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I'm trying to prove that the set of even-length strings with the two middle symbols being equal cannot be accepted by finite automata. I can explain why it cannot be accepted intuitively, but I'm having trouble with the proof. Our symbols are {a, b}.

I allowed L = $\{(ab)^{*{\frac{n}{2} - 1}} aa (ab)^{*{\frac{n}{2} - 1}}\}$. I know the format of the language is wonky, and will be talking to my professor about it tomorrow. For the proof, I allowed $\frac{n}{2} - 1$ to be the combination of symbols before and after the two elements. So, using the Pumping Lemma's condition that |uv|≤ n, I allowed $u = \frac{n}{2} -1$ and $v = n^2$ (for aa); this is obviously greater than n, but I'm having trouble understanding how to choose $u$ and $v$. Is my assignment for these parameters correct?

Raphael
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user3304718
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1 Answers1

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$u$ and $v$ are supposed to be strings such that $uv$ is a prefix of a string in the language longer than $p$ symbols, $p$ being the pumping length. Furthermore, you do not get to choose them. All you know is that their total length does not exceed $p$ and they are prefix of a word $x$ in the regular language, of length $n$.

You do not get to choose them because they are existentially quantified in the statement of the pumping lemma ... you could choose them if they were universally quantified. If you do not understand this last remark, just ignore it.

What you can choose is the value of $n$ and the string $x=uvw$ of size $n$ that will be used for pumping. You choose $n$ in relation to the pumping length $p$ which depends only on the regular language.

I would suggest to choose $n$ quite big, so that you make sure $uv$ is in the first half of the string $x$, since $| uv| ≤ p$.

I would also choose my string $x$ with an appropriate pattern of a and b ... and then check what happens when I pump. It may seem odd, but you have to get even with that problem.

babou
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