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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.
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
how to understand DFA ?
Exercise Show, using Suffix Substitution Closure, that L 3 . L 3 ∈ SL 2 . Explain how it can be the case that L 3 . L 3 ∈ SL 2 , while L 3 . L 3 ⊆ L + 3 and L + 3 ∈ SL
As we are primarily concerned with questions of what is and what is not computable relative to some particular model of computation, we will usually base our explorations of langua
Intuitively, closure of SL 2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a
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
examples of decidable problems
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
This close relationship between the SL2 languages and the recognizable languages lets us use some of what we know about SL 2 to discover properties of the recognizable languages.
draw pda for l={an,bm,an/m,n>=0} n is in superscript
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