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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.
Exercise: Give a construction that converts a strictly 2-local automaton for a language L into one that recognizes the language L r . Justify the correctness of your construction.
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The Universality Problem is the dual of the emptiness problem: is L(A) = Σ∗? It can be solved by minor variations of any one of the algorithms for Emptiness or (with a little le
We will assume that the string has been augmented by marking the beginning and the end with the symbols ‘?' and ‘?' respectively and that these symbols do not occur in the input al
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The Myhill-Nerode Theorem provided us with an algorithm for minimizing DFAs. Moreover, the DFA the algorithm produces is unique up to isomorphism: every minimal DFA that recognizes
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
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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