I'm following an online course which has the following (multiple-choice) quiz question:

Which of the following statements cannot be true, given the current state of knowledge?

  1. Some NP-complete problems are polynomial-time solvable, and some NP-complete problems are not polynomial-time solvable.
  2. There is an NP-complete problem that is polynomial-time solvable.
  3. There is an NP-complete problem that can be solved in $ O(n^{\log n}) $ time, where $ n $ is the size of the input.
  4. There is no NP-complete problem that can be solved in $ O(n^{\log n}) $ time, where $ n $ is the size of the input.

It seems to me that (1) and (2) are false since it has not been proven that any NP-complete problem is polynomial-time solvable. So I'm vacillating between (3) and (4).

In particular, I'm a bit nonplussed where the $O(n^{\log n})$ comes from; I haven't seen this expression in any of the lecture notes. However, I do know about the traveling salesman problem which can be solved in $O(n^2 2^n)$ time using the Held-Karp algorithm. It would seem to me like this function grows less faster asymptotically than $O(n^{\log n})$, since it is usually the expression in the exponent that matters the most, in which case the answer would be (3).

Is this correct? Also, how would I show this formally?

  • 3
    $\begingroup$ I think you need to read the question again. It asks what cannot be true. $\endgroup$ – D.W. Sep 21 '17 at 17:31
  • 5
    $\begingroup$ I had missed that indeed. It turns out that neither (3) or (4) are the correct answer; I believe that (1) is the correct answer (though I haven't yet verified this) because by definition if one NP-complete problem is solvable in polynomial time, then all of them are. $\endgroup$ – Kurt Peek Sep 22 '17 at 11:00

The Exponential Time Hypothesis (ETH) states that SAT cannot be solved in time $2^{o(n)}$. Now suppose that $A$ is some NP-complete problem which can be solved in time $O(n^{\log n})$. Since $A$ is NP-hard, there is a polytime reduction from SAT to $A$, say running in time $O(n^k)$. Using this reduction, we can solve SAT in time $O(n^{k^2 \log n})$, contradicting ETH.

Since ETH is still standing, it is safe to assume that no $O(n^{\log n})$ algorithm is known for any NP-complete problem. We don't know for sure that (3) is true, i.e. we don't know for sure that (4) is false.

Conversely, we can't prove that no $O(n^{\log n})$ algorithm is known for any NP-complete problem. So we don't know for sure that (3) is false either.

  • 1
    $\begingroup$ Most of the time, the $n$ in ETH stands for the number of variables, while the $n$ in the running time of an np hardness reduction stands for the total input size, right? Can you clarify? $\endgroup$ – user53923 Sep 21 '17 at 22:33
  • $\begingroup$ Yes, you are right. But that shouldn't make a big difference here (exercise). $\endgroup$ – Yuval Filmus Sep 22 '17 at 6:47
  • $\begingroup$ I agree that it should not make a big difference $\endgroup$ – user53923 Sep 22 '17 at 7:47
  • 2
    $\begingroup$ OK, but the question asks which of the claims we know to be false. $\endgroup$ – David Richerby Jan 7 '19 at 14:13
  • $\begingroup$ This answer would be better if you also explained why we can't rule out the existence of an $O(n^{\log n})$ algorithm for an NP-complete problem. $\endgroup$ – Gilles 'SO- stop being evil' Jan 7 '19 at 22:23

The accepted answer is incorrect. I've taken the same course, and option 1 is correct. They didn't say why, but I have the following argument.

Quoting Wikipedia:

polynomial time: An algorithm is said to be of polynomial time if its running time is upper bounded by a polynomial expression in the size of the input for the algorithm, i.e., $ T(n) = O(n^k) $ for some positive constant $ k $.

exponential time: An algorithm is said to be exponential time, if $ T(n) $ is upper bounded by $ 2^{poly(n)} $, where $ poly(n) $ is some polynomial in $ n $. More formally, an algorithm is exponential time if $ T(n) $ is bounded by $ O(2^{n^k}) $ for some constant $ k $.

We don't know $ P \neq NP $, so option 2 can't be ruled out.

For all we know, the running time required to solve NP-complete problems could be anywhere between polynomial and exponential (note that $ n^{\log n} $ is more than polynomial but less than exponential). Thus, options 3 and 4 can't be ruled out.

That only leaves option 1. The claim in option 1 is incorrect because if there's a polynomial-time algorithm for any NP-complete problem X, then every other NP problem has a polynomial-time algorithm by reduction to X.

  • $\begingroup$ Yuval's answer is correct. It just focuses on the question, it doesn't solve the whole homework exercise. $\endgroup$ – Gilles 'SO- stop being evil' Jan 7 '19 at 22:23
  • $\begingroup$ @Gilles, It wasn't correct until your edit. The original claim was misleading and led to believe that option 4 was the correct answer. Your comment misrepresents what was; either your edit should stand, or your comment, because making an edit and commenting on what doesn't exist anymore makes no sense. $\endgroup$ – Abhijit Sarkar Jan 7 '19 at 23:52

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