Slightly Superexponential Parameterized Problems

TL;DR

This paper investigates parameterized problems with superexponential running times and proves that for some, the dependence on the parameter cannot be improved to single exponential, highlighting inherent computational complexity limits.

Contribution

The paper demonstrates that for certain parameterized problems with $f(k)=k^{O(k)}$, the superexponential dependence on $k$ is optimal and cannot be reduced to single exponential.

Findings

Proves superexponential dependence is optimal for some problems.

Shows that $f(k)=k^{O(k)}$ cannot be improved to $c^k$ for these problems.

Highlights inherent complexity barriers in parameterized algorithms.

Abstract

A central problem in parameterized algorithms is to obtain algorithms with running time $f(k)\cdot n^{O(1)}$ such that $f$ is as slow growing function of the parameter $k$ as possible. In particular, a large number of basic parameterized problems admit parameterized algorithms where $f(k)$ is single-exponential, that is, $c^k$ for some constant $c$, which makes aiming for such a running time a natural goal for other problems as well. However there are still plenty of problems where the $f(k)$ appearing in the best known running time is worse than single-exponential and it remained ``slightly superexponential'' even after serious attempts to bring it down. A natural question to ask is whether the $f(k)$ appearing in the running time of the best-known algorithms is optimal for any of these problems. In this paper, we examine parameterized problems where $f(k)$ is $k^{O(k)}=2^{O(k\log k)}$ in the best known running time and for a number of such problems, we show that the dependence on $k$ in the running time cannot be improved to single exponential. (See paper for the longer abstract.)

Equations

$$2^k{4}^k\prod _{i=1}^{\ell }(11^{k_i}\log k)\le 2^{O(k)}(\log k{)}^{\ell }\le 2^{O(k\log \log k)}.$$

$$(S,(k_1,\dots ,k_{\ell }),\{((b_1^i,\dots ,b_q^i),(L_1^i,\dots ,L_q^i))|1\le i\le \lceil \log k\rceil \},f_1,\dots ,f_{\ell })$$

$$2^k{4}^k\prod _{i=1}^{\lceil \log k\rceil }(p{)}^{\log k}\prod _{i=1}^{\lceil \log k\rceil }{4}^{{\sum }_{j=1}^{\lceil \log k\rceil }|L_j^i|}\prod _{i=1}^{\ell }(11^{k_i}\log k_i)\le 8^k(2k{)}^{\log k}{4}^{k}12^k\le 2^{O(k+(\log k{)}^3)}\le 2^{O(k)}.$$

$$S_e=\{(i_1,j^{\prime })\mid 1\le j^{\prime }\le k,j^{\prime }\ne j_1\}\cup \{(i_2,j^{\prime })\mid k+1\le j^{\prime }\le 2k,j^{\prime }\ne j_2\}.$$

$$r_1^1,{\bar{r}}_1^1,c_{\delta (1)}^1,{\bar{c}}_{\delta (1)}^1,r_2^1,{\bar{r}}_2^1,c_{\delta (2)}^1,{\bar{c}}_{\delta (2)}^1,\dots ,r_{2k}^1,{\bar{r}}_{2k}^1,c_{\delta (2k)}^1,{\bar{c}}_{\delta (2k)}^1,$$

$$r_1^2,{\bar{r}}_1^2,c_{\delta (1)}^2,{\bar{c}}_{\delta (1)}^2,r_2^2,{\bar{r}}_2^2,c_{\delta (2)}^2,{\bar{c}}_{\delta (2)}^2,\dots ,r_{2k}^2,{\bar{r}}_{2k}^2,c_{\delta (2k)}^2,{\bar{c}}_{\delta (2k)}^2,$$

$$r_1^3,{\bar{r}}_1^3,c_{\delta (1)}^3,{\bar{c}}_{\delta (1)}^3,r_2^3,{\bar{r}}_2^3,c_{\delta (1)}^3,{\bar{c}}_{\delta (2)}^3,\dots ,r_{2k}^3,{\bar{r}}_{2k}^3,c_{\delta (2k)}^3,{\bar{c}}_{\delta (2k)}^3.$$

$$c_a^1,\dots ,c_a^{d+1},v_1,\rho (U_1),v_2,\rho (U_2),v_3,\dots ,v_m,\rho (U_m),v_{m+1},c_b^{d+1},\dots ,c_b^1.$$

Full text

In this paper, we examine parameterized problems where $f(k)$ is $k^{O(k)}=2^{O(k\log k)}$ in the best known running time and for a number of such problems, we show that the dependence on $k$ in the running time cannot be improved to single exponential. More precisely we prove following tight lower bounds, for four natural problems, arising from three different domains:

• •

In the Closest String problem, given strings $s_{1}$, $\dots$, $s_{t}$ over an alphabet $\Sigma$ of length $L$ each, and an integer $d$, the question is whether there exists a string $s$ over $\Sigma$ of length $L$, such that its hamming distance from each of the strings $s_{i}$, $1\leq i\leq t$, is at most $d$. The pattern matching problem Closest String is known to be solvable in time $2^{O(d\log d)}\cdot n^{O(1)}$ and $2^{O(d\log|\Sigma|)}\cdot n^{O(1)}$ . We show that there are no $2^{o(d\log d)}\cdot n^{O(1)}$ or $2^{o(d\log|\Sigma|)}\cdot n^{O(1)}$ time algorithms, unless the Exponential Time Hypothesis (ETH) fails.

• •

The graph embedding problem Distortion, that is, deciding whether a graph $G$ has a metric embedding into the integers with distortion at most $d$ can be solved in time $2^{O(d\log d)}\cdot n^{O(1)}$. We show that there is no $2^{o(d\log d)}\cdot n^{O(1)}$ time algorithm, unless the ETH fails.

• •

The Disjoint Paths problem can be solved in time in time $2^{O(w\log w)}\cdot n^{O(1)}$ on graphs of treewidth at most $w$. We show that there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm, unless the ETH fails.

• •

The Chromatic Number problem can be solved in time in time $2^{O(w\log w)}\cdot n^{O(1)}$ on graphs of treewidth at most $w$. We show that there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm, unless the ETH fails.

To obtain our results, we first prove the lower bound for variants of basic problems: finding cliques, independent sets, and hitting sets. These artificially constrained variants form a good starting point for proving lower bounds on natural problems without any technical restrictions and could be of independent interest. Several follow up works have already obtained tight lower bounds by using our framework, and we believe it will prove useful in obtaining even more lower bounds in the future.

1 Introduction

The goal of parameterized complexity is to find ways of solving NP-hard problems more efficiently than brute force: our aim is to restrict the combinatorial explosion to a parameter that is hopefully much smaller than the input size. Formally, a parameterization of a problem is assigning an integer $k$ to each input instance and we say that a parameterized problem is fixed-parameter tractable (FPT) if there is an algorithm that solves the problem in time $f(k)\cdot|I|^{O(1)}$, where $|I|$ is the size of the input and $f$ is an arbitrary computable function depending on the parameter $k$ only. There is a long list of NP-hard problems that are FPT under various parameterizations: finding a vertex cover of size $k$, finding a cycle of length $k$, finding a maximum independent set in a graph of treewidth at most $k$, etc. For more background, the reader is referred to the monographs [18, 29, 34, 60].

The practical applicability of fixed-parameter tractability results depends very much on the form of the function $f(k)$ in the running time. In some cases, for example in results obtained from Graph Minors theory, the function $f(k)$ is truly horrendous (towers of exponentials), making the result purely of theoretical interest. On the other hand, in many cases $f(k)$ is a moderately growing exponential function: for example, $f(k)$ is $1.2738^{k}$ in the current fastest algorithm for finding a vertex cover of size $k$ [14], which can be further improved to $1.1616^{k}$ in the special case of graphs with maximum degree 3 [67]. For some problems, $f(k)$ can be even subexponential (e.g., $c^{\sqrt{k}}$) [24, 23, 22, 1].

The implicit assumption in the research on fixed-parameter tractability is that whenever a reasonably natural problem turns out to be FPT, then we can improve $f(k)$ to $c^{k}$ with some small $c$ (hopefully $c<2$) if we work on the problem hard enough. Indeed, for some basic problems, the current best running time was obtained after a long sequence of incremental improvements. However, it is very well possible that for some problems there is no algorithm with single-exponential $f(k)$ in the running time.

In this paper, we examine parameterized problems where $f(k)$ is “slightly superexponential” in the best known running time: $f(k)$ is of the form $k^{O(k)}=2^{O(k\log k)}$. Algorithms with this running time naturally occur when a search tree of height at most $k$ and branching factor at most $k$ is explored, or when all possible permutations, partitions, or matchings of a $k$ element set are enumerated. For a number of such problems, we show that the dependence on $k$ in the running time cannot be improved to single exponential. More precisely, we show that a $2^{o(k\log k)}\cdot|I|^{O(1)}$ time algorithm for these problems would violate the Exponential Time Hypothesis (ETH), which is a complexity-theoretic assumption that can be informally stated as saying that there is no $2^{o(n)}$-time algorithm for $n$-variable 3SAT [44].

In the first part of the paper, we prove the lower bound for variants of basic problems: finding cliques, independent sets, and hitting sets. These variants are artificially constrained such that the search space is of size $2^{O(k\log k)}$ and we prove that a $2^{o(k\log k)}\cdot|I|^{O(1)}$ time algorithm would violate the ETH. The results in this section demonstrate that for some problems the natural $2^{O(k\log k)}\cdot|I|^{O(1)}$ upper bound on the search space is actually a tight lower bound on the running time. More importantly, the results on these basic problems form a good starting point for proving lower bounds on natural problems without any technical restrictions.

In the second part of the paper, we use our results on the basic problems to prove tight lower bounds for four natural problems from three different domains:

• •

In the Closest String problem, given strings $s_{1}$, $\dots$, $s_{t}$ over an alphabet $\Sigma$ of length $L$ each, and an integer $d$, the question is whether there exists a string $s$ over $\Sigma$ of length $L$, such that its hamming distance from each of the strings $s_{i}$, $1\leq i\leq t$, is at most $d$. The pattern matching problem Closest String is known to be solvable in time $2^{O(d\log d)}\cdot|I|^{O(1)}$ [40] and $2^{O(d\log|\Sigma|)}\cdot|I|^{O(1)}$ [55]. We show that there are no $2^{o(d\log d)}\cdot n^{O(1)}$ or $2^{o(d\log|\Sigma|)}\cdot n^{O(1)}$ time algorithms, unless the ETH fails.

• •

The graph embedding problem Distortion, that is, deciding whether a $n$ vertex graph $G$ has a metric embedding into the integers with distortion at most $d$ can be done in time $2^{O(d\log d)}\cdot n^{O(1)}$[33]. We show that there is no $2^{o(d\log d)}\cdot n^{O(1)}$ time algorithm, unless the ETH fails.

• •

The Disjoint Paths problem can be solved in time $2^{O(w\log w)}\cdot n^{O(1)}$ on $n$ vertex graphs of treewidth at most $w$ [64]. We show that there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm, unless the ETH fails.

• •

The Chromatic Number problem can be solved in time $2^{O(w\log w)}\cdot n^{O(1)}$ on $n$ vertex graphs of treewidth at most $w$ [46]. We show that there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm, unless the ETH fails.

We remark that the algorithm given in [64] does not mention the running time for Disjoint Paths as $2^{O(w\log w)}\cdot n^{O(1)}$ on graphs of bounded treewidth but a closer look reveals that it is indeed the case. We expect that many further results of this form can be obtained by using the framework of the current paper. Thus parameterized problems requiring “slightly superexponential” time $2^{O(k\log k)}\cdot|I|^{O(1)}$ is not a shortcoming of algorithm design or pathological situations, but an unavoidable feature of the landscape of parameterized complexity.

It is important to point out that it is a real possibility that some $2^{O(k\log k)}\cdot|I|^{O(1)}$ time algorithm can be improved to single-exponential dependence with some work. In fact, there are examples of well-studied problems where the running time was “stuck” at $2^{O(k\log k)}\cdot|I|^{O(1)}$ for several years before some new algorithmic idea arrived that made it possible to reduce the dependence to $2^{O(k)}\cdot|I|^{O(1)}$:

• •

In 1985, Monien [57] gave a $k!\cdot n^{O(1)}$ time algorithm for finding a cycle of length $k$ in a graph on $n$ vertices. Alon, Yuster, and Zwick [2] introduced the color coding technique in 1995 and used it to show that a cycle of length $k$ can be found in time $2^{O(k)}\cdot n^{O(1)}$.

• •

In 1995, Eppstein [31] gave a $O(k^{k}n)$ time algorithm for deciding if a $k$-vertex planar graph $H$ is a subgraph of an $n$-vertex planar graph $G$. Dorn [26] gave an improved algorithm with running time $2^{O(k)}\cdot n$. One of the main technical tools in this result is the use of sphere cut decompositions of planar graphs, which was used earlier to speed up algorithms on planar graphs in a similar way [27].

• •

In 1995, Downey and Fellows [28] gave a $k^{O(k)}\cdot n^{O(1)}$ time algorithm for Feedback Vertex Set (given an undirected graph $G$ on $n$ vertices, delete $k$ vertices to make it acyclic). A randomized $4^{k}\cdot n^{O(1)}$ time algorithm was given in 2000 [6]. The first deterministic $2^{O(k)}\cdot n^{O(1)}$ time algorithms appeared only in 2005 [42, 21], using the technique of iterative compression introduced by Reed et al. [62].

• •

In 2003, Cook and Seymour [17] used standard dynamic programming techniques to give a $2^{O(w\log w)}$ $n^{O(1)}$-time algorithm for Feedback Vertex Set on graphs of treewidth $w$, and it was considered plausible that this is the best possible form of running time. Hence it was a remarkable surprise in 2011 when Cygan et al. [19] presented a $3^{w}n^{O(1)}$ time randomized algorithm by using the so-called Cut & Count technique. Later, Bodlaender et al. [9] and Fomin et al. [36] obtained deterministic single-exponential parameterized algorithms using a different approach.

As we can see in the examples above, achieving single-exponential running time often requires the invention of significant new techniques. Therefore, trying to improve the running time for a problem whose best known parameterized algorithm is slightly superexponential can lead to important new discoveries and developments. However, in this paper we identify problems for which such an improvement is very unlikely. The $2^{O(k\log k)}$ dependence on $f(k)$ seems to be inherent to these problems, or one should realize that in achieving single-exponential dependence one is essentially trying to disprove the ETH.

