# hwsol6 - Theory of Algorithms Spring 2000 Homework 6...

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Theory of Algorithms. Spring 2000. Homework 6 Solutions. Section 4.3 (3) Let L = { w ∈ { a, b } * : n a ( w ) = n b ( w ) } . Then L is not regular. (Since L * = L , L * is also not regular either.) Proof. Assume towards a contradiction that L is regular. Let m > 0 be given by the Pumping Lemma. Then let w = a m b m . Notice that w L and | w | ≥ m . So let w = xyz be the decomposition of w given by the Pumping Lemma. Notice that y = a k for some k with 1 k m . Now let i = 2. Then w i = w 2 = xy 2 z = a m + k b m . So w i / L because m + k 6 = m . This contradicts the Pumping Lemma. So L is not regular. (4a) Let L = a n b l a k : k n + l . 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 = a m b m a 2 m . Notice that w L and | w | ≥ m . So let w = xyz be the decomposition of w given by the Pumping Lemma. Notice that y = a t for some t with 1 t m . Now let i = 2. Then w i = w 2 = xy 2 z = a m + t b m a 2 m . So w i / L because 2 m 6≥ ( m + t ) + m . This contradicts the Pumping Lemma. So L is not regular. (4b) Let L = a n b l a k : k 6 = n + l . Then L is not regular. Proof. Assume towards a contradiction that L is regular. Then ¯ L L ( a * b * a * ) is also regular since the family of regular languages is closed under compliment and intersection. Let us write L 1 for ¯ L L ( a * b * a * ). Notice that L 1 = a n b l a k : k = n + l . We will apply the Pumping Lemma to L 1 . Let m > 0 be given by the Pumping Lemma. Then let w = a m b m a 2 m . Notice that w L 1 and | w | ≥ m . So let w = xyz be the decomposition of w given by the Pumping Lemma. Notice that y = a t for some t with 1 t m . Now let i = 2. Then w i = w 2 = xy 2 z = a m + t b m a 2 m . So w i / L 1 because 2 m 6 = ( m + t ) + m . This contradicts the Pumping Lemma. So L is not regular. (4c) Let L = a n b l a k : n = l or l 6 = k . 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 = a m b m a m . Notice that w L (since n=m=l) and | w | ≥ m . So let w = xyz be the decomposition of w given by the Pumping Lemma. Notice that y = a t for some t with 1 t m . Now let i = 2. Then w i = w 2 = xy 2 z = a m + t b m a m . So w i / L because n = m + t 6 = m = l and l = m = k . This contradicts the Pumping Lemma. So L is not regular. (4d) Let L = a n b l : n l . 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 = a m b m . Notice that w L and | w | ≥ m . So let w = xyz be the decomposition of w given by the Pumping Lemma. Notice that y = a t for some t with 1 t m . Now let i = 2. Then w i = w 2 = xy 2 z = a m + t b m . So w i / L because m + t 6≤ m . This contradicts the Pumping Lemma. So L is not regular.

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• Spring '07
• Staff
• WI, Formal language, Formal languages, Regular expression, Regular language, Pumping lemma for regular languages

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