As I am currently teaching regular languages and pumping lemma, I was searching for nice examples of languages, regular or not, for exercises.

  • $L_1 = \{vv\mid v\in \Sigma^*\}$ is a classic example, as it can be proven to be non regular for $|\Sigma|\geqslant 2$ using the pumping lemma;
  • $L_2 = \{ uvv\mid u,v\in\Sigma^*\}$ looks like $L_1$, however it is a bit of a trap: since $L_2 = \Sigma^*$, it is regular;
  • $L_3 = \{uvv\mid u\in\Sigma^*, v\in\Sigma^+\}$ looks like $L_2$, however it is not regular if $|\Sigma| \geqslant 2$. This can be proven using the pumping lemma: suppose it is regular, and let $n$ be the pumping length. Let $u=a^nba^nb\in L_3$ and $u = xyz$ a decomposition such that $|xy|\leqslant n$ and $|y| > 0$. Then $xz = a^kba^nb$ with $k< n$ and does not end with a (non empty) square, therefore is not in $L_3$ which is not regular;

Things became a bit complicated with $L_4 = \{uuv\mid u\in \Sigma^+, v\in \Sigma^*\}$: after struggling a bit to find a counter-example word to the pumping lemma, I proved $L_4$ to be non regular if $|\Sigma|\geqslant 2$, using the fact that $L_3 = L_4^R$ and that regular languages are closed under mirror image.

However, if $|\Sigma| = 2$, I found out that $L_4$ verifies the conclusion of the pumping lemma, with a pumping length of $5$! Indeed, since any word of length $\geqslant 4$ contains a square, any word $u\in L_4$ with $|u| \geqslant 5$ can be written $u = vwwu'$ with $|vww|\leqslant 5$ and $|v|,|w| > 0$. Therefore $u = xyz$, with $x = \varepsilon$, $y = v$ and $z = wwu'$ verifying:

  • $|xy| = |v| < |vww| \leqslant 5$;
  • $|y| = |v| > 0$;
  • for all $k \geqslant 0$, $xy^kz\in L_4$. Indeed, $xz = z = wwu' \in L_4$ (because $|w| > 0$), $xyz = u \in L_4$ (hypothesis) and if $k>1$, $xy^kz = yyy^{k-2}z \in L_4$ (because $|y| > 0$).

I then tried to find a counter-example to the pumping lemma for $L_4$ when $|\Sigma|>2$, but without success. I found out that in order to not be able to apply the previous reasoning, in any counter-example, the first $n$ (pumping length) letters need to not contain a square, which is possible, but couldn't go further.

My questions are all related to this problem:

  • I don't think I did, but maybe, did I make any mistake?
  • Is there a counter-example word to the pumping lemma for $L_4$ when $|\Sigma| \geqslant 3$?
  • Do you know of any simpler language that is a counter-example to the converse of the pumping lemma (not regular but verifying the conclusion of the pumping lemma)? vonbrand proposed this one which is a tad bit complicated. I also know of $\{ab^iab^jab^j\mid i,j\geqslant 0\}\cup \{uaav\mid u,v\in\Sigma^*\}$ which is not better.
  • 1
    $\begingroup$ As for your last bullet point: Every language of the form $\$L\cup \{\$^k\mid k\ne 1\}\cdot \Sigma^*$ for $L\subseteq \Sigma^*$, $\$\notin\Sigma$, satisfies the regular pumping lemma. cs.stackexchange.com/q/9181/4287 $\endgroup$ Jan 8, 2022 at 1:18
  • $\begingroup$ Thanks for the idea. What I found interesting with the language $\{uuv\mid u\in \Sigma^+, v\in\Sigma^*\}$ is that it is not defined as the union of two other languages (which is the case for your language and the two I gave, and most of the counter-example I found). $\endgroup$
    – Nathaniel
    Jan 8, 2022 at 7:11

