We are going over the pumping lemma in class and we recently went over the following example:

  1. Let $$ L = \{ w \mid w \text{ has a different number of 0s and 1s} \} $$
  2. Consider $$ s = 0^P1^\left(P+P!\right) $$
  3. $ s $ can be divided into $ s = xyz $
  4. Consider $$ y = 0^m; 0 \leq m \leq P $$
  5. Let $ i = \frac{P!}{m} + 1 $
  6. $$ xyz = 0^\left(P + \left(i - 1\right)m\right)1^\left(P + P!\right) \notin L $$

Forgive me if the example is not put together well. This example didn't quite get finished in class. Feel free to expand on this if need be.

However, my question is how does one know to use a factorial in the first place when approaching this proof?

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    $\begingroup$ How does one know how to approach any mathematical proof? Perhaps some intuition, perhaps luck. This is not well-defined. Actually, to show that this language is not regular it is much easier to show that its complement is not regular. $\endgroup$ – Shaull Feb 20 '13 at 7:25
  • $\begingroup$ I assume there is a step 1.5, where you assume that the pumping number of $L$ is $P$? $\endgroup$ – frafl Feb 20 '13 at 16:20
  • $\begingroup$ I would try a different approach. What is the complement of your language? Is that regular? What do you know about complementation of regular languages? $\endgroup$ – Dave Clarke Feb 20 '13 at 16:56
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    $\begingroup$ @DaveClarke: While this is the easier approach for this language, it's not what he asked for. I think the OP wants to learn something about proofs using the pumping lemma and not about the regularity of that specific language. $\endgroup$ – frafl Feb 20 '13 at 17:26

Let's choose a different word in step 2, say $s' = 0^a1^b$, where $a$ and $b$ depend on $P$ and $a \geq P$ (to force an $y$ that has the form $y=0^m$).

It's important to show that for any length $m, 1\leq m \leq P$ there is no valid division $s' = xyz, y = 0^m$ (valid means it fulfils the conditions of the pumping lemma).

So for every such $m$, we need an $i$ such that: $xy^iz = x0^{im}z = 0^{a-m+im}1^b = 0^{a+(i-1)m}1^b {! \atop =} 0^b1^b$. This means that $m$ must divide $b-a$ for every $m$. One number $b-a$ that fulfils this, is $P!$, so e.g. $a=P$, $b=P+P!$, the least common multiple of $1,\dots,m$ is another possible choice for $b-a$.


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