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l3-handout - Outline Proofs of equivalence of DFAs and NFAs...

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Lecture 3: Subset construction and -NFAs David Dill Department of Computer Science 1 Outline Proofs of equivalence of DFAs and NFAs -NFAs Equivalence of -NFAs and DFAs. 2 Subset Construction Given N = ( Q, Σ , q 0 , δ N , F N ) construct D = (2 Q , Σ , { q 0 } , δ D , F D ) such that L ( D ) = L ( N ) . The key is in the definition of δ D and F D : where δ D ( S, a ) = S q S δ N ( q, a ) , for each S 2 Q . The final states of the D are the subsets that contain at least one final state of N F D = { S 2 Q | S F N 6 = ∅} 3 Proof of the subset construction Here is what we want: Theorem For every NFA N , if D is the DFA obtained by the subset construction, L ( N ) = L ( D ) . But it’s actually easier to prove a stronger claim: Lemma For every NFA N , if D is the DFA obtained by the subset construction, ˆ δ D ( { q 0 } , w ) = ˆ δ N ( q 0 , w ) for all w Σ * (I.e., the state you get to by running D on w is the same as the set of states you get to by running N on w .) Reminders: DFA ˆ δ : Base: ˆ δ ( q, ) = q Induction: ˆ δ ( q, xa ) = δ ( ˆ δ ( q, x ) , a ) NFA ˆ δ : Base: ˆ δ ( q, ) = { q }
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