MOS substitution: Difference between revisions

'''MOS substitution''' 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.

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

MOS substitution scales are denoted using the notation "subst ax(bycz)" or just "aX(bYcZ)". Any particular scale of a given MOS substitution type is said to be "a subst ax(bycz)" or "a scale of type ax(bycz)". A specific MOS substitution scale may be denoted {{nowrap|template_MOS_with_slot_letter_X(filling_MOS)}}; to make this notation unique for a particular given MOS-substitution scale, the brightest mode for the template MOS is conventionally used, treating the slot letter X as the smaller step.

|+ style="font-size: 105%;" | The three subst 2'''L'''(1'''m'''2'''s''') scales

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

| LXLXX(mss)

| LXLXX(sms)

| LXLXX(ssm)

== Math notation ==

* ''a'''''X'''''b'''''Y'''(''k'') denotes the mode of ''a'''''X'''''b'''''Y''' which would have [[simplified UDP]] notation <math>k|a+b-1-k</math> under the assumption '''X''' > '''Y''' > '''0'''.

A ternary scale word ''w''('''x'''₁, '''x'''₂, '''x'''₃) is a ''MOS substitution'' scale word if there exists a permutation <math>\pi \in S_3</math> such that the following holds:

* identifying '''x'''_π(1) and '''x'''_π(2) results in a MOS (called the ''template MOS'') and

* deleting all instances of '''x'''_π(3) (called the ''slot letter'') results in a MOS (called the ''filling MOS'').

<math>\displaystyle{\mathsf{subst}\left( a\mathbf{w}(b + c)\mathbf{X}(0) , \mathbf{X}, b\mathbf{y}c\mathbf{z}(k) \right) }</math>

|+ style="font-size: 105%;" | Diaslen as {{nowrap|subst 5'''L'''(2'''m'''4'''s''')}}

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

| 2 || <code>mssmss</code> || 2{{pipe}}0

| 1 || <code>smssms</code> || 1{{pipe}}1

| 0 || <code>ssmssm</code> || 0{{pipe}}2

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

| 3 || <code>msss</code> || 3{{pipe}}0

| 2 || <code>smss</code> || 2{{pipe}}1

| 1 || <code>ssms</code> || 1{{pipe}}2

| 0 || <code>sssm</code> || 0{{pipe}}3

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

| 4 || <code>LsLss</code> || 4{{pipe}}0

| 3 || <code>LssLs</code> || 3{{pipe}}1

| 2 || <code>sLsLs</code> || 2{{pipe}}2

| 1 || <code>sLssL</code> || 1{{pipe}}3

| 0 || <code>ssLsL</code> || 0{{pipe}}4

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.

=== If the template is a primitive MOS, and for some perfect generators <math>p_T, p_F, \ r := \left|p_T\right|_\mathbf{X} = \left|p_F\right|,</math> then MOS substitution yields a parallelogram substring in the lattice ===

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.

Hence these particular MOS substitution scales satisfy a property that we call ''[[parallelogram substring scale|parallelogram substring]]''. An '''e'''-equivalent scale is a ''parallelogram substring'' if there exist integers ''m'' > 0, ''n'' > 0, 0 &le; ''a'' < ''n'', 0 &le; ''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>

\{\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}. % prefix of last row

</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}\}_{i=a}^{n-1}</math> is a (nonempty) suffix of the first row

* <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

* <math>\{\mathbf{a} + i\mathbf{v} + (m-1)\mathbf{w}\}_{i=0}^{b}</math> is a (nonempty) prefix of the last row

* '''v''' and '''w''' are the generator and offset

 

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 a parallelogram substring.

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}}.

 

Given a ternary scale with step signature ''a''L''b''m''c''s with gcd(''a'', ''b'', ''c'') = 1 and a [[JI subgroup]], there exist linearly independent (possibly [[patent val|non-patent]]) [[val]]s ''a'', ''a'' + ''b'', and ''a'' + ''b'' + ''c'' that interpret the scale. Assuming the join is not contorted, a [[rank-3 temperament]] can now be defined as the join of these three vals.

# In UDP, brightness = number of generators up * gcd of the step counts

[[Category:Rank-3 scales| ]]