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Our DFAs are required to have exactly one edge incident from each state for each input symbol so there is a unique next state for every current state and input symbol. Thus, the next state is fully determined by the current state and input symbol. As we saw in the previous section, this simpli?es the proof that the DFA accepts a speci?c language. There are many circumstances, though, in which it will be simpler to de?ne the automaton in the ?rst place if we allow for there to be any one of a number of next states or even no next state at all. Thus there may be many out edges from a given node labeled with a given symbol, or no out edges from that node for that symbol. Such FSA are called non-deterministic because the next step of a computation is not fully determined by the current state and input symbol-we may have a choice of states to move into.
De?nition 1 (NFA without ε-Transitions) A FSA A = (Q,Σ, T, q0, F) is non-deterministic iff either
• there is some q ∈ Q, σ ∈ Σ and p1 = p2 ∈ Q for which hq, p1, σi ∈ T and hq, p2, σi ∈ T,
• or there is some q ∈ Q, σ ∈ Σ for which there is no p ∈ Q such that hq, p, σi ∈ T
For every regular language there is a constant n depending only on L such that, for all strings x ∈ L if |x| ≥ n then there are strings u, v and w such that 1. x = uvw, 2. |u
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
While the SL 2 languages include some surprisingly complex languages, the strictly 2-local automata are, nevertheless, quite limited. In a strong sense, they are almost memoryless
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Exercise: Give a construction that converts a strictly 2-local automaton for a language L into one that recognizes the language L r . Justify the correctness of your construction.
The Emptiness Problem is the problem of deciding if a given regular language is empty (= ∅). Theorem 4 (Emptiness) The Emptiness Problem for Regular Languages is decidable. P
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
spam messages h= 98%, m= 90%, l= 80% non spam h=12%, m = 8%, l= 5% The organization estimates that 75% of all messages it receives are spam messages. If the cost of not blocking a
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
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