<p>Since every planar graph on $n$ vertices has treewidth $O(\sqrt{n})$, all problems which are solvable in $O^*(2^{O(k)})$ time for graphs of treewidth at most~$k$ (there are a LOT of such problems) have subexponential-time algorithms on planar graphs by computing a constant-factor approximation to the treewidth in polynomial-time (for example by computing the branchwidth with the ratcatcher algorithm) and then running the treewidth algorithm, resulting in runtimes of the form $O^*(2^{O(\sqrt{n})})$ for graphs on $n$ vertices. Examples are Planar Independent Set and Planar Dominating Set, which are NP-complete of course.</p>

CC BY-SA 2.5
2010-12-07T12:32:06.167Z
2010-12-07T12:32:06.167Z
Bart Jansen
<p>There is a close connection between <strong>sub-exponential time solvability</strong> (SUBEPT) and <strong>fixed parameter tractability</strong> (FPT). The link between them is provided in the following  paper. </p>

<blockquote>
  <p><a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.92.531&amp;rep=rep1&amp;type=pdf">An isomorphism between subexponential and parameterized complexity theory</a>, Yijia Chen and Martin Grohe, 2006.</p>
</blockquote>

<p>In brief, they introduced a notion called <em>miniaturization mapping</em>, which maps a parameterized problem $(P,\nu)$ into another parameterized problem $(Q,\kappa)$. By viewing a normal problem as a problem parameterized by the input size, we have the following connection. (See theorem 16 in the paper)</p>

<blockquote>
  <p><strong>Theorem</strong>. $(P,\nu)$ is in SUBEPT iff $(Q,\kappa)$ is in FPT.</p>
</blockquote>

<p>Be careful of the definitions here. Normally we view $k$-clique problem as parameterized in $k$, so there is no sub-exponential time algorithm for it assuming Exponential time hypothesis. But here we let the problem be parameterized by the input size $O(m+n)$, thus the problem can be solved in $2^{O(\sqrt{m}\log m)}$, which is a sub-exponential time algorithm. And the theorem tells us that the $k$-clique problem is fixed parameter tractable under the some twist of the parameter $k$, which is reasonable.</p>

<p>In general, problems in SUBEPT under SERF-reductions (sub-exponential reduction families) can be transformed into problems in FPT under FPT-reductions. (Theorem 20 in the paper) Furthermore, the connections are even stronger since they provided an isomorphism theorem between a whole hierarchy of problems in exponential time complexity theory and parameterized complexity theory. (Theorem 25 and 47) Though the isomorphism is not complete (there are some missing links between them), it is still nice to have a clear picture about these problems, and we can study sub-exponential time algorithms via parameterized complexity.</p>

<p>See the <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.142.1009&amp;rep=rep1&amp;type=pdf">survey</a> by Jörg Flum and Martin Grohe, together with Jacobo Torán, the editor of the complexity column, for more information.</p>

CC BY-SA 2.5
2010-12-08T17:05:37.787Z
Yes.  btw, Flum and Grohe wrote the survey; Toran is the Complexity Column editor.
Andy Drucker
CC BY-SA 2.5
2010-12-09T01:00:20.737Z
@Andy: Thank you for the correction. I'll modified the article accordingly.
Hsien-Chih Chang 張顯之
CC BY-SA 2.5
2010-12-08T04:51:51.210Z
2010-12-09T01:02:34.603Z
2010-12-09T01:02:34.603Z
Hsien-Chih Chang 張顯之
<p>another example can be Cop and Robber game, which is NP-hard but solvable in time $2^{o(n)}$ on graphs with n vertices.
BibTeX bibliographical record in XML Fedor V. Fomin, Petr A. Golovach, Jan Kratochvíl, Nicolas Nisse, Karol Suchan: Pursuing a fast robber on a graph. Theor. Comput. Sci. 411(7-9): 1167-1181 (2010)</p>

