Study Notes of Pumping Lemma

Pumping Lemma gives a necessity for regular languages.

Pumping Lemma is not sufficient, that is, even if there is an integer n that satisfies the conditions of Pumping Lemma, the language is not necessarily regular.

Pumping Lemma can not be used to prove the regularity of a language.

It can only show that a language is non-regular.

Example: L = a^kb^k is non-regular, where k is a natural number.

• Suppose that L is regular and let n be the number of states of an FA that accepts L. Consider a string x = aⁿbⁿ for that n.
• Then there must be strings u, v, and w such that
• x = uvw, |uv| ≤ n |v| > 0, and for every m ≥ 0, uv^mw ∈ L.
• Since |v| > 0, v has at least one symbol.
• Also since |uv| ≤ n, v = a^p, for some p > 0,
• Let us now consider the string uv^mw for m = 2.
• Then uv²w = a^n-pa^2pbⁿ = a^n+pbⁿ. Since p > 0 , n + p ≠ n .
• Hence a^n+pbⁿ can not be in the language L represented by a^kb^k.
• This violates the condition that for every m ≥ 0, uv^mw ∈ L.
• Hence L is not a regular language.

Pumping Lemma for CFL’s

• Let L be a CFL. Then there exists a constant N such that if z ∈L s.t. |z|≥N, then we can write z=uvwxy,
• |vwx| ≤ N
• vx ≠ ε
• For all k ≥ 0 : uv^kwx^ky ∈ L

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