# User:Inthar/MOS substitution

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MOS substitution[idiosyncratic term] is a procedure for obtaining a ternary (3 step sizes) scale from two MOS patterns. It consists of substituting the step pattern of one MOS pattern (called the filling MOS), scale step for scale step, for all occurrences of a chosen step size of another MOS pattern (called the template MOS). Unlike MV3 scales, a MOS substitution scale may have any combination of step sizes.

For example, if the template MOS is LXLXX, and the filling MOS is mss, then the resulting MOS substitution scales are LmLss, LsLms, and LsLsm. The first scale is denoted "subst 2L(1m2s) 2|0", using UDP notation for the filling MOS, and is said to be a "subst 2L(1m2s)". We always substitute into the brightest mode of the template MOS, where X is treated as the smaller step.

Under defined conditions, MOS substitution scales have the advantage of having a generator sequence.

The same operation can be done even when the scales involved are not MOS scales or necessarily even binary scales, in which context it may be called letterwise substitution or simply substitution.

## Conventions

• Boldface Latin variables are step sizes, and $\mathbf{L} \gt \mathbf{m} \gt \mathbf{s} \gt \mathbf{0}.$ $\mathbf{0}$ 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 $m, n, \ (m, n) := \gcd(m, n).$
• 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 $\mathsf{subst}(w, \mathbf{X}, u)$ 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 $dk|d(a/d+b/d-1-k)\ (d), \ d = \gcd(a,b)$ 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 $a\mathbf{L}b\mathbf{m}$ (which we write in place of $a\mathbf{L}b\mathbf{s}$ for convenience's sake, since $\mathbf{s}$ 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 $a, c$ or $b, c$ 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). The idea is that modifying the input scales in a sufficiently controlled fashion from the nicest case of MOS template scales and MOS filling scales whose period divides the count of unknown letters in the template will result in a scale that retains some degree of elegance in its lattice structure. However, this condition is not necessary for MOS substitution to result in a binary generator sequence (with two distinct generators), though the generator sequence necessary to generate the scale will be longer.

In the original aberrismic-informed context, say that $d = (a, c) \gt 1.$ Consider the MOS word $(a + c)\mathbf{X}b\mathbf{m}$, which we call the template MOS[idiosyncratic term]. Since the "most even" arrangement (in the sense of distributional evenness) of $a$-many $\mathbf{L}$ steps and $c$-many $\mathbf{s}$ steps is the MOS $a\mathbf{L}b\mathbf{s}$ (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 $\mathbf{X}$ steps. Fixing a choice of which $\mathbf{X}$ in the MOS $(a + c)\mathbf{X}b\mathbf{m}$ you start from, we can choose one of $(a+c)/d$ modes of $a \mathbf{L} c \mathbf{s}.$ 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 $a\mathbf{L}c\mathbf{s}$. Of course, one may do this using template MOS $a\mathbf{L}(b + c)\mathbf{X}$ and the $(b, c)$-multiperiod filling MOS $b\mathbf{m} c\mathbf{s}$ instead. This article denotes the resulting scale $\mathsf{MOS\_subst}(a, b, c; \mathbf{y}, \mathbf{z}; k):$

$\displaystyle{ \mathsf{MOS\_subst}(a, b, c; \mathbf{y}, \mathbf{z}; k) := \mathsf{subst}( a\mathbf{w}(b + c)\mathbf{X}(0) , \mathbf{X}, b\mathbf{y}c\mathbf{z}(k) ) }$

For brevity, this should usually be written:

$\mathsf{subst} \ a\mathbf{x}(b\mathbf{y}c\mathbf{z}) \ k|b+c-1-k\ (p),$

using UDP for the filling MOS.

where $\mathbf{z}$ is the new step size inserted, $\mathbf{y}$ is the step size in the starting MOS identified with $\mathbf{z}$ by the template MOS, and $k$ is the brightness of the mode of the filling MOS used ($k = 0$ corresponds to the darkest mode; the conventional understanding of "brightness" makes sense as $\mathbf{L}$ (resp. $\mathbf{m}$) > $\mathbf{s}$).

