What's the worst-case asymptotic-complexity of the Set-cover problem in Big O notation?

I've been developing some novel techniques to try and solve this problem but am having trouble finding the theoretical limits I need to surpass.

Thanks for all information


1 Answer 1


I don't see how you can say that SCP or any other problem has "worst-case asymptotic complexity", but surely an algorithm solving it can have. However, the problem is NP-complete, and the optimization version is NP-hard.

For an exact algorithm, this is clearly a "theoretical limit". For an exact algorithm, see for example E. Balas and M. C. Carrera, A Dynamic Subgradient-Based Branch-and-Bound Procedure for Set Covering, Operations Research 44, 875-890, 1996. According to A. Caprara et al., Algorithms for the Set Covering Problem, 2000, this algorithm of Balas and Carrera is the fastest known exact algorithm. The result is a bit old already, so I'm not sure if there are newer and faster exact algorithms.

There is a simple greedy polynomial time approximation algorithm with an approximation ratio of $H_n$, where $H_n$ is the $n$th harmonic number. Under reasonable assumptions, this is the best one can do, since if there exists a $c \ln n$-approximation algorithm for the (unweighted) set cover problem for some constant $c < 1$, then there is an $O(n^{O(\log \log n)})$-time deterministic algorithm for each NP-complete problem. I think the Wikipedia page gives the essential references.

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    $\begingroup$ Actually Feige shows if it's possible to approximate Set Cover better than $(1-o(1))\ln n$ then there is an $n^{O(\log\log n)}$ algorithm for NP. Other people have subsequently shown (as byproducts of improved PCPs) that for some $c > 0$, no $c\ln n$ approximation is possible unless P=NP. $\endgroup$ Commented May 7, 2012 at 17:58

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