MOS substitution: Difference between revisions

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Let <math>\mathsf{MOS}(a,b;k)(\mathbf{x}, \mathbf{y})</math> be the mode of <math>a\mathbf{x}b\mathbf{y}</math> that would have brightness k if <math>\mathbf{x}</math> were <math>\mathbf{L}</math> and <math>\mathbf{y}</math> were <math>\mathbf{s}.</math> For example, <math>\mathsf{MOS}(5,2;5)(\mathbf{x},\mathbf{y}) = \mathbf{xxyxxxy}.</math> Let <math> n = a+b+c</math> and <math>q = (a + c)/(a,c)</math>.

# Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{MOS}(a+c,b;0)(\mathbf{X},\mathbf{m}).</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>n,</math> <math>r</math>-steps in the filling MOS <math>F = \mathsf{MOS}(a,c;k)(\mathbf{L},\mathbf{s})</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>(n - 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}) = \mathsf{MOS}(a+c,b;0)(\mathbf{X},\mathbf{m}).</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>n,</math> <math>r</math>-steps in the filling MOS <math>F = \mathsf{MOS}(a,c;k)(\mathbf{L},\mathbf{s})</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>(n - 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.

# Assume that template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{MOS}(a+c, b; 0)(\mathbf{X}, \mathbf{m})</math> is primitive. Suppose that the chunks (maximal substrings of consecutive <math>\mathbf{X}</math> steps) <math>T</math> that we use has either <math>\ell</math> or <math>(\ell + 1)</math>-many <math>\mathbf{X}</math> steps. Suppose the sizes for <math>\ell</math>-steps in <math>F</math> are <math>t\mathbf{L} + u\mathbf{s}</math> and <math>(t-1)\mathbf{L}+(u+1)\mathbf{s}.</math>

#* <math>S</math> becomes a MOS with <math>\mathbf{s} = 0</math> for <math>k \in \{0, ..., q-v-1\},</math> where <math>v</math> is the number of occurrences of the <math>(\ell + 1)</math>-step <math>(t + 1)\mathbf{L} + u\mathbf{s}</math> in <math>F</math>. This is because the numbers of <math>\mathbf{m}</math> steps that occur in chunks of <math>\mathbf{X}</math> steps in <math>T</math> after substitution determine whether the result of substitution and deleting <math>\mathbf{s}</math> steps is a MOS. In particular, if the interval class of <math>(\ell + 1)</math>-steps is <math>\{t\mathbf{L}+(u+1)\mathbf{s},(t-1)\mathbf{L}+(u+2)\mathbf{s}\},</math> then <math>S</math> with <math>\mathbf{s} = 0</math> is a MOS for any <math>k \in \{0, ..., q-1\}</math>. The practical import of this result is that when this holds, the resulting scale may justly be considered the original MOS <math>a\mathbf{L}b\mathbf{m}</math> detempered by inserting <math>\mathbf{s}</math> steps.

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==Examples==

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 Let <math>\mathsf{MOS}(a,b;k)(\mathbf{x}, \mathbf{y})</math> be the mode of <math>a\mathbf{x}b\mathbf{y}</math> that would have brightness k if <math>\mathbf{x}</math> were  <math>\mathbf{L}</math> and <math>\mathbf{y}</math> were  <math>\mathbf{s}.</math> For example, <math>\mathsf{MOS}(5,2;5)(\mathbf{x},\mathbf{y}) = \mathbf{xxyxxxy}.</math> Let <math> n = a+b+c</math> and <math>q = (a + c)/(a,c)</math>.
-# Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{MOS}(a+c,b;0)(\mathbf{X},\mathbf{m}).</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>n,</math> <math>r</math>-steps in the filling MOS <math>F = \mathsf{MOS}(a,c;k)(\mathbf{L},\mathbf{s})</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>(n - 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}) = \mathsf{MOS}(a+c,b;0)(\mathbf{X},\mathbf{m}).</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>n,</math> <math>r</math>-steps in the filling MOS <math>F = \mathsf{MOS}(a,c;k)(\mathbf{L},\mathbf{s})</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>(n - 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.
-# Assume that template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{MOS}(a+c, b; 0)(\mathbf{X}, \mathbf{m})</math> is primitive. Suppose that the chunks (maximal substrings of consecutive <math>\mathbf{X}</math> steps) <math>T</math> that we use has either <math>\ell</math> or <math>(\ell + 1)</math>-many <math>\mathbf{X}</math> steps. Suppose the sizes for <math>\ell</math>-steps in <math>F</math> are <math>t\mathbf{L} + u\mathbf{s}</math> and <math>(t-1)\mathbf{L}+(u+1)\mathbf{s}.</math>
-#* <math>S</math> becomes a MOS with <math>\mathbf{s} = 0</math> for <math>k \in \{0, ..., q-v-1\},</math> where <math>v</math> is the number of occurrences of the <math>(\ell + 1)</math>-step <math>(t + 1)\mathbf{L} + u\mathbf{s}</math> in <math>F</math>. This is because the numbers of <math>\mathbf{m}</math> steps that occur in chunks of <math>\mathbf{X}</math> steps in <math>T</math> after substitution determine whether the result of substitution and deleting <math>\mathbf{s}</math> steps is a MOS. In particular, if the interval class of <math>(\ell + 1)</math>-steps is <math>\{t\mathbf{L}+(u+1)\mathbf{s},(t-1)\mathbf{L}+(u+2)\mathbf{s}\},</math> then <math>S</math> with <math>\mathbf{s} = 0</math>  is a MOS for any <math>k \in \{0, ..., q-1\}</math>. The practical import of this result is that when this holds, the resulting scale may justly be considered the original MOS <math>a\mathbf{L}b\mathbf{m}</math> detempered by inserting <math>\mathbf{s}</math> steps.
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 ==Examples==