Interleaving: Difference between revisions

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Moreover, the offset '''δ''' = '''Z''', i.e. ''S''2 is separated by the interval '''Z''' to the right of ''S''1.

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.

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 ~~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.

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.

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)

@@ Line 44: / Line 44: @@
 Moreover, the offset '''δ''' = '''Z''', i.e. ''S''<sub>2</sub> is separated by the interval '''Z''' to the right of ''S''<sub>1</sub>.
-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.
+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 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.
+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.
 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)