Linear Kernels on Graphs Excluding Topological Minors Research funded by DFG-Project RO 927/12-1 titled Theoretical and Practical Aspects of Kernelization. A simpler proof of the results of this paper along with additional results appears in http://arxiv.org/abs/1207.0835

# Linear Kernels on Graphs Excluding Topological Minors Research funded by DFG-Project RO 927/12-1 titled “Theoretical and Practical Aspects of Kernelization.”A simpler proof of the results of this paper along with additional results appears in http://arxiv.org/abs/1207.0835

## Abstract

We show that problems that have finite integer index and satisfy a requirement we call treewidth-bounding admit linear kernels on the class of -topological-minor free graphs, for an arbitrary fixed graph . This builds on earlier results by Bodlaender et al. on graphs of bounded genus [2] and by Fomin et al. on -minor-free graphs [9]. Our framework encompasses several problems, the prominent ones being Chordal Vertex Deletion, Feedback Vertex Set and Edge Dominating Set.

## 1Introduction

Parameterized complexity deals with algorithms for decision problems whose instances consist of a secondary measurement known as the parameter. A major goal in parameterized complexity is to investigate whether a problem with parameter admits an algorithm with running time . Parameterized problems that admit such algorithms are called fixed-parameter tractable and the class of all such problems is denoted .

A closely related concept is that of kernelization. A kernelization algorithm for a parameterized problem takes as instance of the problem and, in time polynomial in , outputs an equivalent instance such that , for some function . The function is called the size of the kernel and may be viewed as a measure of the “compressibility” of a problem using polynomial-time preprocessing rules. By now it is a folklore result in the area that a decidable problem is fixed-parameter tractable iff it has a kernelization algorithm. What makes kernelization interesting is that many problems have a small kernel, meaning that the function is polynomial or some times even linear.

An important research direction is to investigate the parameterized complexity of problems that are 1 in general in special graph classes. It turns out (not surprisingly) that several problems are not only in in special graph classes but admit linear kernels. A celebrated result is the linear kernel for Dominating Set in planar graphs by Alber, Fellows, and Niedermeier [1]. This paper prompted an explosion of research papers on linear kernels in planar graphs, including Dominating Set [1], Feedback Vertex Set [3], Cycle Packing [4], Induced Matching [19], Full-Degree Spanning Tree [14] and Connected Dominating Set [17].

Guo and Niedermeier showed that several problems that admit a “distance property” admit linear kernels in planar graphs [13]. This result was subsumed by that of Bodlaender, Fomin, Lokshtanov, Penninkx, Saurabh and Thilikos in [2] who provided a meta-theorem for problems to have a linear kernel on graphs of bounded genus (a strictly larger class than planar graphs). Later Fomin, Lokshtanov, Saurabh and Thilikos in [9] extended these results for bidimensional problems to an even larger graph class, namely, -minor-free and apex-minor-free graphs. The last two papers have provided deep insight into the circumstances under which problems admit linear (and polynomial) kernels in sparse graphs. The property of finite integer index, introduced by Bodlaender and van Antwerpen-de Fluiter [5], has emerged to be of central importance to the aforementioned results: it guarantees the existence of small gadgets that “simulate” large portions of the instance satisfying certain properties. Finally note that a recent result by Fomin, Lokshtanov, Saurabh and Thilikos now provides a linear kernel for Dominating Set and Connected Dominating Set on -minor-free graphs [10].

In this paper, we partially extend the results of Fomin et al. in [9] by giving a meta-result for linear kernels on -topological-minor-free graphs. More specifically, we show that any graph problem that has finite integer index and is treewidth-bounding has a linear kernel in -topological-minor-free graphs. Informally, we call a problem treewidth-bounding if there exists a vertex set of small size whose deletion reduces the treewidth of the remaining graph to within a constant.

Its worthwhile to note that Marx and Grohe have recently developed a decomposition theorem for -topological-minor-free graphs along the same lines as the one for -minor-free graphs [12]. As the latter proved to be extremely useful in designing linear kernels for -minor-free graphs, it would be very interesting to see how one can apply this structure theorem to obtain kernels on graphs excluding a fixed topological minor. Note, however, that for the results of this paper we do not make use of this structure theorem.

The rest of the paper is organized as follows: Section 2 contains the basic definitions and some important aspects of -topological-minor-free graphs as well as a key lemma used extensively in the proof of the main result. In Section 3 we present our main result, its implications in Section 4. Finally Section 5 contains the conclusion and some open questions.