## Examples

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

### 5L2m4s

To derive groundfault's diamech scale which has step pattern $5\mathbf{L}2\mathbf{m}4\mathbf{s}$ as $\mathsf{MOS\_subst}(5, 2, 4; \mathbf{m}, \mathbf{s}; k)$, we exploit $(b, c) = 2$ and substitute $2\mathbf{m}4\mathbf{s}$ into the template MOS $5\mathbf{L}6\mathbf{X}$ ($\mathbf{LXLXLXLXLXX}$). Since $2\mathbf{m}4\mathbf{s}$ has three distinct modes ($\mathbf{ssmssm}, \mathbf{smssms}, \mathbf{mssmss}$) and $5\mathbf{L}6\mathbf{X}$ is primitive, we obtain three distinct scales, all of which admit length-3 generator sequences of 2-steps, representing all 3 possible rotations of $(\mathbf{L}+\mathbf{m}, \mathbf{L}+\mathbf{s}, \mathbf{L}+\mathbf{s})$ as displayed in the following table:

Diamech as subst 5L(2m4s)
$k$ filling MOS UDP for filling MOS step pattern generator sequence MOS for $\mathbf{s} = \mathbf{0}$
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

### 5L2m6s

subst 5L(2m6s)
$k$ filling MOS (1 period) UDP for filling MOS step pattern generator sequence MOS for $\mathbf{s} = \mathbf{0}$
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 an optimum for 2.3.5.11, and is given by a GS using three 4/3's and one 5/4.

### 6L7m9s

subst 7m(6L9s)
$k$ filling MOS (1 period) UDP for filling MOS step pattern generator sequence MOS for $\mathbf{s} = \mathbf{0}$
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

## 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:

1. the result of identifying L steps with m steps is a MOS;
2. 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 $S = \mathsf{MOS\_subst}(a, b, c; \mathbf{L}, \mathbf{s}; k)$ (and after switching $\mathbf{L}$ with $\mathbf{m}$ and $a$ with $b,$ for $\mathsf{MOS\_subst}(a, b, c; \mathbf{m}, \mathbf{s}; k)$ as well):

Consider the mode of the template MOS $T = T(\mathbf{m},\mathbf{X}) = (a+c)\mathbf{X}b\mathbf{m}(0).$ This is the mode of $T$ that has the most $\mathbf{X}$ steps near the end. If $T$ is primitive, let $r$ be the count of $\mathbf{X}$ steps in a chosen (reduced) generator of $T.$ Since $r$ must be coprime to $a+c,$ $r$-steps in the filling MOS $F = a\mathbf{L}c\mathbf{s}(k)$ come in exactly 2 sizes, $i\mathbf{L}+j\mathbf{s}$ and $(i-1)\mathbf{L}+(j+1)\mathbf{s}.$ Since the detempering of the imperfect generator of $T$ occurs only once in $S$, $S$ admits a particularly elegant well-formed binary (using two distinct generators) generator sequence of length $q,$ corresponding to the circle of $r$-steps in the filling MOS. Letting $\mathsf{GS}(g_1, ..., g_{q})$ be this generator sequence, $g_j$ is either $p\mathbf{m} + i\mathbf{L} + j\mathbf{s}$ or $p\mathbf{m} + (i-1)\mathbf{L} + (j+1)\mathbf{s},$ according as the $j$-th $r$-step in the sequence of stacked $r$-steps on the chosen mode of $F$ is $i\mathbf{L} + j\mathbf{s}$ or $(i-1)\mathbf{L} + (j+1)\mathbf{s}.$ (We could have chosen to use the mode of $T$ on the other extreme of its generator arc instead, which corresponds to taking the circle of $(a+c - r)$-steps in $F$ 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 $p_T, p_F, \ |p_T|_\mathbf{X} = |p_F|,$ 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

$\{\mathbf{a} + i\mathbf{v}\}_{i=a}^{n-1} \cup \{\mathbf{a} + i\mathbf{v} + j\mathbf{w}\}_{(i,j) \in [n]_0 \times [m-2]_1} \cup \{\mathbf{a} + i\mathbf{v} + (m-1)\mathbf{w}\}_{i=0}^{b}.$

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

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?
3. Call an almost parallelogram scale with a = 1 and b = n − 2 transposable[ad-hoc term]. Classify transposable MOS substitution scales.