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