It is very well known that if the classes $\sf FP$ and $\sf FNP$ are equal, then also the classes $\sf P$ and $\sf NP$ are equal (see e.g. FNP on Wikipedia).

Is it also true that if $\sf FNP=FEXPTIME$ then also $\sf NP=EXPTIME$? (See the exponential time conjecture.) I did find a paper constructing real functions that are in $\sf FEXPTIME$.

  • $\begingroup$ If it was trivial and direct, would you not already have a proof? $\endgroup$
    – Raphael
    Jul 1, 2014 at 12:10
  • $\begingroup$ ... er, FNP is a class of relations and FEXPTIME is a class of functions. $\;$ $\endgroup$
    – user12859
    Jul 1, 2014 at 12:23
  • $\begingroup$ @RickyDemer, aren't FP, FNP, FEXP and Fblah (for any blah) all defined formally as classes of binary relations? Moreover, a function is a binary relation anyway. $\endgroup$ Jul 1, 2014 at 13:06
  • $\begingroup$ If they're defined like that then $\:$FP = FNP$\;$. $\;\;\;$ (How else could FP be defined?) $\hspace{1.67 in}$ $\endgroup$
    – user12859
    Jul 1, 2014 at 13:20
  • $\begingroup$ @RickyDemer, FP is the class of binary relations Pxy for which y can be computed in deterministic polynomial time given x, or equivalently the subclass of FNP that can be computed in polynomial time. $\endgroup$ Jul 2, 2014 at 0:40

1 Answer 1


Yes, vacuously, since it is trivial that $\operatorname{FNP} \neq \operatorname{FEXPTIME}$, because it takes an exponentially long amount of time to give an exponentially long output.

  • $\begingroup$ Presumably FEXPTIME is defined so that $|y|$ is polynomial in $|x|$. $\endgroup$ Jul 2, 2014 at 7:16

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