By Ivanyi A. (ed.)

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Sample text

2-3 Give a DFA to accept respectively the following languages: L1 = {an bm | n ≥ 1, m ≥ 0}, L2 = {an bm | n ≥ 1, m ≥ 1}, n m L3 = {a b | n ≥ 0, m ≥ 0}, L4 = {an bm | n ≥ 0, m ≥ 1}. 2-4 Give an NFA which accepts words containing at least two 0’s and any number of 1’s. Give an equivalent DFA. 2-5 Minimize the DFA’s in Fig. 28. (a) is a minimum state automaton. 2-7 Transform NFA in Fig. (b) in a DFA, and after this minimize it. 58 1. 26 Associating finite automaton to regular expression ε + (0 + 1)∗ 1.

We proceed as in the previous example using here word u = 0n 1n , where n is the natural number associated by lemma to language L2 . 19 We prove, using the pumping lemma, that L3 = uu | u ∈ {a, b}∗ is not a regular language. Let w = an ban b = xyz be, where n here is also the natural number associated to L3 by the pumping lemma. From |xy| ≤ n we have that y contains no other letters than a, but it contains at least one. By lemma we have xz ∈ L3 , that is not possible. Therefore L3 is not regular.

Using this method in the case of Fig. 19, the following equations will be obtained X = ε + X1 + Y 1 Y = X0 + Y 0 Therefore X = ε + (X + Y )1 Y = (X + Y )0. Adding the two equations we will obtain X + Y = ε + (X + Y )(0 + 1), from where (considering ε as β and (0 + 1) as α) we get the result X + Y = (0 + 1)∗ . From here the value of X after the substitution is X = ε + (0 + 1)∗ 1, which is equivalent to the expression obtained using the other methods. 2. 25 Possible transformations to obtain finite automaton associated to a regular expression.

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