# Can the String, $0^p 0^p 0^p$, be Used with the Pumping Lemma to Show that $w^r w w^r$ is Not Context Free?

I'm trying to show that $L=\left\{w^rww^r:w \in \{0,1\}^*\right\}$ is not context free using the pumping lemma.

I thought picking the string, $0^p0^p0^p$, would be a good candidate for this, but someone told me I might want to consider another string. Why is this? Any way in which I divide the string and then pump the string results in the balances between the $w^r$on either side of the string being thrown off, no?

For $p>0$, you can always split $0^{3p}$ into $uvxyz$ such that $u=x=\varepsilon$, $v=0$, $y=0^2$, and $z=0^{3p-3}$. Such a split clearly satisfies $|vxy|\leq p$, $|vy|\geq1$. Finally, for all $n\geq0$, $uv^nxy^nz=0^{3(p-1+n)}$, which is in $L$, since $0^{3m}\in L$ for all $m$.
Since you want to show, that for an arbitrary split satisfying the length conditions, there is some $n$ for which $uv^nxy^nz\notin L$, the choice of $0^{3p}$ is unfortunate.
The string $0^{3p}$ can always be pumped so that it stays in $L$. Recall what the pumping lemma says: there exists a $p \in \mathbb{N}$ such that any $w \in L$ of size at least $p$ can be written as $w=uvxyz$ such that $|vxy| \leq p$, $|vy| \geq 1$ and $uv^ixy^iz \in L$ for all $i \in \mathbb{N}$. In your case, we can choose $u = x = y = \epsilon$ and $v=0^3$, and then $uv^ixy^iz = 0^{3(p-1+i)} \in L$, and so no contradiction can be obtained using the word $0^{3p}$.
In order to prove using the pumping lemma that $L$ is not regular, you need to come up with a word $w$ so that for every possible decomposition $w = uvxyz$ such that $|vxy| \leq p$ and $|vy| \geq 1$, there exists some $i \in \mathbb{N}$ such that $uv^ixy^i z \notin L$. For your word $w$ this is not the case, since the decomposition indicated above is "bad". While you are free to choose $w$, you are not free to choose the decomposition $w = uvxyz$; your argument should work for all possible decompositions.
In fact, the word $w = 0^{3p}$ doesn't even help in showing that $L$ is not regular; we can choose the decomposition $w = xyz$ with $x = \epsilon$ and $y = 0^3$, and then $xy^i z = 0^{3(p-1+i)} \in L$.
The root cause of all these problems is that $L \cap 0^* = \{0^n : 3|n\}$, which is regular. So you have to look at a word which contains both $0$s and $1$s.