There are some lower bound results on FPT problems in the parameterized complexity literature, but not of the form that we are proving here. Cai and Juedes [12] proved that if the parameterized version of a MAXSNP-complete problems (such as Vertex Cover on graphs of maximum degree 3) can be solved in time $2^{o(k)}\cdot|I|^{O(1)}$, then ETH fails. Using parameterized reductions, this result can be transfered to other problems: for example, assuming the ETH, there is a no $2^{o(\sqrt{k})}\cdot|I|^{O(1)}$ time algorithm for planar versions of Vertex Cover, Independent Set, and Dominating Set (and this bound is tight). However, no lower bound above $2^{O(k)}$ was obtained this way for any problem so far.

Flum, Grohe, and Weyer [35] tried to rebuild parameterized complexity by redefining fixed-parameter tractability as $2^{O(k)}\cdot|I|^{O(1)}$ time and introducing appropriate notions of reductions, completeness, and complexity classes. This theory could be potentially used to show that the problems treated in the current paper are hard for certain classes, and therefore they are unlikely to have single-exponential parameterized algorithms. However, we see no reason why these problems would be complete for any of those classes (for example, the only complete problem identified in [35] that is actually FPT is a model checking on problem on words for which it was already known that $f(k)$ cannot even be elementary). Moreover, we are not only giving evidence against single-exponential time algorithms in this paper, but show that the $2^{O(k\log k)}$ dependence is actually tight.

2 Basic problems

In this section, we modify basic problems in such a way that they can be solved in time $2^{O(k\log k)}|I|^{O(1)}$ by brute force, and this is best possible assuming the ETH. In all the problems of this section, the task is to select exactly one element from each row of a $k\times k$ table such that the selected elements satisfy certain constraints. This means that the search space is of size $k^{k}=2^{O(k\log k)}$. We denote by $[k]\times[k]$ the set of elements in a $k\times k$ table, where $(i,j)$ is the element in row $i$ and column $j$. Thus selecting exactly one element from each row gives a set $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ for some mapping $\rho:[k]\to[k]$. In some of the variants, we not only require that exactly one element is selected from each row, but we also require that exactly one element is selected from each column, that is, $\rho$ has to be a permutation. The lower bounds for such permutation problems will be essential for proving hardness results on Closest String (Section 3) and Distortion (Section 4). The key step in obtaining the lower bounds for permutation problems is the randomized reordering argument of Theorem 2.11. The analysis and derandomization of this step is reminiscent of the color coding [2] and chromatic coding [1] techniques.

To prove that a too fast algorithm for a certain problem $P$ contradicts the Exponential Time Hypothesis, we have to reduce $n$-variable 3SAT to problem $P$ and argue that the algorithm would solve 3SAT in time $2^{o(n)}$. It will be somewhat more convenient to do the reduction from 3-Coloring. We use the well-known fact that there is a polynomial-time reduction from 3SAT to 3-Coloring where the number of vertices of the graph is linear in the size formula.

Proposition 2.1**.**

Given a 3SAT formula $\phi$ with $n$-variables and $m$-clauses, it is possible to construct a graph $G$ with $O(n+m)$ vertices in polynomial time such that $G$ is 3-colorable if and only if $\phi$ is satisfiable.

Proposition 2.1 implies that an algorithm for 3-Coloring with running time subexponential in the number of vertices gives an algorithm for 3SAT that is subexponential in the number of clauses. This is sufficient for our purposes, as the Sparsification Lemma of Impagliazzo, Paturi and Zane [44] shows that such an algorithm already violates the ETH.

Lemma 2.2** ([44]).**

Assuming the ETH, there is no $2^{o(m)}$ time algorithm for $m$-clause 3SAT.

Combining Proposition 2.1 and Lemma 2.2 gives the following proposition:

Proposition 2.3**.**

Assuming the ETH, there is no $2^{o(n)}$ time algorithm for deciding whether an $n$-vertex graph is $3$-colorable.

2.1 $k\times k$** Clique**

The first problem we investigate is the variant of the standard clique problem where the vertices are the elements of a $k\times k$ table, and the clique we are looking for has to contain exactly one element from each row.

$k\times k$ Clique

Input:

A graph $G$ over the vertex set $[k]\times[k]$

Parameter:

$k$

Question:

Is there a $k$-clique in $G$ with exactly one element from each row?

Note that the graph $G$ in the $k\times k$ Clique instance has $O(k^{2})$ vertices at most $O(k^{4})$ edges, thus the size of the instance is $O(k^{4})$.

Theorem 2.4**.**

Assuming the ETH, there is no $2^{o(k\log k)}$ time algorithm for $k\times k$ Clique.

Proof.

Suppose that there is an algorithm $\mathbb{A}$ that solves $k\times k$ Clique in $2^{o(k\log k)}$ time. We show that this implies that 3-Coloring on a graph with $n$ vertices can be solved in time $2^{o(n)}$, which contradicts the ETH by Proposition 2.3.

Let $H$ be a graph with $n$ vertices. Let $k$ be the smallest integer such that $3^{n/k+1}\leq k$, or equivalently, $n\leq k\log_{3}k-k$. Note that such a finite $k$ exists for every $n$ and it is easy to see that $k\log k=O(n)$ for the smallest such $k$. Intuitively, it will be useful to think of $k$ as a value somewhat larger than $n/\log n$ (and hence $n/k$ is somewhat less than $\log n$).

Let us partition the vertices of $H$ into $k$ groups $X_{1}$, $\dots$, $X_{k}$, each of size at most $\lceil n/k\rceil$. For every $1\leq i\leq k$, let us fix an enumeration of all the proper 3-colorings of $H[X_{i}]$. Note that there are most $3^{\lceil n/k\rceil}\leq 3^{n/k+1}\leq k$ such 3-colorings for every $i$. We say that a proper 3-coloring $c_{i}$ of $H[X_{i}]$ and a proper 3-coloring $c_{j}$ of $H[X_{j}]$ are compatible if together they form a proper coloring of $H[X_{i}\cup X_{j}]$: for every edge $uv$ with $u\in X_{i}$ and $v\in X_{j}$, we have $c_{i}(u)\neq c_{j}(v)$. Let us construct a graph $G$ over the vertex set $[k]\times[k]$ where vertices $(i_{1},j_{1})$ and $(i_{2},j_{2})$ with $i_{1}\neq i_{2}$ are adjacent if and only if the $j_{1}$-th proper coloring of $H[X_{i_{1}}]$ and the $j_{2}$-th proper coloring of $H[X_{i_{2}}]$ are compatible (this means that if, say, $H[X_{i_{1}}]$ has less than $j_{1}$ proper colorings, then $(i_{1},j_{1})$ is an isolated vertex).

We claim that $G$ has a $k$-clique having exactly one vertex from each row if and only if $H$ is 3-colorable. Indeed, a proper 3-coloring of $H$ induces a proper 3-coloring for each of $H[X_{1}]$, $\dots$, $H[X_{k}]$. Let us select vertex $(i,j)$ if and only if the proper coloring of $H[X_{i}]$ induced by $c$ is the $j$-th proper coloring of $H[X_{i}]$. It is clear that we select exactly one vertex from each row and they form a clique: the proper colorings of $H[X_{i}]$ and $H[X_{j}]$ induced by $c$ are clearly compatible. For the other direction, suppose that $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ form a $k$-clique for some mapping $\rho:[k]\to[k]$. Let $c_{i}$ be the $\rho(i)$-th proper 3-coloring of $H[X_{i}]$. The colorings $c_{1}$, $\dots$, $c_{k}$ together define a coloring $c$ of $H$. This coloring $c$ is a proper 3-coloring: for every edge $uv$ with $u\in X_{i_{1}}$ and $v\in X_{i_{2}}$, the fact that $(i_{1},\rho(i_{1}))$ and $(i_{2},\rho(i_{2}))$ are adjacent means that $c_{i_{1}}$ and $c_{i_{2}}$ are compatible, and hence $c_{i_{1}}(u)\neq c_{i_{2}}(v)$.

Running the assumed algorithm $\mathbb{A}$ on $G$ decides the 3-colorability of $H$. Let us estimate the running time of constructing $G$ and running algorithm $\mathbb{A}$ on $G$. The graph $G$ has $k^{2}$ vertices and the time required to construct $G$ is polynomial in $k$: for each $X_{i}$, we need to enumerate at most $k$ proper 3-colorings of $G[X_{i}]$. Therefore, the total running time is $2^{o(k\log k)}\cdot k^{O(1)}=2^{o(n)}$ (using that $k\log k=O(n)$). It follows that we have a $2^{o(n)}$ time algorithm for 3-Coloring on an $n$-vertex graph, contradicting the ETH. ∎

$k\times k$ Permutation Clique is a more restricted version of $k\times k$ Clique: in addition to requiring that the clique contains exactly one vertex from each row, we also require that it contains exactly one vertex from each column. In other words, the vertices selected in the solution are $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ for some permutation $\rho$ of $[k]$. Given an instance $I$ of $k\times k$ Clique having a solution $S$, if we randomly reorder the vertices in each row, then with some probability the reordered version of solution $S$ contains exactly one vertex from each row and each column of the reordered instance. In Theorem 2.5, we use this argument to show that a $2^{o(k\log k)}$ time algorithm for $k\times k$ Permutation Clique gives a randomized $2^{o(k\log k)}$ time algorithm for $k\times k$ Clique. Section 2.1.1 shows how the proof of Theorem 2.5 can be derandomized.

Theorem 2.5**.**

If there is a $2^{o(k\log k)}$ time algorithm for $k\times k$ Permutation Clique, then there is a randomized $2^{o(m)}$ time algorithm for $m$-clause 3SAT.

Proof.

We show how to transform an instance $I$ of $k\times k$ Clique into an instance $I^{\prime}$ of $k\times k$ Permutation Clique with the following properties: if $I$ is a no-instance, then $I^{\prime}$ is a no-instance, and if $I$ is a yes-instance, then $I^{\prime}$ is a yes-instance with probability at least $2^{-O(k)}$. This means that if we perform this transformation $2^{O(k)}$ times and accept $I$ as a yes-instance if and only at least one of the $2^{O(k)}$ constructed instances is a yes-instance, then the probability of incorrectly rejecting a yes-instance can be reduced to an arbitrary small constant. Therefore, a $2^{o(k\log k)}$ time algorithm for $k\times k$ Permutation Clique implies a randomized $2^{O(k)}\cdot 2^{o(k\log k)}=2^{o(k\log k)}$ time algorithm for $k\times k$ Clique.

Let $c(i,j):[k]\times[k]\to[k]$ be a mapping chosen uniform at random; we can imagine $c$ as a coloring of the $k\times k$ vertices. Let $c^{\prime}(i,j)=\bigstar$ if there is a $j^{\prime}\neq j$ such that $c(i,j)=c(i,j^{\prime})$ and let $c^{\prime}(i,j)=c(i,j)$ otherwise (i.e., if $c(i,j)=x\neq\bigstar$, then no other vertex has color $x$ in row $i$). The instance $I^{\prime}$ of $k\times k$ Permutation Clique is constructed the following way: if there is an edge between $(i_{1},j_{1})$ and $(i_{2},j_{2})$ in instance $I$ and $c^{\prime}(i_{1},j_{1}),c^{\prime}(i_{2},j_{2})\neq\bigstar$, then we add an edge between $(i_{1},c^{\prime}(i_{1},j_{1}))$ and $(i_{2},c^{\prime}(i_{2},j_{2}))$ in instance $I^{\prime}$. That is, we use mapping $c$ to rearrange the vertices in each row. If vertex $(i,j)$ clashes with some other vertex in the same row (that is, $c(i,j)=\bigstar$), then all the edges incident to $(i,j)$ are thrown away.

Suppose that $I^{\prime}$ has a $k$-clique $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ for some permutation $\rho$ of $[k]$. For every $i$, there is a unique $\delta(i)$ such that $c^{\prime}(i,\delta(i))=\rho(i)$: otherwise $(i,\rho(i))$ is an isolated vertex in $I^{\prime}$. It is easy to see that $(1,\delta(i))$, $\dots$, $(k,\delta(k))$ is a clique in $I$: vertices $(i_{1},\delta(i_{1}))$ and $(i_{2},\delta(i_{2}))$ have to be adjacent, otherwise there would be no edge between $(i_{1},\rho(i_{1}))$ and $(i_{2},\rho(i_{2}))$ in $I^{\prime}$. Therefore, if $I$ is a no-instance, then $I^{\prime}$ is a no-instance as well.

Suppose now that $I$ is a yes-instance: there is a clique $(1,\delta(1))$, $\dots$, $(k,\delta(k))$ in $I$. Let us estimate the probability that the following two events occur:

• (1)

For every $1\leq i_{1}<i_{2}\leq k$, $c(i_{1},\delta(i_{1}))\neq c(i_{2},\delta(i_{2}))$.

• (2)

For every $1\leq i\leq k$ and $1\leq j\leq k$ with $j\neq\delta(i)$, $c(i,\delta(i))\neq c(i,j)$.

Event (1) means that $c(1,\delta(1))$, $\dots$, $c(k,\delta(k))$ is a permutation of $[k]$. Therefore, the probability of (1) is $k!/k^{k}=e^{-O(k)}$ (using Stirling’s Formula). For a particular $i$, event (2) holds if $k-1$ randomly chosen values are all different from $c(i,\delta(i))$. Thus the probability that (2) holds for a particular $i$ is $(1-1/k)^{-(k-1)}\geq e^{-1}$ and the probability that (2) holds for every $i$ is at least $e^{-k}$. Furthermore, events (1) and (2) are independent: we can imagine the random choice of the mapping $c$ as first choosing the values $c(1,\delta(1))$, $\dots$, $c(k,\delta(k))$ and then choosing the remaining $k^{2}-k$ values. Event (1) depends only on the first $k$ choices, and for any fixed result of the first $k$ choices, the probability of event (2) is the same. Therefore, the probability that (1) and (2) both hold is $e^{-O(k)}$.

Suppose that (1) and (2) both hold. Event (2) implies that $c(i,\delta(i))=c^{\prime}(i,\delta(i))\neq\bigstar$ for every $1\leq i\leq k$. Event (1) implies that if we set $\rho(i):=c(i,\delta(i))$, then $\rho$ is a permutation of $[k]$. Therefore, the clique $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ is a solution of $I^{\prime}$, as required. ∎

In the next section, we show that instead of random colorings, we can use a certain deterministic family of colorings. This will imply:

Corollary 2.6**.**

Assuming the ETH, there is no $2^{o(k\log k)}$ time algorithm for $k\times k$ Permutation Clique.

