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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 ‘?') and the edges were labeled with individual alphabet symbols. The k-factors of the automaton could be recovered by appending the symbol on an edge to the factor of the node it is incident from. The key value of the graphs is the way that they capture the set of all computations of the automaton in a concise form: every computation of the automaton corresponds to a path through the automaton from ‘?' to ‘?' and vice versa. The su?x substitution closure property is, in essence, a consequence of this fact. All that is signi?cant about the initial portion of a computation is the node it ends on. All strings that lead to the same node are equivalent in the sense that any continuation that extends one of them to form a string that is accepted will extend any of them to form a string that is accepted, and any continuation that leads one of them to be rejected will lead any of them to be rejected.
In adapting this idea for LTk automata, we have to confront the fact that the last k - 1 symbols of the input are no longer enough to characterize the initial portion of a string. We now will also need the record of all k-factors which occurred in that initial portion. To accommodate this, we will extend the labeling of our nodes to include sets of k-factors. The node set will be pairs in which the ?rst component is a k - 1 factor (the last k - 1 symbols of the input) and the second component is a set of k-factors. At the initial node, not having scanned any of the input yet, we have seen no k-factors, that is, the initial set of k-factors is empty (∅). The label of the initial node, then is (?, ∅).
Suppose A = (Q,Σ, T, q 0 , F) is a DFA and that Q = {q 0 , q 1 , . . . , q n-1 } includes n states. Thinking of the automaton in terms of its transition graph, a string x is recogn
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Computation of a DFA or NFA without ε-transitions An ID (q 1 ,w 1 ) computes (qn,wn) in A = (Q,Σ, T, q 0 , F) (in zero or more steps) if there is a sequence of IDs (q 1
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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
For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable. "Positiveness Problem". Note that
Find a regular expression for the regular language L={w | w is decimal notation for an integer that is a multiple of 4}
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