MOS substitution: Difference between revisions

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The following holds for <math>S = \mathsf{MOS\_subst}(a, b, c; \mathbf{L}, \mathbf{s}; k)</math> (and after switching <math>\mathbf{L}</math> with <math>\mathbf{m}</math> and <math>a</math> with <math>b,</math> for <math>\mathsf{MOS\_subst}(a, b, c; \mathbf{m}, \mathbf{s}; k)</math> as well):

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.

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> 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.

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

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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.

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

Consider a MOS substitution scale {{nowrap|a'''X''' (b'''Y''' c'''Z''')}}. It is obvious that '''X''' has [[block balance]] 1, since we can replace the MOS substitution scale with the MOS scale a'''X''' ({{nowrap|b + c}})'''W''' to make this argument. '''Y''' and '''Z''' have block balance at most 2, since we can consider windows of the MOS scale of size ''k'' or {{nowrap|''k'' + 1}}, and the number of times '''Y''' (and also '''Z''') differs by at most 2. This is proved below for '''Y''', but it's exactly the same argument for '''Z''':

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 The following holds for <math>S = \mathsf{MOS\_subst}(a, b, c; \mathbf{L}, \mathbf{s}; k)</math> (and after switching <math>\mathbf{L}</math> with <math>\mathbf{m}</math> and <math>a</math> with <math>b,</math> for <math>\mathsf{MOS\_subst}(a, b, c; \mathbf{m}, \mathbf{s}; k)</math> as well):
-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.
+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> 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.
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 === If the template is a primitive MOS, and for some perfect generators <math>p_T, p_F, \ \left|p_T\right|_\mathbf{X} = \left|p_F\right|,</math> then MOS substitution yields almost parallelograms in the lattice ===
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 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.
 -->
 === MOS substitution scales have block balance at most 2 ===
 Consider a MOS substitution scale {{nowrap|a'''X''' (b'''Y''' c'''Z''')}}. It is obvious that '''X''' has [[block balance]] 1, since we can replace the MOS substitution scale with the MOS scale a'''X''' ({{nowrap|b + c}})'''W''' to make this argument. '''Y''' and '''Z''' have block balance at most 2, since we can consider windows of the MOS scale of size ''k'' or {{nowrap|''k'' + 1}}, and the number of times '''Y''' (and also '''Z''') differs by at most 2. This is proved below for '''Y''', but it's exactly the same argument for '''Z''':