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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.
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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
A.(A+C)=A
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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
We developed the idea of FSA by generalizing LTk transition graphs. Not surprisingly, then, every LTk transition graph is also the transition graph of a FSA (in fact a DFA)-the one
Find a regular expression for the regular language L={w | w is decimal notation for an integer that is a multiple of 4}
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.
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