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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.
Suppose A = (Σ, T) is an SL 2 automaton. Sketch an algorithm for recognizing L(A) by, in essence, implementing the automaton. Your algorithm should work with the particular automa
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prove following function is turing computable? f(m)={m-2,if m>2, {1,if
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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
All that distinguishes the de?nition of the class of Regular languages from that of the class of Star-Free languages is that the former is closed under Kleene closure while the lat
We can then specify any language in the class of languages by specifying a particular automaton in the class of automata. We do that by specifying values for the parameters of the
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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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