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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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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
The Myhill-Nerode Theorem provided us with an algorithm for minimizing DFAs. Moreover, the DFA the algorithm produces is unique up to isomorphism: every minimal DFA that recognizes
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Exercise Show, using Suffix Substitution Closure, that L 3 . L 3 ∈ SL 2 . Explain how it can be the case that L 3 . L 3 ∈ SL 2 , while L 3 . L 3 ⊆ L + 3 and L + 3 ∈ SL
The project 2 involves completing and modifying the C++ program that evaluates statements of an expression language contained in the Expression Interpreter that interprets fully pa
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(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
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