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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.
write grammer to produce all mathematical expressions in c.
State & prove pumping lemma for regular set. Show that for the language L={ap |p is a prime} is not regular
1. Does above all''s properties can be used to prove a language regular? 2..which of the properties can be used to prove a language regular and which of these not? 3..Identify one
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We developed the idea of FSA by generalizing LTk transition graphs. Not surprisingly, then, every LTk transition graph is also the transition graph of a FSA (in fact a DFA)-the one
As we are primarily concerned with questions of what is and what is not computable relative to some particular model of computation, we will usually base our explorations of langua
how many pendulum swings will it take to walk across the classroom?
Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
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
how to understand DFA ?
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