MOS substitution: Difference between revisions

Take for example <math>d = (a, c)</math>, let <math>a^\prime = a/d, c^\prime = c/d.</math> 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.  

We tentatively denote the resulting scale <math>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, x, 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, since '''L''' (or '''m''') > '''s''').

The following holds for <math>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>\mathsf{aberrize\_by\_mos\_subst}(a, b, c, \mathbf{m}, k)</math> as well):

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 &minus; 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'' &minus; 1)'''L''' + (''u'' + 1)'''s'''.

#* <math>S</math> becomes a MOS with '''s''' = 0 for ''k'' in {''v'', ..., ''q'' &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.

To derive 5L2m4s as <math>\mathsf{aberrize\_by\_mos\_subst}(5, 2, 4, \mathbf{m}, k)</math>, we exploit gcd(b, c) = 2 and substitute 2'''m'''4'''s''' into the template MOS 5'''L'''6'''X''' ('''LXLXLXLXLXX'''). Since 2'''m'''4'''s''' has three distinct modes ('''ssmssm''', '''smssms''', and '''mssmss''') and 5'''L'''6'''X''' 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''').

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

!rowspan=2| ''k''

!rowspan=2| MOS for s = 0?

(*) such that the perfect generator has fewer '''X''''s than the imperfect counterpart

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

!rowspan=2| ''k''

!rowspan=2| MOS for s = 0?

 

(*) such that the perfect generator has fewer '''X''''s than the imperfect counterpart