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How would about proving this is not regular with the pumping lemma. Please include all steps and explain all steps. I am really new with this.

$1^{2x}0^y$ and $y>= x$

Does it matter which side you pump? How do you pick which side to pump?

This is what I was thinking:

$$u = 1^{2x}$$ $$v = 0^y$$ $$w = 1^{2x}0^y$$

Am I on the right track? What else do I need to do?

nitishch
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cokedude
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  • Think what pumping lemma says. It says that there exists some $p$ with some conditions. Try to use $p$ in your equations and try to get a contradiction – nitishch Mar 12 '14 at 09:46
  • So change my x to p? What title would you recommend? I thought I was trying to prove the pumping lemma was not regular. – cokedude Mar 12 '14 at 15:44

1 Answers1

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assuming you have a pump length $p$ then there is a string $1^{2x}0^y$ with $p \lt x \le y$ as part part of the language

thus the partition $uvw$ with $\text{length}(uv)<p$ must have $uv$ as all $1$s

this means that your pump string $v$ contains all $1$ which if pumped can create twice as many $1$ than $0$s which is not part of the language. QED

ratchet freak
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  • So is what I did right? I thought I could pump the u. What else do I need? Like I said I am really new with this. – cokedude Mar 12 '14 at 15:41
  • @cokedude You need a string longer than the pump length and $uv$ must be shorter than the pump length, this meant that I could construct a string where the number of $1$s was greater than the pump length which forced $uv$ to be all $1$s as well. This in turn meant that you had to pump $1$. You could use $vw$ shorter than $p$ and pump $v$; this relies on the fact that the reverse of a regular languages is also regular. – ratchet freak Mar 12 '14 at 15:48