By Ivanyi A. (ed.)

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It is not necessary to consider all subset of the set of states of NFA. The states of DFA A can be obtained successively. Begin with the state q 0 = I and determine the states δ(q 0 , a) for all a ∈ Σ. For the newly obtained states we determine the states accessible from them. This can be continued until no new states arise. In our previous example q 0 := {q0 , q1 } is the initial state. 2. Finite automata and regular languages δ(q 0 , 0) = {q1 }, where q 1 := {q1 }, δ(q 1 , 0) = ∅, δ(q 2 , 0) = {q 2 }, 33 δ(q 0 , 1) = {q 2 }, where q 2 := {q2 }, δ(q 1 , 1) = {q 2 }, δ(q 2 , 1) = {q 2 }.

Qn the k states of the automaton A. Dene languages Rij for all −1 ≤ k ≤ n and 0 ≤ i, j ≤ n. k Rij is the set of words, for which automaton A goes from state qi to state qj without using any state with index greater than k . Using transition graph we can say: a word k is in Rij , if from state qi we arrive to state qj following the edges of the graph, and concatenating the corresponding labels on edges we get exactly that word, not using k any state qk+1 , . . qn . Sets Rij can be done also formally: −1 Rij = {a ∈ Σ | (qi , a, qj ) ∈ E}, if i = j , −1 Rii = {a ∈ Σ | (qi , a, qi ) ∈ E} ∪ {ε}, k−1 k−1 k−1 k Rij = Rij ∪ Rik Rkk ∗ k−1 Rkj for all i, j, k ∈ {0, 1, .

24 A language L is accepted by a nondeterministic pushdown automaton V1 by empty stack if and only if it can be accepted by a nondeterministic pushdown automaton V2 by nal state. Proof a) Let V1 = (Q, Σ, W, E, q0 , z0 , ∅) be the pushdown automaton which accepts by empty stack language L. Dene pushdown automaton V2 = (Q ∪ {p0 , p}, Σ, W ∪ {x}, E , p0 , x, {p}), where p, p0 ∈ Q, , x ∈ W and E = E ∪ p0 , (ε, x/xz0 ), q0 ∪ q, (ε, x/ε), p q ∈ Q .