CC BY-SA 2.5
2010-12-07T13:24:47.530Z
Oops, this may be shameful, but I had a long time believing $\mathsf{NP}$-hard problems do not have sub-exponential time algorithms, just because the Exponential Time Hypothesis. :(
Hsien-Chih Chang 張顯之
CC BY-SA 2.5
2010-12-08T17:03:06.900Z
No shame... but, one easy way to see this is not true is to take any $NP$-hard language $L \in NPTIME(n^k)$, and then form a 'padded' version $L'$ in which the 'yes' instances are of form $(x, 1^{|x|^c})$, with $x \in L$, for some fixed $c > k$.  Then $L'$ is $NP$, but has a deterministic algorithm running in time essentially $2^{n^{k/c}}$.
Andy Drucker
CC BY-SA 2.5
2010-12-07T13:08:48.100Z
2010-12-07T13:08:48.100Z
XXYYXX
<p>The best approximation algorithm for clique gives an unbelievably bad approximation factor $n/\text{polylog } n$ (recall that approximation factor of $n$ is trivial).</p>

<p>There are hardness of approximation results under various hardness assumptions that don't quite match this, but still give hardness of $n^{1-o(1)}$. Personally, I believe that $n/\text{polylog } n$ approximation for clique is as good as polynomial-time algorithms would ever do. </p>

<p>But approximation of $n/\text{polylog } n$ for clique can easily be done in quasi-polynomial time.</p>

<hr>

<p>An NP-hard problem is a problem that has a polynomial-time reduction from SAT.
Even if SAT needs time $2^{\Omega(n)}$, this may translate to time $2^{\Omega(N^\epsilon)}$ for the problem we reduce to. If the latter has input size N, it may be the case that $N=n^{1/\epsilon}$ for a small constant $\epsilon$.</p>

CC BY-SA 2.5
2010-12-07T20:35:42.520Z
2010-12-07T20:43:16.347Z
2010-12-07T20:43:16.347Z
Dana Moshkovitz
<p>By the way the Max Clique problem, in full generality, can be solved in time $2^{\tilde O(\sqrt N)}$ where $N$ is the size of the input.</p>

<p>This is trivial if the graph is represented via an adjacency matrix, because then $N=|V|^2$, and a brute force search will take time $2^{O(|V|)}$.</p>

<p>But we can get the same bound even if the graph is represented by adjacency lists, via an algorithm of running time $2^{\tilde O(\sqrt{|V| + |E|})}$. To see how, let's get a $2^{\tilde O(\sqrt{|V| + |E|})}$-time algorithm for the NP-complete decision problem in which we are given a graph $G=(V,E)$ and $k$ and we want to know if there is a clique of size $\geq k$.</p>

<p>The algorithm simply removes all vertices of degree $&lt; k$ and the edges incident on them, then does it again, and so on, until we are left with a vertex-induced subgraph over a subset $V&#39;$ of vertices, each of degree $\geq k$, or with an empty graph. In the latter case, we know that no clique of size $\geq k$ can exist. In the former case, we do a brute-force search running in time roughly $|V&#39;|^k$. Note that  $|E| \geq k\cdot |V&#39;| /2$ and $k\leq |V&#39;|$, so that that $|E| \geq k^2/2$, and so a brute-force search running in time $|V&#39;|^k$ is actually running in time $2^{O(\sqrt{|E|} \cdot \log |V|)}$.</p>

CC BY-SA 2.5
2010-12-13T03:02:33.897Z
Indeed, for these kinds of reasons Impagliazzo, Paturi and Zane argued that when asking about $2^{\Omega(n)}$ vs $2^{o(n)}$ complexity you need to set $n$ to be the size of the witness (which you need to define as part of the problem). In the $k$-clique case the witness is of size $\log \binom{|V|}{k} \sim k\log |V|$ for small $k$, while, as you say, you can assume w.l.o.g there are at least $k|V|$ edges and the input size is much larger than the witness size.
Boaz Barak
CC BY-SA 2.5
2010-12-08T16:24:42.947Z
2010-12-08T16:24:42.947Z
Luca Trevisan
<p>In light of the recent result of Arora, Barak, and Steurer, <a href="http://www.cs.princeton.edu/~dsteurer/subexpug.pdf">Subexponential Algorithms for Unique Games and Related Problems</a>, I'm interested in graph problems that have subexponential time algorithms but believed not to be polynomially solvable. A famous example is graph isomorphism which has subexponential algorithm of $2^{O(n^{1/2} \log n)}$ run-time. Another example is log-Clique problem which is solvable in quasi-polynomial time ($n^{O(\log n)}$).</p>

<blockquote>
  <p>I'm looking for interesting examples and preferably a reference to surveys of subexponential hard graph problems (not necessarily $NP$-complete). Also, Are there any $NP$-complete graph problems with subexponential time algorithms?</p>
</blockquote>

