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28+Slides--Finite+Automata+II

28+Slides--Finite+Automata+II - CS103 HO#28 Finite Automata...

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CS103 HO#28 Finite Automata II 4/22/11 1 Note: It is important to do the reading in the Sipser text. You should try to understand it line by line. Come to office hours , write us with questions, or talk to your classmates! Suppose L = {w { a , b } * | no two consecutive characters are the same} q 1 a b b a a, b q 2 q 3 q 4 a b last character a, only b OK reject input last character b, only a OK Q = {q 1 , q 2 , q 3 , q 4 } = {a, b} q 1 is the start state F = {q 1 , q 2 , q 3 } a b q 1 q 2 q 3 q 2 q 4 q 3 q 3 q 2 q 4 q 4 q 4 q 4 Suppose L = {w | w = a m b n for m, n > 0} q 1 b a q 2 q 4 q 3 a a b b a, b Suppose L = {w | w is the string representation of a floating point number } d . d +, – E d d d d d +, d E +2.0 532.67 0.3E123 1.5006E+27 -0.4E-2 -6E8 d stands for any decimal digit (transitions to non-accepting state for illegal inputs not shown) Let = {a, b, c, d} Let L Missing = {w | there is a symbol from not in w} Start state: all letters missing After one character, the state could be a read, b, c, d still missing b read, a, c, d still missing c read, a, b, d still missing d read, a, b, c still missing After two characters, the state could be any of the previous, or a, b read, c, d still missing a, c read, b, d still missing ... After three characters, ... Some Important Definitions Let M = (Q, , , q 0 , F) be a finite automaton and let w = w 1 , w 2 , ..., w n be a string where each w i   . Then m accepts w if there exists a sequence of states r 0 , r 1 , ..., r n in Q such that: 1. r 0 = q 0 , 2. (r i , w i+1 ) = r i+1 for i = 0, 1, ..., n – 1, and 3. r n F. q 1 a b b a a, b q 2 q 3 q 4 a b Q = {q 1 , q 2 , q 3 , q 4 } = {a, b} q 1 is the start state : Q    Q F = {q 1 , q 2 , q 3 } a b q 1 q 2 q 3 q 2 q 4 q 3 q 3 q 2 q 4 q 4 q 4 q 4
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