Interleaving: Difference between revisions

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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 either ''w''1 or ''w''2 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.

If ''k'' > 1, stack the word of ''k''-steps in the scale, yielding a circular word ''T''. 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)

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)

Suppose that '''δ''' = ''m'''''X''' + ''n'''''Y''' + ''p'''''Z''', ''m'' + ''n'' + ''p'' = ''k''.

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 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 either ''w''<sub>1</sub> or ''w''<sub>2</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.
-If ''k'' > 1, stack the word of ''k''-steps in the scale, yielding a circular word ''T''. 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)
+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)
 Suppose that '''δ''' = ''m'''''X''' + ''n'''''Y''' + ''p'''''Z''', ''m'' + ''n'' + ''p'' = ''k''.