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Unformatted text preview: CSE 105: Automata and Computability Theory Winter 2012 Problem Set #1
Due: Monday, January 30th, 2012 Problem 1 Let M = (Q, , , q0 , F ) be a DFA. Let w = w1 w2 wn be a string accepted by M , for which let r0 , r1 , . . . , rn be the corresponding accepting path. (See the section titled "Formal Definition of Computation," page 40, in Sipser.) Suppose that ri = rj for some i and j such that i < j. Prove that, in addition to accepting w, M also accepts some string w that is shorter than w, i.e., such that |w | < |w|. Problem 2 In class, we showed that swapping the accepting and nonaccepting states of a DFA whose language is L gives a DFA whose language is L = \ L. a. Show (by construction) that swapping the accepting and nonaccepting states of an NFA whose language is L does not necessarily give an NFA whose language is L. Hint: There are examples with a very small number of states. b. Explain, given an NFA whose language is L, how to construct another NFA whose language is L. Problem 3 In this problem, we consider a generalization of DFAs called second-order deterministic finite automata, or DFA(2) s for short. The transition function of a DFA(2) depends not only on the current symbol and the current state, but also the previous state. That is, the transition function is a function (2) : QQ Q, that maps every pair of states (qprev , qcurr ) and every input symbol a to a next state qnext = (2) (qprev , qcurr , a). There is a special case for processing the very first input symbol, when there is no previous state; we define the next state to be (2) (q0 , q0 , a), where q0 is the start state and a is the first input symbol. Clearly any DFA can be transformed into an equivalent DFA(2) in which the transition function does not depend on the previous state, i.e., (2) (qprev , qcurr , a) = (qcurr , a) regardless of qprev . Show that every DFA(2) can be transformed into an equivalent DFA. Hint: Recall the transformation from NFA to DFA. 1 ...
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