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Let :

$ L = \{ w = \{0,1\}^* | w = 1^k \textrm{ or } w = 0^j {1^k}^2, j >= 1, k>=0 \} $

So i split that language into L1 and L2 languages and treated them separately.The first one $(1^k)$ is for sure regular. Second language is a bit of a problem for me. Let $j,k = p$. So it looks like it : $0^p{1^p}^2$

Word $|w|$ should be longer or equal p so it checked.

Then $|xy| <= p$ and y not equal of empty word. I divided it like that :

$x = 0^a, a>=0$

$y = 0^b, b>=1$

$a+b=p$

$z =0^{p-a-b}{1^p}^2$

And lastly : $xyyz = 0^a0^b0^b0^{p-a-b}{1^p}^2 = 0^{p+b}{1^p}^2$

I feel like this led to me to nowhere, could you enlight me what should I do next?

Eri
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  • "$w=1^k w$" probably should be "$w=1^k \textbf{ or } w=0^j...$", considering the text afterwards. – Al.G. Oct 25 '24 at 18:38
  • @Al.G. That's true, I will edit it right away. – Eri Oct 25 '24 at 18:50
  • Thanks for editing! You can type that with 1^k \textrm{ or } w= to get the text straight and spaced. – Al.G. Oct 25 '24 at 18:56
  • And by the way, I'd advise you to include the language in the title. This will make it more attractive to people interested in such problems. – Al.G. Oct 25 '24 at 18:58

1 Answers1

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Hint: Note that $\{0^j: j\ge 1\}$ is regular, so the problem is not with the zeroes and the pumping lemma won't help. That is, pumping zeroes into the word will not take the word outside the language.

The problem is with the $k^2$ in $\{1^{k^2}:k\ge 0\}$. Apply the lemma there, and pump 1's instead!

Al.G.
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  • Ah yes! Thank you for your answer, don't know why i pump 0's here, maybe because of my little understanding of that topic. I guess now the 1's will resolve to $$ x = 0^p1^a, y = 1^b, z = 1^c $$ where $$ a + b + c = p^2 $$ – Eri Oct 25 '24 at 19:11
  • Yeah, no problem. Try to answer the question "which part of the language is problematic with respect to regularity" and focus there. – Al.G. Oct 25 '24 at 19:18
  • But, not so fast! Your comment is not really an improvement: you are still using the same word as before $0^p 1^{p^2}$. This word will not contradict the pumping's lemma, because the lemma applies only to a $p$-sized prefix of it. Since the p-long prefix of your initial w contains only zeroes and pumping them won't help, you should try to think of another word. :) – Al.G. Oct 25 '24 at 19:19