Algorithms of informatics. Foundations by Ivanyi A. (ed.)

By Ivanyi A. (ed.)

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Remark. The above algorithm can be used also in the case of an DFA which is not complete, that is there are states for which does not exist transition. Then a pair ∅, {q} may occur, and if q is a nal state, consider this pair marked. 16 Let be the DFA in Fig. 15. We will use a table for marking pairs with a star. Marking pair {p, q} means putting a star in the cell corresponding to row p and column q (or row q and column p). 2. Finite automata and regular languages 47 First we mark pairs {q2 , q0 }, {q2 , q1 }, {q2 , q3 }, {q2 , q4 } and {q2 , q5 } (because q2 is the single nal state).

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 . e. there exists qk ∈ q k ∩F , that is by the denitions of qk ∈ F and q k there is qk−1 such that (qk−1 , ak , qk ) ∈ E . Similarly, there are the states qk−2 , .

Q, q ) elem δ(q, a) , elem δ (q , a) ... In the rst column of the rst row we put (q0 , q0 ) and complete the rst row using the above method. If in the rst row in any column there occur a pair of states from which one is a nal state and the other not then the algorithm ends, the two automata are not equivalent . If there is no such a pair of states, every new pair is written in the rst column. The algorithm continues with the next unlled row. If no new pair of states occurs in the table and for each pair both of states are nal or both are not, then the algorithm ends and the two DFA are equivalent .

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