<p><a href="http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=743516">Impagliazzo, Paturi and Zane</a> showed that the Exponential Time Hypothesis implies that Clique, k-Colorability, and Vertex Cover need $2^{\Omega(n)}$ time.</p>

CC BY-SA 2.5
2010-12-08T00:32:51.180Z
Just for completeness: **log-CLIQUE** = $\{(G, k) | G \text{ has }n\text{ vertices, }k = \log n\text{ and }G\text{ has a clique of size }k\}$
Sadeq Dousti
<cc.complexity-theory><reference-request>
Subexponentially solvable hard graph problems
Mohammad Al-Turkistany
# Subexponentially solvable hard graph problems

(Asked by: username_0 on 2010-12-07)

In light of the recent result of Arora, Barak, and Steurer, [Subexponential Algorithms for Unique Games and Related Problems](http://www.cs.princeton.edu/~dsteurer/subexpug.pdf), I'm interested in graph problems that have subexponential time algorithms but believed not to be polynomially solvable. A famous example is graph isomorphism which has subexponential algorithm of $2^{O(n^{1/2} \log n)}$ run-time. Another example is log-Clique problem which is solvable in quasi-polynomial time ($n^{O(\log n)}$).

> I'm looking for interesting examples and preferably a reference to surveys of subexponential hard graph problems (not necessarily $NP$-complete). Also, Are there any $NP$-complete graph problems with subexponential time algorithms?

[Impagliazzo, Paturi and Zane](http://ieeexplore.ieee.org/xpl/freeabs_all.jsp?arnumber=743516) showed that the Exponential Time Hypothesis implies that Clique, k-Colorability, and Vertex Cover need $2^{\Omega(n)}$ time.

**Tags:** cc.complexity-theory, reference-request
> **Comments:**
> **username_1**: Just for completeness: **log-CLIQUE** = $\{(G, k) | G \text{ has }n\text{ vertices, }k = \log n\text{ and }G\text{ has a clique of size }k\}$  

---

## 5 answers

### Answer by username_7 on 2010-12-08. Score: 20 | Accepted answer

By the way the Max Clique problem, in full generality, can be solved in time $2^{\tilde O(\sqrt N)}$ where $N$ is the size of the input.

This is trivial if the graph is represented via an adjacency matrix, because then $N=|V|^2$, and a brute force search will take time $2^{O(|V|)}$.

But we can get the same bound even if the graph is represented by adjacency lists, via an algorithm of running time $2^{\tilde O(\sqrt{|V| + |E|})}$. To see how, let's get a $2^{\tilde O(\sqrt{|V| + |E|})}$-time algorithm for the NP-complete decision problem in which we are given a graph $G=(V,E)$ and $k$ and we want to know if there is a clique of size $\geq k$.

The algorithm simply removes all vertices of degree $< k$ and the edges incident on them, then does it again, and so on, until we are left with a vertex-induced subgraph over a subset $V'$ of vertices, each of degree $\geq k$, or with an empty graph. In the latter case, we know that no clique of size $\geq k$ can exist. In the former case, we do a brute-force search running in time roughly $|V'|^k$. Note that $|E| \geq k\cdot |V'| /2$ and $k\leq |V'|$, so that that $|E| \geq k^2/2$, and so a brute-force search running in time $|V'|^k$ is actually running in time $2^{O(\sqrt{|E|} \cdot \log |V|)}$.

> **Comments:**
> **username_8**: Indeed, for these kinds of reasons Impagliazzo, Paturi and Zane argued that when asking about $2^{\Omega(n)}$ vs $2^{o(n)}$ complexity you need to set $n$ to be the size of the witness (which you need to define as part of the problem). In the $k$-clique case the witness is of size $\log \binom{|V|}{k} \sim k\log |V|$ for small $k$, while, as you say, you can assume w.l.o.g there are at least $k|V|$ edges and the input size is much larger than the witness size.  

---

### Answer by username_2 on 2010-12-07. Score: 22

Since every planar graph on $n$ vertices has treewidth $O(\sqrt{n})$, all problems which are solvable in $O^\*(2^{O(k)})$ time for graphs of treewidth at most~$k$ (there are a LOT of such problems) have subexponential-time algorithms on planar graphs by computing a constant-factor approximation to the treewidth in polynomial-time (for example by computing the branchwidth with the ratcatcher algorithm) and then running the treewidth algorithm, resulting in runtimes of the form $O^\*(2^{O(\sqrt{n})})$ for graphs on $n$ vertices. Examples are Planar Independent Set and Planar Dominating Set, which are NP-complete of course.


