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To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the Myhill graph is acyclic, then no path from x to x can be longer than card(Σ) + 2, since otherwise some node would have to occur at least twice in the path.
The question of finiteness of L(A), then, can be reduced to the question of acyclicity of the corresponding Myhill graph. And we established that there is an algorithm for testing acyclicity of graphs in Algorithms and Data Structures. Our algorithm for deciding finiteness of L(A) just interprets A as a graph and calls the algorithm for deciding acyclicity as a subroutine.
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Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about
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Let L 3 = {a i bc j | i, j ≥ 0}. Give a strictly 2-local automaton that recognizes L 3 . Use the construction of the proof to extend the automaton to one that recognizes L 3 . Gi
Generate 100 random numbers with the exponential distribution lambda=5.0.What is the probability that the largest of them is less than 1.0?
De?nition Deterministic Finite State Automaton: For any state set Q and alphabet Σ, both ?nite, a ?nite state automaton (FSA) over Q and Σ is a ?ve-tuple (Q,Σ, T, q 0 , F), w
Prepare the consolidated financial statements for the year ended 30 June 2011. On 1 July 2006, Mark Ltd acquired all the share capitall of john Ltd for $700,000. At the date , J
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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