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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.
We represented SLk automata as Myhill graphs, directed graphs in which the nodes were labeled with (k-1)-factors of alphabet symbols (along with a node labeled ‘?' and one labeled
Let L 3 = {a i bc j | i, j ≥ 0}. Give a strictly 2-local automaton that recognizes L 3 . Use the construction of the proof to extend the automaton to one that recognizes L 3 . Gi
A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note
RESEARCH POSTER FOR MEALY MACHINE
In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no
DEGENERATE OF THE INITIAL SOLUTION
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 .
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