2.1.1 Derandomization

In this section, we give a coloring family that can be used instead of the random coloring in the proof of Theorem 2.5. We call a graph $G$ to be a cactus-grid graph if the vertices are elements of a $k\times k$ table and the graph precisely consists of a clique containing exactly one vertex from each row and each vertex in the clique is adjacent to every other vertex in its row. There are no other edges in the graph, thus the graph has exactly $\binom{k}{2}+k(k-1)$ edges. We are interested in a coloring family ${\cal F}=\left\{f:[k]\times[k]\rightarrow[k+1]\right\}$ with the property that for any cactus-grid graph $G$ with vertices from $k\times k$ table, there exists a function $f\in{\cal F}$ such that $f$ properly colors the vertices of $G$. We call such a $\cal F$ as a coloring family for cactus-grid graphs.

Before we proceed to construct a coloring family $\cal F$ of size $2^{O(k\log\log k)}$, we explain how this can be used to obtain the derandomized version of Theorem 2.5, the Corollary 2.6. Suppose that the instance $I$ of $k\times k$ Clique is a yes-instance. Then there is a clique $(1,\delta(1))$, $\dots$, $(k,\delta(k))$ in $I$. Consider the cactus-grid graph $G$ consisting of clique $(1,\delta(1))$, $\dots$, $(k,\delta(k))$ and for each $1\leq i\leq k$, the edges between $(i,\delta(i))$ and $(i,j)$ for every $j\neq\delta(i)$. Let $f\in{\cal F}$ be a proper coloring of $G$. Now since $(1,\delta(1))$, $\dots$, $(k,\delta(k))$ is a clique in $G$ they get distinct colors by $f$ and since all the vertices in the row $i$, $(i,j)$, $j\neq\delta(i)$, are adjacent to $(i,\delta(i))$ we have that $f((i,j))\neq f(i,\delta(i))$. So if we use this $f$ in place of $c(i,j)$, the random coloring used in the proof of Theorem 2.5, then events (1) and (2) hold and we know that the instance $I^{\prime}$ obtained using $f$ is a yes-instance of $k\times k$ Permutation Clique. Thus we know that an instance $I$ of $k\times k$ Clique has a clique of size $k$ containing exactly one element from each row if and only if there exists an $f\in{\cal F}$ such that the corresponding instance $I^{\prime}$ of $k\times k$ Permutation Clique has a clique of size $k$ such that it contains exactly one element from each row and column. This together with the fact that the size of ${\cal F}$ is bounded by $2^{O(k\log\log k)}$ imply the Corollary 2.6.

To construct our deterministic coloring family we also need a few known results on perfect hash families. Let ${\cal H}=\{f:[n]\rightarrow[k]\}$ be a set of functions such that for all subsets $S$ of size $k$ there is a $h\in{\cal H}$ such that it is one-to-one on $S$. The set $\cal H$ is called $(n,k)$-family of perfect hash functions. There are some known constructions for set $\cal H$. We summarize them below.

Proposition 2.7** ([2, 59]).**

There exists explicit construction $\cal H$ of $(n,k)$-family of perfect hash functions of size $O(11^{k}\log n)$. There is also another explicit construction $\cal H$ of $(n,k)$-family of perfect hash functions of size $O(e^{k}k^{O(\log k)}\log n)$.

Now we are ready to state the main lemma of this section.

Lemma 2.8**.**

There exists explicit construction of coloring family $\cal F$ for cactus-grid graphs of size $2^{O(k\log\log k)}$.

Proof.

Our idea for deterministic coloring family $\cal F$ for cactus-grid graphs is to keep $k$ functions $f_{1},\ldots,f_{k}$ where each $f_{i}$ is an element of a $(k,k^{\prime})$-family of perfect hash functions for some $k^{\prime}$ and use it to map the elements of $\{i\}\times k$ (the column $i$). We guess the number of vertices of $G$ that appear in each column, and we reserve that many private colors for the column so that these colors are not used on the vertices of any other columns. This will ensure that we get the desired coloring family. We make our intuitive idea more precise below. A description of a function $f\in{\cal F}$ consists of a tuple having

• •

a set $S\subseteq[k]$;

• •

a tuple $(k_{1},k_{2},\ldots,k_{\ell})$ where $k_{i}\geq 1$, $\ell=|S|$ and $\sum_{i=1}^{\ell}k_{i}=k$;

• •

$\ell$ functions $f_{1},\ldots,f_{\ell}$ where $f_{i}\in{\cal H}_{i}$ and ${\cal H}_{i}$ is a $(k,k_{i})$-family of perfect hash functions.

The set $S$ tells us which columns the clique intersects. Let the elements of $S=\{s_{1},\ldots,s_{\ell}\}$ be sorted in increasing order, say $s_{1}<s_{2}<\dots<s_{\ell}$. Then the tuple $(k_{1},k_{2},\ldots,k_{\ell})$ tells us that the column $s_{j}$, $1\leq j\leq\ell$, contains $k_{j}$ vertices from the clique. Hence with this interpretation, given a tuple $(S,(k_{1},\ldots,k_{\ell}),f_{1},\ldots,f_{\ell})$ we define the coloring function $g:[k]\times[k]\rightarrow[k]$ as follows. Every element in $[k]\times\{1,\ldots,k\}\setminus S$ is mapped to $k+1$. Now for vertices in $[k]\times\{s_{j}\}$ (vertices in column $s_{j}$), we define $g(i,s_{j})=f_{j}(i)+\sum_{1\leq i<j}k_{i}$. We do this for every $j$ between $1$ and $\ell$. This concludes the description. Now we show that it is indeed a coloring family for cactus-grid graphs. Given a cactus grid graph $G$, we first look at the columns it intersects and that forms our set $S$ and then the number of vertices it intersects in each column makes the the tuple $(k_{1},k_{2},\ldots,k_{\ell})$. Finally for each of the columns there exists a function $h$ in the perfect $(k,k_{i})$-hash family that maps the elements of clique in this column one to one with $[k_{i}]$; we store this function corresponding to this column. Now we show that the function $g$ corresponding to this tuple properly colors $G$. The function $g$ assigns different values from $[k]$ to the columns in $S$ and hence we have that the vertices of clique gets distinct colors as in each column we have a function $f_{i}$ that is one-to-one on the vertices of $S$. Now we look at the edge with both end-points in the same row. If any of the end-point occurs in column that is not in $S$, then we know that it has been assigned $k+1$ while the vertex from the clique has been assigned color from $[k]$. If both end-points are from $S$, then the offset we use to give different colors to vertices in these columns ensures that these end-points get different colors. This shows that $g$ is indeed a proper coloring of $G$. This shows that for every cactus-grid graph we have a function $g\in\cal F$. Finally, the bound on the size of $\cal F$ is as follows,

[TABLE]

This concludes the proof. ∎

The bound achieved in Equation 1 on the size of $\cal F$ is sufficient for our purpose but it is not as small as $2^{O(k)}$ that one can obtain using a simple application of probabilistic methods. We provide a family $\cal F$ of size $2^{O(k)}$ below which could be of independent algorithmic interest.

Lemma 2.9**.**

There exists explicit construction of coloring family $\cal F$ for cactus-grid graphs of size $2^{O(k)}$.

Proof.

We incurred a factor of $(\log k)^{\ell}$ in the construction given in Lemma 2.8 because for every column we applied hash functions from $[k]\rightarrow[k_{i}]$. Loosely speaking, if we could replace these by $[k_{i}^{2}]\rightarrow[k_{i}]$, then the size of family will be $11^{k_{i}}\log k_{i}\leq 12^{k_{i}}$ and then $\prod_{i=1}^{\ell}11^{k_{i}}\log k_{i}\leq 12^{k}$. Next we describe a procedure to do this by incurring an extra cost of $2^{O(\log^{3}k)}$. To do this we use the following classical lemma proved by Fredman, Komlós and Szemerédi [38].

Lemma 2.10** ([38]).**

Let $W\subseteq[n]$ with $|W|=r$. The mapping $f:[n]\rightarrow[2r^{2}]$ such that $f(x)=(tx\mod p)\mod~{}2r^{2}$ is one-to-one when restricted to $W$ for at least half of the values $t\in[p]$. Here $p$ is any prime between $n$ and $2n$.

The idea is to use Lemma 2.10 to choose multipliers ($t$ in the above description) appropriately. Let us fix a prime $p$ between $k$ and $2k$. Given a set $S$ and a tuple $(k_{1},k_{2},\ldots,k_{\ell})$ we make a partition of set $S$ as follows $S_{i}=\{s_{j}~{}|s_{j}\in S,2^{i-1}<k_{j}\leq 2^{i}\}$ for $i\in\{0,\ldots,\lceil\log k\rceil\}$. Now let us fix a set $S_{i}$, by our construction we know that the size of intersection of the clique with each of the columns in $S_{i}$ is roughly same. For simplicity of argument, let us fix a clique $W$ of some cactus grid graph $G$. Consider a bipartite graph $(A,B)$ where $A$ contains a vertex for each column in $S_{i}$ and $B$ consists of numbers from $[p]$. Now we give an edge between vertex $a\in A$ and $b\in B$ if we can use $b$ as a multiplier in Lemma 2.10, that is, the map $f(x)=(bx~{}\mod~{}p)\mod~{}2^{2i+1}$ is one-to-one when restricted to the vertices of the clique $W$ to the column $a$.

Observe that because of Lemma 2.10, every vertex in $A$ has degree at least $p/2$ and hence there exists a vertex $b\in B$ that can be used as a multiplier for at least half of the elements in the set $A$. We can repeat this argument by removing a vertex $b\in B$, that could be used as a multiplier for half of the vertices in $A$, and all the columns for which it can be multiplier. This implies that there exits a set $X_{i}\subseteq[p]$ of size $\log|A|\leq\log k$ that could be used as a multiplier for every column in $A$. Now we give a description of a function $f\in{\cal F}$ that consists of a tuple having

• •

a set $S\subseteq[k]$;

• •

a tuple $(k_{1},k_{2},\ldots,k_{\ell})$ where $k_{i}\geq 1$, $\ell=|S|$ and $\sum_{i=1}^{\ell}k_{i}=k$;

• •

$((b^{i}_{1},\ldots,b^{i}_{q}),(L^{i}_{1},\ldots,L^{i}_{q}))$, $1\leq i\leq\lceil\log k\rceil$, $q=\lceil\log k\rceil$; Here $(L^{i}_{1},\ldots,L^{i}_{q})$ is a partition of $S_{i}$ and the interpretation is that for every column in $L_{j}^{i}$ we will use $b^{i}_{j}$ as a multiplier for range reduction;

• •

$\ell$ functions $f_{1},\ldots,f_{\ell}$ where $f_{i}\in{\cal H}_{i}$ and ${\cal H}_{i}$ is a $(8k_{i}^{2},k_{i})$-family of perfect hash functions.

This completes the description. Now given a tuple

[TABLE]

we define the coloring function $g:[k]\times[k]\rightarrow[k]$ as follows. Every element in $[k]\times\{1,\ldots,k\}\setminus S$ is mapped to $k+1$. Now for vertices in $[k]\times\{s_{j}\}$ (vertices in column $s_{j}$), we do as follows. Suppose $s_{j}\in L_{\alpha}^{\beta}$ then we define $g(i,s_{j})=(\sum_{1\leq i<j}k_{i})+f_{j}(((b_{\alpha}^{\beta}s_{j})~{}\mod~{}p)\mod~{}ck_{j}^{2})$. We do this for every $j$ between $1$ and $\ell$. This concludes the description for $g$. Observe that given a vertex in column $s_{j}$ we first use the function in Lemma 2.10 to reduce its range to roughly $O(k_{j}^{2})$ and still preserving that for every subset $[k]$ of size at most $2k_{j}$ there is some multiplier which maps it injective. It is evident from the above description that this is indeed a coloring family of cactus grid graphs. The range of any function in $\cal F$ is $k+1$ and the size of this family is

[TABLE]

The last assertion follows from the fact that $\sum_{i=1}^{\lceil\log k\rceil}\sum_{j=1}^{\lceil\log k\rceil}|L_{j}^{i}|\leq k$ and $\sum_{i=1}^{\ell}k_{i}=k$. This concludes the proof. ∎

2.2 $k\times k$** Independent Set**

The lower bounds in Section 2.4 for $k\times k$ (Permutation) Clique obviously hold for the analogous $k\times k$ (Permutation) Independent Set problem: by taking the complement of the graph, we can reduce one problem to the other. We state here a version of the independent set problem that will be a convenient starting point for reductions in later sections:

$2k\times 2k$ Bipartite Permutation Independent Set

Input:

A graph $G$ over the vertex set $[2k]\times[2k]$ where every edge is between $I_{1}=\{(i,j)\mid i,j\leq k\}$ and $I_{2}=\{(i,j)\mid i,j\geq k+1\}$.

Parameter:

$k$

Question:

Is there an independent set $(1,\rho(1))$, $\dots$, $(2k,\rho(2k))\subseteq I_{1}\cup I_{2}$ in $G$ for some permutation $\rho$ of $[2k]$?

That is, the upper left quadrant $I_{1}$ and the lower right quadrant $I_{2}$ induce independent sets, and every edge is between these two independent sets. The requirement that the solution is a subset of $I_{1}\cup I_{2}$ means that $\rho(i)\leq k$ for $1\leq i\leq k$ and $\rho(i)\geq k+1$ for $k+1\leq i\leq 2k$.

Theorem 2.11**.**

Assuming the ETH, there is no $2^{o(k\log k)}$ time algorithm for $2k\times 2k$ Bipartite Permutation Independent Set.

Proof.

Given an instance $I$ of $k\times k$ Permutation Independent Set, we construct an equivalent instance $I^{\prime}$ of $2k\times 2k$ Bipartite Permutation Independent Set the following way. For every $1\leq i\leq k$ and $1\leq j,j^{\prime}\leq k$, $j\neq j^{\prime}$, we add an edge between $(i,j)$ and $(i+k,j^{\prime}+k)$ in $I^{\prime}$. If there is an edge between $(i_{1},j_{1})$ and $(i_{2},j_{2})$ in $I$, then we add an edge between $(i_{1},j_{1})$ and $(i_{2}+k,j_{2}+k)$ in $I^{\prime}$. This completes the description of $I^{\prime}$.

