MOS substitution: Difference between revisions
→Facts: This statement wasn't quite right; it's according to how many times the (r+1)-step occurs. |
|||
| Line 12: | Line 12: | ||
# 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. | # 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. | ||
#* If the interval class of (''r'' + 1)-steps has ''t''L + (''u'' + 1)s and (''t'' − 1)''L'' + (''u'' + 2)''s'', ''S'' becomes a mos after deleting s steps for any ''k'' in {0, ..., ''q'' − 1}.<!-- | #* If the interval class of (''r'' + 1)-steps has ''t''L + (''u'' + 1)s and (''t'' − 1)''L'' + (''u'' + 2)''s'', ''S'' becomes a mos after deleting s steps for any ''k'' in {0, ..., ''q'' − 1}.<!-- | ||
#* If the interval class of (''r'' + 1)-steps has ''t''L + (''u'' + 1)''s'' and (''t'' + 1)L + ''u''s, ''S'' becomes a mos after deleting s steps for k in {0, ..., ''q'' − ''v'' − 1}, where ''v'' is the number of occurrences of (''t'' + 1)L + ''u''s in ''F'' | #* If the interval class of (''r'' + 1)-steps has ''t''L + (''u'' + 1)''s'' and (''t'' + 1)L + ''u''s, ''S'' becomes a mos after deleting s steps for k in {0, ..., ''q'' − ''v'' − 1}, where ''v'' is the number of occurrences of (''t'' + 1)L + ''u''s in ''F''. | ||
== Examples == | == Examples == | ||