1 Answer 1


When the alphabet is of size $4$ or more, it is easy to prove that $L_4$ is not regular using the pumping lemma. Given a pumping length $n$, let $s$ be a square-free word over $\{0,1,2\}$ of length $n$; such a word exists since there is an infinite square-free word over $\{0,1,2\}$. We choose $w = 3s33s3 \in L_4$. Suppose that we could write $v = xyz$ so that $|xy| \leq n$, $y \neq \epsilon$, and $xy^iz \in L_4$ for all $i \geq 0$. In particular, $xz \in L_4$. We consider two cases:

  1. $x \neq \epsilon$. In this case, $xz = 3t 33s3$, where $t$ is a word over $\{0,1,2\}$ satisfying $0 < |t| < s$. If $xz$ started with a square $u^2$, then $u$ starts with a $3$, and so the second occurrence of $u$ must also start with a $3$. The second letter cannot be a $3$ (since $t$ doesn't start with $3$), hence $u = 3t3$, but since $|t| < |s|$, the word $u^2$ is not a prefix of $xz$. Hence this case is impossible.

  2. $x = \epsilon$. In this case, $xz = t33s3$, where $t$ is a non-empty suffix of $s$. Since $xz$ contains only three $3$s, if $xz$ starts with a square $u^2$, then either $u$ contains no $3$ or it contains a single $3$. It $u$ contains no $3$ then $u^2$ is a prefix of $t$, which is impossible since $s$ is square-free. If $u$ contains a single $3$ then $u = t3$, which is impossible since $t$ doesn't start with a $3$ whereas the second occurrence of $u$ does. Hence this case is also impossible.

In the ternary case, it suffices to find a collection or arbitrarily long square-free ternary words $w$ such that for any decomposition $w = xyz$ with $y \neq \epsilon$, we have $xy^i z \notin L_4$ for some $i$. I conjecture that prefixes of the Leech word satisfy this condition (even if we restrict $i$ to be either $0$ or $2$).

Suppose that arbitrarily long words exist satisfying the above constraints. Given a pumping constant $n$, choose such a word $w_k$ of length at least $n+2$ which starts and ends with the same letter $\sigma$, and let $w = w_k \sigma w_k \sigma \in L_4$. (Note that since each word satisfying the constraints is square-free, every subword of length 5 contains all different letters; hence we can always truncate a long enough word to one which ends with the same letter with which it starts.)

Suppose that $w = xyz'$, where $y \neq \epsilon$ and $|xy| \leq n \leq |w_k| - 2$. Then we can write $z' = z\sigma w_k \sigma$, where $w_k = xyz$; note that $|z| \geq 2$ and $z$ ends with $\sigma$.

According to the conjecture above, we can find $i \neq 1$ such that $xy^i z$ does not have a square prefix. If $xy^i z'$ does have a square prefix $u^2$, then $u^2$ must reach the first $\sigma$; in particular, $|u| \geq 2$.

If the first copy of $u$ doesn't reach the first $\sigma$ then then $u$ doesn't contain $\sigma^2$, but then its second copy contains both $\sigma$ and either the last letter of $z$ or the first letter of $w_k$, and so does contain $\sigma^2$. Thus the first copy of $u$ does reach the first $\sigma$.

If the first copy of $u$ doesn't end at the first $\sigma$ then it contains $\sigma^3$. Since this is the unique copy of $\sigma^3$ in $w$, this is impossible. Hence $u = xy^iz\sigma$, and so $u$ is a prefix of $w_k \sigma$. In particular, $i = 0$, since otherwise $|u| > |w_k \sigma|$. But then $u$ is a proper prefix of $w_k \sigma$, implying that $w_k$ contains a copy of $\sigma^2$, which is false.

  • $\begingroup$ Nice counter-example. Did you find anything for $|\Sigma| = 3$? $\endgroup$
    – Nathaniel
    Jan 8, 2022 at 11:03
  • $\begingroup$ I have a proposed construction using the Leech word. $\endgroup$ Jan 8, 2022 at 12:27
  • 1
    $\begingroup$ Thanks for the construction. I have verified that the conjecture is true for Leech words up to ~2000 letters, but I am not so sure as to how to prove it for the general case. $\endgroup$
    – Nathaniel
    Jan 9, 2022 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.