This must be known, but somehow I can't locate a reference about this. Let $A$ be a nondeterministic finite automaton (NFA) over words of an alphabet $\Sigma$. I say that $A$ is unambigous if, for every word $w \in \Sigma^*$, then $A$ has at most one accepting run over $w$. I say that $A$ is 2-ambiguous if, for every word $w \in \Sigma^*$, then $A$ has at most two accepting runs over $w$.
Can I transform in polynomial time an input 2-ambiguous NFA into an unambiguous NFA which is equivalent? (i.e., recognizes the same language) Or is it known that it is not possible? e.g., there is a family of 2-ambiguous NFAs where the number of states of any equivalent unambiguous NFAs must be exponential?
In terms of related work, I found the 1989 SIAM J. Comput. paper by Ravikumar and Ibarra Relating the type of ambiguity of finite automata to the succinctness of their representation (unfortunately I didn't find an open-access version) which shows that, given a $k$-ambiguous NFA, there is an exponential blowup in $k$ to convert it to an unambiguous NFA. But this does not imply anything for $k=2$. I also found lots of works about more exotic automata models. But I guess the result on vanilla NFAs must be known (or simple to show)?