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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 path function δ : Q × Σ* → P(Q) is the extension of δ to strings: This just says that the path labeled ε from any given state q goes only to q itself (or rather never l
Lemma 1 A string w ∈ Σ* is accepted by an LTk automaton iff w is the concatenation of the symbols labeling the edges of a path through the LTk transition graph of A from h?, ∅i to
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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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Can you say that B is decidable? If you somehow know that A is decidable, what can you say about B?
Differentiate between DFA and NFA. Convert the following Regular Expression into DFA. (0+1)*(01*+10*)*(0+1)*. Also write a regular grammar for this DFA.
The Recognition Problem for a class of languages is the question of whether a given string is a member of a given language. An instance consists of a string and a (?nite) speci?cat
The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
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