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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.
The generalization of the interpretation of strictly local automata as generators is similar, in some respects, to the generalization of Myhill graphs. Again, the set of possible s
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Give DFA''s accepting the following languages over the alphabet {0,1}: i. The set of all strings beginning with a 1 that, when interpreted as a binary integer, is a multiple of 5.
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
Construct a PDA that accepts { x#y | x, y in {a, b}* such that x ? y and xi = yi for some i, 1 = i = min(|x|, |y|) }. For your PDA to work correctly it will need to be non-determin
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
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#can you solve a problem of palindrome using turing machine with explanation and diagrams?
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