MOS substitution: Difference between revisions

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'''MOS substitution''' is a procedure for obtaining a ternary scale with arbitrary scale signature ~~aLbmcs~~. Originally developed by Inthar for the purpose of adding aberrisma steps in an orderly manner to a MOS pattern ~~aLbm~~ (which we write in place of ~~aLbs~~ for convenience's sake, since s denotes the new steps added to the MOS) in the context of groundfault's aberrismic theory, MOS substitution is intended to take advantage of extra symmetry when a, c or b, c is not a coprime pair and generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales.

'''MOS substitution''' is a procedure for obtaining a ternary scale with arbitrary scale signature a'''L'''b'''m'''c'''s'''. Originally developed by Inthar for the purpose of adding aberrisma steps in an orderly manner to a MOS pattern a'''L'''b'''m''' (which we write in place of a'''L'''b'''s''' for convenience's sake, since s denotes the new steps added to the MOS) in the context of groundfault's aberrismic theory, MOS substitution is intended to take advantage of extra symmetry when a, c or b, c is not a coprime pair and generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales. (This article bolds steps '''L''', '''m''', '''s''', and '''X'''.)

Take for example d = (a, c) (:= gcd(a, c)), let a' = a/d and c' = c/d. Consider the MOS word (a + c)'''X'''b'''m''', which we call the ''template MOS''. The most even arrangement of a'-many '''L''' steps and c'-many '''s''' steps is the MOS a'Lc's, so this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the '''X''''s. Fixing a choice of which '''X''' in (a + c)'''X'''b'''m''' you start from, you have to choose a mode of a'Lc's. (Todo: count the distinct choices.) If a' = c' = 1 (equivalently if a = c), we obtain a balanced (thus MV3) ternary scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'Lc's. Of course, one may do this using template MOS a'''L'''(b + c)'''X''' and filling MOS (b/(b, c))'''m''' (c/(b, c))'''s''' instead.

Take for example d = (a, c) (:= gcd(a, c)), let a' = a/d and c' = c/d. Consider the MOS word (a + c)Xbm, which we call the ''template MOS''. The most even arrangement of a'-many L steps and c'-many s steps is the MOS a'Lc's, so this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the X's. Fixing a choice of which X in (a + c)Xbm you start from, you have to choose a mode of a'Lc's. (Todo: count the distinct choices.) If a' = c' = 1 (equivalently if a = c), we obtain a balanced (thus MV3) ternary scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'Lc's. Of course, one may do this using template MOS aL(b + c)X and filling MOS (b/(b, c))m (c/(b, c))s instead.

We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{\mathbf{L}, \mathbf{m}\}</math> is the step size identified with s by the template MOS and k is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).

We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{L, m\}</math> is the step size identified with s by the template MOS and k is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).

== Facts ==

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

The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, \mathbf{L}, c, k)</math> (and mutatis mutandis, 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>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>.

# If the template MOS <math>T = T(m,X) = M_{b,a+c}(m,X;n-1)</math> is [[primitive]], let <math>r</math> 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}(L,s;k)</math> come in exactly 2 sizes, <math>iL+js</math> and <math>(i-1)L+(j+1)s.</math> Since the detempering of the imperfect generator of ''T'' occurs only once in ''S'', we have a well-formed [[generator sequence]] of length <math>q</math> for ''S''. Letting <math>\mathsf{GS}(g_1, ..., g_{q})</math> be this generator sequence, <math>g_j</math> is either <math>pm + iL + js</math> or <math>pm + (i-1)L + (j+1)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>iL + js</math> or <math>(i-1)L + (j+1)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.)

# If the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = M_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> is [[primitive]], let <math>r</math> 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 [[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(m,X) = M_{b,a+c}(m,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}) = M_{b,a+c}(\mathbf{m},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'''.

#* ''S'' 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'', ''S'' becomes a MOS with s = 0 for any ''k'' in {0, ..., ''q'' − 1}.

