Algorithms of informatics, vol. 1 by Ivanyi A. (ed.)

By Ivanyi A. (ed.)

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Qn−2 → an−1 qn−1 , qn−1 → an were used, which by definition means that in DFA A there is a walk a an−1 a a a n 3 2 1 qn , · · · −→ qn−1 −→ q2 −→ q1 −→ q0 −→ and since qn is a final state, u ∈ L(A) \ {ε} . If the DFA accepts also the empty word ε, then in the above grammar we introduce a new start symbol q0 instead of q0 , consider the new production q0 → ε and for each production q0 → α introduce also q0 → α. 13 Let A = ({q0 , q1 , q2 }, {a, b}, E, {q0 }, {q2 }) be a DFA, where E = (q0 , a, q0 ), (q0 , b, q1 ), (q1 , b, q2 ), (q2 , a, q2 ) .

R0i , R0i from expressions representing languages R0i p 2 1 Further on we give some procedures which associate DFA to regular expressions and conversely regular expression to DFA. Associating regular expressions to finite automata. We present here three methods, each of which associate to a DFA the corresponding regular expression. Method 1. Using the result of the theorem of Kleene, we will construct the sets k n n Rij , and write a regular expression which represents the language L = R0i ∪ R0i ∪ 1 2 n .

Qk ∈ Sk and since qk ∈ F we get Sk ∩ F = ∅, so Sk ∈ F . Thus, there exists a walk a a a ak−1 a k 1 2 3 S0 −→ S1 −→ S2 −→ · · · −→ Sk−1 −→ Sk , S0 ⊆ I, Sk ∈ F . There are sets S0 , . . , Sk for which S0 = I, and for i = 0, 1, . . , k we have Si ⊆ Si , and ak−1 ak a1 a2 a3 S0 −→ S1 −→ S2 −→ · · · −→ Sk−1 −→ Sk is a productive walk. Therefore w ∈ L(A). That is L(A) ⊆ L(A). b) Now we show that L(A) ⊆ L(A). Let w = a1 a2 . . ak ∈ L(A). Then there is a walk ak−1 ak a1 a2 a3 q0 −→ q1 −→ q2 −→ · · · −→ q k−1 −→ q k , q 0 ∈ I, q k ∈ F .

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