Generator-offset property: Difference between revisions

Line 86:

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''X(''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''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 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. 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 scale ''U'' has ''k'' X's and ''k'' Y's, gcd(''j'', 2''k'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''k'' Y's and ''k'' Z's alternate.

'''Claim 2''': If a scale ''U'' has ''b'' X's and ''b'' Y'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.

Proof: write η and ζ for the two sizes of ''j''-steps. Since gcd(''j'', 2''k'') = 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.

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.

These two claims prove that ''E''X(S) = (YZ)''k'', which also proves that the two alternants' sizes differ by replacing one Y for a Z.

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

===== Statement (5) =====

@@ Line 86: / Line 86: @@
 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. 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 scale ''U'' has ''k'' X's and ''k'' Y's, gcd(''j'', 2''k'') = 1, and consecutively stacked ''j''-steps occur in 2 alternating sizes, the ''k'' Y's and ''k'' Z's alternate.
+'''Claim 2''': If a scale ''U'' has ''b'' X's and ''b'' Y'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.
-Proof: write η and ζ for the two sizes of ''j''-steps. Since gcd(''j'', 2''k'') = 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.
+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.
-These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''k''</sup>, which also proves that the two alternants' sizes differ by replacing one Y for a Z.
+These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''b''</sup>, which also proves that the two alternants' sizes differ by replacing one Y for a Z.
 ===== Statement (5) =====