Suppose that $I$ has a solution $(1,\delta(1))$, $\dots$, $(k,\delta(k))$ for some permutation $\delta$ of $[2k]$. Then it is obvious from the construction of $I^{\prime}$ that $(1,\delta(1))$, $\dots$, $(k,\delta(k))$, $(1+k,\delta(1)+k)$, $\dots$, $(2k,\delta(k)+k)$ is an independent set of $I^{\prime}$ and $\delta(1)$, $\dots$, $\delta(k)$, $\delta(1)+k$, $\dots$, $\delta(k)+k$ is clearly a permutation of $[2k]$. Suppose that $(1,\rho(1))$, $\dots$, $(2k,\rho(2k))$ is solution of $I^{\prime}$ for some permutation $\rho$ of $[2k]$. By definition, $\rho(i)\leq k$ for $1\leq i\leq k$. We claim that $(1,\rho(k))$, $\dots$, $(k,\rho(k))$ is an independent set of $I$. Observe first that $\rho(i+k)=\rho(i)+k$ for every $1\leq i\leq k$: otherwise there is an edge between $(i,\rho(i))$ and $(i+k,\rho(i+k))$ in $I^{\prime}$. If there is an edge between $(i_{1},\rho(i_{1}))$ and $(i_{2},\rho(i_{2}))$ in $I$, then by construction there is an edge between $(i_{1},\rho(i_{1}))$ and $(i_{2}+k,\rho(i_{2})+k)=(i_{2}+k,\rho(i_{2}+k))$ in $I^{\prime}$, contradicting the assumption that $(1,\rho(k))$, $\dots$, $(2k,\rho(2k))$ is an independent set in $I^{\prime}$. ∎

2.3 $k\times k$** Hitting Set**

Hitting Set is a W[2]-complete problem, but if we restrict the universe to a $k\times k$ table where only one element can be selected from each row, then it can be solved in time $O^{*}(k^{k})$ by brute force.

$k\times k$ Hitting Set

Input:

Sets $S_{1},\dots,S_{m}\subseteq[k]\times[k]$.

Parameter:

$k$

Question:

Is there a set $S$ containing exactly one element from each row such that $S\cap S_{i}\neq\emptyset$ for any $1\leq i\leq m$?

We say that the mapping $\rho$ hits a set $S\subseteq[k]\times[k]$, if $(i,\rho(i))\in m$ for some $1\leq i\leq S$. Note that unlike for $k\times k$ Clique and $k\times k$ Independent Set, the size of the $k\times k$ Hitting Set instance cannot be bounded by a function of $k$.

It is quite easy to reduce $k\times k$ Independent Set to $k\times k$ Hitting Set: for every pair $(i_{1},j_{1})$, $(i_{2},j_{2})$ of adjacent vertices, we need to ensure that they are not selected simultaneously, which can be forced by a set that contains every element of rows $i_{1}$ and $i_{2}$, except $(i_{1},j_{1})$ and $(i_{2},j_{2})$. However, in Section 3.1 we prove the lower bound for Closest String by reduction from a restricted form of $k\times k$ Hitting Set where each set contains at most one element from each row. The following theorem proves the lower bound for this variant of $k\times k$ Hitting Set. The basic idea is that an instance of $2k\times 2k$ Bipartite Permutation Independent Set can be transformed in an easy way into an instance of Hitting Set where each set contains at most one element from each column and we want to select exactly one element from each row and each column. By adding each row as a new set, we can forget about the restriction that we want to select exactly one element from each row: this restriction will be automatically satisfied by any solution. Therefore, we have a Hitting Set instance where we have to select exactly one element from each column and each set contains at most one element from each column. By changing the role of rows and columns, we arrive to a problem of the required form.

Theorem 2.12**.**

Assuming the ETH, there is no $2^{o(k\log k)}\cdot n^{O(1)}$ time algorithm for $k\times k$ Hitting Set, even in the special case when each set contains at most one element from each row.

Proof.

To make the notation in the proof less confusing, we introduce a transposed variant of the problem (denote by $k\times k$ Hitting SetT), where exactly one element has to be selected from each column. We prove the lower bound for $k\times k$ Hitting SetT with the additional restriction that each set contains at most one element from each column; this obviously implies the theorem.

Given an instance $I$ of $2k\times 2k$ Bipartite Permutation Independent Set, we construct an equivalent $2k\times 2k$ Hitting SetT instance $I^{\prime}$ on the universe $[2k]\times[2k]$. For $1\leq i\leq k$, let set $S_{i}$ contain the first $k$ elements of row $i$ and for $k+1\leq i\leq 2k$, let set $S_{i}$ contain the last $k$ elements of row $i$. For every edge $e$ in instance $I$, we construct a set $S_{e}$ the following way. By the way $2k\times 2k$ Bipartite Permutation Independent Set is defined, we need to consider only edges connecting some $(i_{1},j_{1})$ and $(i_{2},j_{2})$ with $i_{1},j_{1}\leq k$ and $i_{2},j_{2}\geq k+1$. For such an edge $e$, let us define

[TABLE]

Suppose that $(1,\delta(1))$, $\dots$, $(2k,\delta(2k))$ is a solution of $I$ for some permutation $\rho$ of $[2k]$. We claim that it is a solution of $I^{\prime}$. As $\rho$ is a permutation, the set satisfies the requirement that it contains exactly one element from each column. As $\delta(i)\leq k$ if and only if $i\leq k$, the set $S_{i}$ is hit for every $1\leq i\leq 2k$. Suppose that there is an edge $e$ connecting $(i_{1},j_{1})$ and $(i_{2},j_{2})$ such that set $S_{e}$ of $I^{\prime}$ is not hit by this solution. Elements $(i_{1},\delta(i_{1}))$ and $(i_{2},\delta(i_{2}))$ are selected and we have $1\leq\delta(i_{1})\leq k$ and $k+1\leq\delta(i_{2})\leq 2k$. Thus if these two elements do not hit $S_{e}$, then this is only possible if $\delta(i_{1})=j_{1}$ and $\delta(i_{2})=j_{2}$. However, this means that the solution for $I$ contains the two adjacent vertices $(i_{1},j_{1})$ and $(i_{2},j_{2})$, a contradiction.

Suppose now that $(\rho(1),1)$, $\dots$, $(\rho(2k),2k)$ is a solution for $I^{\prime}$. Because of the sets $S_{i}$, $1\leq i\leq 2k$, the solution contains exactly one element from each row, i.e., $\rho$ is a permutation of $2k$. Moreover, the sets $S_{1}$, $\dots$, $S_{k}$ have to be hit by the $k$ elements in the first $k$ columns. This means that $\rho(i)\leq k$ if $i\leq k$ and consequently $\rho(i)>k$ if $i>k$. We claim that $(\rho(1),1)$, $\dots$, $(\rho(2k),2k)$ is also a solution of $I$. It is clear that the only thing that has to be verified is that these $2k$ vertices form an independent set. Suppose that $(\rho(j_{1}),j_{1})$ and $(\rho(j_{2}),j_{2})$ are connected by an edge $e$. We can assume that $\rho(j_{1})\leq k$ and $\rho(j_{2})>k$, which implies $j_{1}\leq k$ and $j_{2}>k$. The solution for $I^{\prime}$ hits set $S_{e}$, which means that either the solution selects an element $(\rho(j_{1}),j^{\prime})$ or an element $(\rho(j_{2}),j^{\prime})$. Elements $(\rho(j_{1}),j_{1})$ and $(\rho(j_{2}),j_{2})$ are the only elements of this form in the solution, but neither of them appears in $S_{e}$. Thus $(\rho(1),1)$, $\dots$, $(\rho(2k),2k)$ is indeed a solution of $I$ ∎

3 Closest String

Computational biology applications often involve long sequences that have to be analyzed in a certain way. One core problem is finding a “consensus” of a given set of strings: a string that is close to every string in the input. The Closest String problem defined below formalizes this task.

Closest String

Input:

Strings $s_{1}$, $\dots$, $s_{t}$ over an alphabet $\Sigma$ of length $L$ each, an integer $d$

Parameter:

$d$

Question:

Is there a string $s$ of length $L$ such $d(s,s_{i})\leq d$ for every $1\leq i\leq t$?

We denote by $d(s,s_{i})$ the Hamming distance of the strings $s$ and $s_{i}$, that is, the number of positions where they have different characters. The solution $s$ will be called the center string.

Closest String and its generalizations (Closest Substring, Distinguishing (Sub)string Selection, Consensus Patterns) have been thoroughly explored both from the viewpoint of approximation algorithms and fixed-parameter tractability [55, 66, 56, 40, 51, 16, 32, 39, 49, 25]. In particular, Gramm et al. [40] showed that Closest String is fixed-parameter tractable parameterized by $d$: they gave an algorithm with running time $O(d^{d}\cdot|I|)$. The algorithm works over an arbitrary alphabet $\Sigma$ (i.e., the size of the alphabet is part of the input). It is an obvious question whether the dependence on $d$ can be reduced to single exponential, i.e., whether the running time can be improved to $2^{O(d)}\cdot|I|^{O(1)}$. For small fixed alphabets, Ma and Sun [55] achieved single-exponential dependence on $d$: the running time of their algorithm is $|\Sigma|^{O(d)}\cdot|I|^{O(1)}$. Improved algorithms with running time of this form, but with better constants in the exponent were given in [66, 16]. We show here that the $d^{d}$ and $|\Sigma|^{d}$ dependence are best possible (assuming the ETH): the dependence cannot be improved to $2^{o(d\log d)}$ or to $2^{o(d\log|\Sigma|)}$. More precisely, what our proof actually shows is that $2^{o(t\log t)}$ dependence is not possible for the parameter $t=\max\{d,|\Sigma|\}$. In particular, single exponential dependence on $d$ cannot be achieved if the alphabet size is unbounded.

Theorem 3.1**.**

Assuming the ETH, there is no $2^{o(d\log d)}\cdot|I|^{O(1)}$ or $2^{o(d\log|\Sigma|)}\cdot|I|^{O(1)}$ time algorithm for Closest String.

Proof.

We prove the theorem by a reduction from the Hitting Set problem considered in Theorem 2.12. Let $I$ be an instance of $k\times k$ Hitting Set with sets $S_{1}$, $\dots$, $S_{m}$; each set contains at most one element from each row. We construct an instance $I^{\prime}$ of Closest String as follows. Let $\Sigma=[2k+1]$, $L=k$, and $d=k-1$ (this means that the center string has to have at least one character common with every input string). Instance $I^{\prime}$ contains $(k+1)m$ input strings $s_{x,y}$ ($1\leq x\leq m$, $1\leq y\leq k+1$). If set $S_{x}$ contains element $(i,j)$ from row $i$, then the $i$-th character of $s_{x,y}$ is $j$; if $S_{x}$ contains no element of row $i$, then the $i$-th character of $s_{x,y}$ is $y+k$. Thus string $s_{x,y}$ describes the elements of set $S_{x}$, using a certain dummy value between $k+1$ and $2k+1$ to mark the rows disjoint from $S_{x}$. The strings $s_{x,1}$, $\dots$, $s_{x,k+1}$ differ only in the choice of the dummy values.

We claim that $I^{\prime}$ has a solution if and only if $I$ has. Suppose that $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ is a solution of $I$ for some mapping $\rho:[k]\to[k]$. Then the center string $s=\rho(1)\dots\rho(k)$ is a solution of $I^{\prime}$: if element $(i,\rho(i))$ of the solution hits set $S_{x}$ of $I$, then both $s$ and $s_{x,y}$ have character $\rho(i)$ at the $i$-th position. For the other direction, suppose that center string $s$ is a solution of $I^{\prime}$. As the length of $s$ is $k$, there is a $k+1\leq y\leq 2k+1$ that does not appear in $s$. If the $i$-th character of $s$ is some $1\leq c\leq k$, then let $\rho(i)=c$; otherwise, let $\rho(i)=1$ (or any other arbitrary value). We claim that $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ is a solution of $I$, i.e., it hits every set $S_{x}$ of $I$. To see this, consider the string $s_{x,y}$, which has at least one character common with $s$. Suppose that character $c$ appears at the $i$-th position in both $s$ and $s_{x,y}$. It is not possible that $c>k$: character $y$ is the only character larger than $k$ that appears in $s_{x,y}$, but $y$ does not appear in $s$. Therefore, we have $1\leq c\leq k$ and $\rho(i)=c$, which means that element $(i,\rho(i))=(i,c)$ of the solution hits $S_{x}$.

The claim in the previous paragraph shows that solving instance $I^{\prime}$ using an algorithm for Closest String solves the $k\times k$ Hitting Set instance $I$. Note that the size $n$ of the instance $I^{\prime}$ is polynomial in $k$ and $m$. Therefore, a $2^{o(d\log d)}\cdot|I|^{O(1)}$ or a $2^{o(d\log|\Sigma|)}\cdot|I|^{O(1)}$ algorithm for Closest String would give a $2^{o(k\log k)}\cdot(km)^{O(1)}$ time algorithm for $k\times k$ Hitting Set, violating the ETH (by Theorem 2.12). ∎

4 Distortion

Given an undirected graph $G$ with the vertex set $V(G)$ and the edge set $E(G)$, a metric associated with $G$ is $M(G)=(V(G),D)$, where the distance function $D$ is the shortest path distance between $u$ and $v$ for each pair of vertices $u,v\in V(G)$. We refer to $M(G)$ as to the graph metric of $G$. Given a graph metric $M$ and another metric space $M^{\prime}$ with distance functions $D$ and $D^{\prime}$, a mapping $f:M\rightarrow M^{\prime}$ is called an embedding of $M$ into $M^{\prime}$. The mapping $f$ has contraction $c_{f}$ and expansion $e_{f}$ if for every pair of points $p,q$ in $M$, $D(p,q)\leq D^{\prime}(f(p),f(q))\cdot c_{f}$ and $D(p,q)\cdot e_{f}\geq D^{\prime}(f(p),f(q))$ respectively. We say that $f$ is non-contracting if $c_{f}$ is at most $1$. A non-contracting mapping $f$ has distortion $d$ if $e_{f}$ is at most $d$. One of the most well studied case of graph embedding is when the host metric $M^{\prime}$ is $\mathbb{R}^{1}$ and $D^{\prime}$ is the Euclidean distance. This is also called embedding the graph into integers or line. Formally, the problem of Distortion is defined as follows.

