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By Ivanyi A. (ed.)

Ivanyi A. (ed.) Algorithms of informatics, vol.1.. foundations (2007)(ISBN 9638759615)

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For example instead of ((x∗ )(x + y)) we can consider x∗ (x + y). Two regular expressions are equivalent if they represent the same language, that is x ≡ y if X = Y , where X and Y are the languages represented by regular expressions x and y respectively. 18 shows some equivalent expressions. We show that to any nite language L can be associated a regular expression x which represent language L. If L = ∅, then x = ∅. If L = {w1 , w2 , . . , wn }, then x = x1 + x2 + . . + xn , where for any i = 1, 2, .

If |Q| = n, |Q | = n and |Σ| = m then taking into account that in worst case loop repeat is executed nn times, loop for m times, the running time of the algorithm in worst case will be O(nn m), or if n = n then O(n2 m). Our algorithm was described to determine the equivalence of two complete DFA's. If we have to determine the equivalence of two NFA's, rst we transform them into complete DFA's and after this we can apply the above algorithm. 2. 11 Determine if the two DFA's in Fig. 6 are equivalent or not.

An . Therefore, u ∈ L(G). 2. 8. 13. 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 denition means that in DFA A there is a walk a a an−1 a a 1 2 3 n q0 −→ q1 −→ q2 −→ · · · −→ qn−1 −→ qn , and since qn is a nal 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.

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