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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 be a graph with n > 2 vertices with (n2 - 3n + 4)/2 edges. Prove that G is connected.
Another way of representing a strictly 2-local automaton is with a Myhill graph. These are directed graphs in which the vertices are labeled with symbols from the input alphabet of
how to prove he extended transition function is derived from part 2 and 3
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We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
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spam messages h= 98%, m= 90%, l= 80% non spam h=12%, m = 8%, l= 5% The organization estimates that 75% of all messages it receives are spam messages. If the cost of not blocking a
Let there L1 and L2 . We show that L1 ∩ L2 is CFG . 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 second
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
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