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Sample text

We are interested to determine if they are or not equivalent. We construct a table with elements of form (q, q ), where q ∈ Q and q ∈ Q . Beginning with the second column of the table, we associate a column to each letter of the alphabet Σ. If the first element of the ith row is (q, q ) then at the cross of ith row and the column associated to letter a will be the pair elem δ(q, a) , elem δ (q , a) . a ... ... (q, q ) elem δ(q, a) , elem δ (q , a) ... 2. Finite automata and regular languages 35 In the first column of the first row we put (q0 , q0 ) and complete the first row using the above method.

States S1 , S4 , S5 , S6 are inaccessible and can be removed from the DFA. 2. Finite automata and regular languages 33 δ 0 1 S0 {S2 } {S3 } S2 S3 ∅ {S3 } {S3 } {S3 } The corresponding transition graph is in Fig. 5. 10 can be simplified. 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.

Marking pair {p, q} means putting a star in the cell corresponding to row p and column q (or row q and column p). First we mark pairs {q2 , q0 }, {q2 , q1 }, {q2 , q3 }, {q2 , q4 } and {q2 , q5 } (because q2 is the single final state). Then consider all unmarked pairs and examine them as the algorithm requires. Let us begin with pair {q0 , q1 }. Associate with it pairs {elem δ(q0 , 0) , elem δ(q1 , 0) }, {elem δ(q0 , 1) , elem δ(q1 , 1) }, that is {q1 , q4 }, {q4 , q2 }. Because pair {q4 , q2 } is already marked, mark also pair {q0 , q1 }.

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