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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.
how to find whether the language is cfl or not?
While the SL 2 languages include some surprisingly complex languages, the strictly 2-local automata are, nevertheless, quite limited. In a strong sense, they are almost memoryless
For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable. "Positiveness Problem". Note that
Lemma 1 A string w ∈ Σ* is accepted by an LTk automaton iff w is the concatenation of the symbols labeling the edges of a path through the LTk transition graph of A from h?, ∅i to
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
1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization
We can then specify any language in the class of languages by specifying a particular automaton in the class of automata. We do that by specifying values for the parameters of the
turing machine for prime numbers
If the first three words are the boys down,what are the last three words??
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
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