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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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Theorem The class of ?nite languages is a proper subclass of SL. Note that the class of ?nite languages is closed under union and concatenation but SL is not closed under either. N
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Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
Intuitively, closure of SL 2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a
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Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
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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