A 13k-kernel for Planar Feedback Vertex Setvia Region DecompositionA preliminary version (with a slightly weaker result) was presented at IPEC 2014, Wrocław, Poland. Work partially supported by the ANR Grant EGOS (2012-2015) 12 JS02 002 01 (MB) and by the National Science Centre of Poland, grant number UMO-2013/09/B/ST6/03136 (ŁK).

A -kernel for Planar Feedback Vertex Set
via Region Decompositionthanks: A preliminary version (with a slightly weaker result) was presented at IPEC 2014, Wrocław, Poland. Work partially supported by the ANR Grant EGOS (2012-2015) 12 JS02 002 01 (MB) and by the National Science Centre of Poland, grant number UMO-2013/09/B/ST6/03136 (ŁK).

Marthe Bonamy LIRMM, France    Łukasz Kowalik University of Warsaw, Poland
Abstract

We show a kernel of at most vertices for the Feedback Vertex Set problem restricted to planar graphs, i.e., a polynomial-time algorithm that transforms an input instance to an equivalent instance with at most vertices. To this end we introduce a few new reduction rules. However, our main contribution is an application of the region decomposition technique in the analysis of the kernel size. We show that our analysis is tight, up to a constant additive term.

1 Introduction

A feedback vertex set in a graph is a set of vertices such that is a forest. In the Feedback Vertex Set problem, given a graph and integer one has to decide whether has a feedback vertex set of size . This is one of the fundamental NP-complete problems, in particular it is among the 21 problems considered by Karp [11]. It has applications e.g. in operating systems (see [15]), VLSI design, synchronous systems and artificial intelligence (see [8]).

In this paper we study kernelization algorithms, i.e., polynomial-time algorithms which, for an input instance either conclude that has no feedback vertex set of size or return an equivalent instance , called kernel. In this paper, by the size of the kernel we mean the number of vertices of . Burrage et al. [5] showed that Feedback Vertex Set has a kernel of size , which was next improved to by Bodlaender [3] and to by Thomassé [16]. Actually, as argued by Dell and van Melkebeek [7] the kernel of Thomassé can be easily tuned to have the number of edges bounded by . This cannot be improved to for any , unless  [7].

In this paper we study Planar Feedback Vertex Set problem, i.e., Feedback Vertex Set restricted to planar graphs. Planar versions of NP-complete graph problems often enjoy kernels with vertices. Since an -vertex planar graph has edges, this implies they have edges, and hence are called linear kernels. The first nontrivial result of that kind was presented in the seminal work of Alber, Fellows and Niedermeier [2] who showed a kernel of size for Planar Dominating Set. One of the key concepts of their paper was the region decomposition technique in the analysis of the kernel size. Roughly, in this method the reduced plane instance is decomposed into regions (i.e. subsets of the plane) such that every region contains vertices of the graph. It was next applied by Guo and Niedermeier to a few more graph problems [10]. In fact it turns out that for a number of problems on planar graphs, including Planar Dominating Set and Planar Feedback Vertex Set, one can get a kernel of size by general method of protrusion decomposition [9]. However, in this general algorithm the constants hidden in the notation are very large, and researchers keep working on problem-specific linear kernels with the constants as small as possible [6, 14, 17, 13, 12].

In the case of Planar Feedback Vertex Set, Bodlaender and Penninkx [4] gave an algorithm which outputs a kernel of size at most . This was next improved by Abu-Khzam and Khuzam [1] to . Very recently, and independently of our work, Xiao [18] has presented an improved kernel of vertices. However, neither of these papers uses the region decomposition. Indeed, it seems non-obvious how the regions of the region decomposition can be defined for Planar Feedback Vertex Set. Instead, the authors of the previous works cleverly apply simple bounds on the number of edges in general and bipartite planar graphs. Moreover, for certain problems these methods turned out to give better results and simpler proofs than those based on region decomposition, see e.g., the work of Wang, Yang, Guo and Chen [17] on Connected Vertex Cover, Edge Dominating Set, and Triangle Packing in planar graphs improving previous results of Guo and Niedermeier [10].

Somewhat surprisingly, in this work we show that region decomposition can be successfully applied to Planar Feedback Vertex Set, and moreover it gives much tighter bounds than the previous methods. Furthermore, we add a few new reduction rules to improve the bound even further, to . More precisely, we show the following result.

Theorem 1.

There is an algorithm that, given an instance of Planar Feedback Vertex Set, either reports that has no feedback vertex set of size or produces an equivalent instance with at most vertices. The algorithm runs in expected time, where is the number of vertices of .

We use the region decomposition approach in a slightly relaxed way: the regions are the faces of a -vertex plane graph and the number of vertices of the reduced graph in each region is linear in the length of the corresponding face. We show that this gives a tight bound, i.e., we present a family of graphs which can be returned by our algorithm and have vertices.

