MOS substitution
MOS substitution is a procedure for obtaining a ternary scale with arbitrary scale signature aLbmcs. Originally developed by Inthar for the purpose of adding aberrisma steps in an orderly manner to a MOS pattern aLbm (which we write in place of aLbs for convenience's sake, since s denotes the new steps added to the MOS) in the context of groundfault's aberrismic theory, MOS substitution is intended to take advantage of extra symmetry when a, c or b, c is not a coprime pair and generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales.
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]\displaystyle{ \mathsf{mos\_subst\_aberrize}(a, b, x, c, k), }[/math] where [math]\displaystyle{ x \in \{L, m\} }[/math] is the step size identified with s by the template MOS and [math]\displaystyle{ 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).
Facts
Let [math]\displaystyle{ M_{a,b}(x,y;k) }[/math] be the mode of axby that would have brightness k if x were L and y were s. For example, [math]\displaystyle{ M_{a,b}(5,2;5)(x,y) = xxyxxxy. }[/math] Let [math]\displaystyle{ n = a+b+c }[/math] and [math]\displaystyle{ q = a + c }[/math] (resp. [math]\displaystyle{ b + c }[/math]).
The following holds for [math]\displaystyle{ S = \mathsf{mos\_subst\_aberrize}(a, b, L, c, k) }[/math] (and mutatis mutandis, for [math]\displaystyle{ \mathsf{mos\_subst\_aberrize}(a, b, m, c, k) }[/math] as well):
- If the template MOS [math]\displaystyle{ T = T(m,X) = M_{b,a+c}(m,X;n-1) }[/math] is primitive, let [math]\displaystyle{ r }[/math] the count of X steps in a chosen (reduced) generator of [math]\displaystyle{ T. }[/math] Since [math]\displaystyle{ r }[/math] must be coprime to [math]\displaystyle{ n }[/math], [math]\displaystyle{ r }[/math]-steps in the filling MOS [math]\displaystyle{ F = M_{a,c}(L,s;k) }[/math] come in exactly 2 sizes, [math]\displaystyle{ iL+js }[/math] and [math]\displaystyle{ (i-1)L+(j+1)s }[/math], and taking this generator of [math]\displaystyle{ T }[/math] results in a generator sequence of length [math]\displaystyle{ q }[/math]. Letting [math]\displaystyle{ \mathsf{GS}(g_1, ..., g_{q}) }[/math] be this generator sequence, [math]\displaystyle{ g_j }[/math] is either [math]\displaystyle{ pm + iL + js }[/math] or [math]\displaystyle{ pm + (i-1)L + (j+1)s, }[/math] according as the j-th r-step in the sequence of stacked [math]\displaystyle{ r }[/math]-steps in the chosen mode of [math]\displaystyle{ F }[/math] is [math]\displaystyle{ iL + js }[/math] or [math]\displaystyle{ (i-1)L + (j+1)s }[/math] (We could have chosen to use the "darkest" mode of [math]\displaystyle{ T }[/math] instead, which corresponds to taking the circle of (n − r)-steps in F and is thus also valid.)
- Assume that template MOS [math]\displaystyle{ T = T(m,X) = M_{b,a+c}(m,X;n-1) }[/math] is primitive. Suppose that the perfect generator of T that we use has r-many X steps and that the imperfect generator has (r + 1)-many X steps. Suppose the sizes for r-steps in F are tL + us and (t − 1)L + (u + 1)s. Then the interval class of (r + 1)-steps has either (a) tL + (u + 1)s and (t − 1)L + (u + 2)s, or (b) tL + (u + 1)s and (t + 1)L + us.
- In case (a), S becomes a mos after deleting s steps for any k in {0, ..., q − 1}.
- In case (b), S becomes a mos after deleting s steps for k in {0, ..., v − 1}, where v is the number of generators stacked to obtain (t + 1)L + us in the filling MOS F.
Example
For 5L2m4s, we exploit gcd(b, c) = 2 and substitute 2m4s into the template MOS 5L6X (LXLXLXLXLXX). Since 2m4s has three distinct modes (ssmssm, smssms, and mssmss) and 5L6X is primitive, we obtain three distinct scales: LsLsLmLsLsm, LsLmLsLsLms, and LmLsLsLmLss. The first two are a chiral pair of billiard scales, and the last is achiral but not deletion-MOS. All three scales admit short generator sequences of 2-steps, respectively GS(L+s, L+s, L+m), GS(L+s, L+m, L+s), and GS(L+m, L+s, L+s), notably representing all 3 possible rotations of (L+s, L+m, L+s).