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### Answer by username_3 on 2010-12-08. Score: 15

There is a close connection between **sub-exponential time solvability** (SUBEPT) and **fixed parameter tractability** (FPT). The link between them is provided in the following paper.

> [An isomorphism between subexponential and parameterized complexity theory](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.92.531&rep=rep1&type=pdf), Yijia Chen and Martin Grohe, 2006.

In brief, they introduced a notion called *miniaturization mapping*, which maps a parameterized problem $(P,\nu)$ into another parameterized problem $(Q,\kappa)$. By viewing a normal problem as a problem parameterized by the input size, we have the following connection. (See theorem 16 in the paper)

> **Theorem**. $(P,\nu)$ is in SUBEPT iff $(Q,\kappa)$ is in FPT.

Be careful of the definitions here. Normally we view $k$-clique problem as parameterized in $k$, so there is no sub-exponential time algorithm for it assuming Exponential time hypothesis. But here we let the problem be parameterized by the input size $O(m+n)$, thus the problem can be solved in $2^{O(\sqrt{m}\log m)}$, which is a sub-exponential time algorithm. And the theorem tells us that the $k$-clique problem is fixed parameter tractable under the some twist of the parameter $k$, which is reasonable.

In general, problems in SUBEPT under SERF-reductions (sub-exponential reduction families) can be transformed into problems in FPT under FPT-reductions. (Theorem 20 in the paper) Furthermore, the connections are even stronger since they provided an isomorphism theorem between a whole hierarchy of problems in exponential time complexity theory and parameterized complexity theory. (Theorem 25 and 47) Though the isomorphism is not complete (there are some missing links between them), it is still nice to have a clear picture about these problems, and we can study sub-exponential time algorithms via parameterized complexity.

See the [survey](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.142.1009&rep=rep1&type=pdf) by Jörg Flum and Martin Grohe, together with Jacobo Torán, the editor of the complexity column, for more information.

> **Comments:**
> **username_4**: Yes.  btw, Flum and Grohe wrote the survey; Toran is the Complexity Column editor.  
> **username_3**: @Andy: Thank you for the correction. I'll modified the article accordingly.  

---

### Answer by username_5 on 2010-12-07. Score: 12

another example can be Cop and Robber game, which is NP-hard but solvable in time $2^{o(n)}$ on graphs with n vertices.
BibTeX bibliographical record in XML Fedor V. Fomin, Petr A. Golovach, Jan Kratochvíl, Nicolas Nisse, Karol Suchan: Pursuing a fast robber on a graph. Theor. Comput. Sci. 411(7-9): 1167-1181 (2010)

> **Comments:**
> **username_3**: Oops, this may be shameful, but I had a long time believing $\mathsf{NP}$-hard problems do not have sub-exponential time algorithms, just because the Exponential Time Hypothesis. :(  
> **username_4**: No shame... but, one easy way to see this is not true is to take any $NP$-hard language $L \in NPTIME(n^k)$, and then form a 'padded' version $L'$ in which the 'yes' instances are of form $(x, 1^{|x|^c})$, with $x \in L$, for some fixed $c > k$.  Then $L'$ is $NP$, but has a deterministic algorithm running in time essentially $2^{n^{k/c}}$.  

---

### Answer by username_6 on 2010-12-07. Score: 8

The best approximation algorithm for clique gives an unbelievably bad approximation factor $n/\text{polylog } n$ (recall that approximation factor of $n$ is trivial).

There are hardness of approximation results under various hardness assumptions that don't quite match this, but still give hardness of $n^{1-o(1)}$. Personally, I believe that $n/\text{polylog } n$ approximation for clique is as good as polynomial-time algorithms would ever do.

But approximation of $n/\text{polylog } n$ for clique can easily be done in quasi-polynomial time.

---

An NP-hard problem is a problem that has a polynomial-time reduction from SAT.
Even if SAT needs time $2^{\Omega(n)}$, this may translate to time $2^{\Omega(N^\epsilon)}$ for the problem we reduce to. If the latter has input size N, it may be the case that $N=n^{1/\epsilon}$ for a small constant $\epsilon$.


---

