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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.
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
Give the Myhill graph of your automaton. (You may use a single node to represent the entire set of symbols of the English alphabet, another to represent the entire set of decima
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What is the arbwnememmsmdbdbfbfjmfksmjejfnfnfnnrndmnfjfjfnrnkrkfjfnfmkrjrbfbbfjfnfjruhrvrjkgktithhrbenfkiffnbr ki rnrjjdjrnrk bd n FBC..jcb?????????????????????????????????????????
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