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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.
State & prove pumping lemma for regular set. Show that for the language L={ap |p is a prime} is not regular
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Suppose G = (N, Σ, P, S) is a reduced grammar (we can certainly reduce G if we haven't already). Our algorithm is as follows: 1. Define maxrhs(G) to be the maximum length of the
The fact that the Recognition Problem is decidable gives us another algorithm for deciding Emptiness. The pumping lemma tells us that if every string x ∈ L(A) which has length grea
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As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
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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
Automaton (NFA) (with ε-transitions) is a 5-tuple: (Q,Σ, δ, q 0 , F i where Q, Σ, q 0 and F are as in a DFA and T ⊆ Q × Q × (Σ ∪ {ε}). We must also modify the de?nitions of th
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