# Suppose we are given a DFA D = (Q,S,d,Q0,F), and we are asked to design an NFA N such that it…

Suppose we are given a DFA D = (Q,S,d,Q0,F), and we are asked to design an NFA N such that it recognizes the middle-third of every string recognized by D. More formally, we want the language of the NFA N, L (N), to be L (N) = {y : ?x, y, z ? S * : len(x) = len(y) = len(z)? x yz ? L (D)} Of course, for every string in L (D) whose length is not divisible by 3, there will be no corresponding middle-third string contributed to N. Build the NFA using the mk_nfa call provided within Def_NFA, and then produce the minimized DFA using Joves suite of functions. Hint: The idea is to imagine keeping a

»Suppose we are given a DFA D = (Q,S,d,Q0,F), and we are asked to design an NFA N such that it recognizes the middle-third of every string recognized by D. More formally, we want the language of the NFA N, L (N), to be L (N) = {y : ?x, y, z ? S * : len(x) = len(y) = len(z)? x yz ? L (D)} Of course, for every string in L (D) whose length is not divisible by 3, there will be no corresponding middle-third string contributed to N. Build the NFA using the mk_nfa call provided within Def_NFA, and then produce the minimized DFA using Joves suite of functions. Hint: The idea is to imagine keeping a nondeterministically chosen milepost a third of the way and another nondeterministically chosen milepost two-thirds of the way. Model the NFA state as Q 5 (the Q being the set of states of the DFA), i.e., Q ×Q ×Q ×Q ×Q. Let the initial state be (q0, q1, q1, q2, q2). It is as if we placed one mile-post token at some state q1, another at q2, and let a token from q0 seek the one at q1, let another token at q1 seek q2, and a final token at q2 seek qf , a state in F. We move three tokens with equal velocities, and if/when they manage to hit a magical configuration, we have found our middle-third string. More specifically, the idea is then to consider the following types of state transitions (notice that the second and fourth components of the state tuple stay put, serving as fixed mileposts): (q0, q1, q1, q2, q2) ? (q01, q1, q12, q2, q21) ? (q02, q1, q12, q2, q22) ? (q03, q1, q13, q2, q23) ? … (q1, q1, q2, q2, qf ) In this scenario, the last state is our magical configuration. Putting these ideas to work, we write down the following components of our NFA: Q N = Q 5 S N = S Q N 0 = {(q0, q1, q1, q2, q2) : q1, q2 ? Q} (i.e., start the NFA from the set of these initial states). F N = {(q1, q1, q2, q2, qf ) : qf ? F}. d N ((qa, q1, qb, q2, qc), y) = {(d(qa, c1), q1,d(qb, y), q2,d(qc, c2)) : c1, c2 ? S} One can see how only the middle token listens to y; the other two tokens chaotically wander. They seek to find the magical configuration in all possible ways. Try animating this on a piece of paper using colored beads in place of tokens.

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