Pumping lemma, L={WW^R | W can be {1}+}

Question

im trying to find out, if L is regular or not using pumping lemma.

I have L={WW^R | W can be {1}+}

So possible strings would be 11, 1111, 111111. In every cases i have googled so far are examples with alphabet greater then 1, so there is understandable, that WW^R is not regular, but in this case, im lost.

Let's have a string w=111111, p=5 so

x=11 y=111 z=1

|xy|<=p p>=1

xy^iz where i=0 i would have w=111 which is not WW^R and that would mean L is not regular.

But... Honesly, im not sure, whether am i doing it right or wrong.

Can some body help me out? And... Is that L realy NOT REGULAR?

Answer 1

Observe that $L=\{1^{2n}\mid n\in\Bbb N,\, n>0\}$ which is regular. Can you find a finite state automata (with three states) that accepts exactly $L$?

Answer 2

The pumping lemma states that there exists a factorization $xyz$ that satisfies the pumping condition, not that every such factorization fulfills the lemma. The factorization you are looking at indeed does not fulfill the pumping condition.

In this case, any factorization with $y$ of an even length (and fulfilling the conditions of the lemma) will do. For example $$x= \lambda, y=11, z = 1111$$ should do, if the lemma's formulation allows empty $x$, which it usually does.