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There exists a variation of the pumping lemma with necessary and sufficient conditions for a language to be Regular.

According to that lemma:

A language $L$ is regular iff $\exists k$, $\forall x\in \Sigma^k$, $\exists u,v,w\in \Sigma^*$, $ x=uvw \cap |v|\ge 1$ such that:

$$\forall i \ge 0,\ \forall z\in \Sigma^*: uvwz\in L \iff uv^iwz\in L.$$

My question to you is: is there any way changing the for all $i \ge 0$ condition to for all $ 0\le i\le N$ for some $N$ - and the lemma will still be correct?

That $N$ may be constant, depend on the lemma's k, and so on.

I can't find an approach to prove it, any ideas?

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  • $\begingroup$ Try constructing a counterexample first. $\endgroup$ Commented Jun 11, 2020 at 7:34
  • $\begingroup$ In the proof of the lemma in the attached paper, it seems that they only use N=1, when proving that a language is regular. When assuming a language is regular, it's easy to say that it works with i={0,1}, because then there are less things to prove. So i seems the answer to my question is true, for N=1. Have I got it wrong somewhere? $\endgroup$
    – Tom
    Commented Jun 11, 2020 at 8:43
  • $\begingroup$ It’s hard to say without seeing the proof. If the proof still works even when $N=1$, then the lemma holds even when $N=1$. $\endgroup$ Commented Jun 11, 2020 at 9:35

1 Answer 1

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Yes, what you have observed is correct. We can, in fact, always take $N=0$. Here is the variant of Jeffrey Jaffe's pumping lemma where strings are pumped up or down exactly once.


A language $L$ is regular iff $\exists k$, $\forall x\in \Sigma^k$, $\exists u,v,w\in \Sigma^*$ such that $ x=uvw$, $|v|\ge 1$ and $\forall z\in \Sigma^*$, $$uvwz\in L \iff uwz\in L.$$


As you have observed, the article has basically proved the above fact.

Here is a simpler proof of the "$\Longleftarrow$" direction. Let $[y]$ be the Myhill-Nerode equivalence class represented by string $y$. Suppose $|y|\ge k$.

Let $x$ be the first $k$ symbols of $y$ and $t$ be the rest of $y$, i.e., $y=xt$ and $|x|=k$. By assumption, there exists $u,v,w\in\Sigma^*$ such that $x=uvw$, $|v|=1$ and $$yz=x(tz)=uvw(tz)\in L\iff uw(tz)=(uwt)z\in L.$$ The equivalence above means $[y]=[uwt]$. Since $uwt$ is $y$ with $v$ deleted, $|uwt|\lt |y|$. That means the shortest string in a Myhill-Nerode equivalence class must be shorter than $k$.

Since there are finitely many strings that are shorter than $k$ symbols, there are only finitely many Myhill-Nerode equivalence classes. By the celebrated Myhill-Nerode theorem, $L$ is regular.

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  • $\begingroup$ 'the "⟸" direction' means the "only if" part of that "iff". $\endgroup$
    – John L.
    Commented Jun 15, 2020 at 16:59
  • $\begingroup$ Loved the answer, thank you! $\endgroup$
    – Tom
    Commented Jun 16, 2020 at 9:03
  • $\begingroup$ I have a Further question: can the "forall z" be restricted to "forall z upto length K" or any significant limitation to it (meaning that the number of possible z will be finite), and it will still imply that L is regular? $\endgroup$
    – Tom
    Commented Jun 16, 2020 at 9:14
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    $\begingroup$ That sounds like an interesting question. $\endgroup$
    – John L.
    Commented Jun 16, 2020 at 13:18

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