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Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about how to do this. For this claim the assumption that the solution of each instance is unique is not necessary; but both of the others are. If you had a program that checks whether a proposed solution to an instance of a problem is correct and another that systematically generates every instance of the problem along with every possible solution, how could you use them (as subroutines) to build a program that, when given an instance, was guaranteed to ?nd a correct solution to that problem under the assumption that such a solution always exists?
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
Suppose A = (Q,Σ, T, q 0 , F) is a DFA and that Q = {q 0 , q 1 , . . . , q n-1 } includes n states. Thinking of the automaton in terms of its transition graph, a string x is recogn
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The Emptiness Problem is the problem of deciding if a given regular language is empty (= ∅). Theorem 4 (Emptiness) The Emptiness Problem for Regular Languages is decidable. P
what is regular expression?
prove following function is turing computable? f(m)={m-2,if m>2, {1,if
We have now de?ned classes of k-local languages for all k ≥ 2. Together, these classes form the Strictly Local Languages in general. De?nition (Strictly Local Languages) A langu
Let L 3 = {a i bc j | i, j ≥ 0}. Give a strictly 2-local automaton that recognizes L 3 . Use the construction of the proof to extend the automaton to one that recognizes L 3 . Gi
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
Our DFAs are required to have exactly one edge incident from each state for each input symbol so there is a unique next state for every current state and input symbol. Thus, the ne
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