MOS substitution: Difference between revisions

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# 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>, <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>, 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 ''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.)

# Suppose that the perfect generator of ''T'' that we use has r-many X steps~~, where~~ r ~~< |T|/2~~. Suppose the ~~perfect generator is tL~~ + us and ~~the imperfect generator is~~ (t - 1)L + (u + 1)s. Then the interval class of (r + 1)-steps has either (a) tL + (u + 1)s and (t - 1)L + (u + 2)s, or (b) tL + (u + 1)s and (t + 1)L + us.

# 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. Then the interval class of (''r'' + 1)-steps has either (a) ''t''L + (''u'' + 1)s and (''t'' − 1)''L'' + (''u'' + 2)''s'', or (b) ''t''L + (''u'' + 1)''s'' and (''t'' + 1)L + ''u''s.

#* In case (a), S becomes a mos after deleting s steps for any k in {0, ..., q-1}.

#* In case (a), ''S'' becomes a mos after deleting s steps for any ''k'' in {0, ..., ''q'' − 1}.

#* In case (b), S becomes a mos after deleting s steps for k = 0, ..., v, where v is the number of generators stacked to obtain (t + 1)L + us in the filling MOS F.

#* In case (b), ''S'' becomes a mos after deleting s steps for k in {0, ..., ''v''}, where ''v'' is the number of generators stacked to obtain (''t'' + 1)L + ''u''s in the filling MOS F.

== 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).

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 # 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>, <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>, 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 ''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.)
-# Suppose that the perfect generator of ''T'' that we use has r-many X steps, where r < |T|/2. Suppose the perfect generator is tL + us and the imperfect generator is (t - 1)L + (u + 1)s. Then the interval class of (r + 1)-steps has either (a) tL + (u + 1)s and (t - 1)L + (u + 2)s, or (b) tL + (u + 1)s and (t + 1)L + us.
+# 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. Then the interval class of (''r'' + 1)-steps has either (a) ''t''L + (''u'' + 1)s and (''t'' &minus; 1)''L'' + (''u'' + 2)''s'', or (b) ''t''L + (''u'' + 1)''s'' and (''t'' + 1)L + ''u''s.
-#* In case (a), S becomes a mos after deleting s steps for any k in {0, ..., q-1}.
+#* In case (a), ''S'' becomes a mos after deleting s steps for any ''k'' in {0, ..., ''q'' &minus; 1}.
-#* In case (b), S becomes a mos after deleting s steps for k = 0, ..., v, where v is the number of generators stacked to obtain (t + 1)L + us in the filling MOS F.
+#* In case (b), ''S'' becomes a mos after deleting s steps for k in {0, ..., ''v''}, where ''v'' is the number of generators stacked to obtain (''t'' + 1)L + ''u''s in the filling MOS F.
 == 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).