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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.
The computation of an SL 2 automaton A = ( Σ, T) on a string w is the maximal sequence of IDs in which each sequential pair of IDs is related by |- A and which starts with the in
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Another striking aspect of LTk transition graphs is that they are generally extremely ine?cient. All we really care about is whether a path through the graph leads to an accepting
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The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automa
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As we are primarily concerned with questions of what is and what is not computable relative to some particular model of computation, we will usually base our explorations of langua
The fact that the Recognition Problem is decidable gives us another algorithm for deciding Emptiness. The pumping lemma tells us that if every string x ∈ L(A) which has length grea
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
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