CODEWITHSUNDEEP
computer-sciencecontext-free-grammarpumping-lemma
Neal
Neal

Reputation: 6992

Pumping lemma for CFLs

My question is:

Let L = { x in {a,b}* | x has an equal number of a's and b's}

I know this is a context free language because I can create a grammar for it (e is epsilon):

S -> aX | bY | e

X -> bS | aXX

Y -> aS | bYY

You can also prove it is context free by using the fact that a context free language intersected with a regular language is context free.

Since it is a context free language, according to the pumping lemma for CFL's, any string longer than the pumping length p should be able to be pumped. However, if I choose the string s = a^p b^p a^p b^p, this string cannot be pumped, so the language should not be context free.

Where am I going wrong?

Upvotes: 3

Views: 1576

Answers (2)

Jim Lewis
Jim Lewis

Reputation: 45115

Let u = a^p, v = b^(p-1), x = ba, y = a^(p-1), z = b^p, so that your string s = uvxyz.

Then any string of the form u v^i x y^i z is in the language, so the conditions of the CFL pumping lemma are satisfied.

The pumping length isn't "p" for your example...maybe that's where you're getting confused?

Edit: sepp2k correctly points out that my choice of vxy violates the condition that |vxy| < = p, the pumping length of the language. His solution v=b, x=e, y=a is correct. For this language, any string of length 2 or greater will pump -- "ab" or "ba" must appear somewhere, so vy = ab or vy = ba will always work.

Upvotes: 1

sepp2k
sepp2k

Reputation: 370425

Sure the string can be pumped. Let u = a^p b^(p-1), v = b, x = e, y = a, z=a^(p-1) b^p. Now uvxyz = s and for any i u v^i x y^i z has an equal amount of as and bs.

Upvotes: 4

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