hwsoln03 - CS 341: Foundations of Computer Science II Prof....

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Unformatted text preview: CS 341: Foundations of Computer Science II Prof. Marvin Nakayama Homework 3 Solutions 1. Give NFAs with the specified number of states recognizing each of the following lan- guages. In all cases, the alphabet is = { , 1 } . (a) The language { w | w ends with 00 } with three states. 1 2 3 , 1 (b) The language { w | w contains the substring 0101 , i.e., w = x 0101 y for some x,y } with five states. 1 2 3 4 5 , 1 1 1 , 1 (c) The language { w | w contains at least two s, or exactly two 1 s } with six states. 1 2 5 3 4 6 , 1 1 , 1 1 1 (d) The language { } with one state. 1 1 (e) The language 1 with three states. 1 2 3 1 2. (a) Show by giving an example that, if M is an NFA that recognizes language C , swapping the accept and non-accept states in M doesnt necessarily yield a new NFA that recognizes C . Answer: The NFA M below recognizes the language C = { w | w ends with 00 } , where = { , 1 } . 1 2 3 , 1 Swapping the accept and non-accept states of M gives the following NFA M : 1 2 3 , 1 Note that M accepts the string 100 negationslash C = { w | w does not end with 00 } , so M does not recognize the language C . (b) Is the class of languages recognized by NFAs closed under complement? Explain your answer. Answer: The class of languages recognized by NFAs is closed under complement, which we can prove as follows. Suppose that C is a language recognized by some NFA M , i.e., C = L ( M ) . Since every NFA has an equivalent DFA (Theorem 1.19), there is a DFA D such that L ( D ) = L ( M ) = C . By problem 3 on Homework 2, we then know there is another DFA D that recognizes the language L ( D ) . Since 2 every DFA is also an NFA, this then shows that there is an NFA, in particular D , that recognizes the language C = L ( D ) . Thus, the class of languages recognized by NFAs is closed under complement. 3. Use the construction given in Theorem 1.39 to convert the following NFA N into an equivalent DFA. 1 2 3 a a a,b b Answer: Let NFA N = ( Q, ,, 1 ,F ) , where Q = { 1 , 2 , 3) , = { a,b } , 1 is the start state, F = { 2 } , and the transition function as in the diagram of N ....
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hwsoln03 - CS 341: Foundations of Computer Science II Prof....

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