#* <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 be considered an extension of the original MOS a'''L'''b'''m''' with '''s''' steps.

==Examples==

=== 5L2m4s ===

To derive 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, m, 4, k)</math>, we exploit gcd(b, c) = 2 and substitute ~~2m4s~~ into the template MOS ~~5L6X~~ (LXLXLXLXLXX). Since 2m4s has three distinct modes (ssmssm, smssms, and mssmss) and ~~5L6X~~ is primitive, we obtain three distinct scales: LsLsLmLsLsm, LsLmLsLsLms, and LmLsLsLmLss~~. The first two are a chiral pair of billiard scales, and the last is achiral but not deletion-MOS~~. All three scales admit short generator sequences of 2-steps, respectively GS(L+s, L+s, L+m), GS(L+s, L+m, L+s), and GS(L+m, L+s, L+s), ~~notably~~ representing all 3 possible rotations of (L+s, L+m, L+s).

To derive 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, \mathbf{m}, 4, k)</math>, we exploit gcd(b, c) = 2 and substitute 2'''m'''4'''s''' into the template MOS 5'''L'''6'''X''' ('''LXLXLXLXLXX'''). Since 2m4s has three distinct modes ('''ssmssm''', '''smssms''', and '''mssmss''') and 5'''L'''6X is primitive, we obtain three distinct scales: '''LsLsLmLsLsm''', '''LsLmLsLsLms''', and '''LmLsLsLmLss'''. All three scales admit short generator sequences of 2-steps, respectively GS('''L'''+'''s''', '''L'''+'''s''', '''L'''+'''m'''), GS('''L'''+'''s''', '''L'''+'''m''', '''L'''+'''s'''), and GS('''L'''+'''m''', '''L'''+'''s''', '''L'''+'''s'''), representing all 3 possible rotations of ('''L'''+'''s''', '''L'''+'''m''', '''L'''+'''s''').

{| class="wikitable"

|+ 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, m, 4, k)</math>

|+ 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, \mathbf{m}, 4, k)</math>

|-

!rowspan=2| ''k''

Line 33:

| 2 || <code>mssmss</code> || 4|0(2)

|colspan=2 style="text-align:right;"| <code>LmLsLsLmLss</code>

|colspan=2| GS(L+m, L+s, L+s) || yes

|colspan=2| GS('''L'''+'''m''', '''L'''+'''s''', '''L'''+'''s''') || yes

|-

| 1 || <code>smssms</code> || 2|2(2)

|colspan=2 style="text-align:right;"| <code>LsLmLsLsLms</code>

|colspan=2| GS(L+s, L+m, L+s) || yes

|colspan=2| GS('''L'''+'''s''', '''L'''+'''m''', '''L'''+'''s''') || yes

|-

| 0 || <code>ssmssm</code> || 0|4(2)

|colspan=2 style="text-align:right;"| <code>LsLsLmLsLsm</code>

|colspan=2| GS(L+s, L+s, L+m) || yes

|colspan=2| GS('''L'''+'''s''', '''L'''+'''s''', '''L'''+'''m''') || yes

|}

=== 6L7m9s ===

{| class="wikitable"

|+ 6L7m9s as <math>\mathsf{mos\_subst\_aberrize}(6, 7, L, 9, k)</math>

|+ 6L7m9s as <math>\mathsf{mos\_subst\_aberrize}(6, 7, \mathbf{L}, 9, k)</math>

|-

!rowspan=2| ''k''