## 2Preliminaries

We use standard graph-theoretic notation (see [8] for any undefined terminology). Let be an edge in a graph . By , we denote the graph obtained by contracting the edge into a new vertex , and making it adjacent to all the former neighbors of and . A minor of is a graph obtained from a subgraph of by contracting zero or more edges. A family of graphs is said to be minor-closed if for all , every minor of is contained in . A graph is said to be -minor-free if no minor of is isomorphic to . The class of -minor-free graphs can be easily seen to be minor-closed. Note that if is -minor-free then it is also -minor-free, where . Therefore no -minor-free graph contains a clique with or more vertices. If a chordal graph is -minor-free, then every bag of the natural tree decomposition of is a maximal clique of size at most .

Given a graph , a tree-decomposition of is a pair , where is a tree and is a collection of vertex sets of with one set for each node of the tree such that the following hold:

1. ;

2. for every edge in , there exists such that ;

3. for each vertex , the set of nodes induces a subtree.

The vertices of the tree are usually referred to as nodes and the sets are called bags. The width of a tree-decomposition is the size of the largest bag minus one. The treewidth of , denoted , is the smallest width of a tree-decomposition of .

Given a subtree of a tree-decomposition of a graph , the bags of refer to the bags in that correspond to the nodes in . We let denote the graph induced by the vertices that occur in the bags of .

### 2.1Protrusions, t-Boundaried Graphs and Finite Integer Index

In this subsection, we restate the definitions and results required for using the protrusion machinery used extensively in [2].

Given a graph and a set , we define as the set of vertices in that have a neighbor in . For a set the neighborhood of is . Subscripts are omitted when it is clear which graph is being referred to.

If is an -protrusion, the vertex set is the restricted protrusion of .

A -boundaried graph is a graph with distinguished vertices labeled through . The set of labeled vertices is denoted by and is called the boundary or the terminals of . For -boundaried graphs and , we let denote the graph obtained by taking the disjoint union of and and identifying each vertex in with the vertex in with the same label. This operation is called gluing.

We now restate the definition of one of the most important notions used in this paper.

To test whether a parameterized problem has finite integer index on a graph class, one can use the sufficiency test introduced in [9] called strong monotonicity. We restate its definition for parameterized vertex deletion problems. An instance of a parameterized vertex deletion problem consists of a graph and a parameter , and the question is whether there exists a vertex set of size at most whose deletion results in a graph with some pre-specified property. Fix a vertex-deletion parameterized problem (analogous for an edge-deletion problem). Given -boundaried graphs and , we let denote the size of the smallest vertex set such that is a solution to for the problem . If no such exists, we define .

Informally, a parameterized problem is strongly monotone if for every -boundaried graph , a local solution for has nearly the same size as a global solution for restricted to for every -boundaried graph .

It turns out that any graph-theoretic optimization problem where the objective is to find a maximum or minimum sized vertex or edge set satisfying a (counting) MSO-predicate has finite integer index if it is strongly monotone.

We adapt the notion of quasi-compact problems introduced in [2] for graphs of bounded genus to that of treewidth bounding problems by removing the radial distance, which is not applicable to the more general class of graphs excluding a fixed topological minor.

For problems whose solution is a vertex subset, the set will be the solution set of . For simplicity we will call the set the solution in the following.

### 2.2Properties of H-topological-minor-free graphs

In this section we list some properties of -topological-minor-free graphs that we use in the proofs to follow. We use to denote .

The first property states that graphs that exclude a fixed graph as a topological minor are sparse in some sense.

As a corollary, a graph with average degree larger than contains as a topological minor.

It is clear that if a graph excludes as a topological minor, then it does not have as a topological minor. What is also true is that the total number of cliques (not necessarily maximal) is linear in the number of vertices.

We will sometimes be sloppy with our notation and, for a subgraph of , write instead of .

One technique frequently used in the proofs that follow is embodied in the proof of the following lemma.

We construct a topological minor such that each edge in corresponds to a subgraph . The construction works as follows. Delete all edges in the graph . For each connected subgraph , choose distinct vertices such that is not an edge and both and are adjacent to and in , respectively. We explicitly allow the case .

Next choose a path from to in and delete all vertices of save those from . Finally contract the path to an edge between and . This sequence of operations clearly produces a topological minor since the only edges that were contracted had at least one endpoint with degree at most two.

