TITLE: Emptiness and determinization of NFAs
QUESTION [7 upvotes]: Consider an NFA on n states. Is it possible to determine whether it accepts all strings in poly(n) time?
Suppose the NFA above has an equivalent DFA on d states. Is it possible to construct this DFA in poly(n,d) time?

REPLY [2 votes]: Your second question is a little ambiguous, because it admits the following cheating affirmative answer, which is probably not what you intend. 
Namely, every NFA with n states is equivalent to a DFA with d states for d sufficiently large. And for large $d$, allowing poly(n,d) steps is plenty of time. The well-known equivalent DFA, such as the one provided in Sipser's book, has $2^n$ states (plus a constant), and this example can be constructed in poly(n,2^n) steps, simply because $d=2^n$ is already so large here. More generally, for even larger $d$, we can simply pad this one with extra irrelevant states, and build it also in poly(n,d) steps. 
Perhaps you mean to ask about the optimal $d$? Or do you want to ask for all $d$ for which there is an equivalent DFA with $d$ states? Or do you also want to ask whether the optimal $d$ itself is poly(n)?