1. CSE322 PUMPING LEMMA FOR REGULAR SETS AND ITS APPLICATIONS
Lecture #11

2. Non-regular languages

3. Problem: this is not easy to prove
How can we prove that a language
is not regular?
Prove that there is no DFA that accepts
Problem: this is not easy to prove
Solution: the Pumping Lemma !!!

4. The Pigeonhole Principle

5. pigeons
pigeonholes

6. A pigeonhole must
contain at least two pigeons

7. pigeons
pigeonholes

8. The Pigeonhole Principle
pigeons
pigeonholes
There is a pigeonhole
with at least 2 pigeons

9. The Pigeonhole Principle and DFAs

10. DFA with states

11. In walks of strings:
no state
is repeated

12. In walks of strings:
a state
is repeated

13. If string has length :
Then the transitions of string
are more than the states of the DFA
Thus, a state must be repeated

14. In general, for any DFA:
String has length number of states
A state must be repeated in the walk of
walk of
......
......
Repeated state

15. In other words for a string :
transitions are pigeons
states are pigeonholes
walk of
......
......
Repeated state

16. The Pumping Lemma

17. Take an infinite regular language
There exists a DFA that accepts
states

18. Take string with
There is a walk with label :
walk

19. then, from the pigeonhole principle: a state is repeated in the walk
If string has length
(number
of states
of DFA)
then, from the pigeonhole principle:
a state is repeated in the walk
......
......
walk

20. Let be the first state repeated in the
walk of
......
......
walk

21. Write
......
......

22. Observations:
length
number
of states
of DFA
length
......
......

23. Observation:
The string
is accepted
......
......

24. Observation:
The string
is accepted
......
......

25. Observation:
The string
is accepted
......
......

26. In General:
The string
is accepted
......
......

27. In General:
Language accepted by the DFA
......
......

28. In other words, we described:
The Pumping Lemma !!!

29. The Pumping Lemma: Given a infinite regular language
there exists an integer
for any string with length
we can write
with and
such that:

30. Applications of the Pumping Lemma

31. Theorem:
The language
is not regular
Proof:
Use the Pumping Lemma

32. Assume for contradiction
that is a regular language
Since is infinite
we can apply the Pumping Lemma

33. Let be the integer in the Pumping Lemma
Pick a string such that:
length
We pick

34. Write:
From the Pumping Lemma
it must be that length
Thus:

35. From the Pumping Lemma:
Thus:

36. From the Pumping Lemma:
Thus:

37. BUT:
CONTRADICTION!!!

38. Conclusion: Therefore: Our assumption that
is a regular language is not true
Conclusion:
is not a regular language

39. Non-regular languages

40. More Applications of the Pumping Lemma

41. The Pumping Lemma: Given a infinite regular language
there exists an integer
for any string with length
we can write
with and
such that:

42. Non-regular languages

43. Theorem:
The language
is not regular
Proof:
Use the Pumping Lemma

44. Assume for contradiction
that is a regular language
Since is infinite
we can apply the Pumping Lemma

45. Let be the integer in the Pumping Lemma
Pick a string such that:
and
length
We pick

46. Write
From the Pumping Lemma
it must be that length
Thus:

47. From the Pumping Lemma:
Thus:

48. From the Pumping Lemma:
Thus:

49. BUT:
CONTRADICTION!!!

50. Conclusion: Therefore: Our assumption that
is a regular language is not true
Conclusion:
is not a regular language

51. Non-regular languages

52. Theorem:
The language
is not regular
Proof:
Use the Pumping Lemma

53. Assume for contradiction
that is a regular language
Since is infinite
we can apply the Pumping Lemma

54. Let be the integer in the Pumping Lemma
Pick a string such that:
and
length
We pick

55. Write
From the Pumping Lemma
it must be that length
Thus:

56. From the Pumping Lemma:
Thus:

57. From the Pumping Lemma:
Thus:

58. BUT:
CONTRADICTION!!!

59. Conclusion: Therefore: Our assumption that
is a regular language is not true
Conclusion:
is not a regular language

60. Non-regular languages

61. Theorem:
The language
is not regular
Proof:
Use the Pumping Lemma

62. Assume for contradiction
that is a regular language
Since is infinite
we can apply the Pumping Lemma

63. Let be the integer in the Pumping Lemma
Pick a string such that:
length
We pick

64. Write
From the Pumping Lemma
it must be that length
Thus:

65. From the Pumping Lemma:
Thus:

66. From the Pumping Lemma:
Thus:

67. Since:
There must exist such that:

68. However:
for
for any

69. BUT:
CONTRADICTION!!!

70. Conclusion: Therefore: Our assumption that
is a regular language is not true
Conclusion:
is not a regular language