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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.
how many pendulum swings will it take to walk across the classroom?
Generate 100 random numbers with the exponential distribution lambda=5.0.What is the probability that the largest of them is less than 1.0?
Kleene called this the Synthesis theorem because his (and your) proof gives an effective procedure for synthesizing an automaton that recognizes the language denoted by any given r
We now add an additional degree of non-determinism and allow transitions that can be taken independent of the input-ε-transitions. Here whenever the automaton is in state 1
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
Sketch an algorithm for the universal recognition problem for SL 2 . This takes an automaton and a string and returns TRUE if the string is accepted by the automaton, FALSE otherwi
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
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Perfect shuffle permutation
. On July 1, 2010, Harris Co. issued 6,000 bonds at $1,000 each. The bonds paid interest semiannually at 5%. The bonds had a term of 20 years. At the time of issuance, the market r
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