By Ivanyi A. (ed.)

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**Extra info for Algorithms of informatics. Foundations**

**Sample text**

10 can be simplied. 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 }.

The transition table is δ 0 1 q0 q1 q2 {q 1 } ∅ {q 2 } {q 2 } {q 2 } {q 2 } which is the same (excepted the notation) as before. The next algorithm will construct for an NFA A = (Q, Σ, E, I, F ) the transition table M of the equivalent DFA A = (Q, Σ, E, I, F ), but without to determine the nal states (which can easily be included). Value of IsIn(q, Q) in the algorithm is true if state q is already in Q and is false otherwise. Let a1 , a2 , . . , am be an ordered list of the letters of Σ. Nfa-Dfa(A) 1 2 3 4 5 6 7 q0 ← I Q ← {q 0 } i←0 k←0 repeat ✄ i counts the rows.

Using the transition function δ of NFA A we construct the sets S0 = {q0 }, δ(S0 , a1 ) = S1 , . . δ(Sk−1 , ak ) = Sk . Then q1 ∈ S1 , . . , qk ∈ Sk and since qk ∈ F we get Sk ∩ F = ∅, so Sk ∈ F . Thus, there exists a walk a a ak−1 a a 1 2 3 k S0 −→ S1 −→ S2 −→ · · · −→ Sk−1 −→ Sk , S0 ⊆ I, Sk ∈ F . There are sets S0 , . . , 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).