problem set 1

problem set 1 - CSE 105: Automata and Computability Theory...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online