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The Emptiness Problem is the problem of deciding if a given regular language is empty (= ∅).
Theorem 4 (Emptiness) The Emptiness Problem for Regular Languages is decidable.
Proof: We'll sketch three different algorithms for deciding the Emptiness Problem, given some DFA A = (Q,Σ, T, q0, F).
(Emptiness 1) A string w is in L(A) iff it labels a path through the transition graph of A from q0 to an accepting state. Thus, the language will be non-empty iff there is some such path. So the question of Emptiness reduces to the question of connectivity: the language recognized by A is empty iff there is no accepting state in the connected component of its transition graph that is rooted at q0. The problem of determining connected components of directed graphs is algorithmically solvable,by Depth-First Search, for instance (and solvable in time linear in the number of nodes). So, given A, we just do a depth-?rst search of the transition graph rooted at the start state keeping track of whether we encounter any accepting state. We return "True" iff we ?nd none.
A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note
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The Equivalence Problem is the question of whether two languages are equal (in the sense of being the same set of strings). An instance is a pair of ?nite speci?cations of regular
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Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with
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It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ v) directly computes another (p, v) via
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