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Another way of representing a strictly 2-local automaton is with a Myhill graph. These are directed graphs in which the vertices are labeled with symbols from the input alphabet of the automaton (plus {x,x}), with an edge from a vertex labeled σ1 to a vertex labeled σ2 ix the pair σ1σ2 is included in T. (Note that if we interpret the strings in T as pairs of symbols, then the Myhill graph of A = (Σ, T) is just G = (Σ+, T).) The Myhill graph of the automaton of Figure 2 is given in Figure. For consistency with the graphs we will use later, the entry point to the graph is indicated with an edge "from nowhere" and the exit point is indicated by circling it.
The key property of Myhill graphs is that every path through the graph from the ‘x' node to the ‘x' node corresponds to a computation of the automaton and every computation of the automaton corresponds to such a path. So we can reason about the strings that are accepted by the automaton by reasoning about the sequences of nodes that occur on paths from ‘x' to ‘x'. (For simplicity, we will refer to paths from ‘x' to ‘x' as "paths through the graph".)
For example, the shortest strings in the language recognized by the automaton will those labeling the shortest paths through the graph, which is to say, the acyclic paths from ‘x' to ‘x'. In this particular case, these are the paths (x,x) and (x, a, b,x), corresponding to the strings ε and ab.
proof of arden''s theoram
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Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second
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
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
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