Organization of the paper.  In Section 2 we present a kernelization algorithm which is obtained from the algorithms in [4, 1] by generalizing a few reduction rules, and adding some completely new rules. In Section 3 we present an analysis of the size of the kernel obtained by our algorithm. In the analysis we assume that in the reduced graph, for every induced path with internal vertices, the internal vertices have at least three neighbors outside the path. Based on this, we get the bound of for the number of vertices in the kernel. In Section 2 we present reduction rules which guarantee that in the kernel , resulting in the kernel size bound of . To get the claimed bound of vertices in Section 4 we present a complex set of reduction rules, which allow us to conclude that . In Section 5 we discuss the running time of the algorithm. Finally, in Section 6 we discuss possibilities of further research.

Notation.  In this paper we deal with multigraphs, though for simplicity we refer to them as graphs. (Even if the input graph is simple, our algorithm may introduce multiple edges.) By the degree of a vertex in a multigraph , denoted by , we mean the number of edges incident to in . By , or shortly , we denote the set of neighbors of , while is the closed neighborhood of . Note that in a multigraph , but the equality does not need to hold. The neighborhood of a set of vertices is defined as , while the closed neighborhood of is . For a face in a plane graph, a facial walk of is the shortest closed walk induced by all edges incident with . The length of , denoted by is the length of its facial walk.

2 Our kernelization algorithm

In this section we describe our algorithm which outputs a kernel for Planar Feedback Vertex Set. The algorithm exhaustively applies reduction rules. Each reduction rule is a subroutine which finds in polynomial time a certain structure in the graph and replaces it by another structure, so that the resulting instance is equivalent to the original one. More precisely, we say that a reduction rule for parameterized graph problem is correct when for every instance of it returns an instance such that:

  1. is an instance of ,

  2. is a yes-instance of iff is a yes-instance of , and

  3. .

Below we state the rules we use. The rules are applied in the given order, i.e., in each rule we assume that the earlier rules do not apply. We begin with some rules used in the previous works [1, 4].

(a)

(b)

(c)

(d)

(e)

(f)
Figure 7: Reduction rules 22. Dashed edges are optional. We draw in black the vertices whose incident edges are all already drawn (as solid or dashed edges), in white the vertices which might be incident to other edges. Regardless of their color, vertices in the figures may not coincide.

Rule 2 If there is a loop at a vertex , remove and decrease by one.

Rule 2 Delete vertices of degree at most one.

Rule 2 If a vertex is of degree two, with incident edges and , then delete and add the edge . (Note that if then a loop is added.)

Rule 2 If a vertex has exactly two neighbors and , edge is double, and edge is simple, then delete and and decrease by one.

Rule 2 If there are at least three edges between a pair of vertices, remove all but two of the edges. 

Rule 2 Assume that there are five vertices such that 1) both and are neighbors of each of and 2) each vertex is incident with at most one edge such that . Then remove all the five vertices and decrease by two.

The correctness of the above reduction rules was proven in [1]. (In [1], Rule 2 is formulated in a slightly less general way which forbids multiplicity of some edges, but the correctness proof stays the same.) Now we introduce a few new rules.

Rule 2 If a vertex has exactly three neighbors , and , is also adjacent to and , and both edges and are simple, then contract and add an edge (increasing its multiplicity if it already exists). If edge was not simple, add a loop at .

Lemma 2.

Rule 2 is correct.

Proof.

Let be the graph obtained from a graph by a single application of Rule 2. Let be a feedback vertex set of size in . We claim is a feedback vertex set in too. Assume for a contradiction that there is a cycle in . Then , for otherwise . If then and is a cycle in , a contradiction. If , then and hence is the only neighbor of in , so is the 2-cycle . But then contains a loop at , a contradiction.

Let be a feedback vertex set of size in . If , then is a feedback vertex set of size in . Assume . Then we can assume for otherwise we replace by , which is also a feedback vertex set in . If there is a cycle in , then , for otherwise . But then is a cycle in , a contradiction. Finally, if then both and are in , so is also a feedback vertex set in . ∎

The graph modification in Rule 2 is an example of a gadget replacement, i.e., a subgraph of is replaced by another subgraph in such a way that the answer to the Feedback Vertex Set problem does not change. We will use many rules of this kind, and their correctness proofs all use similar arguments. In order to make our proofs more compact, we define gadget replacement formally below, and prove a technical lemma (Lemma 3 below) which will be used in many rule correctness proofs.

Gadget replacement in graph is a triple , where , is a set of vertices disjoint with , and is a set of edges with both endpoints in . The result of gadget replacement is a new graph , obtained from by deleting and and inserting and .

For an example, in Rule 2, is a result of gadget replacement . Note that if is a gadget replacement in that results in then is a gadget replacement in and its result is .

Lemma 3.

Let be a gadget replacement in graph , and let be its result. Let and . Let be a feedback vertex set in . Let be a subset of vertices of such that and is a forest. Finally, assume that for every pair if there is a -path in then there is a -path in . Then is a feedback vertex set of .

Proof.

Assume for a contradiction that there is a cycle in . Since is a forest, has at least one vertex outside . Assume has all vertices outside . But then and since we also have , so is not a feedback vertex set of , a contradiction.

