Interleaving: Difference between revisions

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 Moreover, the offset '''δ''' = '''Z''', i.e. ''S''<sub>2</sub> is separated '''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, ''w''<sub>1</sub> with consecutive '''Z''''s and ''w''<sub>2</sub> with consecutive non-'''Z''' letters. Consecutive non-'''Z''' letters contradict '''δ''' = '''Z''': Let ''r'' be the length of a maximal subword of non-'''Z''' letters in ''w''<sub>2</sub>.
+If any maximal subword of consecutive '''Z'''s has ''q'' > 1, then the scale can be split into two subwords of the same length, ''w''<sub>1</sub> with the maximal number of consecutive '''Z''''s and ''w''<sub>2</sub> with the minimal number of '''Z'''s.
-'''Z''' '''W'''<sub>1</sub>...'''W'''<sub>''r''</sub> '''Z''' ('''W'''<sub>''i''</sub> are all non-'''Z'''s)
+Scoot ''w''<sub>1</sub> to the right until it loses one '''Z'''. Because of the offset, this proves that a non-'''Z''' letter is equal to '''Z'''. Hence ''q'' = 1, as desired.
-(Assume WOLOG) '''W'''<sub>1</sub> + '''W'''<sub>2</sub> (in strand ''S''<sub>2</sub>) = '''Z''' + '''W'''<sub>1</sub> (in strand ''S''<sub>1</sub>)
-'''W'''<sub>2</sub> = '''Z''' (contradiction)
-Hence ''q'' = 1, as desired.
 Case 2: ''k'' &ge; 3.