Distortion

Input:

A graph $G$, and a positive integer $d$

Parameter:

$d$

Question:

Is there an embedding $g:V(G)\rightarrow Z$ such that for all $u,v\in V(G)$, $D(u,v)\leq|g(u)-g(v)|\leq d\cdot D(u,v)$?

The problem of finding embedding with good distortion between metric spaces is a fundamental mathematical problem [45, 52] that has been studied intensively [3, 4, 5, 48]. Embedding a graph metric into a simple low-dimensional metric space like the real line has proved to be a useful algorithmic tool in various fields (for an example see [43] for a long list of applications). Bădoiu et al. [4] studied Distortion from the viewpoint of approximation algorithms and exact algorithms. They showed that there is a constant $a>1$, such that $a$-approximation of the minimum distortion of embedding into the line, is NP-hard and provided an exact algorithm computing embedding of a $n$ vertex graph into line with distortion $d$ in time $n^{O(d)}$. Subsequently, Fellows et al. [33] improved the running time of their algorithm to $d^{O}(d)\cdot n$ and thus proved Distortion to be fixed parameter tractable parameterized by $d$. We show here that the $d^{O(d)}$ dependence in the running time of Distortion algorithm is optimal (assuming the ETH). To achieve this we first obtain a lower bound on an intermediate problem called Constrained Permutation, then give a reduction that transfers the lower bound from Constrained Permutation to Distortion. The superexponential dependence on $d$ is particularly interesting, as $c^{n}$ time algorithms for finding a minimum distortion embedding of a graph on $n$ vertices into line have been given by Fomin et al.  [37] and Cygan and Pilipczuk [20].

Constrained Permutation

Input:

Subsets $S_{1}$, $\dots$, $S_{m}$ of $[k]$

Parameter:

$k$

Question:

A permutation $\rho$ of $[k]$ such that for every $1\leq i\leq m$, there is a $1\leq j<k$ such that $\rho(j),\rho(j+1)\in S_{i}$.

Given a permutation $\rho$ of $[k]$, we say that $x$ and $y$ are neighbors if $\{x,y\}=\{\rho(i),\rho(i+1)\}$ for some $1\leq i<k$. In the Constrained Permutation problem the task is to find a permutation that hits every set $S_{i}$ in the sense that there is a pair $x,y\in S_{i}$ that are neighbors in $\rho$.

Theorem 4.1**.**

Assuming the ETH, there is no $2^{o(k\log k)}m^{O(1)}$ time algorithm for Constrained Permutation.

Proof.

Given an instance $I$ of $2k\times 2k$ Bipartite Permutation Independent Set, we construct an equivalent instance $I^{\prime}$ of Constrained Permutation. Let $k^{\prime}=24k$ and for ease of notation, let us identify the numbers in $[k^{\prime}]$ with the elements $r^{\ell}_{i}$, $\bar{r}^{\ell}_{i}$, $c^{\ell}_{j}$, $\bar{c}^{\ell}_{j}$ for $1\leq\ell\leq 3$, $1\leq i,j\leq 2k$. The values $r^{\ell}_{i}$ represent the rows and the values $c^{\ell}_{j}$ represent the columns. If $\bar{r}^{\ell}_{i}$ and $c^{\ell}_{j}$ are neighbors in $\rho$, then we interpret it as selecting element $j$ from row $i$. More precisely, we want to construct the sets $S_{1}$, $\dots$, $S_{m}$ in such a way that if $(1,\delta(1))$, $\dots$, $(2k,\delta(2k))$ is a solution of $I$, then the following permutation $\rho$ of $[k^{\prime}]$ is a solution of $I^{\prime}$:

[TABLE]

The first property that we want to ensure is that every solution of $I^{\prime}$ looks roughly like $\rho$ above: pairs $r^{\ell}_{i}\bar{r}^{\ell}_{i}$ and pairs $c^{\ell}_{j}\bar{c}^{\ell}_{j}$ alternate in some order. Then we can define a permutation $\delta$ such that $\delta(i)=j$ if $r^{1}_{i}\bar{r}^{1}_{i}$ is followed by the pair $c^{1}_{j}\bar{c}^{1}_{j}$. The sets in instance $I^{\prime}$ will ensure that this permutation $\delta$ is a solution of $I$. Let instance $I^{\prime}$ contain the following groups of sets:

For every $1\leq\ell\leq 3$ and $1\leq i\leq 2k$, there is a set $\{r^{\ell}_{i},\bar{r}^{\ell}_{i}\}$ , 2. 2.

For every $1\leq\ell\leq 3$ and $1\leq j\leq 2k$, there is a set $\{c^{\ell}_{j},\bar{c}^{\ell}_{j}\}$, 3. 3.

For every $1\leq\ell^{\prime}<\ell^{\prime\prime}\leq 3$, $1\leq i\leq 2k$, $X\subseteq[2k]$, there is a set $\{\bar{r}^{\ell^{\prime}}_{i},\bar{r}^{\ell^{\prime\prime}}_{i}\}\cup\{c^{\ell^{\prime}}_{j}\mid j\in X\}\cup\{c^{\ell^{\prime\prime}}_{j}\mid j\not\in X\}$, 4. 4.

For every $1\leq i\leq k$, there is a set $\{\bar{r}^{1}_{i}\}\cup\{c^{1}_{j}\mid 1\leq j\leq k\}$, 5. 5.

For every $k+1\leq i\leq 2k$, there is a set $\{\bar{r}^{1}_{i}\}\cup\{c^{1}_{j}\mid k+1\leq j\leq 2k\}$, 6. 6.

For every two adjacent vertices $(i_{1},j_{1})\in I_{1}$ and $(i_{2},j_{2})\in I_{2}$, there is a set $\{\bar{r}^{1}_{i_{1}},\bar{r}^{1}_{i_{2}}\}\cup\{c^{1}_{j}\mid 1\leq j\leq k,j\neq j_{1}\}\cup\{c^{1}_{j}\mid k+1\leq j\leq 2k,j\neq j_{2}\}$.

Recall that every edge of instance $I$ goes between the independent sets $I_{1}=\{(i,j)\mid i,j\leq k\}$ and $I_{2}=\{(i,j)\mid i,j\geq k+1\}$. Let us verify first that if $\delta$ is a solution of $I$, then the permutation $\rho$ described above satisfies every set. It is clear that sets in the first two groups are satisfied. To see that every set in group 3 is satisfied, consider a set corresponding to a particular $1\leq\ell^{\prime}<\ell^{\prime\prime}\leq 3$, $1\leq i\leq 2k$, $X\subseteq[2k]$. If $\delta(i)\in X$, then $\bar{r}^{\ell^{\prime}}_{i}$ and $c^{\ell^{\prime}}_{\delta(i)}$ are neighbors and both appear in the set; if $\delta(i)\not\in X$, then $\bar{r}^{\ell^{\prime\prime}}_{i}$ and $c^{\ell^{\prime\prime}}_{\delta(i)}$ are neighbors and both appear in the set. Sets in group 4 and 5 are satisfied because $\delta(i)\leq k$ for $1\leq i\leq k$ and $\delta(i)\geq k+1$ for $k+1\leq i\leq 2k$. Finally, let $(i_{1},j_{1})\in V_{1}$ and $(i_{2},j_{2})\in V_{2}$ be two adjacent vertices and consider the corresponding set in group 6. As the solution of $I$ is an independent set, either $\delta(i_{1})\neq j_{1}$ or $\delta(i_{2})\neq j_{2}$. In the first case, $\bar{r}^{1}_{i_{1}}$ and $c^{1}_{\delta(i_{1})}$ are neighbors and both appear in the set; in the second case, $\bar{r}^{1}_{i_{2}}$ and $c^{1}_{\delta(i_{2})}$ are neighbors and both appear in the set.

Next we show that if $\rho$ is a solution of $I^{\prime}$, then a solution for $I$ exists. We say that an element $\bar{r}^{\ell}_{i}$ is good if its neighbors are $r^{\ell}_{i}$ and $c^{\ell^{\prime}}_{j}$ for some $1\leq\ell^{\prime}\leq 3$ and $1\leq j\leq 2k$. Similarly, an element $c^{\ell}_{j}$ is good if its neighbors are $\bar{c}^{\ell}_{j}$ and $\bar{r}^{\ell^{\prime}}_{i}$ for some $1\leq\ell^{\prime}\leq 3$ and $1\leq i\leq 2k$. Our first goal is to show that every $\bar{r}^{\ell}_{i}$ and $c^{\ell}_{j}$ is good. The sets in group 1 and 2 ensure that $r^{\ell}_{i}$ and $\bar{r}^{\ell}_{i}$ are neighbors, and $c^{\ell}_{j}$ and $\bar{c}^{\ell}_{j}$ are neighbors.

We claim that for every $1\leq\ell^{\prime}<\ell^{\prime\prime}\leq 3$, and $1\leq i\leq 2k$, if elements $\bar{r}^{\ell^{\prime}}_{i}$ and $\bar{r}^{\ell^{\prime\prime}}_{i}$ are not neighbors, then both of them are good. Let us build a $4k$-vertex graph $B$ whose vertices are $c^{\ell^{\prime}}_{j}$, $c^{\ell^{\prime\prime}}_{j}$ ($1\leq j\leq 2k$). Let us connect by an edge those vertices that are neighbors in $\rho$. Moreover, let us make $c^{\ell^{\prime}}_{j}$ and $c^{\ell^{\prime\prime}}_{j}$ adjacent for every $1\leq j\leq 2k$. Observe that the degree of every vertex is at most 2 (as $c^{\ell^{\prime}}_{j}$ has only one neighbor besides $\bar{c}^{\ell^{\prime}}_{j}$). Moreover, $B$ is bipartite: in every cycle, edges of the form $c^{\ell^{\prime}}_{j}c^{\ell^{\prime\prime}}_{j}$ alternate with edges not of this form. Therefore, there is a bipartition $(Y,\bar{Y})$ of $B$ such that the set $Y$ (and hence $\bar{Y}$) contains exactly one of $c^{\ell^{\prime}}_{j}$ and $c^{\ell^{\prime\prime}}_{j}$ for every $1\leq j\leq 2k$. Group 3 contains a set $S_{Y}=\{\bar{r}^{\ell^{\prime}}_{i},\bar{r}^{\ell^{\prime\prime}}_{i}\}\cup Y$ and a set $S_{\bar{Y}}=\{\bar{r}^{\ell^{\prime}}_{i},\bar{r}^{\ell^{\prime\prime}}_{i}\}\cup\bar{Y}$: as $Y$ contains exactly one of $c^{\ell^{\prime}}_{j}$ and $c^{\ell^{\prime\prime}}_{j}$, there is a choice of $X$ that yields these sets. Permutation $\rho$ satisfies $S_{Y}$ and $S_{\bar{Y}}$, thus each of $S_{Y}$ and $S_{\bar{Y}}$ contains a pair of neighboring elements. By assumption, this pair cannot be $\bar{r}^{\ell^{\prime}}_{i}$ and $\bar{r}^{\ell^{\prime\prime}}_{i}$. As $Y$ induces an independent set of $B$, this pair cannot be contained in $Y$ either. Thus the only possibility is that one of $\bar{r}^{\ell^{\prime}}_{j}$ and $\bar{r}^{\ell^{\prime\prime}}_{j}$ is the neighbor of an element of $Y$. If, say, $\bar{r}^{\ell^{\prime}}_{j}$ is a neighbor of an element $y\in Y$, then $\bar{r}^{\ell^{\prime}}_{j}$ is good. In this case, $\bar{r}^{\ell^{\prime}}_{j}$ is not the neighbor of any element of $\bar{Y}$, which means that the only way two members of $S_{\bar{Y}}$ are neighbors if $\bar{r}^{\ell^{\prime\prime}}_{j}$ is a neighbor of a member of $\bar{Y}$, i.e., $\bar{r}^{\ell^{\prime\prime}}_{j}$ is also good.

At most one of $\bar{r}^{2}_{i}$ and $\bar{r}^{3}_{i}$ can be the neighbor of $\bar{r}^{1}_{i}$, thus we can assume that $\bar{r}^{1}_{i}$ and $\bar{r}^{\ell}_{i}$ are not neighbors for some $\ell\in\{2,3\}$. By the claim in the previous paragraph, $\bar{r}^{1}_{i}$ and $\bar{r}^{\ell}_{i}$ are both good. In particular, this means that $\bar{r}^{1}_{i}$ is not the neighbor of $\bar{r}^{2}_{i}$ and $\bar{r}^{3}_{i}$, hence applying again the claim, it follows that $\bar{r}^{2}_{i}$ and $\bar{r}^{3}_{i}$ are both good. Thus $\bar{r}^{\ell}_{i}$ is good for every $1\leq\ell\leq 3$ and $1\leq i\leq 2k$, and the pigeonhole principle implies that $c^{\ell}_{j}$ is good for every $1\leq\ell\leq 3$ and $1\leq i\leq 2k$.

As every $c^{1}_{j}$ is good, the sets in groups 4 and 5 can be satisfied only if every $\bar{r}^{1}_{i}$ has a neighbor $c^{1}_{j}$. Let $\delta(i)=j$ if $c^{1}_{j}$ is the neighbor of $\bar{r}^{1}_{i}$; clearly $\delta$ is a permutation of $[2k]$. We claim that $\delta$ is a solution of $I$. The sets in group 4 and 5 ensure that $\delta(i)\leq k$ for every $1\leq i\leq k$ and $\delta(i)\geq k+1$ if $k+1\leq i\leq 2k$. To see that $(1,\delta(i))$, $\dots$, $(2k,\delta(2k))$ is an independent set, consider two adjacent vertices $(i_{1},j_{1})\in I_{1}$ and $(i_{2},j_{2})\in I_{2}$. We show that it is not possible that $\delta(i_{1})=j_{1}$ and $\delta(i_{2})=j_{2}$. Consider the set $S$ in group 6 corresponding to the edge connecting $(i_{1},j_{1})$ and $(i_{2},j_{2})$. As $\bar{r}^{1}_{i_{1}}$, $\bar{r}^{1}_{i_{2}}$, and every $c^{1}_{j}$ is good, then only way $S$ is can be satisfied is that $\bar{r}^{1}_{i_{1}}$ or $\bar{r}^{1}_{i_{2}}$ is the neighbor of some $c^{1}_{j}$ appearing in $S$. If $\delta(i_{1})=j_{1}$ and $\delta(i_{2})=j_{2}$, then the $c^{1}_{j_{1}}$ and $c^{1}_{j_{2}}$ are the neighbors of $\bar{r}^{1}_{i_{1}}$ and $\bar{r}^{1}_{i_{2}}$, respectively, but $c^{1}_{j_{1}}$ and $c^{1}_{j_{2}}$ do not appear in $S$. This shows that if there is a solution for $I^{\prime}$, then there is a solution for $I$ as well.

