Interleaving: Difference between revisions

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

Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''~~. We get a linear relation probably, but I haven't checked~~.

Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''.

We have '''V'''w'''W''' where '''V''' != '''W''', ‖'''V'''''w''‖ = '''δ''', ‖''w'''''W'''‖ = ‖''s''‖ - '''δ'''

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=> '''δ''' - '''V''' = ‖''w''‖ = ‖''s''‖ - '''δ''' - '''W'''

=> 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'''}.

=> 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.

== Generalizations ==

@@ Line 104: / Line 104: @@
 Case 3: gcd(2(''a'' + ''b''), ''k'') > 1. ''k'' being even contradicts the interleaving property, hence ''k'' = ''a'' + ''b'' which must be odd.
-Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''. We get a linear relation probably, but I haven't checked.
+Scoot the (''a'' + ''b'')-letter subword across ''s''. On ''a'' + ''b'' consecutive notes, the interval subtended by this word is '''δ''', and on the other ''a'' + ''b'' notes, the interval subtended by this word is ‖''s''‖ - '''δ'''.
 We have '''V'''w'''W''' where '''V''' != '''W''', ‖'''V'''''w''‖ = '''δ''', ‖''w'''''W'''‖ = ‖''s''‖ - '''δ'''
@@ Line 110: / Line 110: @@
 => '''δ''' - '''V''' = ‖''w''‖ = ‖''s''‖ - '''δ''' - '''W'''
-=> 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'''}.
+=> 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.
 == Generalizations ==