3
$\begingroup$

When applying the pumping lemma to $L = \{ 0^n \mid n>0\}$ I do the following:

$S = 0^p$
$x = \varepsilon$
$y = 0^p$
$z = \varepsilon$

so $S = xyz = (\varepsilon)0^p(\varepsilon)$

For $x y^i z$ if i choose $i = 0$ (pumping down) i get $S = \varepsilon$, which is not an element of $L$.

Therefore according to the pumping lemma this language is not regular. But since I am able to draw a DFA for $L$, I know it is regular.

I would like to know what rules I am violating by trying to apply the pumping lemma to $L$ in this manner.

$\endgroup$
1
1
$\begingroup$

The pumping lemma says that you can split the word into $xyz$. It doesn't say that every such split works. For instance, if you take $|x|=1$ in your example, the pumping works up and down. This would be a "correct split" that makes the pumping lemma hold.

As a reference, recall the definition of the pumping language for regular languages (from Wikipedia):

Let $L$ be a regular language. Then there exists an integer $p\ge 1$ (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. $|y| \ge 1$
  2. $|xy| \le p$ and
  3. for all $i \ge 0$, $xy^iz \in L$.
    $y$ is the substring that can be pumped (removed or repeated any number of times, and the resulting string is always in $L$).
$\endgroup$
4
  • $\begingroup$ Thank you for the quick reply. Did you reformat the post using Latex? $\endgroup$
    – BrentW
    Oct 29 '14 at 4:30
  • $\begingroup$ Yes, see a short guide in meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – Ran G.
    Oct 29 '14 at 14:16
  • $\begingroup$ But wait -- if n = 1 in the op's L, you can't take |x| = 1, because |y| has to be >= 1. That leaves $\varepsilon$ for x, which is |x|=0. How can it be pumped down then? $\endgroup$ May 20 at 19:05
  • $\begingroup$ Ahh, all words longer than p can be pumped, so in the case we need to omit n=1 and start at n=2 cs.stackexchange.com/a/1848/24350 $\endgroup$ May 20 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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