The size of the constructed instance $I^{\prime}$ is polynomial in $2^{k}$. Thus if $I^{\prime}$ can be solved in time $2^{o(k^{\prime}\log k^{\prime})}\cdot|I^{\prime}|=2^{o(k\log k)}\cdot 2^{O(k)}=2^{o(k\log k)}$, then this gives a $2^{o(k\log k)}$ time algorithm for $2k\times 2k$ Bipartite Permutation Independent Set. ∎

Theorem 4.2**.**

Assuming the ETH, there is no $2^{o(d\log d)}\cdot n^{O(1)}$ time algorithm for Distortion.

Proof.

We prove the theorem by a reduction from the Constrained Permutation problem. Let $I$ be an instance of Constrained Permutation  consisting of subsets $S_{1}$, $\dots$, $S_{m}$ of $[k]$. Now we show how to construct the graph $G$, an input to Distortion corresponding to $I$. For an ease of presentation we identify $[k]$ with vertices $u_{1},\ldots,u_{k}$. We also set $U=\{u_{1},\ldots,u_{k}\}$ and $d=2k$. The vertex set of $G$ consists of the following set of vertices.

• •

A vertex $u^{i}_{j}$ for every $1\leq i\leq m$ and $1\leq j\leq k$. We also denote the set $\{u_{1}^{i},\ldots,u_{k}^{i}\}$ by $U_{i}$.

• •

A vertex $s_{i}$ for each set $S_{i}$.

• •

Two cliques $C_{a}$ and $C_{b}$ of size $d+1$ consisting of vertices $c_{a}^{1},\ldots,c_{a}^{d+1}$ and $c_{b}^{1},\ldots,c_{b}^{d+1}$ respectively.

• •

A path $P$ of length $m$ (number of edges) consisting of vertices $v_{1},\ldots,v_{m+1}$.

We add the following more edges among these vertices. We add edges from all the vertices in clique $C_{a}$ but $c_{a}^{1}$ to $v_{1}$ and add edges from all the vertices in clique $C_{b}$ but $c_{b}^{1}$ to $v_{m+1}$. For all $1\leq i<m$ and $1\leq j\leq k$, make $u^{i}_{j}$ adjacent to $v_{i}$, $v_{i+1}$ and $u^{i+1}_{j}$. For $1\leq j\leq k$, make $u^{m}_{j}$ adjacent to $v_{m}$, $v_{m+1}$. Finally make $s_{i}$ adjacent to $u_{j}^{i}$ if $u_{j}\in S_{i}$. This concludes the construction. A figure corresponding to the construction can be found in Figure 1.

For our proof of correctness we also need the following known facts about distortion $d$ embedding of a graph into integers. For an embedding $g$, let $v_{1}$, $v_{2},\dots,v_{q}$ be an ordering of the vertices such that $g(v_{1})<g(v_{2})<\dots<g(v_{n})$. If $g$ is such that for all $1\leq i<q$, $D(v_{i},v_{i+1})=|g(v_{i})-g(v_{i+1})|$, then the mapping $g$ is called pushing embedding. It is known that pushing embeddings are always non-contracting and if $G$ can be embedded into integers with distortion $d$, then there is a pushing embedding of $G$ into integers with distortion $d$ [33].

Let a permutation $\rho$ of $[k]=U$ be a solution to $I$, an instance of Constrained Permutation. This automatically leads to a permutation on $U$ that we represent by $\rho(U)$. There is a natural bijection between $U$ and $U_{i}$ with $u_{j}\in U$ being mapped to $u_{j}^{i}$. So when we write $\rho(U_{i})$ then this means that the vertices of $U$ are permuted with respect to $\rho$ and being identified with its counterpart in $U_{i}$. Now we give a pushing embedding for the vertices in $G$ with $c_{a}^{1}$ being placed at [math]. All the vertices except the set vertices $s_{i}$ appear in the following order

[TABLE]

Since $\rho$ is a solution to $I$ we know that for every $S_{i}$ there exists a $1\leq j<k$ such that $\rho(j)\rho(j+1)\in S_{i}$. We place $s_{i}$ between $\rho(u_{j}^{i})$ and $\rho(u_{j+1}^{i})$. By our construction the given embedding is pushing and hence non-contracting. To show that for every pair of vertices $u,v\in V(G)$, $|g(u)-g(v)|\leq d\cdot D(u,v)$, we only have to show that for every edge $uv\in E(G)$, $|g(u)-g(v)|\leq d$. This can be readily checked from the construction. What needs to be verified is that for any two adjacent vertices $u$ and $v$, the sequence of vertices between $u$ and $v$ in the pushing embedding give a total distance at most $d\cdot D(u,v)$. The cruical observation is that the distance between two consecutive vertices from $U_{i}$ is $2$, and hence it must be at least distance $2$ apart on the line. If $s_{i}$ is adjacent to two consecutive vertices in $U_{i}$ we can “squeeze” in $s_{i}$ between those two vertices without disturbing the rest of the construction.

In the reverse direction, assume that we start with a distortion $d$ pushing embedding of $G$. Consider the layout of the graph induced on $C_{a}$ and the vertex $v_{1}$. This is a clique of size $d+2$ minus an edge and hence $C_{a}\cup\{v_{1}\}$ can be layed out in two ways: $c_{a}^{1},C_{a}\setminus\{c_{a}^{1}\},v_{1}$ or $v_{1},C_{a}\setminus\{c_{a}^{1}\},c_{a}^{1}$. Since we can reverse the layout, we can assume without loss of generality that it is $c_{a}^{1},C_{a}\setminus\{c_{a}^{1}\},v_{1}$. Without loss of generality we can also assume that $v_{1}$ is placed on position [math]. Since every vertex in $U_{1}$ is adjacent to $v_{1}$ and the negative positions are taken by the vertices in $C_{a}$, the $k=d/2$ vertices of $U_{1}$ must lie on the positions $\{1,\ldots,d\}$. We first argue that no vertex of $U_{1}$ occupies the position $d$. Suppose it does. Then the rightmost vertex of $U_{2}$ (to the right of $v_{1}$ in the embedding) must be on position at least $2d$. Simultaneously $v_{2}$ must be on position at most $d-1$ since $d$ is already occupied and $v_{2}$ is adjacent to $v_{1}$. But $v_{2}$ is adjacent to the rightmost vertex of $U_{2}$ and hence the distance on the line between them becomes at least $d+1$, a contradiction. So $U_{1}$ must use only positions in $\{1,\ldots,d-1\}$. Since the distance between two consecutive vertices in $U_{1}$ is $2$ together with the fact that we started with a pushing embedding imply that the vertices of $U_{1}$ occupy all odd positions of $\{1,\ldots..,d-1\}$. Now, $U_{2}$ must be on the positions in $\{d+1,\ldots,2d\}$ with the rightmost vertex in $U_{2}$ being on at least $2d-1$. Since $d-1$ is occupied by someone in $U_{1}$ and $v_{2}$ is adjacent to both $v_{1}$ and the rightmost vertex of $U_{2}$ it follows that $v_{2}$ must be on position $d$.

We can now argue similarly to the previous paragraph that $U_{2}$ does not use position $2d$, and hence $v_{3}$ is on position $2d$ while $U_{2}$ must use the odd positions of $\{d+1,\ldots,2d-1\}$. We can repeat this argument for all $i$ and position the vertex $v_{i}$ of the path at $d(i-1)$ and place the vertices of $U_{i}$ at odd positions between $d(i-1)$ and $di$. Of course, all the vertices of the clique $C_{b}$ will come after $v_{m+1}$.

Consider the order in which the embedding puts the vertices of $U_{1}$. We claim that it must put the vertices of $U_{2}$ in the same order. Look at the embedding of $U_{1}$ and $U_{2}$ from left to right and let $j$ be the first index where $u_{\alpha}^{1}$ of $U_{1}$ is placed between [math] and $d$ while $u_{\beta}^{2}$ of $U_{2}$ is placed between $d$ and $2d$ and $\alpha\neq\beta$. This implies that $u_{\alpha}^{2}$ appears further back in the permutation of $U_{2}$ and hence the distance between the positions of $u_{\alpha}^{1}$ and $u_{\alpha}^{2}$ in $U_{1}$ is more than $d$ while $u_{\alpha}^{1}$ and $u_{\alpha}^{2}$ are adjacent to each other in the graph. By repeating this argument for all $i$ and $i+1$ we can show that order of all $U_{i}$’s is the same. Consider $s_{i}$. It must be put on some even position, with some vertices of $U_{j}$ coming before and after $s_{i}$. But then, because we started with pushing embedding we have that $s_{i}$ is adjacent to both those vertices, and hence $i=j$ as $s_{i}$ is adjacent to only the vertices in $U_{i}$.

Now we take the permutation $\rho$ for $[k]$, imposed by the ordering of $U_{1}$, as a solution to the instance $I$ of Constrained Permutation. For every set $S_{i}$ we need to show that there exists a $1\leq j<k$ such that $\rho(j),\rho(j+1)\in S_{i}$. Consider the corresponding $s_{i}$ in the embedding and look at the vertices that are placed left and right of it. Let these be $u_{\alpha}^{i}$ and $u_{\beta}^{i}$. Then by construction $\alpha$ and $\beta$ are neighbors to $s_{i}$ in $G$ and hence $\alpha$ and $\beta$ belong to $S_{i}$. Now since the ordering of $U_{i}$’s are same we have that they are consecutive in the permutation $\rho$. This concludes the proof in the reverse direction.

The claim in the previous paragraph shows that an algorithm finding a distortion $d$ embedding of $G$ into line solves the instance $I$ of Constrained Permutation. Note the number of vertices in $G$ is bounded by a polynomial in $k$ and $m$. Therefore a $2^{o(d\log d)}\cdot|V(G)|^{O(1)}$ algorithm for Distortion would give a $2^{o(k\log k)}\cdot(km)^{O(1)}$ algorithm for Constrained Permutation, violating the ETH by Theorem 4.1. ∎

5 Disjoint Paths

There are many natural graph problems that are fixed-parameter tractable parameterized by the treewidth of the input graph. In most cases, these results can be obtained by well-understood dynamic programming techniques. In fact, Courcelle’s Theorem provide a clean way of obtaining such results. If the dynamic programming needs to keep track of a permutation, partition, or a matching at each node, then running time of such an algorithm is typically of the form $w^{O(w)}\cdot n^{O(1)}$ on graphs with treewidth $w$ [64]. We demonstrate a problem where this form of running time is necessary for the solution and it cannot be improved to $2^{o(w\log w)}\cdot n^{O(1)}$. We start with definitions of treewidth and pathwidth.

Definitions of Treewidth and Pathwidth.

A tree decomposition of a graph $G$ is a pair $(\mathcal{X},T)$ where $T$ is a tree and ${\cal X}=\{X_{i}\mid i\in V(T)\}$ is a collection of subsets of $V$ such that:

$\bigcup_{i\in V(T)}X_{i}=V$, 2. 2.

for each edge $xy\in E$, $\{x,y\}\subseteq X_{i}$ for some $i\in V(T)$; 3. 3.

for each $x\in V$ the set $\{i\mid x\in X_{i}\}$ induces a connected subtree of $T$.

The width of the tree decomposition is $\max_{i\in V(T)}\{|X_{i}|-1\}$. The treewidth of a graph $G$ is the minimum width over all tree decompositions of $G$. We denote by ${\textbf{tw}}(G)$ the treewidth of graph $G$. If in the definition of treewidth we restrict the tree $T$ to be a path then we get the notion of pathwidth and denote it by ${\textbf{pw}}(G)$.

Now we return to our problem. Given an undirected graph $G$ and $p$ vertex pairs $(s_{i},t_{i})$, the Disjoint Paths problem asks whether there exists $p$ mutually vertex disjoint paths in $G$ linking these pairs. This is one of the classic problems in combinatorial optimization and algorithmic graph theory, and has many applications, for example in transportation networks, VLSI layout, and virtual circuits routing in high-speed networks. The problem is NP-complete if $p$ is part of the input and remains so even if restrict the input graph to be planar [47, 54]. However if $p$ is fixed then the problem is famously fixed-parameter tractable as a consequence of the seminal Graph Minors theory of Robertson and Seymour [63]. A basic building block in their algorithm for Disjoint Paths is an algorithm for Disjoint Paths on graphs of bounded treewidth. To our interest is the following parameterization of Disjoint Paths.

Disjoint Paths

Input:

A graph $G$ together with a tree-decomposition of width $w$, and $p$ vertex pairs $(s_{i},t_{i})$.

Parameter:

$w$

Question:

Does there exist $p$ mutually vertex disjoint paths in $G$ linking $s_{i}$ to $t_{i}$?

The best known algorithm for this problem runs in time $2^{O(w\log w)}\cdot n$ [64] and here we show that this is indeed optimal. To get this lower bound we first give a linear parameter reduction from $k\times k$ Hitting Set to Directed Disjoint Paths, a variant of Disjoint Paths where the input is a directed graph, parameterized by pathwidth of the underlying undirected graph. Finally we obtain a lower bound of $2^{o(k\log k)}|V(G)|^{O(1)}$ on Disjoint Paths parameterized by pathwidth under the ETH, by giving a linear parameter reduction from Directed Disjoint Paths  parameterized by pathwidth to Disjoint Paths  parameterized by pathwidth. Obviously, this proves the same lower bound under the (potentially much smaller) parameter treewidth as well.

Theorem 5.1**.**

Assuming the ETH, there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm for Directed Disjoint Paths.

Proof.

The key tool in the reduction from $k\times k$ Hitting Set to Directed Disjoint Paths is the following gadget. For every $k\geq 1$ and set $S\subseteq[k]\times[k]$, we construct the gadget $G_{k,S}$ the following way (see Figure 2 for illustration).

• •

For every $1\leq i\leq k$, it contains vertices $a_{i}$, $b_{i}$.

• •

For every $1\leq i,j\leq k$, it contains a vertex $v_{i,j}$ and edges $\overrightarrow{a_{i}v_{i,j}}$, $\overrightarrow{v_{i,j}b_{j}}$.

