We know under $ETH$ we cannot solve $K$-SUM in $f(K)poly(nK)$ time under any function $f(K)$ (usually $2^{O(K)}$).

Is there any conjecture that prevents a $(\log n)^{O(K)}$ complexity (this is entirely consistent with possibility as $K=\Omega(n)$ we need exponential time for subset sum) or is such possibility allowed?


1 Answer 1


ETH itself precludes this possibility.

In https://people.csail.mit.edu/rrw/cnf-sat-feasible.pdf we show that any $n^{O(1)} n^{k/\alpha(k)}$ time algorithm for k-SUM, for any monotone nondecreasing unbounded function $\alpha$, would imply ETH is false.

  • 3
    $\begingroup$ Do you mean that $\alpha$ is strictly increasing, or at least goes to infinity? $\endgroup$ Commented Apr 30, 2019 at 5:46
  • $\begingroup$ @RyanWilliams Similar in spirit to ETH like obstruction. Is there anything that would prevent $O((\log n)^{O(k)})$ complexity with polynomial size advice or a PPAD oracle? $\endgroup$
    – Turbo
    Commented May 1, 2019 at 3:29
  • $\begingroup$ Added "unbounded" :) $\endgroup$ Commented May 11, 2019 at 5:15
  • $\begingroup$ @Brout Note that (log(n))^k is an FPT function, so yeah, ETH rules that out. With poly size advice it would mean subexponential size circuits for 3sat. With a PPAD oracle it would seem to imply that ETH implies PPAD not in P. To me that would be a breakthrough, I don't know of much corroborating evidence that PPAD isn't in P $\endgroup$ Commented May 11, 2019 at 5:19

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