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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 scanning a string in L1 . L2, for instance, when to switch from keeping track of factors for L1 to keeping track of factors from L2.
Assuming that the alphabets were not disjoint, there is (evidently, since LT is not closed under concatenation) no way, in general, to know that. For the recognizable languages, on the other hand, we have the convenience of being able to work with non-determinism. We don't actually have to know when to switch from one automaton to the next. Whenever we get to a point in the string that could possibly be the end of the pre?x that is in L1 we can just allow for a non-deterministic choice of whether to continue scanning for A1 (the machine recognizing L1) or to switch to scanning for A2. Since whenever the string is in L1 . L2 there will be some correct place to switch and since acceptance by a NFA requires only that there some accepting computation, the combined automaton will accept every string in L1 . L2. Moreover, the combined automaton will accept a string iff there is some point at which it can be split into a string accepted by A1 followed by one accepted by A2: it accepts all and only the strings in L1 . L2.
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?
#can you solve a problem of palindrome using turing machine with explanation and diagrams?
Construct a Mealy machine that can output EVEN or ODD According to the total no. of 1''s encountered is even or odd.
PROPERTIES OF Ardens therom
design a tuning machine for penidrome
The objective of the remainder of this assignment is to get you thinking about the problem of recognizing strings given various restrictions to your model of computation. We will w
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
designing DFA
conversion from nfa to dfa 0 | 1 ___________________ p |{q,s}|{q} *q|{r} |{q,r} r |(s) |{p} *s|null |{p}
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
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