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].

Conventions

• Boldface Latin variables are step sizes, and [math]\displaystyle{ \mathbf{L} > \mathbf{m} > \mathbf{s} > \mathbf{0}. }[/math] [math]\displaystyle{ \mathbf{0} }[/math] denotes the zero step (0 cents).
• Italic lowercase Latin variables are integers.
• Italic uppercase Latin variables are scale words.
• Function names in sans serif font are scale constructions.
• For integers [math]\displaystyle{ m, n, \ (m, n) := \gcd(m, n). }[/math]

Motivation

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 aberrisma 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 mildly generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales with simple generator sequence expressions (in the sense of being binary, i.e. using only two distinct generators).

In the original aberrismic-informed context, 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. This article denotes the resulting scale [math]\displaystyle{ \mathsf{MOS\_subst}(a, b, c; \mathbf{x}, \mathbf{y}; k), }[/math] where [math]\displaystyle{ \mathbf{y} }[/math] is the new step size inserted, [math]\displaystyle{ \mathbf{x} }[/math] is the step size in the starting MOS identified with [math]\displaystyle{ \mathbf{y} }[/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]).

Examples

In the following tables, the interval class of the generators stacked in the generator sequence is such that the perfect generator has fewer [math]\displaystyle{ \mathbf{X} }[/math] steps than the imperfect counterpart.

5L2m4s

To derive groundfault's diamech scale which has step pattern [math]\displaystyle{ 5\mathbf{L}2\mathbf{m}4\mathbf{s} }[/math] as [math]\displaystyle{ \mathsf{MOS\_subst}(5, 2, 4; \mathbf{m}, \mathbf{s}; 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 length-3 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{MOS\_subst}(5, 2, 4; \mathbf{m}, \mathbf{s}; k) }[/math]

[math]\displaystyle{ k }[/math]	filling MOS	UDP for filling MOS	step pattern		generator sequence		MOS for [math]\displaystyle{ \mathbf{s} = \mathbf{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

5L2m6s

[math]\displaystyle{ 5\mathbf{L}2\mathbf{m}6\mathbf{s} }[/math] as [math]\displaystyle{ \mathsf{MOS\_subst}(5, 2, 6; \mathbf{m}, \mathbf{s}; k) }[/math]

[math]\displaystyle{ k }[/math]	filling MOS (1 period)	UDP for filling MOS	step pattern		generator sequence		MOS for [math]\displaystyle{ \mathbf{s} = \mathbf{0} }[/math]
			template MOS:	LXLXXLXLXXLXX	intvl. class of gen.:	5-steps
3	msss	6|0(2)	LmLssLsLmsLss		GS((2L+m+2s)3, 2L+3s)		yes
2	smss	4|2(2)	LsLmsLsLsmLss		GS((2L+m+2s)2, 2L+3s, 2L+m+2s)		yes
1	ssms	2|4(2)	LsLsmLsLssLms		GS(2L+m+2s, 2L+3s, (2L+m+2s)2)		yes
0	sssm	0|6(2)	LsLssLmLssLsm		GS(2L+3s, (2L+m+2s)3)		yes

Here the notation G^k denotes repeating the generator G k times in the generator sequence.

These are four of the 8 billiard words that have pattern 5L2m6s. The other four billiard words have length-3 subwords of non-X letters, unlike the MOS substitution scales.

This scale pattern is available in 37edo with step ratio 5:3:1; the generator sequence in the tuning has 2L+m+2s = 486.5 (~4/3) and 2L+3s = 421.6 (~14/11), and notably this tuning represents all primes from 3 to 13 with only 3 being inaccurate. 65edo's 9:7:1 is another optimum for 2.3.5.11.13, and is given by a GS using three 4/3's and one 5/4.

6L7m9s

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

[math]\displaystyle{ k }[/math]	filling MOS (1 period)	UDP for filling MOS	step pattern		generator sequence		MOS for [math]\displaystyle{ \mathbf{s} = \mathbf{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

Facts

The following holds for [math]\displaystyle{ S = \mathsf{MOS\_subst}(a, b, c; \mathbf{L}, \mathbf{s}; k) }[/math] (and after switching [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{MOS\_subst}(a, b, c; \mathbf{m}, \mathbf{s}; k) }[/math] as well):

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

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] [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 on 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.) The generator of the template MOS serves as the "guide generator" for this generator sequence.

The generator sequence thus yields q parallel chains C₁, ..., C_q of the aggregate generator, and the offset between C_i and C_i+1 is equal to the result of substituting the perfect generator of the filling MOS into the generator of the template MOS. Hence in the GS,

• the perfect generator of the filling MOS corresponds to advancing from C_i to C_i+1;
• the imperfect generator of the filling MOS corresponds to looping back to C₁ but on the next note of C₁, so it and the q − 1 notes thereafter are advanced by 1 note from any predecessor notes in the chains.

In other words, there exist non-negative integers m, n, a < n, b < n, and vectors a, v and w in the lattice such that the set of notes in the scale is

[math]\displaystyle{ \{\mathbf{a} + i\mathbf{v}\}_{i=a}^{n-1} \cup \{\mathbf{a} + i\mathbf{v} + j\mathbf{w}\}_{(i,j) \in [n]_0 \times [m-1]_1} \cup \{\mathbf{a} + i\mathbf{v} + m\mathbf{w}\}_{i=0}^{b}. }[/math]

Let us call rank-3 scales with such lattice shapes almost parallelogram-shaped. MOS substitution scales are thus almost parallelogram-shaped, but the converse is false, as the scale in 5 letters [9/8 28/27 9/8 64/63 9/8 28/27 243/224 28/27 64/63 567/512 64/63] is almost parallelogram-shaped.

Open questions

1. Is there a simple answer for when a MOS substitution scale becomes a MOS after deleting the "added" steps?
2. For an arbitrary ternary scale that results from MOS substitution, when are the GSes obtained via the procedure the shortest possible GSes?