1. Pumping Lemma Examples

2. L> is not regular. We prove it using the Pumping Lemma.
L> = {aibj : i > j}
L> is not regular. We prove it using the Pumping Lemma.

3. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.

4. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.

5. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
|s|≥ n

6. L> = {aibj : i > j} aaa…aabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
aaa…aabb…b
n+1 n

7. L> = {aibj : i > j} aaa…aabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties.
aaa…aabb…b
n+1 n

8. L> = {aibj : i > j} aaa…aabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties.
|xy|≤ n
|y|≥ 1
aaa…aabb…b
n+1 n

9. L> = {aibj : i > j} aaa…aabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties.
|xy|≤ n
|y|≥ 1
aaa…aabb…b
n+1 n

10. Y L> = {aibj : i > j} aaa…aabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties.
|xy|≤ n
|y|≥ 1 Y
aaa…aabb…b
n+1 n

11. Y L> = {aibj : i > j} aaa…aabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties: y = am, 1 ≤ m ≤ n. Y
aaa…aabb…b
n+1 n

12. L> = {aibj : i > j} aaabb…b L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties: y = am, 1 ≤ m ≤ n.
xz =an+1-mbn ∉ L>.
aaabb…b
n+1-m n n

13. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.
s = an+1bn ϵ L>.
Consider all possible splittings of s in x,y,z with the desired properties: y = am, 1 ≤ m ≤ n.
xz =an+1-mbn ∉ L>.
So L> is not regular!

14. L={ww : w in {a,b}*}
First, figure out what this language is.

15. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language?

16. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab

17. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language?

18. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa

19. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language?

20. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb

21. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb
Is ε in the language?

22. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb
Is ε in the language? YES! (ε = εε)

23. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb
Is ε in the language? YES! (ε = εε)
Is aa in the language?

24. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb
Is ε in the language? YES! (ε = εε)
Is aa in the language? YES!

25. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb
Is ε in the language? YES! (ε = εε)
Is aa in the language? YES!
Is a in the language?

26. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language? aaaaaa
A string not in the language? abbb
Is ε in the language? YES! (ε = εε)
Is aa in the language? YES!
Is a in the language? NO!

27. abaabba|abaabba L={ww : w in {a,b}*}
First, figure out what this language is.
L = {ε, aa, bb, aaaa, abab, baba, bbbb, aaaaaa …}
abaabba|abaabba

28. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.

29. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
First fix an arbitrary number n>0 to be the pumping length.

30. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language

31. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Choose wisely!!!

32. L={ww : w in {a,b}*} aaa…aaa|aaa…aaa
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
aaa…aaa|aaa…aaa n n

33. L={ww : w in {a,b}*} aaa…aaa|aaa…aaa
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
For x = ε, y = a2, z = a2n-2 y z
aaa…aaa|aaa…aaa n n

34. L={ww : w in {a,b}*} aaaaa…aa|aaaa…aaa ϵ L
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
For x = ε, y = a2, z = a2n-2 y y z
aaaaa…aa|aaaa…aaa
ϵ L
n+1
n+1

35. L={ww : w in {a,b}*} aaaaaaa…a|aaaaa…aaa ϵ L
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
For x = ε, y = a2, z = a2n-2 y y y z
aaaaaaa…a|aaaaa…aaa
ϵ L
n+2
n+2

36. L={ww : w in {a,b}*} a…aaaa|aa…aaa ϵ L
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
For x = ε, y = a2, z = a2n-2 z
a…aaaa|aa…aaa
ϵ L
n-1
n-1

37. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
For x = ε, y = a2, z = a2n-2, there is no i: xyiz ∉ L!

38. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = a2n
For x = ε, y = a2, z = a2n-2, there is no i: xyiz ∉ L!
s = a2n doesn’t work!!!

39. L={ww : w in {a,b}*} abab…abab|abab…abab
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = (ab)2n
abab…abab|abab…abab n n

40. L={ww : w in {a,b}*} abab…abab|abab…abab
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = (ab)2n
For x = ε, y = abab, z = (ab)2n-2 y z
abab…abab|abab…abab n n

41. abababab…ab|ababab…abab
L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = (ab)2n
For x = ε, y = abab, z = (ab)2n-2 y y z
abababab…ab|ababab…abab
ϵ L
n+1
n+1

42. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = (ab)2n
For x = ε, y = abab, z = (ab)2n-2
For any i, xyiz = (ab)2i(ab)2n-2 = (ab)2(i-n-2) ϵ L!

43. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language. Example: For s = (ab)2n
For x = ε, y = abab, z = (ab)2n-2
For any i, xyiz = (ab)2i(ab)2n-2 = (ab)2(i-n-2) ϵ L!
s = (ab)2n doesn’t work!

44. L={ww : w in {a,b}*} aaaa…aab|aaaa...aab
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language.
Use s = anbanb
aaaa…aab|aaaa...aab n n

45. y L={ww : w in {a,b}*} aaaa…aab|aaaa...aab
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language.
Use s = anbanb
For any splitting of s in x,y,z with the desired properties: y
aaaa…aab|aaaa...aab n n

46. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language
Use s = anbanb
For any splitting of s in x,y,z with the desired properties: y = am with 1 ≤ m ≤ n.

47. L={ww : w in {a,b}*}
We prove that L is not regular by using the pumping lemma.
Pumping length: n
Choose a proper string in the language
Use s = anbanb
For any splitting of s in x,y,z with the desired properties: y = am with 1 ≤ m ≤ n.
Observe that xy2z = am+nbanb is not in L
QED

48. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?

49. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
A first attempt to design a FA
q10
a,b
q11
a,b
q12
a,b
q13
a,b
...
q1n ε
q2n
a,b
a,b
a,b
a,b
...
q20
q2n-1
q2n-2
q2n-3

50. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
A first attempt to design a FA fails!
q10
a,b
q11
a,b
q12
a,b
q13
a,b
...
q1n
Works for string
sizes up to n! ε
q2n
a,b
a,b
a,b
a,b
...
q20
q2n-1
q2n-2
q2n-3

51. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.

52. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!

53. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length k=2.

54. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length k=2.
For every proper string s in L’,
2n≥2
abbba…abb|bbaba…aaa n n

55. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length k=2.
For every proper string s in L’,
split s in x, y, z with the desired properties.
|y|≥1 and |xy|≤ 2 y z
abbba…abb|bbaba…aaa n n

56. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length k=2.
For every proper string s in L’,
split s in x = ε ,y = first two symbols of s, z = rest. y z
abbba…abb|bbaba…aaa n n

57. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length k=2.
For every proper string s in L’,
split s in x = ε ,y = first two symbols of s, z = rest.
xy2z in L’. y y z
ababbba…ab|bbbaba…aaa
ϵ L’
n+1
n+1

58. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length k=2.
For every proper string s in L’,
split s in x = ε ,y = first two symbols of s, z = rest.
xy3z in L’. y y y z
abababbba…a|bbbbaba…aaa
ϵ L’
n+2
n+2

59. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length n=2.
For every proper string s in L’,
split s in x = ε ,y = first two symbols of s, z = rest.
xy0z in L’. z
bba…abbb|baba…aaa
ϵ L’
n-1
n-1

60. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.
But the pumping lemma holds!
Fix pumping length n=2.
For every proper string s in L’,
split s in x = ε ,y = first two symbols of s, z = rest.
For every i ≥ 0, xyiz in L’.

61. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Consider L’’ = {w : w has even length}.

62. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Consider L’’ = {w : w has even length}.
Every string of even length
abbbaabb….…bbabaaaa 2n

63. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Consider L’’ = {w : w has even length}.
Every string of even length
can be split into two parts of equal length
abbbaabb… …bbabaaaa | n n

64. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Consider L’’ = {w : w has even length}.
Every string of even length
can be split into two parts of equal length
and vice versa.
abbbaabb….…bbabaaaa 2n

65. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Consider L’’ = {w : w has even length}.
L’ = L’’
Every string of even length
can be split into two parts of equal length
and vice versa.

66. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
Consider L’’ = {w : w has even length}
L’ = L’’
A DFA for L’’:
a,b
even
odd
a,b

67. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|}
Is it regular?
YES!!!
L’ = L’’
A DFA for L’:
a,b
even
odd
a,b