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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.
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG too. 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 sec
write grammer to produce all mathematical expressions in c.
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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
The Equivalence Problem is the question of whether two languages are equal (in the sense of being the same set of strings). An instance is a pair of ?nite speci?cations of regular
prove following function is turing computable? f(m)={m-2,if m>2, {1,if
We will assume that the string has been augmented by marking the beginning and the end with the symbols ‘?' and ‘?' respectively and that these symbols do not occur in the input al
Application of the general suffix substitution closure theorem is slightly more complicated than application of the specific k-local versions. In the specific versions, all we had
value chain
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