MOS substitution
MOS substitution[idiosyncratic term] is a procedure for obtaining a ternary scale with arbitrary scale signature [math]\displaystyle{ a\mathbf{L}b\mathbf{m}c\mathbf{s} }[/math].
Todo
Add code and lattice diagrams.
Make explanations more readable.
Conventions
- For more information on the idiosyncratic math notation on this page, see User:Inthar/Notation.
- 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]
- If w is a word (in a specific rotation) in X and possibly other letters, and u is a circular word in a specific modal rotation, then [math]\displaystyle{ \mathsf{subst}(w, \mathbf{X}, u) }[/math] denotes the word w but with the ith occurrence of X replaced with u[i] (for i ≥ 0).
- aXbY(k) denotes the mode of aXbY which would have UDP notation [math]\displaystyle{ dk|d(a/d+b/d-1-k)\ (d), \ d = \gcd(a,b) }[/math] under the assumption X > Y > 0.
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[idiosyncratic term]. 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[idiosyncratic term], 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{y}, \mathbf{z}; k): }[/math]
[math]\displaystyle{ \displaystyle{ \mathsf{MOS\_subst}(a, b, c; \mathbf{y}, \mathbf{z}; k) := \mathsf{subst}( a\mathbf{w}(b + c)\mathbf{X}(0) , \mathbf{X}, b\mathbf{y}c\mathbf{z}(k) ) } }[/math]
where [math]\displaystyle{ \mathbf{z} }[/math] is the new step size inserted, [math]\displaystyle{ \mathbf{y} }[/math] is the step size in the starting MOS identified with [math]\displaystyle{ \mathbf{z} }[/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{ 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{ 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 Gk denotes repeating the generator G k times in the generator sequence.
These are four of the 8 billiard scales 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{ 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 | ||
Mathematical facts
A ternary scale whose L = m and s = 0 temperings are MOS comes from MOS substitution
If a ternary scale with step signature aLbmcs satisfies:
- the result of identifying L steps with m steps is a MOS;
- the result of deleting all s steps is a MOS,
then it is a MOS substitution scale, namely subst((a+b)Xcs(i), X, aLbm(j)) for some brightnesses i and j.
In particular, all monotone-MOS[idiosyncratic term] scales (i.e. such that the results of L = m, m = s, and s = 0 temperings are MOSes) arise from MOS substitution in this way.
If the template MOS is primitive, MOS substitution yields binary well-formed generator sequences
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):
Consider the mode of the template MOS [math]\displaystyle{ T = T(\mathbf{m},\mathbf{X}) = (a+c)\mathbf{X}b\mathbf{m}(0). }[/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{ a+c, }[/math] [math]\displaystyle{ r }[/math]-steps in the filling MOS [math]\displaystyle{ F = a\mathbf{L}c\mathbf{s}(k) }[/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{ (a+c - 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.
If the template is a primitive MOS, and for some perfect generators [math]\displaystyle{ p_T, p_F, \ |p_T|_\mathbf{X} = |p_F|, }[/math] then MOS substitution yields almost parallelograms in the lattice
With the additional assumption that the number of X's in a perfect generator pT of the template MOS be a generator class of the filling MOS, the generator sequence yields q parallel chains C1, ..., Cq of the aggregate generator. The offset between Ci and Ci+1 is equal to subst(pT, X, pF), where pT and pF are perfect generators (of appropriate lengths) of the template and filling MOSes, respectively. The aggregate generator is subst((pT)r, X, Fr), where F is the filling MOS.
Hence in the GS,
- the perfect generator of the filling MOS corresponds to advancing from Ci to Ci+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.
Hence these particular MOS substitution scales satisfy a property that we call almost parallelogram[ad-hoc term]. An e-equivalent scale is almost a parallelogram if there exist non-negative integers m, n, 0 < a < n, 0 < b < n, a vector a, and two linearly independent vectors v and w such that the set of notes in the scale as a subset of the lattice of e-equivalent pitches 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-2]_1} \cup \{\mathbf{a} + i\mathbf{v} + (m-1)\mathbf{w}\}_{i=0}^{b}. }[/math]
In the above case, n = q, v = subst(pT, X, pF), and w = subst((pT)r, X, Fr).
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 a parallelogram.
Open questions
- Is there a simple answer for when a MOS substitution scale becomes a MOS after deleting the "added" steps?
- For an arbitrary ternary scale that results from MOS substitution, when are the GSes obtained via the procedure the shortest possible GSes?
- Call an almost parallelogram scale with a = 1 and b = n − 2 transposable[ad-hoc term]. Classify transposable MOS substitution scales.