MOS substitution

MOS substitution is a procedure for obtaining a ternary scale with arbitrary scale signature [math]\displaystyle{ a\mathbf{L}b\mathbf{m}c\mathbf{s} }[/math]. Originally developed by Inthar for the purpose of adding aberrisma steps in an orderly manner to a MOS pattern [math]\displaystyle{ a\mathbf{L}b\mathbf{m} }[/math] (which we write in place of [math]\displaystyle{ a\mathbf{L}b\mathbf{s} }[/math] for convenience's sake, since [math]\displaystyle{ \mathbf{s} }[/math] 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 potential symmetry when [math]\displaystyle{ a, c }[/math] or [math]\displaystyle{ b, c }[/math] is not a coprime pair and generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales and simple generator sequence expressions (in the sense of using only two distinct generators) for them.

(Note: This article bolds steps [math]\displaystyle{ \mathbf{L}, \mathbf{m}, \mathbf{s}, \mathbf{x}. }[/math] For integers [math]\displaystyle{ m, n, \ (m, n) := \gcd(m, n). }[/math])

Take for example [math]\displaystyle{ d = (a, c) }[/math], let [math]\displaystyle{ a^\prime = a/d, c^\prime = c/d. }[/math] 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{aberrize\_by\_mos\_subst}(a, b, c, x, k), }[/math] where [math]\displaystyle{ 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 darkest mode, since L (or m) > s).

Facts

The following holds for [math]\displaystyle{ 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]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{m}, k) }[/math] as well):

Let [math]\displaystyle{ \mathsf{mos}(a,b;k) }[/math] be the mode of axby that would have brightness k if x were L and y were s. For example, [math]\displaystyle{ \mathsf{mos}(5,2;5)(x,y) = xxyxxxy. }[/math] Let [math]\displaystyle{ n = a+b+c }[/math] and [math]\displaystyle{ q = (a + c)/(a,c) }[/math].

1. Consider the mode of the template MOS [math]\displaystyle{ T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}(b,a+c;n-1)(\mathbf{m},\mathbf{X}). }[/math] This is the mode of T that has the most Xs near the end. If T is primitive, let [math]\displaystyle{ r }[/math] be 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] (the reader is encouraged to check this), [math]\displaystyle{ r }[/math]-steps in the filling MOS [math]\displaystyle{ F = \mathsf{mos}(a,c;k)(\mathbf{L},\mathbf{s}) }[/math] come in exactly 2 sizes, [math]\displaystyle{ i\mathbf{L}+j\mathbf{s} }[/math] and [math]\displaystyle{ (i-1)\mathbf{L}+(j+1)\mathbf{s}. }[/math] Since the detempering of the imperfect generator of [math]\displaystyle{ T }[/math] occurs only once in [math]\displaystyle{ S }[/math], [math]\displaystyle{ S }[/math] admits a particularly elegant well-formed binary generator sequence of length [math]\displaystyle{ q, }[/math] corresponding to the circle of r-steps in the filling MOS. Letting [math]\displaystyle{ \mathsf{GS}(g_1, ..., g_{q}) }[/math] be this generator sequence, [math]\displaystyle{ g_j }[/math] is either [math]\displaystyle{ p\mathbf{m} + i\mathbf{L} + j\mathbf{s} }[/math] or [math]\displaystyle{ p\mathbf{m} + (i-1)\mathbf{L} + (j+1)\mathbf{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{ i\mathbf{L} + j\mathbf{s} }[/math] or [math]\displaystyle{ (i-1)\mathbf{L} + (j+1)\mathbf{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.)
2. Assume that template MOS [math]\displaystyle{ T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}(b,a+c;n-1)(\mathbf{m},\mathbf{X}) }[/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.
   • [math]\displaystyle{ S }[/math] becomes a MOS with s = 0 for k in {v, ..., q − 1}, where v is the number of occurrences of the (r + 1)-step (t + 1)L + us in F. In particular, if the interval class of (r + 1)-steps consists of tL + (u + 1)s and (t − 1)L + (u + 2)s, [math]\displaystyle{ S }[/math] becomes a MOS with s = 0 for any k in {0, ..., q − 1}. When this holds, the "aberrized" scale may rightly be considered a detempering of the original MOS aLbm with additional s steps.

Examples

5L2m4s

To derive 5L2m4s as [math]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(5, 2, 4, \mathbf{m}, k) }[/math], 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. 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), representing all 3 possible rotations of (L+s, L+m, L+s).

5L2m4s as [math]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(5, 2, 4, \mathbf{m}, k) }[/math]

k	filling MOS	UDP for filling MOS	step pattern		generator sequence		MOS for s = 0?
			template MOS:	LXLXLXLXLXX	intvl. class of gen.:(*)	2-steps
2	mssmss	4|0(2)	LmLsLsLmLss		GS(L+m, L+s, L+s)		yes
1	smssms	2|2(2)	LsLmLsLsLms		GS(L+s, L+m, L+s)		yes
0	ssmssm	0|4(2)	LsLsLmLsLsm		GS(L+s, L+s, L+m)		yes

(*) such that the perfect generator has fewer X's than the imperfect counterpart

6L7m9s

6L7m9s as [math]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(6, 7, 9, \mathbf{L}, k) }[/math]

k	filling MOS (1 period)	UDP for filling MOS	step pattern		generator sequence		MOS for s = 0?
			template MOS:	mXXmXXmXXmXXmXXmXXmXXX	intvl. class of gen.:(*)	3-steps
4	LsLss	12|0(3)	mLsmLsmsLmsLmssmLsmLss		GS(L+m+s, L+m+s, L+m+s, L+m+s, m+2s)		yes
3	LssLs	9|3(3)	mLsmsLmsLmssmLsmLsmsLs		GS(L+m+s, L+m+s, L+m+s, m+2s, L+m+s)		yes
2	sLsLs	6|6(3)	msLmsLmssmLsmLsmsLmsLs		GS(L+m+s, L+m+s, m+2s, L+m+s, L+m+s)		yes
1	sLssL	3|9(3)	msLmssmLsmLsmsLmsLmssL		GS(L+m+s, m+2s, L+m+s, L+m+s, L+m+s)		yes
0	ssLsL	0|12(3)	mssmLsmLsmsLmsLmssmLsL		GS(m+2s, L+m+s, L+m+s, L+m+s, L+m+s)		no

(*) such that the perfect generator has fewer X's than the imperfect counterpart

Open questions

1. When are the GSes obtained via this procedure the shortest possible GSes?