tut03sol - CSCI 3130 Formal Languages and Automata Theory...

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CSCI 3130 Formal Languages and Automata Theory Fall 2015 Week 3 Tutorial Session Solution 1. (a) Write down a regular expression for the following NFA. For this problem, you do not have to go through the procedure described in class. (b) Convert the following NFA into a DFA. A B C D 0 , 1 0 0 1 1 0 , 1 Solution: (a) By inspection, the NFA accepts all strings that contain 00 or 11 somewhere. A corresponding regular expression is ( 0 + 1 ) * ( 00 + 11 )( 0 + 1 ) * (b) One can check that the set of active states of the NFA changes as follows: inputs 0 1 { A } { A, B } { A, C } { A, B } { A, B, D } { A, C } { A, C } { A, B } { A, C, D } { A, B, D } { A, B, D } { A, C, D } { A, C, D } { A, B, D } { A, C, D } Here is the corresponding DFA. { A } { A, B } { A, C } { A, B, D } { A, C, D } 0 1 0 1 1 0 0 1 1 0 2. (a) Write down the definition of regular expressions over an alphabet Σ. (b) Given a string w , define w R as the string w in reverse order. That is, if w = w 1 w 2 . . . w n , then w R = w n w n - 1 . . . w 1 . For example, if w = live , then w R = evil . Given a language L , define its reversal L R as the set of strings in L in reverse. More precisely, L R = { w R | w L } . For example, if L = { live , raw , level } , then L R = { evil , war , level } . If L is a regular language, prove that L R as also regular. Solution: (a) A regular expression E is defined recursively as follows: 1
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The “atomic” expressions , ε , and a (for any symbol a Σ) are all regular expressions If an expression E can be written as E = R + S or E = RS or E = R * for some regular expression, then E is also a regular expression (b) Proof 1
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