1. Solution Exercise 1.43 a A r r s q q > b b e b s’ q r q’ b r’ 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 a A r r s q q
> b b e b s’ q r q’ b r’ a A’
and removing from the list of favorables all states in A

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

5. 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: s1 1 s2 2 … m sm
Where each si are states in D D:
>

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