Interleaving: Difference between revisions
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=== Attempted proof of Conjecture === | === Attempted proof of Conjecture === | ||
{{proof|Suppose the scale is made of two interleaved subsets offset by the abstract interval '''δ'''. | |||
Case 1: gcd(2(''a'' + ''b''), ''k'') = 1. If ''k'' = 1, then any '''Z'''<sup>''q''</sup> a maximal subword of consecutive '''Z'''s has ''q'' odd, and they all must be the same length and separated by one non-'''Z''' letter '''W''': | Case 1: gcd(2(''a'' + ''b''), ''k'') = 1. If ''k'' = 1, then any '''Z'''<sup>''q''</sup> a maximal subword of consecutive '''Z'''s has ''q'' odd, and they all must be the same length and separated by one non-'''Z''' letter '''W''': | ||
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=> 2'''δ''' + '''W''' = ‖''s''‖ + '''V'''. If this linear relation is nontrivial, it's a contradiction. If this linear relation is trivial, the third letter '''U''' occurs an even number of times in ''s'', '''V''' occurs an odd number of times in ''s'', and '''W''' occurs an even number of times. Since gcd(''a'', ''b'') = 1, {'''U''', '''W'''} != {'''X''', '''Y'''}, and so either {'''U''', '''W'''} = {'''X''', '''Z'''} or {'''U''', '''W'''} = {'''Y''', '''Z'''}. But this is again a contradiction since gcd(''a'' + ''b'', ''b'') = gcd(''a'', ''a'' + ''b'') = gcd(''a'', ''b'') = 1. | => 2'''δ''' + '''W''' = ‖''s''‖ + '''V'''. If this linear relation is nontrivial, it's a contradiction. If this linear relation is trivial, the third letter '''U''' occurs an even number of times in ''s'', '''V''' occurs an odd number of times in ''s'', and '''W''' occurs an even number of times. Since gcd(''a'', ''b'') = 1, {'''U''', '''W'''} != {'''X''', '''Y'''}, and so either {'''U''', '''W'''} = {'''X''', '''Z'''} or {'''U''', '''W'''} = {'''Y''', '''Z'''}. But this is again a contradiction since gcd(''a'' + ''b'', ''b'') = gcd(''a'', ''a'' + ''b'') = gcd(''a'', ''b'') = 1. | ||
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== Generalizations == | == Generalizations == | ||