The question asks to write a NFA for the following language $L$ above $\Sigma = \left \{0,1 \right \}$.

$L = \{\sigma_1 u \sigma_2 v \sigma_3 \mid (\sigma_1, \sigma_2, \sigma_3 \in \Sigma, u, v \in \Sigma^*, |u| = |v|) \text{ and } (\sigma_1 = \sigma_2 \text{ or } \sigma_3 = \sigma_2 \text{ but not both})\}$

The NFA must contain at the most 7 states.

To be honest, I spent a lot of time with the pumping lemma in order to disprove the claim that the $L$ is regular but it didn't work.

I came up to the conclusion that a word $w \in L$ must hold that $|w| = 2n +3 : n \in \mathbb{N}\cup\left \{ 0 \right \}$ and it led me to the following NFA:

enter image description here

This NFA doesn't cover the cases where $\sigma_1 = \sigma_2 = 0 , \sigma_3 = 1$ and $\sigma_1 = 0 , \sigma_2 = \sigma_3 = 1$.

The constrain of the 7 states prevents me from thinking furthermore.

The question wasn't designed to implement Myhill–Nerode theorem to come up with the solution.

I would like to know what can be added to the NFA I draw, or perhaps I'm not even in the right direction.

  • $\begingroup$ Another issue that I can observe is that your NFA does not require $|u|=|v|$, but that $|u|=2x$ and $|y|=2y$ for some $x,y \in \mathbb{N} \cup \{0\}$ (i.e. that each string contains an even number of characters). $\endgroup$ Commented Apr 18 at 14:18

1 Answer 1


It is indeed regular. Here, your language is basically $1(0+1)^{2m+1}0 + 0(0+1)^{2m+1}1$ for all integers $m \ge 0$. We are basically fixing $\sigma_1 \ne \sigma_3$, and then let $|u\sigma_2v|$ be an odd cycle. This will automatically ensure the rest of the constraints are met. This is easily realizable by a NFA with 8 states, as shown below.8 state NFA

PS: I hope this simplification might lead to your required 7 state solution, which I cannot visualize at this moment.

  • $\begingroup$ Your automaton recognizes $011111$ which is not in the language: using three states for a loop will recognize words like $01^{3m+1}1$, not $01^{2m+1}1$. $\endgroup$
    – Nathaniel
    Commented Apr 18 at 14:59
  • $\begingroup$ Sorry my mistake, I am updating the answer $\endgroup$
    – codeR
    Commented Apr 18 at 15:11
  • $\begingroup$ Stings like 011111 are not possible. All accepted strings are of odd length, and they start and end with different symbols. $\endgroup$
    – codeR
    Commented Apr 18 at 15:15
  • $\begingroup$ Now you just need to delete $q_3$ and $q_6$ and put the corresponding transitions from $q_2$ and $q_5$! $\endgroup$
    – Nathaniel
    Commented Apr 18 at 16:32
  • $\begingroup$ @codeR Thank you very much. I also can't visualize 7 states but what you came with is very useful !! $\endgroup$
    – Daniel
    Commented Apr 18 at 20:48

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