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)'''X'''b'''m''', 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)'''X'''b'''m''' 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 a'''L'''(b + c)'''X''' and filling MOS (b/(b, c))'''m''' (c/(b, c))'''s''' instead.

We tentatively denote the resulting scale <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, x, k),</math> where <math>x \in \{\mathbf{L}, \mathbf{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 mode with the most '''X'''s at the ~~beginning~~).

We tentatively denote the resulting scale <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, x, k),</math> where <math>x \in \{\mathbf{L}, \mathbf{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 mode with the most '''s'''s at the end).

== Facts ==

The following holds for <math>S = \mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{L}, k)</math> (and after replacing '''L''' with '''m''' and a with b, for <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{m}, 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)'''X'''b'''m''', which we call the ''template MOS''. The most even arrangement of a'-many '''L''' steps and c'-many '''s''' steps is the MOS a'<b>L</b>c'<b>s</b>, 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)'''X'''b'''m''' you start from, you have to choose a mode of a'<b>L</b>c'<b>s</b>. (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'<b>L</b>c'<b>s</b>. Of course, one may do this using template MOS a'''L'''(b + c)'''X''' and filling MOS (b/(b, c))'''m''' (c/(b, c))'''s''' instead.
-We tentatively denote the resulting scale <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, x, k),</math> where <math>x \in \{\mathbf{L}, \mathbf{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 mode with the most '''X'''s at the beginning).
+We tentatively denote the resulting scale <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, x, k),</math> where <math>x \in \{\mathbf{L}, \mathbf{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 mode with the most '''s'''s at the end).
 == Facts ==
 The following holds for <math>S = \mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{L}, k)</math> (and after replacing '''L''' with '''m''' and a with b, for <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{m}, k)</math> as well):