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Do we know any problem in NP which has a super-linear time complexity lower bound? Ideally, we would like to show that 3SAT has super-polynomial lower bounds, but I guess we're far away from that. I'd just like to know any examples of super-linear lower bounds.

I know that the time hierarchy theorem gives us problems which can be solved in O(n^3) but not in O(n^2), etc. Thus I put the word "natural" in the question.

I ask for problems in NP, because otherwise someone would give examples of EXP-complete problems.

I know there are time-space tradeoffs for some problems in NP. I don't know if any of them imply a super-linear time complexity lower bound though.

(To address a question below about machine models, consider either multitape Turing machines or the RAM model.)

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4 Answers 4

up vote 20 down vote accepted

Sorry I am so late to the discussion, but I just registered...

There are non-linear time lower bounds on multitape Turing machines for NP-complete problems. These lower bounds follow from the fact that the class of problems solvable in nondeterministic linear time is not equal to the class of those solvable in (deterministic) linear time, in the multitape Turing machine setting. This is proved in:

Wolfgang J. Paul, Nicholas Pippenger, Endre Szemerédi, William T. Trotter: On Determinism versus Non-Determinism and Related Problems (Preliminary Version) FOCS 1983: 429-438

In fact, unraveling the proof shows that there must be some problem solvable in nondeterministic linear time that is not solvable in $o(n \cdot (\log^* n)^{1/4})$ time (again, on a multitape Turing machine). Note the * in the logarithm; this is just "barely" above linear. One known application of this result is that a natural NP-complete problem in automata theory cannot be solved in $o(n \cdot (\log^* n)^{1/4})$ time:

Etienne Grandjean: A Nontrivial Lower Bound for an NP Problem on Automata. SIAM J. Comput. 19(3): 438-451 (1990)

Unfortunately the lower bound of Paul et al. relies crucially on the geometry that arises from accessing one-dimensional tapes. We don't know how to prove a non-linear lower bound even if you allow the Turing machine to have a constant number of two-dimensional tapes. We can prove time lower bounds for NP problems on general computational models if you severely restrict the extra workspace used by the machine. (This is getting into my own work so I won't say more unless you're truly interested.)

As for the comment above me: the sorting lower bound holds only in a comparison-based model, which is extremely restricted. The claim that sorting requires Omega(n log n) time on general computational models is false. There are faster algorithms for sorting integers. See for example:

Yijie Han: Deterministic sorting in O(n log logn) time and linear space. J. Algorithms 50(1): 96-105 (2004)

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That's very interesting. Thanks. I keep hearing these lower bounds for problems in NP which establish bounds like 4n or 5n, perhaps that's a different model then. I guess the model you mention is weaker than the random-access Turing machine model, right? When you speak of restricting extra work space, is this related to the time-space trade-off results, like the paper by Fortnow? –  Rune Dec 18 '09 at 4:34
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The bounds of the form 4n and 5n are for Boolean circuits with AND, OR, and NOT gates. Those lower bounds are much harder to come by. One big reason is that the Boolean circuit model is <i>non-uniform</u>, meaning that for each input length n, one has a different "program" which works for all strings of length n. Such a model can compute some non-recursive functions! The multi-tape Turing machine model certainly seems weaker than the random-access Turing machine, but this isn't known for sure. My other comment was indeed referring to the work on time-space tradeoffs. –  Ryan Williams Dec 18 '09 at 9:07
    
I understand, thanks! –  Rune Dec 18 '09 at 20:23
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If you restrict space usage to be sublinear then you can prove superlinear time lower bounds on SAT. See this article on Lipton's blog for a nice exposition.

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The short answer is NO. The best general lower bounds are linear.

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Could you flesh this out? Or give a link to a reference? Why is finding super-linear lower bounds hard? If it is obvious, I apologize in advance. –  Sam Nead Nov 14 '09 at 22:23
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Here is a citation and a link : Uri Zwick, A 4n lower bound on the combinatorial complexity of certain symmetric Boolean functions over the basis of unate dyadic Boolean functions SIAM Journal on Computing 20, 499-505 (1991) An explicit lower bound of 5n-o (n) for boolean circuits, K Iwama, O. Lachish, H Morizumi, and R. Raz springerlink.com/index/XCP1TCRY1C236RDT.pdf The introduction of the second paper and the slow incremental improvements may give some idea on the difficulty of the problem. –  Gil Kalai Nov 21 '09 at 14:20
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Running time depends on the model of computation. Eg the problem "Given (m, n, p) does m + n = p?" takes at least quadratic time on a one-tape Turing machine. So you can give a more reasonable model of computation... and the next person to come along will explain how it is much too strong, etc.

Hmmm. Perhaps your question might be rephrased in terms of using some other resource (which in turn requires at least a set amount of time). So perhaps invent a decision problem around sorting, and appeal to the nlog(n) lower bound...

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I wouldn't mind multitape Turing machine, or even the RAM model. The sorting lower bound isn't really a time complexity lower bound. It's a decision tree complexity lower bound. –  Rune Nov 11 '09 at 1:00
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