Wikipedia has the following definition of the pumping lemma for regular langauges...

Let $L$ be a regular language. Then there exists an integer $p$ ≥ 1 depending only on $L$ such that every string $w$ in $L$ of length at least $p$ ($p$ is called the "pumping length") can be written as $w$ = $xyz$ (i.e., $w$ can be divided into three substrings), satisfying the following conditions:

  1. |$y$| ≥ 1
  2. |$xy$| ≤ $p$
  3. for all $i$ ≥ 0, $xy^iz$ ∈ $L$

I do not see how this is satisfied for a simple finite regular language. If I have an alphabet of {$a,b$} and regular expression $ab$ then $L$ consists of just the one word which is $a$ followed by $b$. I now want to see if my regular language satisfies the pumping lemma...

As nothing repeats in my regular expression the value of $y$ must be empty so that condition 3 is satisifed for all $i$. But if so then it fails condition 1 which says $y$ must be at least 1 in length!

If instead I let $y$ be either $a$, $b$ or $ab$ then it will satisfy condition 1 but fail condition 3 because it never actually repeats itself.

I am obviously missing something mind blowingly obvious. Which is?

  • $\begingroup$ You are missing that the lemma only states that strings at least $p$ long can be pumped. If there are no such strings, nothing can be pumped. In particular, for a finite language $p$ (the number of states in a DFA accepting the language, check the proof) is larger than the longest string in the language. $\endgroup$ – vonbrand Feb 9 at 16:01

You are right - we cannot allow "pumping" words of a finite $L$. The thing you are missing is that the lemma says there exists a number $p$, but does not tell us the number.

All words longer than $p$ can be pumped, by the lemma. For a finite $L$, it happens so that $p$ is larger than the length of the longest word in $L$. Thus, the lemma only holds vacuously, and cannot be applied to any word in $L$, i.e., any word in $L$ does not satisfy the condition of "having length at least $p$" as the lemma requires.

A corollary: if $L$ has pumping length $p$, and there exists some word $w\in L$ of length at least $p$, then $L$ is infinite.

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  • 3
    $\begingroup$ A nice instance of the empty set fulfilling $\forall$-statements. $\endgroup$ – Raphael May 15 '12 at 9:26

Pumping lemma is usually used on infinite languages, i.e. languages that contain infinite number of word. For any finite language $L$, since it can always be accepted by an DFA with finite number of state, $L$ must be regular.

According to wikipedia (http://en.wikipedia.org/wiki/Pumping_lemma_for_regular_languages#Formal_statement), pumping lemma says: $\begin{array}{l} (\forall L\subseteq \Sigma^*) \\ \quad (\mbox{regular}(L) \Rightarrow \\ \quad ((\exists p\geq 1) ( (\forall w\in L) ((|w|\geq p) \Rightarrow \\ \quad\quad ((\exists x,y,z \in \Sigma^*) (w=xyz \land (|y|\geq 1 \land |xy|\leq p \land (\forall i\geq 0)(xy^iz\in L)))))))) \end{array}$

For any finite language $L$, let $l_{max}$ be the max length of words in $L$, and let $p$ in pumping lemma be $l_{max}+1$. The pumping lemma holds since there are no words in $L$ whose length $\geq l_{max}+1$.

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  • $\begingroup$ It does apply to finite languages just fine: The "pumping length" $p$ in those cases is just larger than the longest string in the language, thus there are no strings susceptible to pumping in the language. $\endgroup$ – vonbrand Feb 15 at 16:25

One way to formalize the core part of the Pumping lemma is this, using $L^{\geq k} = \{ w \in L \mid |w| \geq k\}$:

If $L$ is regular, there exists $p \in \mathbb{N}$ so that

$\qquad \displaystyle \forall w \in L^{\geq p}.\ \exists x,y,z \dots$ (*).

For all finite $L$ and $p > \max \{ |w| \mid w \in L\}$, we have obviously that $L^{\geq p} = \emptyset$. Therefore (*) is (vacuously) true for such $p$.

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