MOS substitution: Difference between revisions
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Let <math>\mathsf{mos}(a,b;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}(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)</math> be the mode of axby that would have brightness k if x were '''L''' and y were '''s'''. 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>. | ||
# Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}(b,a+c;n-1)(\mathbf{m},\mathbf{X}).</math> 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;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 [[generator sequence]] of length <math>q,</math> corresponding to the circle of ''r''-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 ''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;n-1)(\mathbf{m},\mathbf{X}).</math> 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;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 [[generator sequence]] of length <math>q,</math> corresponding to the circle of ''r''-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 ''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}) = \mathsf{mos}(b,a+c;n-1)(\mathbf{m},\mathbf{X})</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;n-1)(\mathbf{m},\mathbf{X})</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. | #* <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. | ||
Revision as of 15:49, 24 January 2024
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 and simple generator sequence expressions (in the sense of using only two distinct generators) for them.
Note: 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)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]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(a, b, c, x, k), }[/math] where [math]\displaystyle{ 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).
Facts
The following holds for [math]\displaystyle{ S = \mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{L}, k) }[/math] (and after replacing L with m and a with b, for [math]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{m}, k) }[/math] as well):
Let [math]\displaystyle{ \mathsf{mos}(a,b;k) }[/math] be the mode of axby that would have brightness k if x were L and y were s. For example, [math]\displaystyle{ \mathsf{mos}(5,2;5)(x,y) = xxyxxxy. }[/math] Let [math]\displaystyle{ n = a+b+c }[/math] and [math]\displaystyle{ q = (a + c)/(a,c) }[/math].
- Consider the mode of the template MOS [math]\displaystyle{ T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}(b,a+c;n-1)(\mathbf{m},\mathbf{X}). }[/math] This is the mode of T that has the most Xs near the end. If T is primitive, let [math]\displaystyle{ r }[/math] be the count of X steps in a chosen (reduced) generator of [math]\displaystyle{ T. }[/math] Since [math]\displaystyle{ r }[/math] must be coprime to [math]\displaystyle{ n }[/math] (the reader is encouraged to check this), [math]\displaystyle{ r }[/math]-steps in the filling MOS [math]\displaystyle{ F = \mathsf{mos}(a,c;k)(\mathbf{L},\mathbf{s}) }[/math] come in exactly 2 sizes, [math]\displaystyle{ i\mathbf{L}+j\mathbf{s} }[/math] and [math]\displaystyle{ (i-1)\mathbf{L}+(j+1)\mathbf{s}. }[/math] Since the detempering of the imperfect generator of [math]\displaystyle{ T }[/math] occurs only once in [math]\displaystyle{ S }[/math], [math]\displaystyle{ S }[/math] admits a particularly elegant well-formed binary generator sequence of length [math]\displaystyle{ q, }[/math] corresponding to the circle of r-steps in the filling MOS. Letting [math]\displaystyle{ \mathsf{GS}(g_1, ..., g_{q}) }[/math] be this generator sequence, [math]\displaystyle{ g_j }[/math] is either [math]\displaystyle{ p\mathbf{m} + i\mathbf{L} + j\mathbf{s} }[/math] or [math]\displaystyle{ 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]\displaystyle{ r }[/math]-steps in the chosen mode of [math]\displaystyle{ F }[/math] is [math]\displaystyle{ i\mathbf{L} + j\mathbf{s} }[/math] or [math]\displaystyle{ (i-1)\mathbf{L} + (j+1)\mathbf{s}. }[/math] (We could have chosen to use the "darkest" mode of [math]\displaystyle{ 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]\displaystyle{ T = T(\mathbf{m},\mathbf{X}) = \mathsf{mos}(b,a+c;n-1)(\mathbf{m},\mathbf{X}) }[/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 tL + us and (t − 1)L + (u + 1)s.
- [math]\displaystyle{ 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 + us in F. In particular, if the interval class of (r + 1)-steps consists of tL + (u + 1)s and (t − 1)L + (u + 2)s, [math]\displaystyle{ 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 aLbm with additional s steps.
Examples
5L2m4s
To derive 5L2m4s as [math]\displaystyle{ \mathsf{aberrize\_by\_mos\_subst}(5, 2, 4, \mathbf{m}, 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. 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).
| k | filling MOS | UDP for filling MOS | step pattern | generator sequence | MOS for s = 0? | ||
|---|---|---|---|---|---|---|---|
| template MOS: | LXLXLXLXLXX
|
intvl. class of gen.:(*) | 2-steps | ||||
| 2 | mssmss |
4|0(2) | LmLsLsLmLss
|
GS(L+m, L+s, L+s) | yes | ||
| 1 | smssms |
2|2(2) | LsLmLsLsLms
|
GS(L+s, L+m, L+s) | yes | ||
| 0 | ssmssm |
0|4(2) | LsLsLmLsLsm
|
GS(L+s, L+s, L+m) | yes | ||
(*) such that the perfect generator has fewer X's than the imperfect counterpart
6L7m9s
| k | filling MOS (1 period) | UDP for filling MOS | step pattern | generator sequence | MOS for s = 0? | ||
|---|---|---|---|---|---|---|---|
| template MOS: | mXXmXXmXXmXXmXXmXXmXXX
|
intvl. class of gen.:(*) | 3-steps | ||||
| 4 | LsLss |
12|0(3) | mLsmLsmsLmsLmssmLsmLss
|
GS(L+m+s, L+m+s, L+m+s, L+m+s, m+2s) | yes | ||
| 3 | LssLs |
9|3(3) | mLsmsLmsLmssmLsmLsmsLs
|
GS(L+m+s, L+m+s, L+m+s, m+2s, L+m+s) | yes | ||
| 2 | sLsLs |
6|6(3) | msLmsLmssmLsmLsmsLmsLs
|
GS(L+m+s, L+m+s, m+2s, L+m+s, L+m+s) | yes | ||
| 1 | sLssL |
3|9(3) | msLmssmLsmLsmsLmsLmssL
|
GS(L+m+s, m+2s, L+m+s, L+m+s, L+m+s) | yes | ||
| 0 | ssLsL |
0|12(3) | mssmLsmLsmsLmsLmssmLsL
|
GS(m+2s, L+m+s, L+m+s, L+m+s, L+m+s) | no | ||
(*) such that the perfect generator has fewer X's than the imperfect counterpart
Open questions
- When are the GSes obtained via this procedure the shortest possible GSes?