• •

For every $1\leq i\leq k$, it contains a directed path $P_{i}=c_{i,0}d_{i,1}v^{*}_{i,1}c_{i,1}\dots d_{i,k}v^{*}_{i,k}c_{i,k}$.

• •

For every $1\leq i,j\leq k$, it contains vertices $f_{i,j}$, $f^{1}_{i,j}$, $f^{2}_{i,j}$ and edges $\overrightarrow{b_{j}f_{i,j}}$, $\overrightarrow{f_{i,j}c_{i,j}}$, $\overrightarrow{f^{1}_{i,j}f_{i,j}}$, $\overrightarrow{f_{i,j}f^{2}_{i,j}}$, $\overrightarrow{f^{1}_{i,j}c_{i,0}}$, $\overrightarrow{c_{i,j-1}f^{2}_{i,j}}$.

• •

It contains two vertices $s$ and $t$, and for every $(i,j)\in S$, there are two edges $\overrightarrow{sd_{i,j}}$, $\overrightarrow{d_{i,j}t}$.

The demand pairs in the gadget are as follows:

• •

For every $1\leq i\leq k$, there is a demand $(a_{i},c_{i,k})$.

• •

For every $1\leq i,j\leq k$, there is a demand $(f^{1}_{i,j},f^{2}_{i,j})$.

• •

There is a demand $(s,t)$.

This completes the description of the gadget. The intuition behind the construction is the following. To satisfy the demand $(a_{i},c_{i,k})$, the path needs to leave $a_{i}$ to $v_{i,j}$ for some $1\leq j\leq k$. Thus if a collection of paths form a solution for the gadget, then for every $1\leq i\leq k$, exactly one of the vertices $v_{i,1}$, $\dots$, $v_{i,k}$ is used by the paths. We say that a solution represents the mapping $\rho:[k]\to[k]$ if for every $1\leq i\leq k$, vertex $v_{i,\rho(i)}$ is used by the paths in the solution. Moreover, if the path satisfying $(a_{i},c_{i,k})$ leaves $a_{i}$ to $v_{i,j}$, then it enters the path $P_{i}$ via the vertex $f_{i,j}$, and reaches $c_{i,k}$ on the path $P_{i}$. In this case, the demand $(f^{1}_{i,j},f^{2}_{i,j})$ cannot use vertex $f_{i,j}$, and has to use the part of $P_{i}$ from $c_{i,0}$ to $c_{i,j-1}$. Then these two paths leave free only vertex $v^{*}_{i,j}$ of $P_{i}$ and no other $v^{*}$. This means that the $v_{i,j}$ and $v^{*}_{i,j}$ vertices behave exactly the opposite way: if $v_{i,\rho(i)}$ is used by the solution, then every vertex $v^{*}_{i,1}$, $\dots$, $v^{*}_{i,k}$ is used, with the exception of $v^{*}_{i,j}$. The following claim formalizes this important property of the gadget.

Claim 5.2**.**

For every $k\geq 1$ and $S\subseteq[k]\times[k]$, gadget $G_{k,S}$ has the following properties:

For every $\rho:[k]\to[k]$ that hits $S$, gadget $G_{k,S}$ has a solution that represents $\rho$, and $v^{*}_{i,\rho(i)}$ is not used by the paths in the solution for any $1\leq i\leq k$. 2. 2.

If $G_{k,S}$ has a solution that represents $\rho$, then $\rho$ hits $S$ and vertex $v^{*}_{i,j}$ is used by the paths in the solution for every $1\leq i\leq k$ and $j\neq\rho(i)$.

Proof.

To prove the first statement, we construct a solution the following way. Demand $(a_{i},c_{i,j})$ is satisfied by the path $a_{i}v_{i,\rho(i)}b_{\rho(i)}f_{i,\rho(i)}c_{i,\rho(i)}\dots c_{i,k}$, where we use a subpath of $P_{i}$ to go from $c_{i,\rho(i)}$ to $c_{i,k}$. For every $1\leq i,j\leq k$, if $j\neq\rho(i)$, then the demand $(f^{1}_{i,j},f^{2}_{i,j})$ is satisfied by the path $f^{1}_{i,j}f_{i,j}f^{2}_{i,j}$. If $j=\rho(i)$, then vertex $f_{i,j}$ is already used by the demand $(a_{i},c_{i,j})$. In this case demand $(f^{1}_{i,j},f^{2}_{i,j})$ is satisfied by the path $f^{1}_{i,j}c_{i,0}\dots c_{i,j-1}f^{2}_{i,j}$. Finally, as $\rho$ hits $S$, there is a $1\leq i\leq k$ such that $(i,\rho(i))\in S$ and hence the edges $\overrightarrow{sd_{i,\rho(i)}}$ and $\overrightarrow{d_{i,\rho(i)}t}$ exist. Therefore, we can satisfy the demand $(s,t)$ via $d_{i,\rho(i)}$. Note that this vertex is not used by the other paths: the path satisfying demand $(a_{i},c_{i,k})$ uses $P_{i}$ only from $c_{i,\rho(i)}$ to $c_{i,k}$, the path satisfying demand $(f^{1}_{i,\rho(i)},f^{2}_{i,\rho(i)})$ uses $P_{i}$ from $c_{i,0}$ to $c_{i,\rho(i)-1}$, and no other path reaches $P_{i}$. This also implies that $v^{*}_{i,\rho(i)}$ is used by none of the paths, as required.

For the second part, consider a solution of $G_{k,S}$ representing some mapping $\rho$. This means that the path of demand $(a_{i},c_{i,k})$ uses vertex $v_{i,\rho(i)}$ and hence $b_{i,\rho(i)}$. The only way to reach $c_{i,k}$ from $b_{i,\rho(i)}$ without going through any other terminal vertex is using the path $f_{i,\rho(i)}c_{i,\rho(i)}\dots c_{i,k}$. This means that demand $(f^{1}_{i,\rho(i)},f^{2}_{i,\rho(i)})$ cannot use vertex $f_{i,\rho(i)}$, hence it has to use the path $f^{1}_{i,\rho(i)}c_{i,0}\dots c_{i,\rho(i)-1}f^{2}_{i,\rho(i)}$. It follows that for every $1\leq i\leq k$ and $j\neq\rho(i)$, vertices $d_{i,j}$ and $v^{*}_{i,j}$ are used by the paths satisfying demands $(a_{i},c_{i,k})$ and $(f^{1}_{i,\rho(i)},f^{2}_{i,\rho(i)})$. This shows that every $v^{*}_{i,j}$ with $j\neq\rho(i)$ is used by the paths in the solution. Moreover, the path satisfying $(s,t)$ has to go through vertex $d_{i,\rho(i)}$ for some $i$. By the way the edges incident to $s$ and $t$ are defined, this is only possible if $\rho(i)\in S$, that is, $\rho$ hits $S$. ∎

Let $S_{1}$, $\dots$, $S_{m}$ be the sets appearing in the $k\times k$ Hitting Set instance $I$. We construct an instance $\vec{I}$ of Directed Disjoint Paths consisting of $m$ gadgets $G_{1}$, $\dots$, $G_{m}$, where gadget $G_{t}$ $(1\leq t\leq m)$ is a copy of the gadget $G_{k,S_{i}}$ defined above. For every $1\leq t<m$ and every $1\leq i,j\leq k$, we identify vertex $v^{*}_{i,j}$ of $G_{t}$ and vertex $v_{i,j}$ of $G_{t+1}$. This completes the description of the instance $\vec{I}$ of Directed Disjoint Paths.

We have to show that the pathwidth of the constructed graph $\vec{G}$ of $\vec{I}$ is $O(k)$ and that $\vec{I}$ has a solution if and only if $I$ has. To bound the pathwidth of $\vec{G}$, for every $0\leq t\leq m$, $1\leq i,j\leq k$, let us define the bag $B_{t,i,j}$ such that it contains vertices $a_{1}$, $\dots$, $a_{k}$, $b_{1}$, $\dots$, $b_{k}$, $s$, $t$, $f_{i,j}$, $f^{1}_{i,j}$, $f^{2}_{i,j}$, and the path $P_{i}$ of gadget $G_{t}$ (unless $t=0$), and vertices $a_{1}$, $\dots$, $a_{k}$, $b_{1}$, $\dots$, $b_{k}$ of gadget $G_{t+1}$ (unless $t=m$). It can be easily verified that the size of each bag is $O(k)$ and if two vertices are adjacent, then they appear together in some bag. Furthermore, if we order the bags lexicographically according to $(t,i,j)$, then each vertex appears precisely in an interval of the bags. This shows that the pathwidth of $\vec{G}$ is $O(k)$.

Next we show that if $I$ has a solution $\rho:[k]\to[k]$, then $\vec{I}$ also has a solution. As $\rho$ hits every $S_{t}$, by the first part of the Claim, each gadget $G_{t}$ has a solution representing $\rho$. To combine these solutions into a solution for $\vec{I}$, we have to make sure that the vertices $v_{i,j}$, $v^{*}_{i,j}$ that were identified are used only in one gadget. Since the solution for gadget $G_{t}$ represents $\rho$, it uses vertices $v_{1,\rho(i)}$, $\dots$, $v_{k,\rho(k)}$, but no other $v_{i,j}$ vertex. As vertex $v_{i,j}$ of gadget $G_{t}$ was identified with vertex $v^{*}_{i,j}$ of gadget $G_{t-1}$, these vertices might be used by the solution of $G_{t-1}$ as well. However, the solution of $G_{t-1}$ also represents $\rho$ and as claimed in the first part of the Claim, the solution does not use vertices $v^{*}_{1,\rho(1)}$, $\dots$, $v^{*}_{k,\rho(k)}$. Therefore, no conflict arises between the solutions of $G_{t}$ and $G_{t-1}$.

Finally, we have to show that a solution for $\vec{I}$ implies that a solution for $I$ exists. We say that a solution for $\vec{I}$ is normal with respect to $G_{t}$ if the paths satisfying the demands in $G_{t}$ do not leave $G_{t}$ (the vertices $v_{i,j}$, $v^{*}_{i,j}$ that were identified are considered as part of both gadgets, so we allow the paths to go through these vertices). We show by induction that the solution for $\vec{I}$ is normal for every $G_{t}$. Suppose that this is true for $G_{t-1}$. If some path $P$ satisfying a demand in $G_{t}$ leaves $G_{t}$, then it has to enter either $G_{t-1}$ or $G_{t+1}$. If $P$ enters a vertex of $G_{t+1}$ that is not in $G_{t}$, then it cannot go back to $G_{t}$: the only way to reach a vertex $v_{i,j}$ of $G_{t+1}$ is from vertex $a_{i}$, which has indegree 0. Therefore, let us suppose that $P$ enters $G_{t-1}$ at some vertex $v^{*}_{i,j}$ of $G_{t-1}$. The only way the path can return to $G_{t}$ is via some vertex $v^{*}_{i,j^{\prime}}$ of $G_{t-1}$ with $j^{\prime}\geq j$. By the induction hypothesis, the solution is normal with respect to $G_{t-1}$, thus the second part of the Claim implies that there is a unique $j$ such that $v^{*}_{i,j}$ is not used by the paths satisfying the demands in $G_{t-1}$. As $P$ can use only this vertex $v^{*}_{i,j}$, it follows that $j^{\prime}=j$ and hence path $P$ does not use any vertex of $G_{t-1}$ not in $G_{t}$, In other words, $P$ does not leave $G_{t}$.

Suppose now that the solution is normal with respect to every $G_{t}$, which means that it induces a solution for every gadget. Suppose that the solution of gadget $G_{t}$ represents mapping $\rho_{t}$. We claim that every $\rho_{t}$ is the same. Indeed, if $\rho_{t}(i)=j$, then the solution of $G_{t}$ uses vertex $v_{i,j}$ of $G_{t}$, which is identical to vertex $v^{*}_{i,j}$ of $G_{t-1}$. This means that the solution of $G_{t-1}$ does not use $v^{*}_{i,j}$, and by the second part of the Claim, this is only possible if $\rho_{t-1}(i)=j$. Thus $\rho_{t-1}=\rho_{t}$ for every $1<i\leq m$, let $\rho$ be this mapping. Again by the claim, $\rho$ hits every set $S_{t}$ in instance $I$, thus $\rho$ is a solution of $I$. ∎

For our main proof we will also need the following lemma.

Lemma 5.3** ([7]).**

Let $G$ be a graph (possibly with parallel edges) having pathwidth at most $w$. Let $G^{\prime}$ be obtained from $G$ by subdividing some of the edges. Then the pathwidth of $G^{\prime}$ is at most $w+1$.

Theorem 5.4**.**

Assuming the ETH, there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm for Disjoint Paths.

Proof.

Let $\vec{I}$ be a instance of Directed Disjoint Paths on a directed graph $D$ having pathwidth $w$. We transform $D$ into an undirected graph $G$, where two adjacent vertices $v_{\text{in}}$, $v_{\text{out}}$ correspond to each vertex $v$ of $D$, and if $\overrightarrow{uv}$ is an edge of $D$, then we introduce a new vertex $e_{uv}$ that is adjacent to both $u_{\text{out}}$ and $v_{\text{in}}$. It is not difficult to see that the pathwidth of $G$ is at most $2w+2=O(w)$: $G$ can be obtained from the underlying graph of $D$ by duplicating vertices (which at most doubles the size of each bag) and subdividing edges (which increases pathwidth at most by one).

Let $I$ be an instance of Disjoint Paths on $G$ where there is a demand $(v_{\text{out}},u_{\text{in}})$ corresponding to every demand of $(v,u)$ of $\vec{I}$. It is clear that if $\vec{I}$ has a solution, then $I$ has a solution as well: every directed path from $u$ to $v$ in $D$ can be turned into a path connecting $u_{\text{out}}$ and $v_{\text{in}}$ in $G$. However, the converse is not true: it is possible that an undirected path $P$ in $G$ reaches $v_{\text{in}}$ from $e_{uv}$ and instead of continuing to $v_{\text{out}}$, it continues to some $e_{wv}$. In this case, there is no directed path corresponding to $P$ in $D$. We add further edges and demands to forbid such paths.

Let $B_{1}$, $\dots$, $B_{n}$ be a path decomposition of $G$ having width $w^{\prime}=O(w)$. For every vertex $x$ of $G$, let $\ell(x)$ and $r(x)$ be the index of the first and last bags, respectively, were $x$ appears. It is well-known that the decomposition can be chosen such that $r(x)\neq r(y)$ for any two vertices $x$ and $y$.

