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What are examples of problems that are known to be sub-exponential, but

  1. are known to be non-polynomial, or
  2. are not known to be polynomial?

EDIT. Here is what I mean by sub-exponential (apologies for leaving this open). A function $f:\mathbb{N}\to\mathbb{N}$ is subexponential if $$f(n) = 2^{n^{o(1)}}.$$ (Thanks to Emil Jerabek for his help in the comment section below!)

Dominic van der Zypen
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    It's good to clarify what you mean by sub-exponential. See also https://cs.stackexchange.com/questions/9813/are-there-subexponential-time-algorithms-for-np-complete-problems – usul Jun 21 '20 at 04:20
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    If the author means subexponential as in III. from the resource in the above comment, then I have a feeling some studied Exponential-time Complete problems would also meet the criteria. – Michael Wehar Jun 21 '20 at 04:23
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    Yes, this is very nonstandard. This makes $2^n$ subexponential. – Emil Jeřábek Jun 23 '20 at 12:08
  • @DominicvanderZypen $\frac1me^n=\Omega(2^n)$. – VS. Jun 23 '20 at 12:46
  • @EmilJeřábek Sorry, right, tried to amend it – Dominic van der Zypen Jun 24 '20 at 06:44
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    The new definition is hard to decipher, but is actually equivalent to just $f(n)=2^{o(n)}$. Why don’t you write it that way? (And while this is not nonstandard, in complexity theory this definition is less common and less convenient than the more natural definition $f(n)=2^{n^{o(1)}}$, which is robust wrt polynomial changes in the representation of the input or in the parameter to specify the problem. E.g., according to your definition, brute force $2^{O(|V|)}$ algorithms for the maximum independent set and similar graph problems are “subexponential” as the size of the input is $|V|^2$.) – Emil Jeřábek Jun 24 '20 at 06:59
  • OK - having digested the definition of $o(\cdot)$ I see your point and will make the post more readable. – Dominic van der Zypen Jun 24 '20 at 07:01

7 Answers7

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Graph Isomorphism is known to be in quasipolynomial time but not known to be in polynomial time: https://arxiv.org/abs/1512.03547

You can also come up with synthetic problems by the time hierarchy theorem, which essentially states that there are problems at any (deterministic) time complexity but not below: https://en.wikipedia.org/wiki/Time_hierarchy_theorem

Mahdi Cheraghchi
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16

VC-dimension of set systems is known to be quasipolynomial but not known to be polynomial, with some evidence (based on the strong exponential time hypothesis) that it is not polynomial.

See:

Linial, Nathan, Mansour, Yishay, and Rivest, Ronald L. (1991), “Results on learnability and the Vapnik–Chervonenkis dimension”, Inf. Comput. 90 (1): 33–49.

Manurangsi, Pasin, and Rubinstein, Aviad (2017), “Inapproximability of VC dimension and Littlestone’s dimension”, Proc. 2017 Conf. Learning Theory (COLT 2017), Proceedings of Machine Learning Research 65, pp. 1432–1460.

Papadimitriou, Christos H., and Yannakakis, Mihalis (1996), “On limited nondeterminism and the complexity of the V–C dimension”, J. Comput. Syst. Sci. 53 (2): 161–170.

David Eppstein
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Minimum dominating set in tournaments problem has quasi-polynomial time algorithm (hence sub-exponential). There is no known polynomial time algorithm for it. Tournament dominating set has polynomial-time algorithm if and only if SAT has sub-exponential time algorithm. Therefore, the exponential time hypothesis (ETH) implies that tournament dominating set can not have polynomial time algorithm. Furthermore, ETH implies that tournament dominating set is in $NP$-intermediate.

Mohammad Al-Turkistany
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Parity games have recently been shown to be solvable in quasipolynomial time: https://doi.org/10.1137/17M1145288

Hermann Gruber
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Factoring integers

The General Number Field Sieve is known to be sub-exponential in the size of the input, for example see this quote from the link I just gave:

The running time of the number field sieve is super-polynomial but sub-exponential in the size of the input.

However no polynomial-time algorithm for factoring integers is known (yet).

user1271772
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    The running time of the algorithm is (up to less significant factors) $2^{n^\epsilon}$ for some constant $\epsilon>0$. More often than not, bounds of this form are considered exponential in complexity theory. See also https://cs.stackexchange.com/a/9814. – Emil Jeřábek Jun 23 '20 at 06:46
1

Discrete log in multiplicative groups of finite fields of small characteristic has quasi-polynomial running time. Specifically, for $\mathbb{F}_{p^n}^\times$ (where inputs have size $\log_2(p^n) = n\log_2(p)$) the (expected) running time is $(pn)^{2\log_2(n) + O(1)} = 2^{2\log_2(p)\log_2(n) + O(1) + 2 \log_2(n)^2 + O(\log_2(n))} = 2^{O(\log_2(n)^2)}$

Mark Schultz-Wu
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  1. Problems in $NTIME[f(n)]$ where $f(n)=\omega(\log n)$ and $f(n)=o(n)$ are not known to be in $DTIME[n^c]$ at any $c\geq1$ while known to be in $DTIME[2^{o(n)}]$.

  2. Conversely there are problems not known to be in $NTIME[f(n)]$ where $f(n)=\omega(\log n)$ and $f(n)=o(n)$ and also not known to be in $DTIME[n^c]$ at any $c\geq1$ while known to be in $DTIME[2^{o(n)}]$.

Examples of $2$ include factoring and discrete logarithm.

VS.
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  • For item 1, if $f(n)=o(n)$, then it seems like the machine does not have enough time to read all of its input bits. Perhaps I am missing something? Maybe the model is something other than a standard Turing Machine, e.g. a word RAM Turing Machine? – Lieuwe Vinkhuijzen Jun 23 '20 at 20:16
  • In computational complexity theory, the complexity class $NTIME(f(n))$ is the set of decision problems that can be solved by a non-deterministic Turing machine which runs in time $O(f(n))$. Nothing more. – VS. Jun 30 '20 at 01:21