales (WIP) =====
 Every mos is slope-LQ. This follows from the fact that every k-step is of the form pL qs or (p+1)L (q-1)s, where p/q <= a/b <= (p+1)/(q-1). This in turn follows from the fact that a mos has a generator, since every k-step is an octave reduction of either k perfect generators or k-1 perfect generators + 1 imperfect generator, and the perfect generator and the imperfect generator differ by changing one L to an s. If both the perfect generator and the imperfect generator had #L/#s > b/a (it must definitely be smaller than ceil(b/a) + 1, since the perfect generator is a collection of L...Ls chunks, and the size of such a chunk is <= ceil(b/a)), this would imply that n-1 perfect generators + 1 imperfect generator would not close at a period multiple. It is obvious that slope-LQ implies MV2.