We modify $G$ to obtain a graph $G^{\prime}$ the following way. If vertex $v$ has $d$ inneighbors $u_{1}$, $\dots$, $u_{d}$ in $D$, then $v_{\text{in}}$ has $d+1$ neighbors in $G$: $v_{\text{out}}$ and $d$ vertices $e_{u_{1}v}$, $\dots$, $e_{u_{d}v}$. Suppose that the neighbors of $v$ are ordered such that $r(e_{u_{1}v})<\dots<r(e_{u_{d}v})$. We introduce $2d-2$ new vertices $v^{s}_{1}$, $\dots$, $v^{s}_{d-1}$, $v^{t}_{1}$, $\dots$, $v^{t}_{d-1}$ such that $v^{s}_{i}$ and $v^{t}_{i}$ are both adjacent to $e_{u_{i}v}$ and $e_{u_{i+1}v}$. For every $1\leq i\leq d-1$, we introduce a new demand $(v^{s}_{i},v^{t}_{i})$. Repeating this procedure for every vertex $v$ of $D$ creates an instance $I^{\prime}$ of undirected Disjoint Paths on a graph $G^{\prime}$.

We show that these new vertices and edges increase the pathwidth at most by a constant factor. Observe that $G^{\prime}$ can be obtained from $G$ by adding two parallel edges between $e_{u_{i}v}$ and $e_{u_{i+1}v}$ and subdividing them. Thus by Lemma 5.3, all we need to show is that adding these new edges increases pathwidth only by a constant factor. If $r(e_{u_{i}v})\geq\ell(e_{u_{i+1}v})$, then the parallel edges between $e_{u_{i}v}$ and $e_{u_{i+1}v}$ can be added without changing the path decomposition: bag $B_{r(e_{iv})}$ contains both vertices. If $r(e_{u_{i}v})<\ell(e_{u_{i+1}v})$, then let us insert vertex $e_{u_{i}v}$ into every bag $B_{j}$ for $r(e_{u_{i}v})<j\leq\ell(e_{u_{i+1}v})$. Now bag $B_{\ell(e_{u_{i+1}v})}$ contains both $e_{u_{i}v}$ and $e_{u_{i+1}v}$, thus we can add two parallel edges between them. Note that vertex $v_{\text{in}}$ appears in every bag where $e_{u_{i}v}$ is inserted: if not, then either $v_{\text{in}}$ does not appear in bags with index at most $r(e_{u_{i}v})$, or it does not appear in bags with index at least $\ell(e_{u_{i+1}v})$, contradicting the fact that $v_{\text{in}}$ is adjacent to both $e_{u_{i}v}$ and $e_{u_{i+1}v}$. Furthermore, vertices $e_{u_{i}v}$ and $e_{u_{j}v}$ are not inserted into the same bag for any $i\neq j$: if $j>i$, then $r(e_{u_{j}v})>r(e_{u_{i+1}v})\geq\ell(e_{u_{i+1}v})$. Therefore, the number of new vertices in each bag is at most the original size of the bag, i.e., the size of each bag increases by at most a factor of 2.

We claim that $I^{\prime}$ has a solution if and only if $\vec{I}$ has. If $\vec{I}$ has a solution, then the directed path satisfying demand $(u,v)$ gives in a natural way an undirected path in $G^{\prime}$ that satisfies demand $(u_{\text{out}},v_{\text{in}})$. Thus we can obtain a pairwise disjoint collection of paths that satisfy the demands of the form $(u_{\text{out}},v_{\text{in}})$. Note that if $v_{\text{out}}$, $e_{u_{1}v}$, $\dots$, $e_{u_{d}v}$ are the neighbors of $v_{\text{in}}$ in $G^{\prime}$, then the paths in this collection use at most one of the vertices $e_{u_{1}v}$, $\dots$, $e_{u_{d}v}$, say, $e_{u_{j}v}$. Now we can satisfy the demands $(v^{s}_{i},v^{t}_{i})$ for every $1\leq i\leq d-1$: for $i<j$, we can use the path $v^{s}_{i}e_{u_{i}v}v^{t}_{i}$, and for $i\geq j$, we can use the path $v^{s}_{i}e_{u_{i+1}v}v^{t}_{i}$. Thus instance $I^{\prime}$ has a solution.

For the other direction, suppose that $I^{\prime}$ has a solution. Let us call a path of this solution a main path if it satisfies a demand of the form $(u_{\text{out}},v_{\text{in}})$. We claim that if $v_{\text{in}}$ is an internal vertex of a main path $P$, then $P$ contains $v_{\text{out}}$ as well. Otherwise, $P$ has to contain at least two of the neighbors $e_{u_{1}v}$, $\dots$, $e_{u_{d}v}$ of $v_{\text{out}}$. In this case, less than $d-1$ vertices out of $e_{u_{1}v}$, $\dots$, $e_{u_{d}v}$ remain available for the $d-1$ demands $(v^{s}_{1},v^{t}_{1})$, $\dots$, $(v^{s}_{d},v^{t}_{d})$, a contradiction.

Consider a main path $P$ that satisfies a demand $(u_{\text{out}},v_{\text{in}})$ of $I^{\prime}$. Clearly, $P$ cannot go through any terminal vertex other than $u_{\text{out}}$ and $v_{\text{in}}$. As $u$ has indegree 0 in $D$, path $P$ has to go to some $e_{uw}$ and then to $w_{\text{in}}$ after starting from $u_{\text{out}}$. By our claim in the previous paragraph, the next vertex has to be $w_{\text{out}}$, then again some $e_{wz}$ and $z_{\text{in}}$ and so on. Thus there is a directed path in $D$ that corresponds to $P$ in $G^{\prime}$. This means that directed paths corresponding to the main paths of the solution for $I^{\prime}$ form a solution for $\vec{I}$. ∎

6 Chromatic Number

In this section, we give another lower bound result for a problem that is known to admit an algorithm with running time $w^{O(w)}\cdot n^{O(1)}$ on graphs with treewidth $w$. In particular we show that the running time cannot be improved to $2^{o(w\log w)}\cdot n^{O(1)}$ unless the ETH collapses. Given a graph $G$, a function $f:V(G)\rightarrow\{1,\ldots,\ell\}$, is called an $\ell$-proper coloring of $G$, if for any edge $uv\in E(G)$, we have that $f(u)\neq f(v)$. The chromatic number of a graph $G$ is the minimum positive integer $\ell$ for which $G$ admits an proper $\ell$-coloring and is denoted by $\chi(G)$. In the Chromatic Number problem, we are given a graph $G$ and objective is to find the value of $\chi(G)$. It is well known that if $G$ has treewidth $w$ then $\chi(G)\leq w+1$. Using, this we can obtain an algorithm for Chromatic Number running in time $w^{O(w)}\cdot n^{O(1)}$ on graphs with treewidth $w$ [46]. We show that in fact this running time is optimal.

In what follows, we give a lower bound for a parameter even larger than the treewidth of the input graph. Given a graph $G$, a subset of vertices $C$ is called vertex cover if for every $uv\in E$, either $u\in C$ or $v\in C$. In other words, $G-C$ is an independent set. In particular, we will study the following parameterization of the problem.

Chromatic Number

Input:

A graph $G$ together with a vertex cover $C$ of size at most $k$ and a positive integer $\ell$.

Parameter:

$k$

Question:

Is $\chi(G)\leq\ell$?

It is well known that if $G$ has a vertex cover of size $k$, then its treewidth is upper bounded by $k+1$ and thus we can test whether $\chi(G)\leq\ell$ in time $k^{O(k)}\cdot n^{O(1)}$.

Theorem 6.1**.**

Assuming the ETH, there is no $2^{o(k\log k)}\cdot n^{O(1)}$ time algorithm for Chromatic Number parameterized by vertex cover number.

Proof.

We prove the theorem by a reduction from the $k\times k$ Permutation Clique  problem. Let $(I,k)$ be an instance of $k\times k$ Permutation Clique consisting of a graph $H$ over the vertex set $[k]\times[k]$ and a positive integer $k$. Recall that in the $k\times k$ Permutation Clique  problem, the goal is to check whether there is a clique containing exactly one vertex from each row, and containing exactly one vertex from each column. In other words, the vertices selected in the solution are $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ for some permutation $\rho$ of $[k]$.

Now we show how to construct the graph $G$, an input to Chromatic Number starting from $H$. The vertex set of $G$ consists of the following set of vertices and edges.

• •

We have two cliques $C_{a}$ and $C_{b}$ of size $k$. The vertex set of $C_{x}$, $x\in\{a,b\}$, consists of $\{x_{1},\ldots,x_{k}\}$.

• •

For every $i,j,x,y\in[k]$, $i\neq j$ and $x\neq y$, for which $(i,x)$ and $(j,y)$ are not adjacent in $H$, we have a new vertex $w_{xy}^{ij}$. We first make $w_{xy}^{ij}$ adjacent to $b_{i}$ and $b_{j}$. Finally, we add edges between $w_{xy}^{ij}$ and $\{a_{1},\ldots,a_{k}\}\setminus\{a_{x},a_{y}\}$.

This concludes the construction.

We now show that $H$ has a permutation clique if and only if $\chi(G)=k$. Let the vertices selected in the permutation clique are $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ for some permutation $\rho$ of $[k]$. Now we define a proper $k$-coloring of $G$. For every $j\in[k]$, we color the vertex $a_{j}$ with $j$ and the vertex $b_{j}$ with $\rho(j)$. The only vertices that are left uncolored are $w_{xy}^{ij}$. Observe that the only colors that we can use for $w_{xy}^{ij}$ are $\{x,y\}$. Thus, if we can show that $Z=\{x,y\}\setminus\{\rho(i),\rho(j)\}$ is non-empty then we can use any color in $Z$ to color $w_{xy}^{ij}$. But that follows since there is an edge between $(i,\rho(i))$ and $(j,\rho(j))$ and there is no edge between $(i,x)$ and $(j,y)$ by definition of $w_{xy}^{ij}$.

Next we show the reverse direction. Let $f$ be a proper $k$-coloring function for $G$. Without loss of generality we can assume that $f(a_{j})=j$. For every $i\in[k]$, define $\rho(i)=f(b_{i})$. Observe that since $C_{b}$ is a clique and $f$ is a proper $k$-coloring for $H$ and hence in particular for $C_{b}$, we have that $\rho$ is a permutation of $[k]$. We claim that $(1,\rho(1))$, $\dots$, $(k,\rho(k))$ forms a permutation clique of $H$. Towards this we only need to show that there is an edge between every $(i,\rho(i))$ and $(j,\rho(j))$ in $H$. For contradiction assume that $(i,\rho(i))$ and $(j,\rho(j))$ are not adjacent in $H$. Consider, the vertex $w^{ij}_{\rho(i)\rho(j)}$ in $G$. It can only be colored with either $\rho(i)$ or $\rho(j)$. However, $f(b_{i})=\rho(i)$ and $f(b_{j})=\rho(j)$. This contradicts the fact that $f$ is a proper $k$-coloring of $G$. This concludes the proof in the reverse direction.

Finally, observe that the vertices of $C_{a}$ and $C_{b}$ form a vertex cover for $G$ of size $2k$. The claim in the previous paragraph shows that an algorithm finding a $\chi(G)$ solves the instance $(I,k)$ of $k\times k$ Permutation Clique. Note the number of vertices in $G$ is bounded by a polynomial in $k$ and the vertex cover of $G$ is bounded by $2k$. Therefore a $2^{o(k\log k)}n^{O(1)}$ algorithm for Chromatic Number would give a $2^{o(k\log k)}$ algorithm for $k\times k$ Permutation Clique, violating the ETH by Theorem 2.5. ∎

7 Conclusion

In this paper we showed that several parameterized problems have slightly superexponential running time unless the ETH fails. In particular we showed for four well-studied problems arising in three different domains that the known superexponential algorithms are optimal: assuming the ETH, there is no $2^{o(d\log d)}\cdot|I|^{O(1)}$ or $2^{o(d\log|\Sigma|)}\cdot|I|^{O(1)}$ time algorithm for Closest String, $2^{o(d\log d)}\cdot|I|^{O(1)}$ time algorithm for Distortion, and $2^{o(w\log w)}\cdot|I|^{O(1)}$ time algorithm for Disjoint Paths and Chromatic Number parameterized by treewidth. We believe that many further results of this form can be obtained by using the framework of the current paper. Two concrete problems that might be amenable to our framework are:

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Are the known parameterized algorithms for Point Line Cover [50, 41] and Directed Feedback Vertex Set [15], parameterized by the solution size, running in time $2^{O(k\log k)}\cdot|I|^{O(1)}$ optimal?

In the conference version of this paper [53], we asked further questions of this form, which have been answered by now.

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Is the $2^{O(k\log k)}\cdot|I|^{O(1)}$ time parameterized algorithm for Interval Completion [65] optimal? In 2016, Cao [13] showed that this is not the case: the problem can be solved in single-exponential time $6^{k}\cdot n^{O(1)}$. In fact, recently Bliznets et al. [8] obtained an algorithm with running time $k^{O(\sqrt{k})}\cdot n^{O(1)}$ for Interval Completion.

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Are the known parameterized algorithms for Hamiltonian Path [34], Connected Vertex Cover [58] and Connected Dominating Set [24], parameterized by the treewidth $w$ of the input graph, running in time $2^{O(w\log w)}\cdot|I|^{O(1)}$ optimal? In 2011, Cygan et al. introduced the technique of Cut & Count [19], which is able to give $2^{O(w)}\cdot|I|^{O(1)}$ time randomized algorithms for all these problems. Later, deterministic algorithms with this running time were found [9, 36]. Cygan et al. showed also that, assuming the ETH, there is no $2^{o(w\log w)}\cdot n^{O(1)}$ time algorithm for Cycle Packing on graphs of treewidth $w$.

It seems that our paper raised awareness in the field of parameterized algorithms that tight lower bounds are possible even for running times that may look somewhat unnatural, and in particular if a problem can be solved in time $2^{O(k\log k)}\cdot n^{O(1)}$, then it is worth exploring whether this can be improved to single-exponential or a lower bound can be proved. The invention of the Cut & Count technique and the related results of Cygan et al. [19] seem to be influenced by this realization. By now, there are other papers building on our work and investigating the optimality of $2^{O(k\log k)}\cdot n^{O(1)}$ time algorithms in the context of bounded-treewidth graphs or graph modification problems [10, 11, 61, 30]