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Equivalence of NFAs & DFAs. Extended transition f n NFAs. Proving equivalence. Construct DFA D from NFA N. N = (Q N , , N ,q 0,N ,F N ) Construct D= (Q D , , D ,q 0,D ,F D ) such that, for any string w *, w is accepted by N iff w is accepted by M. Construction continued.
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Construct DFA Dfrom NFA N • N = (QN,,N,q0,N,FN) • Construct D= (QD,,D,q0,D,FD) such that, • for any string w *, • w is accepted by N iff w is accepted by M
Construction continued • QD = 2QN • q0,D = -close(q0,N) • FD = { q QD | q FN }
Proof that construction works • for any string w *, • w is accepted by N iff w is accepted by D • Or, need to show that:
