Generator-offset property: Difference between revisions

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Proof: Assume 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 note ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, 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'' − 1: ''p'' − 1 + ''i'' + ''j''] and ''S''[''p'' + 1 : ''p'' + 1 + ''i'' + ''j''] are both perfect, 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.

'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''b'' Y's and ''b'' Z's alternate.

'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, the ''b'' Y's and ''b'' Z's alternate.

Proof: Write η and ζ for the two sizes of ''j''-steps. Since gcd(''j'', 2''b'') = 1 there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd. Hence the scale steps of ''U'' are ηζ…ζη mod e and ζη…ηζ mod e, and the step sizes clearly alternate.

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 Proof: Assume 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 note ''p'' of ''E''<sub>X</sub>(''S'') is "contained in" the corresponding "imperfect alternant" of ''S'', the unique (''i'' + ''j'')-step, 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 perfect, 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.
-'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''b'' Y's and ''b'' Z's alternate.
+'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, the ''b'' Y's and ''b'' Z's alternate.
 Proof: Write η and ζ for the two sizes of ''j''-steps. Since gcd(''j'', 2''b'') = 1 there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd. Hence the scale steps of ''U'' are ηζ…ζη mod e and ζη…ηζ mod e, and the step sizes clearly alternate.