Pumping Lemma Examples L> = {aibj : i > j}

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Description: Pumping Lemma Examples L aibj : i j L is not regular. We prove it using the Pumping Lemma. L aibj : i j L is not regular. Fix an arbitrary pumping length n0. L aibj : i j L is not regular. Fix an arbitrary pumping

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slide1. Pumping Lemma Examples<br>
slide2. L> = {aibj : i > j} L> is not regular.
We prove it using the Pumping Lemma.<br>
slide3. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.<br>
slide4. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>.<br>
slide5. L> = {aibj : i > j} L> is not regular.
Fix an arbitrary pumping length n>0.
Choose a proper string s in L>. |s|≥ n<br>
slide6. 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>. aaa…aabb…b n n+1<br>
slide7. 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. aaa…aabb…b n n+1<br>
slide8. 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. |xy|≤ n
|y|≥ 1 aaa…aabb…b n n+1<br>
slide9. 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. |xy|≤ n
|y|≥ 1 aaa…aabb…b n n+1<br>
slide10. 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. |xy|≤ n
|y|≥ 1 aaa…aabb…b Y n n+1<br>
slide11. 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. aaa…aabb…b L> = {aibj : i > j} Y n n+1<br>
slide12. aaabb…b n n+1-m 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>. L> = {aibj : i > j} n<br>
slide13. 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!<br>
slide14. L={ww : w in {a,b}*} First, figure out what this language is.<br>
slide15. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language?<br>
slide16. First, figure out what this language is.
A string in the language? aabaab L={ww : w in {a,b}*}<br>
slide17. L={ww : w in {a,b}*} First, figure out what this language is.
A string in the language? aabaab
Another string in the language?<br>
slide18. 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<br>
slide19. 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?<br>
slide20. 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<br>
slide21. 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?<br>
slide22. 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)<br>
slide23. 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?<br>
slide24. 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!<br>
slide25. 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?<br>
slide26. 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!<br>
slide27. L={ww : w in {a,b}*} First, figure out what this language is.
L = {ε, aa, bb, aaaa, abab, baba, bbbb, aaaaaa …} abaabba|abaabba<br>
slide28. L={ww : w in {a,b}*} We prove that L is not regular by using the pumping lemma.<br>
slide29. 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.<br>
slide30. 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<br>
slide31. 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!!!<br>
slide32. 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 aaa…aaa|aaa…aaa n n<br>
slide33. aaa…aaa|aaa…aaa 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 z y n n<br>
slide34. 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 aaaaa…aa|aaaa…aaa z y n+1 n+1 y ϵ L<br>
slide35. 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 aaaaaaa…a|aaaaa…aaa z y n+2 n+2 y ϵ L y<br>
slide36. 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 a…aaaa|aa…aaa z n-1 n-1 ϵ L<br>
slide37. 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!<br>
slide38. 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!!!<br>
slide39. 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 abab…abab|abab…abab n n<br>
slide40. 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 abab…abab|abab…abab n n z y<br>
slide41. 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 abababab…ab|ababab…abab n+1 n+1 y z y ϵ L<br>
slide42. 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!<br>
slide43. 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!<br>
slide44. 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 aaaa…aab|aaaa...aab n n<br>
slide45. 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: L={ww : w in {a,b}*} y n n<br>
slide46. 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.<br>
slide47. 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<br>
slide48. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} Is it regular?<br>
slide49. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} Is it regular?
A first attempt to design a FA q10 q11 q12 q13 q2n a,b q2n-1 q2n-2 q2n-3 q1n q20 a,b a,b a,b a,b a,b a,b a,b ε ... ...<br>
slide50. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} Is it regular?
A first attempt to design a FA fails! q10 q11 q12 q13 q2n a,b q2n-1 q2n-2 q2n-3 q1n q20 a,b a,b a,b a,b a,b a,b a,b ε ... ... Works for string
sizes up to n!<br>
slide51. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} Is it regular?
Looks similar with L (L = {w1w2 : w1 = w2}.<br>
slide52. 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!<br>
slide53. 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.<br>
slide54. 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’, L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} abbba…abb|bbaba…aaa n n 2n≥2<br>
slide55. abbba…abb|bbaba…aaa 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. z y n n |y|≥1 and |xy|≤ 2<br>
slide56. abbba…abb|bbaba…aaa 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. z y n n<br>
slide57. ababbba…ab|bbbaba…aaa 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’. n+1 ϵ L’ z y y n+1<br>
slide58. abababbba…a|bbbbaba…aaa 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’. n+2 ϵ L’ z y y y n+2<br>
slide59. 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’. bba…abbb|baba…aaa ϵ L’ z n-1 n-1<br>
slide60. 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’.<br>
slide61. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} Is it regular?
Consider L’’ = {w : w has even length}.<br>
slide62. 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<br>
slide63. 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<br>
slide64. 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<br>
slide65. 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.<br>
slide66. 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’’: odd a,b even a,b<br>
slide67. L’ = {w1w2 : w1,w2 ϵ {a,b}*,|w1|=|w2|} Is it regular?
YES!!!
L’ = L’’
A DFA for L’: odd a,b a,b even<br>