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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.
The class of Strictly Local Languages (in general) is closed under • intersection but is not closed under • union • complement • concatenation • Kleene- and positive
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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
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In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
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Normal forms are important because they give us a 'standard' way of rewriting and allow us to compare two apparently different grammars G1 and G2. The two grammars can be shown to
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