Ternary scale theorems: Difference between revisions
| Line 81: | Line 81: | ||
Proof: Consider the "imperfect alternant" of ''S'' on index ''p'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all X's from ''I'' has a j-step subword ''w'' 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 non-X step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-X letter. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' 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'']. | Proof: Consider the "imperfect alternant" of ''S'' on index ''p'', which is ''I'' = ''S''[''p'' : ''p'' + ''i'' + ''j''], and suppose the result of deleting all X's from ''I'' has a j-step subword ''w'' 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 non-X step than ''I'' by our assumption. Thus the word ''I'' must both begin and end in a non-X letter. Removing all the X's from ''I'' results in a word that is ''j'' + 1 letters long and is the ''j''-step word ''w'' 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 | '''Claim 2''': If a binary necklace ''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 u and v 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 (uv)<sup>(''m''−1)/2</sup>u mod e and (vu)<sup>(''m''−1)/2</sup>v mod e, and the step sizes clearly alternate. | Proof: Write u and v 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 (uv)<sup>(''m''−1)/2</sup>u mod e and (vu)<sup>(''m''−1)/2</sup>v mod e, and the step sizes clearly alternate. | ||