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34 1. Automata and Formal Languages 5 repeat 6 for j = 1, 2, . . , m 7 do q ← δ(p, aj ) ✄ j counts the columns. p∈q i if q = ∅ 8 9 then if IsIn(q, Q) 10 then M [i, j] ← {q} 11 else k ← k + 1 12 qk ← q 13 M [i, j] ← {q k } 14 Q ← Q ∪ {q k } 15 else M [i, j] ← ∅ 16 i←i+1 17 until i = k + 1 18 return transition table M of A Since loop repeat is executed as many times as the number of states of new automaton, in worst case the running time can be exponential, because, if the number of states in NFA is n, then DFA can have even 2n − 1 states.

Q1 , q0 such that (qk−2 , ak , qk−1 ) ∈ E, . . , (q0 , a1 , q1 ) ∈ E, where q0 ∈ q 0 = I, thus, there is a walk a a ak−1 a a k 1 2 3 q0 −→ q1 −→ q2 −→ · · · −→ qk−1 −→ qk , q0 ∈ I, qk ∈ F , so L(A) ⊆ L(A). In constructing DFA we can use the corresponding transition function δ:     δ(q, a) = δ(q, a) , ∀q ∈ Q, ∀a ∈ Σ .   q∈q The empty set was excluded from the states, so we used here ∅ instead of {∅}. 10 to transform NFA A in Fig. 3. Introduce the following notation for the states of the DFA: S0 := {q0 , q1 }, S4 := {q0 , q2 }, S1 := {q0 }, S5 := {q1 , q2 }, S2 := {q1 }, S6 := {q0 , q1 , q2 } , S3 := {q2 }, where S0 is the initial state.

2. Finite automata and regular languages 49 Therefore xy i z ∈ L for all i, and this proves the theorem. 17 We use the pumping lemma to show that L1 = {ak bk | k ≥ 1} is not regular. Assume that L1 is regular, and let n be the corresponding natural number given by the pumping lemma. 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.

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