Since topological minor containment is a transitive, the graph obtained by “contracting each connected subgraph into an edge” is also -topological-minor-free. Observe that since we assumed that , for each component , there exists distinct vertices that are adjacent to and which do not yet have an edge between them. If this were not the case, the neighbors of in form a clique of size at least . We would then have an -clique in a topological-minor of , contradicting the fact that it is -topological-minor-free. It now follows that the number of subgraphs is bounded by the number of edges in . By Property ?, the number of edges is linear in the size of and we obtain the following bound:

## 3Main result

In this section we prove our main result.

Let be a yes-instance of , where is -topological-minor-free. Since we assumed that is treewidth-bounding there exists such that , where is a constant that depends only on . Since the problem is assumed to have finite integer index, denote by the size of the largest representative of the equivalence relation , where the representatives are chosen such that they are smallest possible. We use only one reduction rule which is stated below.

Here is the constant in the definition of finite integer index (see Definition ?) that depends on and .

From now on whenever we talk about an instance of the problem , we assume that it is reduced w.r.t. our only reduction rule. In particular, does not contain a -protrusion of size strictly more than .

### 3.1Bounding the size of small components

We first bound the total number of vertices in all components in .

First note that for each , the set of its neighbors in is a separator of in of size at most . Therefore the total number of vertices in all components separated by in is at most , where is the size of the largest representative of the equivalence relation . Thus it is sufficient to show that the number of subsets of that are separators of components in is bounded.

Using the technique outlined in the proof of Lemma ?, we contract the components greedily into edges in . Repeat the following operations for as long as possible. Pick a component arbitrarily and choose two distinct vertices such that is not an edge; create a new edge , and delete from the graph. If there are components which cannot be contracted into edges in this fashion, then it follows that the separators of these components are cliques in . As the subgraph induced by after these operations is a topological minor of , it must be that is -topological minor free. Hence by Property ?, has at most cliques. The number of vertices in all components of separated by such a clique is, as noted before, . Moreover each component that is contracted to an edge also has at most vertices and, by Property ?, has at most edges. Hence the total number of vertices in all components of is bounded from above by .

### 3.2Bounding the size of the large components

Proving that the total number of vertices in all components of is linear in is more involved. As a first step, we use Lemma ? to show that the number of components in is linear in . To bound the total number of vertices in as a linear function of , we propose a technique of decomposing components into connected subgraphs each of bounded size but with a “large” number of neighbors in the set . The following structure plays a crucial role in bounding the size of .

In what follows we let denote a tree-decomposition of that is rooted at some arbitrary bag of degree at least two in the decomposition. We define to be the “forest-decomposition” obtained by taking the disjoint union of all tree-decompositions, that is,

We will employ a marking algorithm that marks bags of to demonstrate that the total number of vertices in all the components is indeed bounded in a reduced instance. We stress however that this algorithm is not efficient, and neither does it have to be, since it is only used to show that the kernel size is small. In what follows, we let denote the set of bags that have already been marked by the algorithm and to be the set of all vertices of the graph which occur in at least one marked bag. Call a subtree of some tree-decomposition in marked if it contains at least one marked bag and unmarked otherwise. Note that an unmarked subtree can contain marked vertices of in its bags, as these vertices could occur in some other marked bag.

The marking algorithm works as follows.

1. Set .

2. Mark bags of the forest decomposition which induce a scrub in the graph that satisfies the following conditions:

• ;

• .

Set .

3. Mark join bags that are parents of unmarked subtrees that induce at least one connected component in with at least neighbors in . Add each such bag to .

4. Iteratively mark the least common ancestor (join) bag of two bags that have already been marked and add it to .

We first point out some features of the marking algorithm.

Suppose that is unmarked but has at least three marked bags as neighbors in . Let be the shortest path from the root of to . If the root of happens to be in , then consists of only the root bag. Now there are at least two marked bags in that are neighbors of that are not on . Clearly one of the bags of must be the least common ancestor of and and the algorithm, in Step 4, would then have marked this bag. This contradicts the hypothesis that has no marked bags.

Since , each bag of the tree-decomposition has size at most (we assume an optimal tree-decomposition). To prove the lemma, it is sufficient to bound the number of bags marked by the algorithm in Steps 2, 3, and 4.

The scrubs that are marked in Step 2 are vertex-disjoint and since each scrub “sees” at least vertices in the set , by Lemma ?, the number of such scrubs is at most . In Step 3, the connected components that are considered are vertex-disjoint and hence the bound of Lemma ? applies again. Finally in Step 4, the number of marked bags doubles in the worst case. This proves the bound on the size of .

