Theory of Computation207910 marksEquivalence of DFA and NFA
Show that, for any NFA N=(Q,Sigma,delta,q0,F) accepting language L=Sigma*, there is a DFA D=(Q',Sigma',q0',delta',F') accepting the same language L.
Show that, for any NFA $N=(Q,\Sigma,\delta,q_0,F)$ accepting language $L=\Sigma^*$, there is a DFA $D=(Q',\Sigma',q_0',\delta',F')$ accepting the same language $L$.[10]
For any NFA N = (Q, Σ, δ, q₀, F) accepting language L, there exists a DFA D = (Q', Σ, δ', q₀', F') such that L(D) = L(N) = L. --- The core insight is that while an NFA can be in multiple states simultaneously, a DFA must be in exactly one state at any point. We simulate the NFA by making each sta...