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I feel that choice should work great for proving non - regularity of the mentioned language.

If $L=\{ww|w \in \{0,1\}^*\}$ and we choose $s=0^p0^p$, meaning s is atleast as long as 'p'. Then we can decompose s as follows $x=empty \ string,\ y = 0^p, \ z = 0^p$

Then we can pump 'y' according to the rules of pumping lemma we will end up with 2p zeroes which is more than the number of zeroes in 'z'

And we have reached a contradiction because this new string is not in the language and according to the pumping lemma it should have, so why does michael sipser say that it's a bad candidate for proving non regularity in this langauge?

Pratik Hadawale
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2 Answers2

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Your proof doesn't work. First of all you can't choose your favorite decomposition of $s$: the pumping lemma just ensures that such a decomposition exists so your argument needs to work for all possible decompositions.

Nevertheless, even with your decomposition it is false that the "pumped" word does not belong to the language. To see why suppose that the pumping length $p$ happens to be $2$. Then $s=0^20^2=0000$, $x=\varepsilon$, $y=0^2=00$, and $z=0^2=00$.

Now let's pump $y=00$ to, e.g., $y^2 = 0000$. The resulting word is $xy^2z = \varepsilon\, 0000\, 0^2 = 000000$. As you can see $000000 \in L$ (by choosing $w=000$).

You run in the same problem regardless of how many times (possibly $0$) you pump $y$.

Steven
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First of all, you don't get to pick the decomposition. So one decomposition you'd need to handle is $x = \varepsilon$, $y = 00$, $z = 0^{2p-2}$. Now if we pump $y$ for $i$ times, we get $xy^iz = 0^{2p + 2i - 2}$, which belongs to $L$.

Arno
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