Question

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?

Answer 1

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.