Line 61:

| 4 || <code>LsLss</code> || 12|0(3)

|colspan=2 style="text-align:right;"| <code>mLsmLsmsLmsLmssmLsmLss</code>

|colspan=2| GS(L+m+s, L+m+s, L+m+s, L+m+s, m+2s) || yes

|colspan=2| GS('''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''m'''+2'''s''') || yes

|-

| 3 || <code>LssLs</code> || 9|3(3)

|colspan=2 style="text-align:right;"| <code>mLsmsLmsLmssmLsmLsmsLs</code>

|colspan=2| GS(L+m+s, L+m+s, L+m+s, m+2s, L+m+s) || yes

|colspan=2| GS('''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''m'''+2'''s''', '''L'''+'''m'''+'''s''') || yes

|-

| 2 || <code>sLsLs</code> || 6|6(3)

|colspan=2 style="text-align:right;"| <code>msLmsLmssmLsmLsmsLmsLs</code>

|colspan=2| GS(L+m+s, L+m+s, m+2s, L+m+s, L+m+s) || yes

|colspan=2| GS('''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''m'''+2'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''') || yes

|-

| 1 || <code>sLssL</code> || 3|9(3)

|colspan=2 style="text-align:right;"| <code>msLmssmLsmLsmsLmsLmssL</code>

|colspan=2| GS(L+m+s, m+2s, L+m+s, L+m+s, L+m+s) || yes

|colspan=2| GS('''L'''+'''m'''+'''s''', '''m'''+2'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''') || yes

|-

| 0 || <code>ssLsL</code> || 0|12(3)

|colspan=2 style="text-align:right;"| <code>mssmLsmLsmsLmsLmssmLsL</code>

|colspan=2| GS(m+2s, L+m+s, L+m+s, L+m+s, L+m+s) || no

|colspan=2| GS('''m'''+2'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''') || no

