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Solution Exercise 1.43

e. e. e. a. e. r’. r. s’. q’. q. b. b. b. A’. Solution Exercise 1.43. a. A. Input : Automaton A accepting L Output : Automaton AD accepting DROP-OUT(L) Process : 1) Let A’ be a duplicate of A

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Solution Exercise 1.43

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  1. e e e a e r’ r s’ q’ q b b b A’ Solution Exercise 1.43 a A • Input: Automaton A accepting L • Output: Automaton AD accepting DROP-OUT(L) • Process: 1) Let A’ be a duplicate of A 2) Let q1, …, qn be the states in A, then rename the states in A’ as q1’, …, qn’ 3) For every transition: ((qk,),qc) in A Construct a new transition ((qk,e),qc’ ) 4) The automation AD is formed by A and A’, plus the transitions in (3), with the starting state of AD being the starting state of A r r s q q > b b b and removing from the list of favorables all states in A

  2. Solution Exercise 1.60 a qk+1 … q q3 q1 q q2 > a,b a,b a a,b b k-1 states

  3. Nonregular Languages L = {akbk : k N} Lets show that is regular: • find a regular expression for L, or • automaton that accepts L How can we proof that L is nonregular?

  4. y has at least 1 letter The substring xy contains at most the first p letters of w (and z contains the rest of letters) Pumping part: xz, xy2z, xy10z and any other string of the form xyiz must be in L Pumping Lemma (Theorem 1.70) • Lemma. Let L be a regular language. There exists an integer p > 0 such that for any string w  L with |w|  p, then there exist strings x, y, and z such that: • w = xyz • y  e • |xy|  p • xyiz L for i = 0, 1, 2, …

  5. D: 1 2 m > s1 s2 … sm Where each si are states in D Sketch of Proof (1) Lets analyze how words are accepted by finite automata. Lets take a generic word: w = 1 2 … m If w is accepted by a DFA D. What does it mean:

  6. same D: 1 m D: 1 2 m > s1 si … … sm sj … s1 s2 … sm x y Where each si are states in D Sketch of Proof (2) > Let p = # of states in D. If m > p, what does this means for these s1, s2, …, sm? z Thus, w can be split into 3 parts x, y and z

  7. Exercises • Lets proof that • L = {akbk : k N} • is nonregular using the pumping lemma • Given a word w, wR denotes the reverse of the word. Lets proof that • Palindromes = {w  * : w = wR} • is nonregular using the pumping lemma ( = {a,b}) • Note: In Example 1.73 of the book (read it!) only case Number 1is necessary to analyze. The other two cases cannot happen because |xy|  p (the text below the cases states it so too but at least one of the selected answers to exercises again analyze more cases than it is necessarily -e.g., Exercise 1.29 (a) -)

  8. Homework (Optional: Wednesday!) • Exercise 1.29 (a), (b) • Problem 1.46 (b) • Problem 1.53 • (Hint: you can pick any numbers x, y and z as long as: • x = y+z, and • Viewed as an string, x = y+z has at least p characters • So what you are trying to do is to take some numbers x, y and z so when you “pump” the summation longer adds up)

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