49 PS6 Solutions

49 PS6 Solutions - Handout #49 May 17, 2010 CS103 Robert...

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Handout #49 CS103 May 17, 2010 Robert Plummer Problem Set #6 Solutions 1. Exercise 1.4, p. 83, part e. Starts with a: At most one b: Just follow the footnote to construct the DFA for the intersection. 2. Exercise 1.5, p. 84, part d. Exchange accepting/non-accepting for complement.
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2 3. Exercise 1.6, p. 84, parts f, i, n. 4. Exercise 1.7, p. 84, parts b and e.
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3 5. Exercise 1.9, p.85, part a. 6. Exercise 1.12, p. 85. D is a subset of b*a*, because it does not contain strings that have ab as a substring. A regular expression is b(bb)*(aa)*. The DFA can be generated from this. 7. Exercise 1.14, p.85. (a) Let N be the DFA with accept and non-accept states swapped. If N accepts a string w, then it ends in an accepting state, which would be a rejecting state of M. So if N accepts w, M rejects it, which means that strings accepted by N are not in the language B. Similarly, if N rejects w, it would be accepted by M and would be in B. So N accepts exactly the strings not accepted by M, and the language of N is the complement of B. Since this argument shows how to build a DFA to recognize the complement of any
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49 PS6 Solutions - Handout #49 May 17, 2010 CS103 Robert...

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