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# Erik Jonsson School of Engineering and Computer Science FEARLESS Engineering CS 5349 – 001 CS 4384 – 001 Automata Theory

## Presentation on theme: "Erik Jonsson School of Engineering and Computer Science FEARLESS Engineering CS 5349 – 001 CS 4384 – 001 Automata Theory"— Presentation transcript:

Erik Jonsson School of Engineering and Computer Science FEARLESS Engineering CS 5349 – 001 CS 4384 – 001 Automata Theory http://www.utdallas.edu/~pervin Thursday: EXAMINATION 1 Tuesday: Context-free Languages Tuesday 9-30-14

TA S. S. Gokhale sxg122830@utdallas.edu ECSS 2.103B1 (West side Open Lab/ TI) MW 1:30-3:30 and other hours possible 2

Extra Assignment Only for CS 5349 Students! 3

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The Pumping Lemma Game We play against an opponent. Our goal is to win the game by establishing a contradiction to the PL, while the opponent tries to foil us. There are four moves in the game. 1) The opponent picks p. 2) Given p we pick a string s(p) in L of length ≥ p. 3) The opponent chooses the decomposition xyz subject to |xy| ≤ p, |y| ≥ 1. We have to assume that the opponent makes the choice that will make it harder for us to win the game. 4) We pick i so that the pumped string is not in L. 5 Don't forget! It will be on the comprehensive final exam!

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9 In class I pointed out that intersecting with the regular language a* makes the problem slightly easier because one would have to pick the s_p we used above.

10 Theorem: Let M be a DFA with p states. (i)L(M) is not empty iff M accepts a string z with |z| < p. (ii) L(M) has an infinite number of members iff M accepts a string z with p <= |z| < 2p. In each case we used the Pumping Lemma to pump “down” to show that the smallest member of the language cannot be of length (i) greater or equal to p; (ii) greater or equal to 2p. Decision Procedures

11Slightly modified

12M&S P. 84 #2.20(4) _

13M&S Problem 2.21 See M&S P. 85 #2.26-2.28

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16Martin, P.120 #6.8b

17 Linz, P. 89 #9b

18Du, P. 53, Example 9.3

19M&S, P. 64 a*b[(b + ab*a)a*b]* BOOK:

20 In class, on the board, I considered the language L = {ww^R | w \in {a,b}*}. (Where w^R is the word w reversed.) I suggested s(p) = a^pbba^p would work for the Pumping Lemma since the two b’s must be and the end of w and the beginning of w^R so our opponent must choose y consisting only of a’s and the b’s would still indicate the middle of the pumped string.

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Example: Sudkamp 2-22 The set of strings over {a,b} with an even number of a’s and an even number of b’s. 23Slide #16 from first class!

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Left-Linear Grammars 26

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