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32. 3. e. for each q ∈ Q, u ∈ Σ∗ and v ∈ W ∗ the triplet (q, u, v) can be a configuration. If u = a1 a2 . . ak and v = x1 x2 . . xm , then the pushdown automaton can change its configuration in two ways: • (q, a1 a2 . . ak , x1 x2 . . xm−1 xm ) =⇒ (p, a2 a3 . . ak , x1 , x2 . . xm−1 w), if q, (a1 , xm /w), p ∈ E • (q, a1 a2 . . ak , x1 x2 . . xm ) =⇒ (p, a1 a2 . . ak , x1 , x2 . . xm−1 w), if q, (ε, xm /w), p ∈ E. ∗ The reflexive and transitive closure of the relation =⇒ will be denoted by =⇒.

11 If L is a language accepted by a DFA, then one may construct a regular grammar which generates language L. Proof Let A = (Q, Σ, E, {q0 }, F ) be the DFA accepting language L, that is L = L(A). 11). 12). 2. 13. • If (p, a, q) ∈ E for p, q ∈ Q and a ∈ Σ, then put production p → aq in P . • If (p, a, q) ∈ E and q ∈ F , then put also production p → a in P . Prove that L(G) = L(A) \ {ε}. Let u = a1 a2 . . an ∈ L(A) and u = ε. Thus, since A accepts word u, there is a walk an−1 a1 a2 a3 an q0 −→ q1 −→ q2 −→ · · · −→ qn−1 −→ q n , qn ∈ F .

It is easy to see that in our case there is only a single possibility: (q0 , abab, z0 ) =⇒ (q1 , bab, z0 z1 ) =⇒ (q2 , ab, z0 ) =⇒ (q0 , ab, ε), but there is no further going, so word abab is not accepted. 27 The transition table ({q0 , q1 }, {0, 1}, {z0 , z1 , z2 }, E, q0 , z0 , ∅) is: Σ ∪ {ε} of the 0 pushdown automaton 1 V2 ε W z0 z1 z2 z0 z1 z2 z0 q0 (z0 z1 , q0 ) (z1 z1 , q0 ) (ε, q1 ) (z2 z1 , q0 ) (z0 z2 , q0 ) (z1 z2 , q0 ) (z2 z2 , q0 ) (ε, q1 ) (ε, q1 ) (ε, q1 ) (ε, q1 ) q1 (ε, q1 ) = The corresponding transition graph can be seen in Fig.

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