By Ivanyi A. (ed.)
Ivanyi A. (ed.) Algorithms of informatics, vol.1.. foundations (2007)(ISBN 9638759615)
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A) Product of two FA. (b) Iteration of an FA. Product. A = A1 · A2 , where Q = Q1 ∪ Q2 , Σ = Σ1 ∪ Σ2 , F = F2 , I = I1 , E = E1 ∪ E2 ∪ p ∈ F1 q ∈ I2 (p, ε, q) For the result automaton see Fig. 14(a). Here also L(A1 · A2 ) = L(A1 )L(A2 ). Iteration. 2. 15. Minimization of DFA. E = E1 ∪ (q0 , ε, p) ∪ p∈I1 (q, ε, p) . q ∈ F1 p ∈ I1 The iteration of an FA can be seen in Fig. 14(b). For this operation it is also ∗ true that L(A∗1 ) = L(A1 ) . The de nition of these tree operations proves again that regular languages are closed under the regular operations.
In our example from the rst equation we get Y = 1. From here Z = 0 + 10 + Z0, or Z = Z0 + (0 + 10), and solving this we get Z = (0 + 10)0∗ . Variable U can be obtained immediately and we obtain U = 11 + (0 + 10)0∗ 1. 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)∗ .
10). Let A = (Q, Σ, E, {q0 }, F ) be this DFA, so L = L(A). Let n be the number of its states, that is |Q| = n. Let u = a1 a2 . . am ∈ L and m ≥ n. Then, because the automaton accepts word u, there are states q0 , q1 , . . , qm and walk am−1 a1 a2 a3 am q0 −→ q1 −→ q2 −→ · · · −→ qm−1 −→ qm , qm ∈ F. Because the number of states is n and m ≥ n, by the pigeonhole principle3 states q0 , q1 , . . , qm can not all be distinct (see Fig. 17), there are at least two of them which are equal. Let qj = qk , where j < k and k is the least such index.