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## Concept of DFA Simulating NFA, Finite Automata and Regular Languages

DFA simulating NFA: the power set constructionTheorem (equivalence NFA-DFA): Any NFA N can be transformed to an equivalent DFA M.

Proof: Thanks to Lemma on ε-transitions, assume with no loss of generality which N = (S, A, f, s0, F) consists of no ε-transitions. Define M = (2S, A, f’, {s0}, F’) as follows. 2S is the power set of S, {s0} the initial state. F’ comprises of all such subsets R ⊆ S which have some final state of N that s, R ∩ F ≠ {}. f’: 2S x A -> 2S is defined as:

For R ∈ 2S and a ∈ A, f’(R, a) = {s ∈ S | s ∈ f(r, a) for some r ∈ R}.

N and M are equal due the given invariant: for all x ∈ A*, f’({s0}, x) = f(s0, x). QED

: Convert the NFA N converted to an ε-free NFA M. E(s1) = {s1, s3}. L = (a ∪ ba* (a ∪ b) a)* to an equivalent DFAIllustrationThe power set construction tends to introduce the unreachable states. Such can be removed by using a transitive closure algorithm. Alternatively, we make states of M only as the corresponding subsets of S are being reached, therefore joining transitive closure with the construction of state space. For simplicity of comparison, let’s redraw the DFA above with no unreachable states, with a layout identical to illustration above. The non-determinism in illustration above: f(s2, a) = {s2, s3} is resolved by introducing the states {s2, s3} and {s1, s2, s3}:

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