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Unformatted text preview: Theory of Algorithms. Spring 2000. Homework 7 Solutions. Section 5.1 (4) Let L = { ab ( bbaa ) n bba ( ba ) n : n ≥ } . Then L is not regular. Proof. Assume towards a contradiction that L is regular. Let m > 0 be given by the Pumping Lemma. Then let w = ab ( bbaa ) m bba ( ba ) m . Notice that w ∈ L and  w  ≥ m . So let w = xyz be the decomposition of w given by the Pumping Lemma. Notice that there are many different possibilities for what y looks like and therefore it would seem as if this proof is going to be difficult. But in fact it will be easy. Let i = 2. I claim that w 2 / ∈ L . To see this, I argue as follows. Notice that bba ( ba ) m is a suffix of w and so it is also a sufix of w 2 . But now notice that our string w is the one and only string in L that has bba ( ba ) m as a suffix. Since w 2 6 = w and yet w 2 has bba ( ba ) m as a suffix, w 2 / ∈ L . This contradicts the Pumping Lemma. So L is not regular. (6) (a) S → aSb  aaa  aa  a  λ (b) S → aSb  A  B  λ A → aA  λ B → bB  bb (c) S → S 1  S 2 S 1 → aaS 1 b  aS 1  a S 2 → AAs 2 b  Ab A → a  λ (d) S → aSbbB  λ B → b  λ As preparation for part (e) let us review exercise 14(b) from section 1.2 (Homework 1.) In that exercise you were asked for a grammar for the language L = { w ∈ { a, b } * : n a ( w ) > n b ( w ) } . My answer was the following grammar G . S → aS 1  aS  S 1 S S 1 → S 1 S 1  λ  aS 1 b  bS 1 a Let me explain why it is true that L ( G ) = L . First notice that by Example 1.12 on page 23, S 1 ⇒ * w , iff n a ( w ) = n b ( w ). It is then clear that L ( G ) ⊆ L , because the only way to eliminate...
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This note was uploaded on 07/17/2011 for the course MAD 3512 taught by Professor Staff during the Spring '07 term at University of Florida.
 Spring '07
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