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If we have a language K, that word w=uv, accepts. $K= \{{w \in \{a, b ,c\}^*: \ \vert u \vert_a \ \text{is not divisible by 3}}\}$. Can we land strings like aaaa in final state? How does NFA decides to break up the string in this case, would it be u = aa, v = aa (w accepted) or u = aaa, v = a (w is not accepted). Or doesn't matter and we can land everything as long as right decomposition exists.

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    $\begingroup$ NFA is, as the name says, non-deterministic. There might be several paths for one input string. As long as the NFA is correct and the input string is an acceptable string, there exists a path to reach the final state. $\endgroup$
    – bigbang
    Commented May 9, 2021 at 10:01
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    $\begingroup$ I suggest looking up the definition of the language accepted by an NFA. The definition contains all the information you need. $\endgroup$ Commented May 9, 2021 at 19:12
  • $\begingroup$ I don't understand the definition of $K$. Do you require that $w$ has a prefix $u$ such that $|u|_a$ is not divisible by $3$? In this case, $K = \{a,b,c\}^*a\{a,b,c\}^*$ $\endgroup$
    – J.-E. Pin
    Commented May 10, 2021 at 8:48

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