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In terms of worst-case asymptotic runtime, which NP-complete problem has the fastest-known (exact) algorithm and what is the algorithm? Is there something known that is faster than $O(n^2*2^n)$?

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  • $\begingroup$ What algorithm has running time $O(n^2 \cdot 2^n)$? EDIT: I assume you mean the Held–Karp algorithm for Traveling Salesman. $\endgroup$ – Guildenstern May 4 '14 at 21:29
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    $\begingroup$ You can take a look at the answers to the question Are there subexponential-time algorithms for NP-complete problems?. $\endgroup$ – Pål GD May 4 '14 at 22:05
  • $\begingroup$ "Faster than $O(\_)$" does not make sense. You mean $\Theta$? Or is the question, "Is there an algorithm with a better proven upper runtime bound than $O(\_)$?" $\endgroup$ – Raphael May 5 '14 at 6:53
  • $\begingroup$ The latter. It's valid point; there could be an algorithm A that's faster than B in practice but not with a tighter upper bound. I'm not sure why it doesn't make sense to say "faster than an upper bound" rather than "faster than a lower AND upper bound"... $\endgroup$ – Wuschelbeutel Kartoffelhuhn May 5 '14 at 13:19
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Vertex Cover has an algorithm running in time $1.2738^k + nk$, and is thus faster than $2^n n^2$, even with $k=n$. You can check out Table of FPT races for a short list of FPT running times of different problems. Here, $n$ is the number of vertices and $k$ is the solution size.

Also, the question Are there subexponential-time algorithms for NP-complete problems? addresses similar questions.

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  • $\begingroup$ The questions asks for fastest known algorithms and the table you link to does have "faster" algorithms than the VC one (in particular subexponential ones), so it's probably not the best to cite. $\endgroup$ – Raphael May 5 '14 at 15:18
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    $\begingroup$ See also this similar question and David Eppstein's answer Best-case Running-time to solve an NP-Complete problem on mathoverflow. $\endgroup$ – Pål GD May 5 '14 at 15:35
  • $\begingroup$ @Raphael Yes, for instance Minimum Fill-In has an algorithm which for every $\epsilon > 0$, runs in $O( (1+\epsilon)^k + \text{poly}(n))$ time. $\endgroup$ – Pål GD Apr 26 '15 at 3:52

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