MOS substitution: Difference between revisions

[[Aberrismic theory]] uses MOS substitution. In fact, groundfault reports having come up with a similar concept but not following up on it.

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{{User:Inthar/Template:Notation}}

* Boldface Latin variables are step sizes, and <math>\mathbf{L} >  \mathbf{m} > \mathbf{s} > \mathbf{0}.</math> <math>\mathbf{0}</math> denotes the zero step (0 cents).

== Formal definition ==

== Original derivation ==

In the original aberrismic-informed context, say that <math>d = (a, c) > 1.</math> Consider the MOS word <math>(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>a</math>-many <math>\mathbf{L}</math> steps and <math>c</math>-many <math>\mathbf{s}</math> steps is the MOS <math>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>\mathbf{X}</math> steps. Fixing a choice of which <math>\mathbf{X}</math> in the MOS <math>(a + c)\mathbf{X}b\mathbf{m}</math> you start from, we can choose one of <math>(a+c)/d</math> modes of <math>a \mathbf{L} c \mathbf{s}.</math> If <math>a = c</math>, we obtain a balanced (thus MV3) ternary scale; when in addition <math>b</math> is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of <math>a\mathbf{L}a\mathbf{s}</math>. Of course, one may do this using template MOS <math>a\mathbf{L}(b + c)\mathbf{X}</math> and the <math>(b, c)</math>-multiperiod filling MOS <math>b\mathbf{m} c\mathbf{s}</math> instead. This article denotes the resulting scale <math>\mathsf{MOS\_subst}(a, b, c; \mathbf{y}, \mathbf{z}; k):</math>  

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

== Examples ==

In particular, all [[monotone-MOS scale]]s (i.e. such that the results of {{nowrap|'''L''' {{=}} '''m''' | '''m''' {{=}} '''s'''}}, and {{nowrap|'''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 ===

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

With the additional assumption that the number of '''X''' letters in a perfect generator ''p''_''T'' of the template MOS be a generator class of the filling MOS, the generator sequence yields ''q'' parallel chains ''C''₁,  

..., ''C''_''q'' of the aggregate generator, the sum of the generators in the GS. The offset between ''C''_''i'' and ''C''_''i''+1 is equal to subst(''p''_''T'', '''X''', ''p''_''F''), where ''p''_''T'' and ''p''_''F'' are perfect generators (of appropriate lengths) of the template and filling MOSes, respectively. The aggregate generator is  subst((''p''_''T'')^''q'', '''X''', ''G''^''r''), where ''G'' is the period of the filling MOS.

* the imperfect generator of the filling MOS corresponds to looping back to ''C''₁ but on the next note of ''C''₁, so it and the ''q'' &minus; 1 notes thereafter are advanced by 1 note from any predecessor notes in the chains.

<math>

In the above case, {{nowrap| ''n'' {{=}} ''q'' | '''v''' {{=}} subst(''p''_''T'', '''X''', ''p''_''F'') | and '''w''' {{=}} subst((''p''_''T'')^''q'', '''X''', ''G''^''r'') (the aggregate generator)}}.

=== MOS substitution scales have block balance at most 2 ===

Case 2: Neither of ''k'' and {{nowrap|''k'' + 1}} equals ({{nowrap|''b'' + ''c''}}). Here, if '''Y''' occurs ''j'' or {{nowrap|''j'' + 1}} times in a window of size ''k'', then '''Y''' occurs {{nowrap|''j'' + 1}} or {{nowrap|''j'' + 2}} times in a window of size {{nowrap|''k'' + 2}}.

== MOS substitution scales and RTT ==

     scale = subst_step_sizes(word, {"X": sizeX, "Y": sizeY, "Z": sizeZ})

     return scale

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