Lemma ? showed that the total number of vertices in marked bags is linearly bounded in . We now go on to show that the total number of vertices in unmarked bags is also linear in . To achieve this goal, we first consider the total size of the scrubs seen by the algorithm in Step 2. Suppose that in this step, the algorithm considers the scrubs in that order while marking bags.

Let and consider a twig for some . By the definition of a scrub, , and hence is separated from the rest of the graph by the set which is of size at most . It follows that in a reduced instance . In fact, the total number of vertices in all twigs of a scrub that are connected to the same set of vertices in is bounded —these twigs share a common separator.

Also note that the scrubs are vertex-disjoint and therefore by Lemma ? it follows that . Therefore in order to bound the total number of vertices in all the scrubs, it is sufficient to bound the total number of twigs. Let . Construct a bipartite graph from with bipartition and edge set as follows:

1. Delete all vertices of that are not in either nor in any scrub .

2. Delete all edges inside the root of scrub , for .

3. Delete all twigs that have no neighbors in .

4. For all twigs in that are connected to the same set in , remove all but one.

5. For each twig , choose arbitrary vertices and . Remove and add the edge to .

Now . For each scrub , the number of vertices in the twigs removed in Step 3 is at most and hence the total number of vertices removed in this step over all scrubs is bounded from above by . For each scrub and each subset , the number of vertices in the twigs removed in Step 4 is bounded from above by .

The bipartite graph is a topological minor of , and since is -topological-minor-free, so is . By Property ?, the number of edges in is at most

The total number of vertices removed in Step 4 is therefore . It follows that the total number of vertices in the scrubs is bounded from above by

At this point, we have accounted for all vertices that occur in a marked bag or a scrub seen by the algorithm in Step 2. We now consider the forest-decomposition obtained from by removing all vertices that occur in marked bags. This corresponds to a forest-decomposition of the graph . Note that we may not remove all the scrub vertices in this process. In order account for the fact that all vertices in the scrubs have been counted, we simplify the forest-decomposition even further. Delete a tree if all its bags only contain scrub vertices from .

The trees in the forest-decomposition can be partitioned into two classes: those that have at most neighbors in and those that have at least neighbors. This motivates us to define and . Define to be the set of trees such that ; is the set of trees such that . By Lemma ?, at most two neighboring bags of a tree are marked.

Let be the vertices of that are contained in the bags of . Note that cannot be the root of a scrub found by the algorithm in Step 2, otherwise would be in some scrub . If , then all must be contained in some scrub . Therefore cannot be empty if . But then is a scrub in not chosen by the algorithm in Step 2. Since the algorithm chooses scrubs of size at least , this implies that .

We next show that the total number of vertices in the trees in is linear in .

By Lemma ?, at most two neighboring bags of a tree are marked. Therefore for each , the number of vertices in the bags of is at most , as the subgraph has a separator of size at most . Moreover the number of trees in that have exactly two marked bags as neighbors is bounded by the number of marked bags (we can simply associate each such tree with one marked bag in the forest ). We therefore have to bound the number of trees that have exactly one marked bag as neighbor. By Lemma ?, the total number of vertices in trees of that are adjacent to exactly one marked bag is at most . Since the number of marked bags is at most , the total number of vertices in bags of is at most which is at most .

All that now remains is to show that the vertices in trees of that have not been accounted for thus far is linear in .

Similar to Lemma ?, we have the following:

If , then as in Lemma ?, we have . Therefore assume that and . Since was not marked in Step 2, for all that induce connected components in , it holds that . By Observation ?, can have at most connected components and hence the claimed bound follows.

Let be the central path of . Construct a path-decomposition of as follows. Take all bags in the path and, for each join bag on this path, add in the vertices of all bags connected to that are not part of . As each such join bag is unmarked, by Lemma ?, the size of such a bag increases by at most . As the size of each bag of is bounded by , the above bound follows.

We next show that if and if the subgraph induced by the vertices in the bags of is large, then we can decompose it into connected subgraphs of constant size such that . Lemma ? assures us that there can be at most such connected subgraphs. Together this would imply a linear bound on the total number of vertices in .

