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

By Ivanyi A. (ed.)

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Sample text

An−1 qn−1 =⇒ a1 a2 . . an . Therefore, u ∈ L(G). Conversely, let u = a1 a2 . . an ∈ L(G) and u = ε. Then there exists a derivation q0 =⇒ a1 q1 =⇒ a1 a2 q2 =⇒ . . =⇒ a1 a2 . . an−1 qn−1 =⇒ a1 a2 . . an , in which productions q0 → a1 q1 , q1 → a2 q2 , . . , 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.

Because the length of the word u = an bn is 2n, this word can be written as in the lemma. We prove that this leads to a contradiction. Let u = xyz be the decomposition as in the lemma. Then |xy| ≤ n, so x and y can contain no other letters than a, and because we must have |y| ≥ 1, word y contains at least one a. Then xy i z for i = 1 will contain a different number of a’s and b’s, therefore xy i z ∈ L1 for any i = 1. This is a contradiction with the third assertion of the lemma, this is why that assumption that L1 is regular, is false.

Initial states are represented by a circle with an arrow, final states by a double circle. Let A1 = (Q1 , Σ1 , E1 , I1 , F1 ) and A2 = (Q2 , Σ2 , E2 , I2 , F2 ) be NFA. The result of any operation is a FA with ε-moves A = (Q, Σ, E, I, F ). Suppose that Q1 ∩Q2 = ∅ always. If not, we can rename the elements of any set of states. Union. A = A1 ∪ A2 , where 44 1. 13 (a) Representation of an NFA. Initial states are represented by a circle with an arrow, final states by a double circle. (b) Union of two NFA’s.

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