Explain Pumping lemma for context free language

Pumping lemma for context free language (CFL) is used to prove that a language is not a Context free language

Assume L is context free language

Then there is a pumping length n such that any string w εL of length>=n can be written as follows −

|w|>=n

We can break w into 5 strings, w=uvxyz, such as the ones given below

• |vxy| >=n
• |vy| # ε
• For all k>=0, the string uv^kxy^yz∈L

The steps to prove that the language is not a context free by using pumping lemma are explained below −

• Assume that L is context free.
• The pumping length is n.
• All strings longer than n can be pumped |w|>=n.
• Now find a string 'w' in L such that |w|>=n.
• Divide w into uvxyz
• Show that uv^kxy^kz ∉L for some k
• Then, consider the ways that w can be divided into uvxyz.
• Show that none of these can satisfy all the 3 pumping conditions at same time.
• w cannot be pumped (contradiction).

Example

Find out whether L={xⁿyn^zn|n>=1} is context free or not

Solution

• Let L be context free.
• L must satisfy pumping length, say n.
• Now we can take a string such that s=xⁿyⁿzⁿ
• We divide s into 5 strings uvxyz.

Let n=4 so, s=x⁴y⁴z⁴

Case 1:

v and y each contain only one type of symbol.

{we are considering only v and y because v and y has power uv²xy²z}

X xx x yyyyz z zz

=uv^kxy^kz when k=2

=uv²xy²z

=xxxxxxyyyyzzzzz

=x⁶y⁴z⁵

(Number of x # number of y #number of z)

Therefore,The resultant string is not satisfying the condition

x⁶y⁴z⁵ ∉ L

If one case fails then no need to check another condition.

Case 2:

Either v or y has more than one kind of symbols

Xx xx yy y y zzzz

=uv^kxy^kz (k=2)

=uv²xy²z

=xxxxyyxxyyyyyzzzz

=x⁴y²x²y⁵z²

This string is not following the pattern of our string xⁿyⁿzⁿ

Bhanu Priya

Updated on: 12-Jun-2021

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