Interleaving: Difference between revisions

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Let ''A'' be the set of all (''k'' - 1)/2-step intervals of ''w''. |''A''| = 1 implies that ''k'' = 1 mod |''w''|, so |''A''| ≥ 2 and contains at least two intervals '''w'''1 and '''w'''2.

Say |''A''| = 2. (If |''A''| ≥ 3 then the proof is easy.) Say ''B'' is the set of all subwords (not intervals) subtending '''w'''1, and ''C'' is the set of all subwords ending in '''Z''' subtending '''w'''2.

Say |''A''| = 2. (If |''A''| ≥ 3 then the proof is easy.) Say ''B'' is the set of all subwords of ''w'' (not intervals) subtending '''w'''1, and ''C'' is the set of all subwords ending in '''Z''' subtending '''w'''2.

Choose ''b'' in ''B'' and ''c'' in ''C''.

Choose ''b'' in ''B'' and ''c'' in ''C'' and consider ''b''('''XZ''', '''YZ''') ''c''('''XZ''', '''YZ''').

''b''('''XZ''', '''YZ''')

''c''('''XZ''', '''YZ''')

If ''k'' = 3, then we can just say ''b'' = ''x'' and ''c'' = ''y''. Suppose '''YZX''' does not occur in ''S''; then ''yx'' does not occur in ''w'' which is a contradiction, so ''YZX'' must occur.

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 Let ''A'' be the set of all (''k'' - 1)/2-step intervals of ''w''. |''A''| = 1 implies that ''k'' = 1 mod |''w''|, so |''A''| &ge; 2 and contains at least two intervals '''w'''<sub>1</sub> and '''w'''<sub>2</sub>.
-Say |''A''| = 2. (If |''A''| &ge; 3 then the proof is easy.) Say ''B'' is the set of all subwords (not intervals) subtending '''w'''<sub>1</sub>, and ''C'' is the set of all subwords ending in '''Z''' subtending '''w'''<sub>2</sub>.
+Say |''A''| = 2. (If |''A''| &ge; 3 then the proof is easy.) Say ''B'' is the set of all subwords of ''w'' (not intervals) subtending '''w'''<sub>1</sub>, and ''C'' is the set of all subwords ending in '''Z''' subtending '''w'''<sub>2</sub>.
-Choose ''b'' in ''B'' and ''c'' in ''C''.
+Choose ''b'' in ''B'' and ''c'' in ''C'' and consider ''b''('''XZ''', '''YZ''') ''c''('''XZ''', '''YZ''').
-''b''('''XZ''', '''YZ''')
-''c''('''XZ''', '''YZ''')
 If ''k'' = 3, then we can just say ''b'' = ''x'' and ''c'' = ''y''. Suppose '''YZX''' does not occur in ''S''; then ''yx'' does not occur in ''w'' which is a contradiction, so ''YZX'' must occur.