MOS substitution: Difference between revisions

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The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, \mathbf{L}, c, k)</math> (and after replacing '''L''' with '''m''' and a with b, for <math>\mathsf{mos\_subst\_aberrize}(a, b, \mathbf{m}, c, k)</math> as well):

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

Let <math>\mathsf{mos}_{a,b}(x,y;k)</math> be the mode of axby that would have brightness k if x were '''L''' and y were '''s'''. For example, <math>\mathsf{mos}_{a,b}(5,2;5)(x,y) = 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}) = M_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> ('''X''' must be the second argument!). This is the mode of ''T'' that has the most X's near the end. If ''T'' is [[primitive]], let <math>r</math> be the count of '''X''' steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>n</math> (the reader is encouraged to check this), <math>r</math>-steps in the filling MOS <math>F = M_{a,c}(\mathbf{L},\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>, we have a well-formed binary [[generator sequence]] of length <math>q</math> for <math>S</math>. 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 ''j''-th ''r''-step in the sequence of stacked <math>r</math>-steps in 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 "darkest" mode of <math>T</math> instead, which corresponds to taking the circle of (''n − r'')-steps in ''F'' and is thus also valid.)

# Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> ('''X''' must be the second argument!). This is the mode of ''T'' that has the most X's near the end. If ''T'' is [[primitive]], let <math>r</math> be the count of '''X''' steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>n</math> (the reader is encouraged to check this), <math>r</math>-steps in the filling MOS <math>F = \mathsf{mos}_{a,c}(\mathbf{L},\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>, we have a well-formed binary [[generator sequence]] of length <math>q</math> for <math>S</math>. 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 ''j''-th ''r''-step in the sequence of stacked <math>r</math>-steps in 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 "darkest" mode of <math>T</math> instead, which corresponds to taking the circle of (''n − r'')-steps in ''F'' and is thus also valid.)

# Assume that template MOS <math>T = T(\mathbf{m},\mathbf{X}) = M_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> is primitive. Suppose that the perfect generator of ''T'' that we use has ''r''-many '''X''' steps and that the imperfect generator has (''r'' + 1)-many '''X''' steps. Suppose the sizes for ''r''-steps in ''F'' are ''t'''''L''' + ''u'''''s''' and (''t'' − 1)'''L''' + (''u'' + 1)'''s'''.

# Assume that template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> is primitive. Suppose that the perfect generator of ''T'' that we use has ''r''-many '''X''' steps and that the imperfect generator has (''r'' + 1)-many '''X''' steps. Suppose the sizes for ''r''-steps in ''F'' are ''t'''''L''' + ''u'''''s''' and (''t'' − 1)'''L''' + (''u'' + 1)'''s'''.

#* <math>S</math> becomes a MOS with '''s''' = 0 for ''k'' in {0, ..., ''q'' − ''v'' − 1}, where ''v'' is the number of occurrences of the (''r'' + 1)-step (''t'' + 1)'''L''' + ''u'''''s''' in ''F''. In particular, if the interval class of (''r'' + 1)-steps consists of ''t'''''L''' + (''u'' + 1)'''s''' and (''t'' − 1)'''L''' + (''u'' + 2)'''s''', <math>S</math> becomes a MOS with '''s''' = 0 for any ''k'' in {0, ..., ''q'' − 1}. When this holds, the "aberrized" scale may rightly be considered a detempering of the original MOS a'''L'''b'''m''' with additional '''s''' steps.

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 The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, \mathbf{L}, c, k)</math> (and after replacing '''L''' with '''m''' and a with b, for <math>\mathsf{mos\_subst\_aberrize}(a, b, \mathbf{m}, c, k)</math> as well):
-Let <math>M_{a,b}(x,y;k)</math> be the mode of axby that would have brightness k if x were '''L''' and y were '''s'''. For example, <math>M_{a,b}(5,2;5)(x,y) = xxyxxxy.</math> Let <math> n = a+b+c</math> and <math>q = (a + c)/(a,c)</math>.
+Let <math>\mathsf{mos}_{a,b}(x,y;k)</math> be the mode of axby that would have brightness k if x were '''L''' and y were '''s'''. For example, <math>\mathsf{mos}_{a,b}(5,2;5)(x,y) = 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}) = M_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> ('''X''' must be the second argument!). This is the mode of ''T'' that has the most X's near the end. If ''T'' is [[primitive]], let <math>r</math> be the count of '''X''' steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>n</math> (the reader is encouraged to check this), <math>r</math>-steps in the filling MOS <math>F = M_{a,c}(\mathbf{L},\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>, we have a well-formed binary [[generator sequence]] of length <math>q</math> for <math>S</math>. 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 ''j''-th ''r''-step in the sequence of stacked <math>r</math>-steps in 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 "darkest" mode of <math>T</math> instead, which corresponds to taking the circle of (''n &minus; r'')-steps in ''F'' and is thus also valid.)
+# Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> ('''X''' must be the second argument!). This is the mode of ''T'' that has the most X's near the end. If ''T'' is [[primitive]], let <math>r</math> be the count of '''X''' steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>n</math> (the reader is encouraged to check this), <math>r</math>-steps in the filling MOS <math>F = \mathsf{mos}_{a,c}(\mathbf{L},\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>, we have a well-formed binary [[generator sequence]] of length <math>q</math> for <math>S</math>. 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 ''j''-th ''r''-step in the sequence of stacked <math>r</math>-steps in 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 "darkest" mode of <math>T</math> instead, which corresponds to taking the circle of (''n &minus; r'')-steps in ''F'' and is thus also valid.)
-# Assume that template MOS <math>T = T(\mathbf{m},\mathbf{X}) = M_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> is primitive. Suppose that the perfect generator of ''T'' that we use has ''r''-many '''X''' steps and that the imperfect generator has (''r'' + 1)-many '''X''' steps. Suppose the sizes for ''r''-steps in ''F'' are ''t'''''L''' + ''u'''''s''' and (''t'' &minus; 1)'''L''' + (''u'' + 1)'''s'''.
+# Assume that template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> is primitive. Suppose that the perfect generator of ''T'' that we use has ''r''-many '''X''' steps and that the imperfect generator has (''r'' + 1)-many '''X''' steps. Suppose the sizes for ''r''-steps in ''F'' are ''t'''''L''' + ''u'''''s''' and (''t'' &minus; 1)'''L''' + (''u'' + 1)'''s'''.
 #* <math>S</math> becomes a MOS with '''s''' = 0 for ''k'' in {0, ..., ''q'' &minus; ''v'' &minus; 1}, where ''v'' is the number of occurrences of the (''r'' + 1)-step (''t'' + 1)'''L''' + ''u'''''s''' in ''F''. In particular, if the interval class of (''r'' + 1)-steps consists of ''t'''''L''' + (''u'' + 1)'''s''' and (''t'' &minus; 1)'''L''' + (''u'' + 2)'''s''', <math>S</math> becomes a MOS with '''s''' = 0 for any ''k'' in {0, ..., ''q'' &minus; 1}. When this holds, the "aberrized" scale may rightly be considered a detempering of the original MOS a'''L'''b'''m''' with additional '''s''' steps.