2. Pumping Lemma Problem: Solution:
Let B be the language of all palindromes over {0,1} containing an equal number of 0s and 1s. Show that B is not CFL
Solution:
We assume that B is a CFL and obtain a contradiction.
Let p be the pumping constant of B that is guaranteed to exist by the pumping lemma. Select the string s = 0p1p1p0p .
Clearly s is a member of B and of length at least p . We can show that no matter how we divide s into uvxyz, one of the conditions of the lemma is violated.
Pumping lemma, | vxy |<= p , we can only place vxy in the following ways:

3. vxy completely falls in the first 0p vxy falls between 0p and 1p
If we pump v and y , then the new string is no longer a palindrome, and the number of 0s will be greater than the number of 1s, which is a contradiction.
vxy falls between 0p and 1p
In this case, v will only contain 0s while y will only contain 1s. So if we pump s , the new string is no longer palindrome, which is a contradiction.
vxy completely falls in the first 1p
Similar to (1), after pumping s , the number of 1s will be greater than the number of 0s, which is a contradiction.
vxy falls between 1p and 1p
After pumping s , the number of 1s will be greater than the number of 0s, which is a contradiction.
vxy completely falls in the second 1p
This case is same with (3).
vxy falls between 1p and 0p
Similar to (2), v will only contain 1s while y will only contain 0s. So if we pump s , the new string is no longer palindrome, which is a contradiction.
vxy completely falls in the second 0p
This case is same with (1).
Hence, B is not context free.

4. Let L = {w  {0; 1}* : w = wR}.
(a) Show that L is context-free by giving a context-free grammar for L.
(b) Show that L is context-free by giving a pushdown automaton for L.
(c) Show that L is not regular.

5. CFG
L is a symmetric language. Consider a context-free grammar for L : (V, Σ, R, S) , where
i. V = {S}
ii. Σ = {0,1}
iii. Rules:
S  0S0 |1S1| 0 |1|ε
iv. S = S V

6. PDA

7. Pumping Lemma
Assume that L regular. Let p be the pumping constant given by the pumping lemma.
Let s be the string 1p-1101p.
s can be split into three pieces, s = xyz , where for any i >= 0 the string xyiz is in L .
According to the pumping lemma condition, we must have | xy |<= p . If this is the case, then y must consist only of 1’s, so xyyz  L . Therefore s cannot be pumped.
This is a contradiction.

9. Pushdown Automaton -- PDA
Input String
Stack
States

10. Initial Stack Symbol Stack Stack stack head top bottom special symbol
Appears at time 0

11. The States
Pop
symbol
Input
symbol
Push
symbol

12. input
stack
top
Replace

13. input
stack
top
Push

14. input
stack
top
Pop

15. input
stack
top
No Change

16. Empty Stack
input
stack
empty
Pop
top
The automaton HALTS
No possible transition after

17. A Possible Transition
input
stack
Pop
top

18. Non-Determinism Allowed non-deterministic transitions
PDAs are non-deterministic
Allowed non-deterministic transitions

19. Example PDA
PDA

20. Basic Idea:
Push the a’s
on the stack
2. Match the b’s on input
with a’s on stack
3. Match
found

21. Input
Stack

22. Input
Stack

23. Input
Stack

24. Input
Stack

25. Input
Stack

26. Input
Stack

27. Input
Stack

28. Input
Stack

29. Input
Stack
accept

30. A string is accepted if there is a computation such that:
All the input is consumed
The last state is an accepting state
At the end of the computation,
we do not care about the stack contents
(the stack can be empty at the last state)

31. is the language accepted by the PDA:

38. reject

39. Another PDA example

40. Push v
on stack
3. Match vR on input
with v on stack
2. Guess
middle
of input
4. Match
found

44. Guess the middle
of string

47. accept

48. Rejection Example:
Input

51. Guess the middle
of string

53. There is no possible transition.
Input is not consumed

54. Another computation on same string:

59. final state
is not reached

61. Another PDA example
aab
abbb (r)

62. Pushing Strings
Pop
symbol
Input
symbol
Push
string

63. Example:
pushed
string
top
Push

64. Another PDA example
abbbaa