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The class of Strictly Local Languages (in general) is closed under
• intersection but is not closed under
• union
• complement
• concatenation
• Kleene- and positive closure
Proof: For intersection, we can adapt the construction and proof for the SL2 case again to get closure under intersection for SLk. This is still not quite enough for SL in general, since one of the languages may be in SLi and the other in SLj for some i = j. Here we can use the hierarchy theorem to show that, supposing i < j, the SLi language is also in SLj . Then the adapted construction will establish that their intersection is in SL .
For non-closure under union (and consequently under complement) we can use the same counterexample as we did in the SL2 case:
To see that this is not in SLk for any k we can use the pair
which will yield abk-1 a under k-local suffix substitution closure.
For non-closure under concatenation we can use the counterexample
The two languages being concatenated are in SL2, hence in SLk for all k ≥ 2 but their concatenation is not in SLk for any k, as we showed in the example above.
Let ? ={0,1} design a Turing machine that accepts L={0^m 1^m 2^m } show using Id that a string from the language is accepted & if not rejected .
The fundamental idea of strictly local languages is that they are speci?ed solely in terms of the blocks of consecutive symbols that occur in a word. We'll start by considering lan
The computation of an SL 2 automaton A = ( Σ, T) on a string w is the maximal sequence of IDs in which each sequential pair of IDs is related by |- A and which starts with the in
The generalization of the interpretation of strictly local automata as generators is similar, in some respects, to the generalization of Myhill graphs. Again, the set of possible s
We can then specify any language in the class of languages by specifying a particular automaton in the class of automata. We do that by specifying values for the parameters of the
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(
To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the
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