MOS substitution: Difference between revisions
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(Note: This article bolds steps <math>\mathbf{L}, \mathbf{m}, \mathbf{s}, \mathbf{x}.</math> For integers <math>m, n, \ (m, n) := \gcd(m, n).</math>) | (Note: This article bolds steps <math>\mathbf{L}, \mathbf{m}, \mathbf{s}, \mathbf{x}.</math> For integers <math>m, n, \ (m, n) := \gcd(m, n).</math>) | ||
In the original aberrismic-informed context, say that <math>d = (a, c) > 1.</math> Consider the MOS word <math>(a + c)\mathbf{X}b\mathbf{m}</math>, which we call the ''template MOS''. Since the "most even" arrangement (in the sense of [[distributional evenness]]) of <math>a</math>-many <math>\mathbf{L}</math> steps and <math>c</math>-many <math>\mathbf{s}</math> steps is the MOS <math>a\mathbf{L}b\mathbf{s}</math> (which will in general be a non-[[primitive]] MOS), this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the <math>\mathbf{X}</math> steps. Fixing a choice of which <math>\mathbf{X}</math> in the MOS <math>(a + c)\mathbf{X}b\mathbf{m}</math> you start from, we can choose one of <math>(a+c)/d</math> modes of <math>a \mathbf{L} c \mathbf{s}.</math> If <math>a = c</math>, we obtain a balanced (thus MV3) ternary scale; when in addition <math>b</math> is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of <math>a\mathbf{L}c\mathbf{s}</math>. Of course, one may do this using template MOS <math>a\mathbf{L}(b + c)\mathbf{X}</math> and the <math>(b, c)</math>-multiperiod filling MOS <math>b\mathbf{m} c\mathbf{s}</math> instead. We tentatively denote the resulting scale <math>\mathsf{ | In the original aberrismic-informed context, say that <math>d = (a, c) > 1.</math> Consider the MOS word <math>(a + c)\mathbf{X}b\mathbf{m}</math>, which we call the ''template MOS''. Since the "most even" arrangement (in the sense of [[distributional evenness]]) of <math>a</math>-many <math>\mathbf{L}</math> steps and <math>c</math>-many <math>\mathbf{s}</math> steps is the MOS <math>a\mathbf{L}b\mathbf{s}</math> (which will in general be a non-[[primitive]] MOS), this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the <math>\mathbf{X}</math> steps. Fixing a choice of which <math>\mathbf{X}</math> in the MOS <math>(a + c)\mathbf{X}b\mathbf{m}</math> you start from, we can choose one of <math>(a+c)/d</math> modes of <math>a \mathbf{L} c \mathbf{s}.</math> If <math>a = c</math>, we obtain a balanced (thus MV3) ternary scale; when in addition <math>b</math> is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of <math>a\mathbf{L}c\mathbf{s}</math>. Of course, one may do this using template MOS <math>a\mathbf{L}(b + c)\mathbf{X}</math> and the <math>(b, c)</math>-multiperiod filling MOS <math>b\mathbf{m} c\mathbf{s}</math> instead. We tentatively denote the resulting scale <math>\mathsf{insert\_by\_MOS\_subst}(a, b, c; x; k),</math> where <math>x \in \{\mathbf{L}, \mathbf{m}\}</math> is the step size identified with <math>\mathbf{s}</math> by the template MOS and <math>k</math> is the brightness of the mode of the filling MOS used (<math>k = 0</math> corresponds to the darkest mode; the conventional understanding of "brightness" makes sense as <math>\mathbf{L}</math> (resp. <math>\mathbf{m}</math>) > <math>\mathbf{s}</math>). | ||
== Facts == | == Facts == | ||
The following holds for <math>S = \mathsf{ | The following holds for <math>S = \mathsf{insert\_by\_MOS\_subst}(a, b, c; \mathbf{L}; k)</math> (and after replacing <math>\mathbf{L}</math> with <math>\mathbf{m}</math> and <math>a</math> with <math>b,</math> for <math>\mathsf{insert\_by\_MOS\_subst}(a, b, c; \mathbf{m}; k)</math> as well): | ||
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)(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;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)(x,y) = xxyxxxy.</math> Let <math> n = a+b+c</math> and <math>q = (a + c)/(a,c)</math>. | ||
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# 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> (which the reader is encouraged to verify), <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 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 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.) | # 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> (which the reader is encouraged to verify), <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 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 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.) | ||
# 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 perfect generator of <math>T</math> that we use has <math>r</math>-many <math>\mathbf{X}</math> steps and that the imperfect generator has <math>(r + 1)</math>-many <math>\mathbf{X}</math> steps. Suppose the sizes for <math>r</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> | # 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 perfect generator of <math>T</math> that we use has <math>r</math>-many <math>\mathbf{X}</math> steps and that the imperfect generator has <math>(r + 1)</math>-many <math>\mathbf{X}</math> steps. Suppose the sizes for <math>r</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>(r + 1)</math>-step <math>(t + 1)\mathbf{L} + u\mathbf{s}</math> in <math>F</math>. In particular, if the interval class of <math>(r + 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 | #* <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>(r + 1)</math>-step <math>(t + 1)\mathbf{L} + u\mathbf{s}</math> in <math>F</math>. In particular, if the interval class of <math>(r + 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. | ||
==Examples== | ==Examples== | ||
=== 5L2m4s === | === 5L2m4s === | ||
To derive <math>5\mathbf{L}2\mathbf{m}4\mathbf{s}</math> as <math>\mathsf{ | To derive <math>5\mathbf{L}2\mathbf{m}4\mathbf{s}</math> as <math>\mathsf{insert\_by\_MOS\_subst}(5, 2, 4; \mathbf{m}; k)</math>, we exploit <math>(b, c) = 2</math> and substitute <math>2\mathbf{m}4\mathbf{s}</math> into the template MOS <math>5\mathbf{L}6\mathbf{X}</math> (<math>\mathbf{LXLXLXLXLXX}</math>). Since <math>2\mathbf{m}4\mathbf{s}</math> has three distinct modes (<math>\mathbf{ssmssm}, \mathbf{smssms}, \mathbf{mssmss}</math>) and <math>5\mathbf{L}6\mathbf{X}</math> is primitive, we obtain three distinct scales, all of which admit short generator sequences of 2-steps, representing all 3 possible rotations of <math>(\mathbf{L}+\mathbf{m}, \mathbf{L}+\mathbf{s}, \mathbf{L}+\mathbf{s})</math> as displayed in the following table: | ||
{| class="wikitable" | {| class="wikitable" | ||
|+ <math>5\mathbf{L}2\mathbf{m}4\mathbf{s}</math> as <math>\mathsf{ | |+ <math>5\mathbf{L}2\mathbf{m}4\mathbf{s}</math> as <math>\mathsf{insert\_by\_MOS\_subst}(5, 2, 4; \mathbf{m}; k)</math> | ||
|- | |- | ||
!rowspan=2| <math>k</math> | !rowspan=2| <math>k</math> | ||
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=== 6L7m9s === | === 6L7m9s === | ||
{| class="wikitable" | {| class="wikitable" | ||
|+ <math>6\mathbf{L}7\mathbf{m}9\mathbf{s}</math> as <math>\mathsf{ | |+ <math>6\mathbf{L}7\mathbf{m}9\mathbf{s}</math> as <math>\mathsf{insert\_by\_MOS\_subst}(6, 7, 9; \mathbf{L}; k)</math> | ||
|- | |- | ||
!rowspan=2| <math>k</math> | !rowspan=2| <math>k</math> | ||