Given the language $ A = \{w \in \{a,b\}^{*} | w = w^{R}\}$ (i.e. palindromes using the symbols $a, b$), I am trying to determine if the Pumping Lemma can be applied to strings of the form $s = a^{2p}$.

From my understanding of the Pumping Lemma, to show it would hold, I need to decompose $ s $ into $ s = xyz $ such that (1) $ xy^{i}z \in A, i \geq 0 $, (2) $ y \neq \epsilon $, and (3) $ |xy| \leq p $.

For cases where $ p \geq 2 $, the decomposition makes sense to me and appears to be quite trivial. For example, when $ p = 2 $, then $ s = aaaa $, and can be decomposed into $xyz$ where $x = a, y = a, z = aa $, which satisfies the conditions above above. Induction could be used to show this holds for larger values of $ p $.

However, I am struggling on the case where $ p = 1 $. Here, $ s = aa $, and it seems like there are not enough symbols to decompose into $ xyz $. My best guess would be to choose $ x = a, y = a, z = \epsilon $, but for some reason this doesn't feel legal -- can you ad-hoc assign pieces of $ xyz $ to be $ \epsilon $? Also, I believe this decomposition would fail to satisfy condition (3) above. Can you offer some guidance on understanding what is happening in the case where $ p = 1 $?

Thanks for your time!

  • 1
    $\begingroup$ Welcome to COMPUTER SCIENCE @SE. There are pumping lemmata for different Chomsky types of languages; you quoted the one for regular languages. Note that $y$ is the only part required to be non-empty. $\endgroup$
    – greybeard
    Oct 6, 2020 at 7:55

1 Answer 1


The empty string is a legitimate string, just like the empty set is a legitimate set. We say that a word $w \in \Sigma^*$ can be pumped with respect to parameter $p \geq 1$ and a language $L$ if there exists a decomposition $w = xyz$, where $x,y,z \in \Sigma^*$ satisfy the following three requirements:

  • $|xy| \leq p$.
  • $y \neq \epsilon$.
  • $xy^iz \in L$ for all $i \geq 0$.

A mathematical definition hides nothing. A word can be pumped if it satisfies the definition as stated. No more, no less.


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