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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.
20*2
Let G be a graph with n > 2 vertices with (n2 - 3n + 4)/2 edges. Prove that G is connected.
The Universality Problem is the dual of the emptiness problem: is L(A) = Σ∗? It can be solved by minor variations of any one of the algorithms for Emptiness or (with a little le
build a TM that enumerate even set of even length string over a
explain turing machine .
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
dfa for (00)*(11)*
The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automa
write short notes on decidable and solvable problem
Applying the pumping lemma is not fundamentally di?erent than applying (general) su?x substitution closure or the non-counting property. The pumping lemma is a little more complica
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