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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 G=(V,T,S,P) where V={a,b,A,B,S}, T={a,b},S the start symbol and P={S->Aba, A->BB, B->ab,AB->b} 1.show the derivation sentence for the string ababba 2. find a sentential form
Can you say that B is decidable? If you somehow know that A is decidable, what can you say about B?
draw pda for l={an,bm,an/m,n>=0} n is in superscript
The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automa
Suppose A = (Q,Σ, T, q 0 , F) is a DFA and that Q = {q 0 , q 1 , . . . , q n-1 } includes n states. Thinking of the automaton in terms of its transition graph, a string x is recogn
Since the signi?cance of the states represented by the nodes of these transition graphs is arbitrary, we will allow ourselves to use any ?nite set (such as {A,B,C,D,E, F,G,H} or ev
Let L 3 = {a i bc j | i, j ≥ 0}. Give a strictly 2-local automaton that recognizes L 3 . Use the construction of the proof to extend the automaton to one that recognizes L 3 . Gi
Perfect shuffle permutation
In general non-determinism, by introducing a degree of parallelism, may increase the accepting power of a model of computation. But if we subject NFAs to the same sort of analysis
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