Interleaving: Difference between revisions

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If any maximal subword of consecutive '''Z'''s has ''q'' > 1, then the scale can be split into two subwords of length ''a'' + ''b'', ''w''1 with the maximal number of consecutive '''Z''''s and ''w''2 with the minimal number of '''Z'''s. We can also choose ''w''1 to begin in ''S''1 (because?)

Scoot ''w''1 to the right one step at a time until it loses ~~one '''Z''', or scoot ''w''2 to the right until it gains~~ one '''Z'''. ~~Because of the offset and because ''w''1 begins in ''S''1 (because ''a'' + ''b'' is odd), this~~ proves that a non-'''Z''' letter is equal to '''Z'''. Hence ''q'' = 1, as desired.

Scoot ''w''1 to the right one step at a time until it loses one '''Z'''. This proves either that a non-'''Z''' letter is equal to '''Z''' or that the offset goes in the wrong direction. Hence ''q'' = 1, as desired.

If ''k'' > 1, stack the word of ''k''-steps in the scale, yielding a circular word ''T'', which traverses all notes of ''S'' since gcd(''k'', 2(''a'' + ''b'')) = 1. Since ''k'' is odd, the letters of this word alternate between beginning in ''S''1 and beginning in ''S''2. By a reasoning similar to the above, ''T'' has a letter '''δ''' between its two mutually interleaved strands. (To be continued)

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 If any maximal subword of consecutive '''Z'''s has ''q'' > 1, then the scale can be split into two subwords of length ''a'' + ''b'', ''w''<sub>1</sub> with the maximal number of consecutive '''Z''''s and ''w''<sub>2</sub> with the minimal number of '''Z'''s. We can also choose ''w''<sub>1</sub> to begin in ''S''<sub>1</sub> (because?)
-Scoot ''w''<sub>1</sub> to the right one step at a time until it loses one '''Z''', or scoot ''w''<sub>2</sub> to the right until it gains one '''Z'''. Because of the offset and because ''w''<sub>1</sub> begins in ''S''<sub>1</sub> (because ''a'' + ''b'' is odd), this proves that a non-'''Z''' letter is equal to '''Z'''. Hence ''q'' = 1, as desired.
+Scoot ''w''<sub>1</sub> to the right one step at a time until it loses one '''Z'''. This proves either that a non-'''Z''' letter is equal to '''Z''' or that the offset goes in the wrong direction. Hence ''q'' = 1, as desired.
 If ''k'' > 1, stack the word of ''k''-steps in the scale, yielding a circular word ''T'', which traverses all notes of ''S'' since gcd(''k'', 2(''a'' + ''b'')) = 1. Since ''k'' is odd, the letters of this word alternate between beginning in ''S''<sub>1</sub> and beginning in ''S''<sub>2</sub>. By a reasoning similar to the above, ''T'' has a letter '''δ''' between its two mutually interleaved strands. (To be continued)