Interleaving: Difference between revisions

Line 100:

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~~)

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. By the lemma this scale has step variety ''r'' > 3, say with letters '''W'''1, ..., '''W'''''r''. Let ''k' '' be the inverse of ''k'' mod 2(''a'' + ''b''). By stacking ''k' ''-step subwords of ''T'', we end up with ''r'' different linear equations with 3 unknowns ''X''', '''Y''', and '''Z''', implying a linear relation between '''X''', '''Y''', and '''Z''' (?). This is a contradiction as '''X''', '''Y''', and '''Z''' are assumed to not have a linear relation.

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

Case 3: gcd(2(''a'' + ''b''), ''k'') > 1. ''k'' being even contradicts the interleaving property, hence ''k'' = ''a'' + ''b''. (To be continued)

@@ Line 100: / Line 100: @@
 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)
+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. By the lemma this scale has step variety ''r'' > 3, say with letters '''W'''<sub>1</sub>, ..., '''W'''<sub>''r''</sub>. Let ''k' '' be the inverse of ''k'' mod 2(''a'' + ''b''). By stacking ''k' ''-step subwords of ''T'', we end up with ''r'' different linear equations with 3 unknowns ''X''', '''Y''', and '''Z''', implying a linear relation between '''X''', '''Y''', and '''Z''' (?). This is a contradiction as '''X''', '''Y''', and '''Z''' are assumed to not have a linear relation.
-Suppose that '''δ''' = ''m'''''X''' + ''n'''''Y''' + ''p'''''Z''', ''m'' + ''n'' + ''p'' = ''k''.
 Case 3: gcd(2(''a'' + ''b''), ''k'') > 1. ''k'' being even contradicts the interleaving property, hence ''k'' = ''a'' + ''b''. (To be continued)