MOS substitution: Difference between revisions

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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,</math> corresponding 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 ~~MOS~~ is primitive and for some perfect generators <math>p_T, p_F, \ |p_T|_\mathbf{X} = |p_F|,</math> then MOS substitution yields almost parallelograms in the lattice ===

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

With the additional ~~condition~~ that the number of X's 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,

With the additional assumption that the number of X's 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 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'')''r'', '''X''', ''f''''r''), where ''f'' is the filling MOS.

..., ''C''''q'' of the aggregate generator. 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'')''r'', '''X''', ''F''''r''), where ''F'' is the filling MOS.

Hence in the GS,

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* 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'' − 1 notes thereafter are advanced by 1 note from any predecessor notes in the chains.

Hence these MOS substitution scales satisfy a property that we call ''almost parallelogram''{{User:Inthar/Template:adhoc}}. 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

Hence these particular MOS substitution scales satisfy a property that we call ''almost parallelogram''{{User:Inthar/Template:adhoc}}. 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>\{\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(''p''''t'', '''X''', ''p''''f''), and '''w''' = subst((''p''''t'')''r'', '''X''', ''f''''r'').

In the above case, ''n'' = ''q'', '''v''' = subst(''p''''T'', '''X''', ''p''''F''), and '''w''' = subst((''p''''T'')''r'', '''X''', ''F''''r'').

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.

@@ Line 144: / Line 144: @@
 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,</math> corresponding 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 MOS is primitive and for some perfect generators <math>p_T, p_F, \ |p_T|_\mathbf{X} = |p_F|,</math> then MOS substitution yields almost parallelograms in the lattice ===
+=== If the template is a primitive MOS, and for some perfect generators <math>p_T, p_F, \ |p_T|_\mathbf{X} = |p_F|,</math> then MOS substitution yields almost parallelograms in the lattice ===
-With the additional condition that the number of X's in a perfect generator ''p''<sub>''T''</sub> of the template MOS be a generator class of the filling MOS, the generator sequence yields ''q'' parallel chains ''C''<sub>1</sub>,
+With the additional assumption that the number of X's in a perfect generator ''p''<sub>''T''</sub> of the template MOS be a generator class of the filling MOS, the generator sequence yields ''q'' parallel chains ''C''<sub>1</sub>,
-..., ''C''<sub>''q''</sub> of the aggregate generator. The offset between ''C''<sub>''i''</sub> and ''C''<sub>''i''+1</sub> is equal to subst(''p''<sub>''t''</sub>, '''X''', ''p''<sub>''f''</sub>), where ''p''<sub>''t''</sub> and ''p''<sub>''f''</sub> are perfect generators (of appropriate lengths) of the template and filling MOSes, respectively. The aggregate generator is  subst((''p''<sub>''t''</sub>)<sup>''r''</sup>, '''X''', ''f''<sup>''r''</sup>), where ''f'' is the filling MOS.
+..., ''C''<sub>''q''</sub> of the aggregate generator. The offset between ''C''<sub>''i''</sub> and ''C''<sub>''i''+1</sub> is equal to subst(''p''<sub>''T''</sub>, '''X''', ''p''<sub>''F''</sub>), where ''p''<sub>''T''</sub> and ''p''<sub>''F''</sub> are perfect generators (of appropriate lengths) of the template and filling MOSes, respectively. The aggregate generator is  subst((''p''<sub>''T''</sub>)<sup>''r''</sup>, '''X''', ''F''<sup>''r''</sup>), where ''F'' is the filling MOS.
 Hence in the GS,
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 * the imperfect generator of the filling MOS corresponds to looping back to ''C''<sub>1</sub> but on the next note of ''C''<sub>1</sub>, so it and the ''q'' &minus; 1 notes thereafter are advanced by 1 note from any predecessor notes in the chains.
-Hence these MOS substitution scales satisfy a property that we call ''almost parallelogram''{{User:Inthar/Template:adhoc}}. 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
+Hence these particular MOS substitution scales satisfy a property that we call ''almost parallelogram''{{User:Inthar/Template:adhoc}}. 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>\{\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(''p''<sub>''t''</sub>, '''X''', ''p''<sub>''f''</sub>), and '''w''' = subst((''p''<sub>''t''</sub>)<sup>''r''</sup>, '''X''', ''f''<sup>''r''</sup>).
+In the above case, ''n'' = ''q'', '''v''' = subst(''p''<sub>''T''</sub>, '''X''', ''p''<sub>''F''</sub>), and '''w''' = subst((''p''<sub>''T''</sub>)<sup>''r''</sup>, '''X''', ''F''<sup>''r''</sup>).
 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.