1.10. pumping lemma for regular sets

Pumping
Lemma for
Regular sets
-Sampath Kumar S,
AP/CSE, SECE

Pumping for Regular Set:
Theorem:
Let L be a regular set. Then there exists a
constant ‘n’ such that for every string W in L −
 |W| ≥ n
We can break W into 3 strings, W = X Y Z, such that:
 |Y| > 0
 |XY| ≤ n
 For all k ≥ 0, the string XYkZ is also in L.
Note: Select k such that the resulting string is not in L.
11/21/2017
Sampath Kumar S, AP/CSE, SECE
2

Applications of Pumping Lemma:
Pumping Lemma is to be applied to show that
certain languages are not regular. It should never be
used to show a language is regular.
 If L is regular, it satisfies Pumping Lemma.
 If L is non-regular, it does not satisfy Pumping
Lemma.
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Method to prove that a language
L is not regular
 At first, we have to assume that L is regular.
 So, the pumping lemma should hold for L.
 Use the pumping lemma to obtain a contradiction
−
 Select W such that |W| ≥ n
 Select Y such that |Y| > 0
 Select X such that |XY| ≤ n
 Assign the remaining string to Z.
 Select k such that the resulting string is not in L.
 Hence L is not regular
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Problems to Discuss:
45. Prove that L = {aibi | i ≥ 0} is not regular.
Solution
 At first, we assume that L is regular and n is the number of
states.
 Let w = anbn. Thus |W| = 2n ≥ n.
 By pumping lemma, let W = XYZ, where |XY|≤ n.
 Let X = ap, Y = aq, and Z = arbn, where p + q + r = n.p ≠ 0, q ≠ 0,
r ≠ 0. Thus |y|≠ 0
 Let k = 2. Then xy2z = apa2qarbn.
 Number of a’s = (p + 2q + r) = (p + q + r) + q = n + q
 Hence, xy2z = an+q bn. Since q ≠ 0, xy2z is not of the form anbn.
 Thus, xy2z is not in L. Hence L is not regular.
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Problems to Discuss:
47. Prove that L = {wwR|w ∈ {a,b}} is not regular.
48. Prove that L = {0n1n2n|n>1} is not regular.
49. Prove that L = {0n1m|m>n & m,n>0} is not
regular
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11/21/2017
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1.10. pumping lemma for regular sets

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