|}

@@ Line 1: / Line 1: @@
-'''MOS substitution''' is a procedure for obtaining a ternary scale with arbitrary scale signature aLbmcs. Originally developed by Inthar for the purpose of adding aberrisma steps in an orderly manner to a MOS pattern aLbm (which we write in place of aLbs for convenience's sake, since s denotes the new steps added to the MOS) in the context of groundfault's aberrismic theory, MOS substitution is intended to take advantage of extra symmetry when a, c or b, c is not a coprime pair and generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales.
+'''MOS substitution''' is a procedure for obtaining a ternary scale with arbitrary scale signature a'''L'''b'''m'''c'''s'''. Originally developed by Inthar for the purpose of adding aberrisma steps in an orderly manner to a MOS pattern a'''L'''b'''m''' (which we write in place of a'''L'''b'''s''' for convenience's sake, since s denotes the new steps added to the MOS) in the context of groundfault's aberrismic theory, MOS substitution is intended to take advantage of extra symmetry when a, c or b, c is not a coprime pair and generalize the congruence substitution procedure for building balanced words to obtain non-balanced but still more "even" scales. (This article bolds steps '''L''', '''m''', '''s''', and '''X'''.)
+<!-- Todo: Remove <math> tags -->
+Take for example d = (a, c) (:= gcd(a, c)), let a' = a/d and c' = c/d. Consider the MOS word (a + c)'''X'''b'''m''', which we call the ''template MOS''. The most even arrangement of a'-many '''L''' steps and c'-many '''s''' steps is the MOS a'<b>L</b>c'<b>s</b>, so this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the '''X''''s. Fixing a choice of which '''X''' in (a + c)'''X'''b'''m''' you start from, you have to choose a mode of a'<b>L</b>c'<b>s</b>. (Todo: count the distinct choices.) If a' = c' = 1 (equivalently if a = c), we obtain a balanced (thus MV3) ternary scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'<b>L</b>c'<b>s</b>. Of course, one may do this using template MOS a'''L'''(b + c)'''X''' and filling MOS (b/(b, c))'''m''' (c/(b, c))'''s''' instead.
-Take for example d = (a, c) (:= gcd(a, c)), let a' = a/d and c' = c/d. Consider the MOS word (a + c)Xbm, which we call the ''template MOS''. The most even arrangement of a'-many L steps and c'-many s steps is the MOS a'Lc's, so this method prescribes following the latter MOS, called the ''filling MOS'', to fill in the X's. Fixing a choice of which X in (a + c)Xbm you start from, you have to choose a mode of a'Lc's. (Todo: count the distinct choices.) If a' = c' = 1 (equivalently if a = c), we obtain a balanced (thus MV3) ternary scale; when in addition b is odd, the scale is also SV3 and chiral, and we recover the two chiralities from the two modes of a'Lc's. Of course, one may do this using template MOS aL(b + c)X and filling MOS (b/(b, c))m (c/(b, c))s instead.
+We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{\mathbf{L}, \mathbf{m}\}</math> is the step size identified with s by the template MOS and k is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).
-We tentatively denote the resulting scale <math>\mathsf{mos\_subst\_aberrize}(a, b, x, c, k),</math> where <math>x \in \{L, m\}</math> is the step size identified with s by the template MOS and k is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).
 == Facts ==
-The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, L, c, k)</math> (and mutatis mutandis, for <math>\mathsf{mos\_subst\_aberrize}(a, b, m, c, k)</math> as well):
+The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, \mathbf{L}, c, k)</math> (and mutatis mutandis, 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>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>.
-# If the template MOS <math>T = T(m,X) = M_{b,a+c}(m,X;n-1)</math> is [[primitive]], let <math>r</math> 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}(L,s;k)</math> come in exactly 2 sizes, <math>iL+js</math> and <math>(i-1)L+(j+1)s.</math> Since the detempering of the imperfect generator of ''T'' occurs only once in ''S'', we have a well-formed [[generator sequence]] of length <math>q</math> for ''S''. Letting <math>\mathsf{GS}(g_1, ..., g_{q})</math> be this generator sequence, <math>g_j</math> is either <math>pm + iL + js</math> or <math>pm + (i-1)L + (j+1)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>iL + js</math> or <math>(i-1)L + (j+1)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.)
