MOS substitution: Difference between revisions

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Take for example d = (a, c) (:= gcd(a, c)), let a' = a/d and c' = c/d. Consider the MOS word (a + c)Xbm, which we call the ''template MOS''. The most even arrangement of a'-many L steps and c'-many s steps is the MOS a'Lc's, so this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the X's. Fixing a choice of which X in (a + c)Xbm you start from, you have to choose a mode of a'Lc's. (Todo: count the distinct choices.) If a' = c' = 1 (equivalently if a = c), we obtain a balanced (thus MV3) ternary scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'Lc's. Of course, one may do this using template MOS aL(b + c)X and filling MOS (b/(b, c))m (c/(b, c))s instead.

We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{L, m\}</math> is the step size identified with s by the template MOS and ~~<math>~~k ~~\in \{0, 1, ..., (b + c)/(b,c)-1\}</math>~~ is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).

We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{L, m\}</math> is the step size identified with s by the template MOS and k is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).

== Facts ==

The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, L, c, k)</math> (and mutatis mutandis, for <math>\mathsf{mos\_subst\_aberrize}(a, b, m, c, k)</math> as well):

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 Take for example d = (a, c) (:= gcd(a, c)), let a' = a/d and c' = c/d. Consider the MOS word (a + c)Xbm, which we call the ''template MOS''. The most even arrangement of a'-many L steps and c'-many s steps is the MOS a'Lc's, so this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the X's. Fixing a choice of which X in (a + c)Xbm you start from, you have to choose a mode of a'Lc's. (Todo: count the distinct choices.) If a' = c' = 1 (equivalently if a = c), we obtain a balanced (thus MV3) ternary scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'Lc's. Of course, one may do this using template MOS aL(b + c)X and filling MOS (b/(b, c))m (c/(b, c))s instead.
-We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{L, m\}</math> is the step size identified with s by the template MOS and <math>k \in \{0, 1, ..., (b + c)/(b,c)-1\}</math> is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).
+We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{L, m\}</math> is the step size identified with s by the template MOS and k is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).
 == Facts ==
 The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, L, c, k)</math> (and mutatis mutandis, for <math>\mathsf{mos\_subst\_aberrize}(a, b, m, c, k)</math> as well):