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Strictly 2-local automata are based on lookup tables that are sets of 2-factors, the pairs of adjacent symbols which are permitted to occur in a word. To generalize, we extend the 2-factors to k-factors. We now have the possibility that the scanning window is actually longer than the augmented string. To accommodate that, we will permit factors of any length up to k as long as they start with ‘x' and end with ‘x' as well as k-factors which may or may not start with ‘x' or end with ‘x'.
So a strictly k-local automaton is just an alphabet and a set of stings of length k in which the ?rst symbol is either x or a symbol of the alphabet and the last is either x or a symbol of the alphabet, plus any number of strings of length no greater than k in which the ?rst and last symbol are x and x, respectively. In scanning strings that are shorter than k - 1, the automaton window will span the entire input (plus the beginning and end symbols). In that case, it will accept i? the sequence of symbols in the window is one of those short strings.
You should verify that this is a generalization of SL2 automata, that if k = 2 the de?nition of SLk automata is the same as the de?nition of SL2 automata.
Theorem (Myhill-Nerode) A language L ⊆ Σ is recognizable iff ≡L partitions Σ* into ?nitely many Nerode equivalence classes. Proof: For the "only if" direction (that every recogn
Question 2 (10 pt): In this question we look at an extension to DFAs. A composable-reset DFA (CR-DFA) is a five-tuple, (Q,S,d,q0,F) where: – Q is the set of states, – S is the alph
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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The Universality Problem is the dual of the emptiness problem: is L(A) = Σ∗? It can be solved by minor variations of any one of the algorithms for Emptiness or (with a little le
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(
These assumptions hold for addition, for instance. Every instance of addition has a unique solution. Each instance is a pair of numbers and the possible solutions include any third
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
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