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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.
Suppose A = (Σ, T) is an SL 2 automaton. Sketch an algorithm for recognizing L(A) by, in essence, implementing the automaton. Your algorithm should work with the particular automa
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
Define the following concept with an example: a. Ambiguity in CFG b. Push-Down Automata c. Turing Machine
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
A common approach in solving problems is to transform them to different problems, solve the new ones, and derive the solutions for the original problems from those for the new ones
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
The class of Strictly Local Languages (in general) is closed under • intersection but is not closed under • union • complement • concatenation • Kleene- and positive
When an FSA is deterministic the set of triples encoding its edges represents a relation that is functional in its ?rst and third components: for every q and σ there is exactly one
Theorem The class of ?nite languages is a proper subclass of SL. Note that the class of ?nite languages is closed under union and concatenation but SL is not closed under either. N
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