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Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B not (there is only one such path). Those leading to region C end in b. Note that once we are in region C the question of whether we have seen bb or not is no longer relevant; in order to accept we must see aa and, since the path has ended with b, we cannot reach aa without ?rst seeing ba (hence, passing through region E). Finally, in region A we have not looked at anything yet. This where the empty string ends up.
Putting this all together, there is no reason to distinguish any of the nodes that share the same region. We could replace them all with a single node. What matters is the information that is relevant to determining if a string should be accepted or can be extended to one that should be. In keeping with this insight, we will generalize our notion of transition graphs to graphs with an arbitrary, ?nite, set of nodes distinguishing the signi?cant states of the computation and edges that represent the transitions the automaton makes from one state to another as it scans the input. Figure 3 represents such a graph for the minimal equivalent of the automaton of Figure 1.
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
Sketch an algorithm for the universal recognition problem for SL 2 . This takes an automaton and a string and returns TRUE if the string is accepted by the automaton, FALSE otherwi
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
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We represented SLk automata as Myhill graphs, directed graphs in which the nodes were labeled with (k-1)-factors of alphabet symbols (along with a node labeled ‘?' and one labeled
Explain the Chomsky's classification of grammar
Question 2 (10 pt): In this question we look at an extension to DFAs. A composable-reset DFA (CR-DFA) is a five-tuple, (Q,S,d,q0,F) where: – Q is the set of states, – S is the alph
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