+# If the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = M_{b,a+c}(\mathbf{m},\mathbf{X};n-1)</math> is [[primitive]], let <math>r</math> 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 [[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(m,X) = M_{b,a+c}(m,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}) = M_{b,a+c}(\mathbf{m},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'''.
-#* ''S'' 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'', ''S'' becomes a MOS with s = 0 for any ''k'' in {0, ..., ''q'' &minus; 1}.
+#* <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 be considered an extension of the original MOS a'''L'''b'''m''' with '''s''' steps.
 ==Examples==
 === 5L2m4s ===
-To derive 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, m, 4, k)</math>, we exploit gcd(b, c) = 2 and substitute 2m4s into the template MOS 5L6X (LXLXLXLXLXX). Since 2m4s has three distinct modes (ssmssm, smssms, and mssmss) and 5L6X is primitive, we obtain three distinct scales: LsLsLmLsLsm, LsLmLsLsLms, and LmLsLsLmLss. The first two are a chiral pair of billiard scales, and the last is achiral but not deletion-MOS. All three scales admit short generator sequences of 2-steps, respectively GS(L+s, L+s, L+m), GS(L+s, L+m, L+s), and GS(L+m, L+s, L+s), notably representing all 3 possible rotations of (L+s, L+m, L+s).
+To derive 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, \mathbf{m}, 4, k)</math>, we exploit gcd(b, c) = 2 and substitute 2'''m'''4'''s''' into the template MOS 5'''L'''6'''X''' ('''LXLXLXLXLXX'''). Since 2m4s has three distinct modes ('''ssmssm''', '''smssms''', and '''mssmss''') and 5'''L'''6X is primitive, we obtain three distinct scales: '''LsLsLmLsLsm''', '''LsLmLsLsLms''', and '''LmLsLsLmLss'''. All three scales admit short generator sequences of 2-steps, respectively GS('''L'''+'''s''', '''L'''+'''s''', '''L'''+'''m'''), GS('''L'''+'''s''', '''L'''+'''m''', '''L'''+'''s'''), and GS('''L'''+'''m''', '''L'''+'''s''', '''L'''+'''s'''), representing all 3 possible rotations of ('''L'''+'''s''', '''L'''+'''m''', '''L'''+'''s''').
 {| class="wikitable"
-|+ 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, m, 4, k)</math>
+|+ 5L2m4s as <math>\mathsf{mos\_subst\_aberrize}(5, 2, \mathbf{m}, 4, k)</math>
 |-
 !rowspan=2| ''k''
@@ Line 33: / Line 33: @@
 | 2 || <code>mssmss</code> || 4&#124;0(2)
 |colspan=2 style="text-align:right;"| <code>LmLsLsLmLss</code>
-|colspan=2| GS(L+m, L+s, L+s) || yes
+|colspan=2| GS('''L'''+'''m''', '''L'''+'''s''', '''L'''+'''s''') || yes
 |-
 | 1 || <code>smssms</code> || 2&#124;2(2)
 |colspan=2 style="text-align:right;"| <code>LsLmLsLsLms</code>
-|colspan=2| GS(L+s, L+m, L+s) || yes
+|colspan=2| GS('''L'''+'''s''', '''L'''+'''m''', '''L'''+'''s''') || yes
 |-
 | 0 || <code>ssmssm</code> || 0&#124;4(2)
 |colspan=2 style="text-align:right;"| <code>LsLsLmLsLsm</code>
-|colspan=2| GS(L+s, L+s, L+m) || yes
+|colspan=2| GS('''L'''+'''s''', '''L'''+'''s''', '''L'''+'''m''') || yes
 |}
 === 6L7m9s ===
 {| class="wikitable"
-|+ 6L7m9s as <math>\mathsf{mos\_subst\_aberrize}(6, 7, L, 9, k)</math>
+|+ 6L7m9s as <math>\mathsf{mos\_subst\_aberrize}(6, 7, \mathbf{L}, 9, k)</math>
 |-
 !rowspan=2| ''k''
@@ Line 61: / Line 61: @@
 | 4 || <code>LsLss</code> || 12&#124;0(3)
 |colspan=2 style="text-align:right;"| <code>mLsmLsmsLmsLmssmLsmLss</code>
-|colspan=2| GS(L+m+s, L+m+s, L+m+s, L+m+s, m+2s) || yes
+|colspan=2| GS('''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''m'''+2'''s''') || yes
 |-
 | 3 || <code>LssLs</code> || 9&#124;3(3)
 |colspan=2 style="text-align:right;"| <code>mLsmsLmsLmssmLsmLsmsLs</code>
-|colspan=2| GS(L+m+s, L+m+s, L+m+s, m+2s, L+m+s) || yes
+|colspan=2| GS('''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''m'''+2'''s''', '''L'''+'''m'''+'''s''') || yes
 |-
 | 2 || <code>sLsLs</code> || 6&#124;6(3)
 |colspan=2 style="text-align:right;"| <code>msLmsLmssmLsmLsmsLmsLs</code>
-|colspan=2| GS(L+m+s, L+m+s, m+2s, L+m+s, L+m+s) || yes
+|colspan=2| GS('''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''m'''+2'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''') || yes
 |-
 | 1 || <code>sLssL</code> || 3&#124;9(3)
 |colspan=2 style="text-align:right;"| <code>msLmssmLsmLsmsLmsLmssL</code>
-|colspan=2| GS(L+m+s, m+2s, L+m+s, L+m+s, L+m+s) || yes
+|colspan=2| GS('''L'''+'''m'''+'''s''', '''m'''+2'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''') || yes
 |-
 | 0 || <code>ssLsL</code> || 0&#124;12(3)
 |colspan=2 style="text-align:right;"| <code>mssmLsmLsmsLmsLmssmLsL</code>
-|colspan=2| GS(m+2s, L+m+s, L+m+s, L+m+s, L+m+s)  || no
+|colspan=2| GS('''m'''+2'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''', '''L'''+'''m'''+'''s''')  || no
 |}