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As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′0. Since they cannot be reached from Q′0 there is no path from Q′0 to a state in F′ which passes through them and they can be deleted from the automaton without changing the language it accepts. In practice it is much easier to build Q′ as needed, only including those state sets that actually are needed.
To see how this works, lets carry out an example. For maximum generality, let's start with the NFA with ε-transitions given above, repeated here:
Because it is simpler to write the transition function (δ) out as a table than it is to write out the transition relation (T) as a set of tuples, we will work with the δ representation. When given a transition graph of an NFA with ε-transitions like this there are 6 steps required to reduce it to a DFA:
1. Write out the transition function and set of ?nal states of the NFA.
2. Convert it to an NFA without ε-transitions.
(a) Compute the ε-Closure of each state in the NFA.
(b) Compute the transition function of the equivalent NFA without ε-transitions.
(c) Compute the set of ?nal states of the equivalent NFA without ε- transitions.
This was one of the ?rst substantial theorems of Formal Language Theory. It's maybe not too surprising to us, as we have already seen a similar equivalence between LTO and SF. But
Kleene called this the Synthesis theorem because his (and your) proof gives an effective procedure for synthesizing an automaton that recognizes the language denoted by any given r
Automaton (NFA) (with ε-transitions) is a 5-tuple: (Q,Σ, δ, q 0 , F i where Q, Σ, q 0 and F are as in a DFA and T ⊆ Q × Q × (Σ ∪ {ε}). We must also modify the de?nitions of th
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We now add an additional degree of non-determinism and allow transitions that can be taken independent of the input-ε-transitions. Here whenever the automaton is in state 1
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