MOS substitution: Difference between revisions

{| class="wikitable"

|-

! rowspan="2" | [[Simplified UDP]] for filling MOS

| style="text-align: right;" | <code>mss</code>

| colspan="2" style="text-align: right;" | <code>LmLss</code>

|-

| 1{{pipe}}1

| style="text-align: right;" | <code>sms</code>

| colspan="2" style="text-align: right;" | <code>LsLms</code>

|-

| 0{{pipe}}2

| style="text-align: right;" | <code>ssm</code>

| colspan="2" style="text-align: right;" | <code>LsLsm</code>

|}

{{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.

Hence in the GS,

* 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>

Here the scale is thought as traversing a series of rows one step of the row at a time, and

* <math>\{\mathbf{a} + i\mathbf{v} + j\mathbf{w}\}_{(i,j) \in [n]_0 \times [m-2]_1}</math> is a (possibly empty) parallelogram where rows are traversed fully

=== 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 ==

     return result

   

# Function returns subst nX X (nY Y nZ Z (brightness_of_filling_mos) | (nY + nZ - gcd(nY, nZ) - brightness_of_filling_mos))

def mos_subst(nX, nY, nZ, sizeX, sizeY, sizeZ, brightness_of_filling_mos):

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

     return scale

</syntaxhighlight>