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Sketch an algorithm for the universal recognition problem for SL2. This takes an automaton and a string and returns TRUE if the string is accepted by the automaton, FALSE otherwise. Assume the automaton is given just as a sequence of pairs of symbols and that the automaton and string are both over the same language. Give an argument justifying the correctness of your algorithm.
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
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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
The initial ID of the automaton given in Figure 3, running on input ‘aabbba' is (A, aabbba) The ID after the ?rst three transitions of the computation is (F, bba) The p
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We now add an additional degree of non-determinism and allow transitions that can be taken independent of the input-ε-transitions. Here whenever the automaton is in state 1
Since the signi?cance of the states represented by the nodes of these transition graphs is arbitrary, we will allow ourselves to use any ?nite set (such as {A,B,C,D,E, F,G,H} or ev
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
The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
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