Consider the problem of computing an exponential sum over a certain function $g(x)=f(x)+h(x)$, that is computing


now if we know that $\sum_{x}f(x)$ and $\sum_{x}h(x)$ are two NP-Hard problems, what can we say about the hardness of $\sum_{x}g(x)$?


1 Answer 1



Lower bound: Suppose $h(x) = -f(x)$. Then $\sum_x g(x) = 0$, which is trivial to compute. If $\sum_x f(x)$ is NP-hard to compute, then $\sum_x h(x)$ will be too, but $\sum_x g(x)$ will be easy to compute.

Upper bound: Suppose $h(x) = f(x)$. Then $\sum_x g(x) = 2 \sum_x f(x)$, which is as hard as computing $\sum_x f(x)$, which is assumed to be NP-hard. Since NP-hard means (roughly speaking) NP-complete or harder, there is no upper bound on the complexity; it could be arbitrarily hard to compute $\sum_x g(x)$.

  • $\begingroup$ It might also be that the sum is not NP-hard. $\endgroup$
    – kutschkem
    Jan 25, 2021 at 15:15
  • 2
    $\begingroup$ @kutschkem: If you can make a nontrivial example I'd like to see it. $\endgroup$
    – Joshua
    Jan 25, 2021 at 16:27
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    $\begingroup$ @Joshua Graph theory is likely full of them - computing the number of graphs with n vertices and a certain property, then computing the number of graphs with n vertices without a certain property. Computing the number of graphs with n vertices is comparatively easy. If that's not non-trivial enough for you, I'm pretty sure computing which graphs have which chromatic number is NP-hard when the chromatic number is >3, but not for <3. So you could do f=>3, h=>4+, and g=>3+. $\endgroup$ Jan 25, 2021 at 19:02
  • $\begingroup$ @Spitemaster Thanks, yes that's the kind of problems I had in mind. $\endgroup$
    – kutschkem
    Jan 26, 2021 at 7:11

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