Generator-offset property: Difference between revisions

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'''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''.

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 note ''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 ~~beginning~~ of one such ''j''-step subword and use S[''q'' : ''q'' + ''i'' + ''j''].

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 note ''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''].

'''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, then ''U'' = (YZ)''b''.

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 because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are ~~ηζ…ζη~~ mod e and ~~ζη…ηζ~~ mod e, and the step sizes clearly 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 because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are (ηζ)(''m''−1)/2η mod e and (ζη)(''m''−1)/2ζ mod e, and the step sizes clearly alternate.

These two claims prove that ''E''X(S) = (YZ)''b'' and that the two alternants' sizes differ by replacing one Y for a Z.

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 '''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''.
-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 note ''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 beginning of one such ''j''-step subword and use S[''q'' : ''q'' + ''i'' + ''j''].
+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 note ''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''].
 '''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, then ''U'' = (YZ)<sup>''b''</sup>.
-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 because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are ηζ…ζη mod e and ζη…ηζ mod e, and the step sizes clearly 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 because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are (ηζ)<sup>(''m''&minus;1)/2</sup>η mod e and (ζη)<sup>(''m''&minus;1)/2</sup>ζ mod e, and the step sizes clearly alternate.
 These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''b''</sup> and that the two alternants' sizes differ by replacing one Y for a Z.