To state this “decomposition lemma,” we introduce additional notation and terminology. Given a path decomposition of a graph and two bags , let denote the graph induced by the vertices in the bags that appear between and in excluding the vertices in and . That is, if denotes the set of bags in the path starting with and ending with , then

The first and last bag of are called its end-bags. Given a bag , we say that is connected to if it either includes a vertex from or is adjacent to a vertex in . Let be a tree-decomposition of a graph and let be bags occurring in . For a subtree , we say that has an -path (or, a path from to ) if there exists a -path in where and . This trivially holds if and contains a vertex of . In the following lemma, we write for the expression as a shorthand.

Let be the width of the path-decomposition . By Lemma ?, this is at most . We first show that for any bag in the decomposition , the graph contains at most connected components. Note that each connected component of is connected to either or . If this were not the case, then the tree-decomposition in of which is a subtree would contain more than one connected component of . This is a contradiction since we assume that each tree in the forest represents a connected component.

The number of connected components of connected to both and is bounded by the width of the decomposition. To see this, simply observe that the graph obtained from by adding edges such that both and induce cliques also has pathwidth at most . If the number of connected components in connected to both and were at least then at least cops would be required to catch a robber in , contradicting the fact that it has pathwidth . The number of components connected exactly to one of or is, by Lemma ?, at most .

Imagine walking along the bags of the decomposition starting at and suppose is the first bag such that . Then contains a connected component with at least vertices. Since our instance is reduced, it must be that . Let be the bag immediately before . Then

and since , an easy calculation shows that proving claim of the lemma. Claim is easier to show. For if , then has a separator of size at most and, since the graph is reduced, has at most vertices.

Finally, we can bound the number of vertices occurring in the bags of trees in .

If has at least vertices then using Lemma ?, we can decompose iteratively into connected components of size at most with . By Lemma ?, the total number of connected components is at most . Finally by Lemma ? the number of vertices in all the bags of trees in is at most , which is at most .

We are now ready to prove the Main Theorem.

Let be a yes-instance of that has been reduced w.r.t. the Protrusion Reduction Rule. Using Lemmas ?, ?, ?, ?, and ? we see that .

This result immediately extends to graphs of bounded degree, as graphs of maximum degree cannot contain as a topological minor.

## 4Implications of the Main Theorem

Some concrete problems that fall under this definition are the following.

This also implies that, by a simple brute force on the kernelized instance, Chordal Vertex Deletion and Interval Vertex Deletion are solvable in time for some constant . On general graphs only a algorithm is known [18].

As an example of a concrete class of problems that satisfy the Main Theorem, consider a hereditary property whose forbidden set contains all holes. Any graph that satisfies must necessarily be chordal. The -Vertex Deletion problem is, given a graph and an integer , to decide whether there exists at most vertices whose deletion results in a graph satisfying . It is easy to show that -Vertex Deletion is both treewidth-bounding and strongly monotone (and hence has finite integer index) on -topological-minor-free graphs and therefore admits a linear kernel on such a graph class.

A natural extension of the problems in Corollary ? is to ask for a connected solution. In many cases, however, the connected version of a problem is not strongly monotone and probably does not have finite integer index. For the following problems, however, strong monotonicity can be shown easily as any solution contains vertices at a constant distance from the boundary.

An interesting property of -topological-minor-free graphs is that the usual width measure are essentially the same.

This entails the following Corollary of the main result.

Finally, we can relate our result to bidimensionality in some natural cases. Consider a vertex-deletion problem -Vertex Deletion for some arbitrary graph property . Then our result entails the following.

Let be a yes-instance with solution set . Then , which entails that, for some constant depending only on , does not contain a -grid as a minor. Otherwise the solution of would be nonempty: if we could contract into a grid that itself is not in , i.e. we need to delete at least one vertex from it to obtain a graph that has the property , this would contradict the assumption that the problem is bidimensional.

Therefore -Vertex Deletion is treewidth-bounding and the above follows.

## 5Conclusion and Open Questions

We have shown that one can obtain linear kernels for a range of problems on graphs excluding a fixed topological minor. This partially extends the results by Bodlaender et al. on graphs of bounded genus [2] and by Fomin et al. on graphs excluding a fixed minor [9].

Two main questions arise: (1) can similar results be obtained for an even larger class of (sparse) graphs and (2) what other problems have linear kernels on -topological-minor free graphs. In particular, does Dominating Set have a linear kernel on graphs excluding a fixed topological minor? It would also be interesting to investigate how the structure theorem by Grohe and Marx can be used in this context [12].

### Footnotes

1. The counterpart of in parameterized complexity.

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