MOS substitution: Difference between revisions

From Xenharmonic Wiki
Jump to navigation Jump to search
Inthar (talk | contribs)
Inthar (talk | contribs)
No edit summary
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 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 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.


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, 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 aL(b + c)X and (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.  
== Worked-out example ==
 
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 <math>k \in \{0, 1, ..., (b + c)/(b,c)-1\}</math> is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).
== Facts ==
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</math> (resp. <math>b + c</math>).
 
The following holds for <math>S = \mathsf{mos\_subst\_aberrize}(a, b, L, c, k)</math> (resp. <math>\mathsf{mos\_subst\_aberrize}(a, b, m, c, k)</math>):
# If the template MOS <math>T = T(m,X) = M_{b,n}(m,X;n-1)</math> (resp. <math>T(L,X)=M_{a, b+c}(L,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>, <math>r</math>-steps in the filling MOS <math>F = M_{a,c}(L,s;k)</math> (resp. <math>M_{b,c}(m,s;k)</math>) come in exactly 2 sizes, <math>iL+js</math> and <math>(i-1)L+(j+1)s</math> (resp. <math>im+js</math> and <math>(i-1)m+(j+1)s</math>), and taking this generator of <math>T</math> results in a [[generator sequence]] of length <math>q</math>. 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 size ''j''-th ''r''-step in the sequence of stacked <math>r</math>-steps in the chosen mode of <math>F</math> is <math>pm + iL + js</math> or <math>pm + (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.)
== Example ==
For 5L2m4s, 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).
For 5L2m4s, 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).
== Open questions ==
# Is the length of the shortest guided generator sequence related to the length of the filler MOS? It could hold when the scale pattern has the divisibilities that this procedure is intended to take advantage of.
# How can we ensure that the deletion of s steps from a scale thus "aberrismized" recovers the MOS version of aLbm?

Revision as of 17:50, 23 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.

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{mos\_subst\_aberrize}(a, b, x, c, k), }[/math] where [math]\displaystyle{ x \in \{L, m\} }[/math] is the step size identified with s by the template MOS and [math]\displaystyle{ k \in \{0, 1, ..., (b + c)/(b,c)-1\} }[/math] is the brightness of the mode of the filling MOS used (0 corresponds to the darkest mode).

Facts

Let [math]\displaystyle{ 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]\displaystyle{ M_{a,b}(5,2;5)(x,y) = xxyxxxy. }[/math] Let [math]\displaystyle{ n = a+b+c }[/math] and [math]\displaystyle{ q = a + c }[/math] (resp. [math]\displaystyle{ b + c }[/math]).

The following holds for [math]\displaystyle{ S = \mathsf{mos\_subst\_aberrize}(a, b, L, c, k) }[/math] (resp. [math]\displaystyle{ \mathsf{mos\_subst\_aberrize}(a, b, m, c, k) }[/math]):

  1. If the template MOS [math]\displaystyle{ T = T(m,X) = M_{b,n}(m,X;n-1) }[/math] (resp. [math]\displaystyle{ T(L,X)=M_{a, b+c}(L,X;n-1) }[/math]) is primitive, let [math]\displaystyle{ r }[/math] 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], [math]\displaystyle{ r }[/math]-steps in the filling MOS [math]\displaystyle{ F = M_{a,c}(L,s;k) }[/math] (resp. [math]\displaystyle{ M_{b,c}(m,s;k) }[/math]) come in exactly 2 sizes, [math]\displaystyle{ iL+js }[/math] and [math]\displaystyle{ (i-1)L+(j+1)s }[/math] (resp. [math]\displaystyle{ im+js }[/math] and [math]\displaystyle{ (i-1)m+(j+1)s }[/math]), and taking this generator of [math]\displaystyle{ T }[/math] results in a generator sequence of length [math]\displaystyle{ q }[/math]. Letting [math]\displaystyle{ \mathsf{GS}(g_1, ..., g_{q}) }[/math] be this generator sequence, [math]\displaystyle{ g_j }[/math] is either [math]\displaystyle{ pm + iL + js }[/math] or [math]\displaystyle{ pm + (i-1)L + (j+1)s, }[/math] according as the size 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{ pm + iL + js }[/math] or [math]\displaystyle{ pm + (i-1)L + (j+1)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.)

Example

For 5L2m4s, 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).