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])

Say that [math]\displaystyle{ d = (a, c) > 1. }[/math] Consider the MOS word [math]\displaystyle{ (a + c)\mathbf{X}b\mathbf{m} }[/math], which we call the template MOS. Since the "most even" arrangement (in the sense of distributional evenness) of [math]\displaystyle{ a }[/math]-many [math]\displaystyle{ \mathbf{L} }[/math] steps and [math]\displaystyle{ c }[/math]-many [math]\displaystyle{ \mathbf{s} }[/math] steps is the MOS [math]\displaystyle{ a\mathbf{L}b\mathbf{s} }[/math] (which will in general be a non-primitive MOS), this method prescribes following the latter MOS, called the filling MOS, to fill in the [math]\displaystyle{ \mathbf{X} }[/math] steps. Fixing a choice of which [math]\displaystyle{ \mathbf{X} }[/math] in the MOS [math]\displaystyle{ (a + c)\mathbf{X}b\mathbf{m} }[/math] you start from, we can choose one of [math]\displaystyle{ (a+c)/d }[/math] modes of [math]\displaystyle{ a \mathbf{L} c \mathbf{s}. }[/math] If [math]\displaystyle{ a = c }[/math], we obtain a balanced (thus MV3) ternary scale; when in addition [math]\displaystyle{ b }[/math] is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of [math]\displaystyle{ a\mathbf{L}c\mathbf{s} }[/math]. Of course, one may do this using template MOS [math]\displaystyle{ a\mathbf{L}(b + c)\mathbf{X} }[/math] and the [math]\displaystyle{ (b, c) }[/math]-multiperiod filling MOS [math]\displaystyle{ b\mathbf{m} c\mathbf{s} }[/math] 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 [math]\displaystyle{ \mathbf{s} }[/math] by the template MOS and [math]\displaystyle{ k }[/math] is the brightness of the mode of the filling MOS used ([math]\displaystyle{ k = 0 }[/math] corresponds to the darkest mode; the conventional understanding of "brightness" makes sense as [math]\displaystyle{ \mathbf{L} }[/math] (resp. [math]\displaystyle{ \mathbf{m} }[/math]) > [math]\displaystyle{ \mathbf{s} }[/math]).

Facts

The following holds for [math]\displaystyle{ S = \mathsf{aberrize\_by\_MOS\_subst}(a, b, c, \mathbf{L}, k) }[/math] (and after replacing [math]\displaystyle{ \mathbf{L} }[/math] with [math]\displaystyle{ \mathbf{m} }[/math] and [math]\displaystyle{ a }[/math] with [math]\displaystyle{ b, }[/math] 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}(a+c,b;0)(\mathbf{X},\mathbf{m}). }[/math] This is the mode of [math]\displaystyle{ T }[/math] that has the most [math]\displaystyle{ \mathbf{X} }[/math] steps near the end. If [math]\displaystyle{ T }[/math] is primitive, let [math]\displaystyle{ r }[/math] be the count of [math]\displaystyle{ \mathbf{X} }[/math] steps in a chosen (reduced) generator of [math]\displaystyle{ T. }[/math] Since [math]\displaystyle{ r }[/math] must be coprime to [math]\displaystyle{ n }[/math] (which the reader is encouraged to verify), [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 (using two distinct generators) generator sequence of length [math]\displaystyle{ q, }[/math] corresponding to the circle of [math]\displaystyle{ r }[/math]-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 [math]\displaystyle{ j }[/math]-th [math]\displaystyle{ r }[/math]-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 mode of [math]\displaystyle{ T }[/math] on the other extreme of its generator arc instead, which corresponds to taking the circle of [math]\displaystyle{ (n - r) }[/math]-steps in [math]\displaystyle{ F }[/math] 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 [math]\displaystyle{ T }[/math] that we use has [math]\displaystyle{ r }[/math]-many [math]\displaystyle{ \mathbf{X} }[/math] steps and that the imperfect generator has [math]\displaystyle{ (r + 1) }[/math]-many [math]\displaystyle{ \mathbf{X} }[/math] steps. Suppose the sizes for [math]\displaystyle{ r }[/math]-steps in [math]\displaystyle{ F }[/math] are [math]\displaystyle{ t\mathbf{L} + u\mathbf{s} }[/math] and [math]\displaystyle{ (t-1)\mathbf{L}+(u+1)\mathbf{s}. }[/math]
   • [math]\displaystyle{ S }[/math] becomes a MOS with [math]\displaystyle{ \mathbf{s} = 0 }[/math] for [math]\displaystyle{ k \in \{0, ..., q-v-1\}, }[/math] where [math]\displaystyle{ v }[/math] is the number of occurrences of the [math]\displaystyle{ (r + 1) }[/math]-step [math]\displaystyle{ (t + 1)\mathbf{L} + u\mathbf{s} }[/math] in [math]\displaystyle{ F }[/math]. In particular, if the interval class of [math]\displaystyle{ (r + 1) }[/math]-steps is [math]\displaystyle{ \{t\mathbf{L}+(u+1)\mathbf{s},(t-1)\mathbf{L}+(u+2)\mathbf{s}\}, }[/math] then [math]\displaystyle{ S }[/math] with [math]\displaystyle{ \mathbf{s} = 0 }[/math] is a MOS for any [math]\displaystyle{ k \in \{0, ..., q-1\} }[/math]. The practical consequence of this result is that when this holds, the "aberrized" scale may justly be considered a detempering of the original MOS [math]\displaystyle{ a\mathbf{L}b\mathbf{m} }[/math] with additional [math]\displaystyle{ \mathbf{s} }[/math] steps.

Examples

5L2m4s

To derive [math]\displaystyle{ 5\mathbf{L}2\mathbf{m}4\mathbf{s} }[/math] as [math]\displaystyle{ \mathsf{aberrize\_by\_MOS\_subst}(5, 2, 4, \mathbf{m}, k) }[/math], we exploit [math]\displaystyle{ (b, c) = 2 }[/math] and substitute [math]\displaystyle{ 2\mathbf{m}4\mathbf{s} }[/math] into the template MOS [math]\displaystyle{ 5\mathbf{L}6\mathbf{X} }[/math] ([math]\displaystyle{ \mathbf{LXLXLXLXLXX} }[/math]). Since [math]\displaystyle{ 2\mathbf{m}4\mathbf{s} }[/math] has three distinct modes ([math]\displaystyle{ \mathbf{ssmssm}, \mathbf{smssms}, \mathbf{mssmss} }[/math]) and [math]\displaystyle{ 5\mathbf{L}6\mathbf{X} }[/math] is primitive, we obtain three distinct scales, all of which admit short generator sequences of 2-steps, representing all 3 possible rotations of [math]\displaystyle{ (\mathbf{L}+\mathbf{m}, \mathbf{L}+\mathbf{s}, \mathbf{L}+\mathbf{s}) }[/math] as displayed in the following table:

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

[math]\displaystyle{ k }[/math]	filling MOS	UDP for filling MOS	step pattern		generator sequence		MOS for [math]\displaystyle{ s = 0 }[/math]
			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 [math]\displaystyle{ \mathbf{X} }[/math] steps than the imperfect counterpart

6L7m9s

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

[math]\displaystyle{ k }[/math]	filling MOS (1 period)	UDP for filling MOS	step pattern		generator sequence		MOS for [math]\displaystyle{ s = 0 }[/math]
			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 [math]\displaystyle{ \mathbf{X} }[/math] steps than the imperfect counterpart

Open questions

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