Generator-offset property: Difference between revisions

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Consider the two alternants, detemperings of the generator ''i''X + ''j''W of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1.

'''Claim 1''': Deleting X's from the ~~alternants~~ of ''S'' gives every ''j''-step subword in the scale ''E''X(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''.

'''Claim 1''': Deleting X's from the alternant subwords of ''S'' gives every ''j''-step subword in the scale ''E''X(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''.

Proof: Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on index ''p'' of ''E''X(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''X(''I'') has ''R'' as a substring. Then ''S''[''p'' − 1: ''p'' − 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' ≠ ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j''].

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 Consider the two alternants, detemperings of the generator ''i''X + ''j''W of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1.
-'''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''.
+'''Claim 1''': Deleting X's from the alternant subwords of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from ''S''.
 Proof: Assume, possibly after inverting the generator, that the imperfect generator of ''T'' has ''j'' + 1 W's and the perfect generator has ''j'' W's. Suppose that one ''j''-step word ''R'' on index ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j'']. By this we mean that ''E''<sub>X</sub>(''I'') has ''R'' as a substring. Then ''S''[''p'' &minus; 1: ''p'' &minus; 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both detemperings of perfect generators of ''T'', and have one fewer step that is Y or Z. Thus the word ''I'' must both begin and end in a letter that is either Y or Z. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step we started with, with just one extra letter appended. Thus one of the two perfect generators above, namely the one that removes the extra letter, must contain this ''j''-step. The rest of the ''j''-step subwords are all contained in "perfect" alternants; take ''q'' &ne; ''p'' to be the index of the first letter of one such ''j''-step subword (as contained in ''S'') and use S[''q'' : ''q'' + ''i'' + ''j''].