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I'm trying to design and DFA that accept string with an odd number of 0`s, but counting only the ones within sub-strings with two or more 0.

So, for example, 011000, will be accepted since it has 4 0`s, but only three or them are inside a sub-string of two or more. But 0001000, wont be accepted.

Xhark
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1 Answers1

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I believe your language is not regular and if I'm not mistaken, the pumping lemma suffices: Let us assume by contradiction your language is regular, thus it should hold all 3 conditions of the pumping lemma.

Let $w=0^{2p+1}111$, notice $w\in L$ because $2n$ is always even, thus by adding $1$ we have an odd number of $0$s.

Now let us take a simple example where $p=1$, but this should work for any $p$.

In this case, $p=1$, notice $|w|>p$, since $|w|=5$, essentially $w=000111$. In our case this means $y=0$ and because $|xy|\le 1$ and because $|y|>0$ where 1 is our pumping length this means $x=\epsilon$.

This leaves $z=0011$. However, for $i=2$, we receive: 000011 which is $y^2z \equiv 000011 $.

Then the number of 0s in your substring is now even, meaning $000011 \notin L$. Again, we can do this for any pumping length $p$ by simply pumping up or pumping down.

Thus there won't be a DFA/NFA for that. You can try using a TM.

For more information: pumping lemma.

Reckless Engineer
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  • Perhaps some of the data was missing, making the language hold the conditions of the pumping lemma, let me know =) – Reckless Engineer Aug 14 '18 at 13:14
  • You cannot assume what $p$ is when using pumping lemma. – xskxzr Aug 14 '18 at 13:39
  • You're correct, thank you for your comment. However, for this is simply an example, I'll edit the answer to emphasize that. Regardless, for such a word, we could always then make the number of 0s be even using pumping. – Reckless Engineer Aug 14 '18 at 14:42