Hence we know that has vertices both inside and outside . It follows that can be divided into subpaths of two kinds: subpaths in and in . Every such subpath in is of the form , where and . Hence, by the assumption of the lemma, there is a -path in . In particular, (we will use this observation later). Let be the closed walk obtained from by replacing every maximal subpath in by a path in . Now consider a maximal subpath of in . It is of the form , where and . Since and are also endpoints of paths in , it holds that as argued above. Since , it holds that . Hence . It follows that is a closed walk in , hence there is a cycle in , a contradiction. ∎

When applying Lemma 3, we will examine reachability relations in and in , in order to check whether the last assumption holds. It will be convenient to introduce the following notation. For a graph and set of vertices , let be the reachability relation in truncated to , i.e., iff and there is an -path in . The set does not need to be a subset of ; for every vertex , forms an equivalence class of .

Figure 8: Reduction rules 2 and 2.

Rule 2 Assume there are six vertices , , , , , , , such that , , , and . Then contract the edge to a new vertex and add an edge , as presented in Figure 8 (left).

Lemma 4.

Rule 2 is correct.

Proof.

Let be the graph obtained from a graph by a single application of Rule 2. Note that is a result of a gadget replacement with and

Let be a feedback vertex set of size in . We claim that there is a feedback vertex set in of size at most . If , then by Lemma 3 we see that works. Hence we can assume . Then or to hit the triangle and the quadrangle . In the prior case we pick . Then and are the same relations since the path in corresponds to the edge in , so Lemma 3 applies. We are left with the case . Then we pick . Again, since the path in corresponds to the edge in , so Lemma 3 applies.

Let be a feedback vertex set of size in . We claim that there is a feedback vertex set in of size at most . If , then by Lemma 3 we see that works. Hence we can assume . Then or to hit the digon . In the prior case we pick . Then since the path in corresponds to the edge in , so Lemma 3 applies. We are left with the case . Then we pick . Again, since the path in corresponds to the edge in , so Lemma 3 applies. ∎

Rule 2 Assume is an induced path such that for two vertices , outside the path, , and , , and . Then replace with the gadget presented in Figure 8 (right), i.e., remove and and add a vertex , edges and , and double edges and .

Lemma 5.

Rule 2 is correct.

Proof.

Let be the graph obtained from a graph by a single application of Rule 2. Note that is a result of a gadget replacement with and

Let be a solution of . If or we proceed as in the proof of Lemma 4. Otherwise, to hit the triangle , equals either or . In both cases, has exactly one equivalence class . We observe that for the relation has also one equivalence class, so by Lemma 3, is a solution of .

Let be a feedback vertex set of size in . If we proceed as in the proof of Lemma 4. Otherwise, . If then we put , and otherwise . Note that is a forest, since . Moreover, and are the same (total) relation, so Lemma 3 applies and is a solution of . ∎

Rule 2 Let and let and be two vertices in , . If no cycle in intersects , and there is a subgraph such for every vertex , we have , then remove and decrease by 1.

Lemma 6.

Rule 2 is correct.

Proof.

Let be the graph obtained from a graph by a single application of Rule 2, i.e., . Let be a feedback vertex set of size in . Then every cycle in contains , so is a feedback vertex set of size in .

Let be a feedback vertex set of size in . If , then clearly is a solution of the instance . Hence assume . We claim that . Assume the contrary, i.e., . Since is a forest,

(1)

On the other hand, by the degree bound, and because and ,

(2)

By (1) and (2), . Since this implies , a contradiction. It follows that . Then is of size at most . Moreover, is a feedback vertex set in , since is a feedback vertex set and by . Again, this implies that is a solution of the instance , as required.

Figure 9: Configurations in lemmas 7 and 8.

Rule 2 is not used directly in our algorithm, because it seems impossible to detect it in time. However, to get the claimed kernel size we need just two special cases of Rule 2, which are stated in lemmas  7 and 8 below.

Lemma 7.

Assume there are five vertices , , , , such that , , there is at most one edge incident to and a vertex outside , and there is at most one edge incident to and a vertex outside . Then Rule 2 applies.

Proof.

It is easy to see that condition of Rule 2 is satisfied. We proceed to condition . Since Rule 2 does not apply, is adjacent to or ; by symmetry assume the former. Let . We build as follows. We start with . Since Rule 2 does not apply, or one of is a double edge. Hence, we add or another copy of one of to , respectively. Note that for every we have , as required. ∎

Lemma 8.

Assume there are five vertices , , , , such that , , and there is at most one edge incident to and a vertex outside . Moreover, the edges and are simple. Then Rule 2 applies.

Proof.

It is easy to see that condition of Rule 2 is satisfied. We proceed to condition . Let . We build as follows. We start with . There are some cases to consider. Since Rule 2 does not apply, or .

CASE 1:  . Then, since Rule 2 does not apply to the cycle, or is a double edge. Moreover, since Rule 2 does not apply, or . We add to edges , , either or (but not both), and the second copy of either or (but not both). Then and , so holds.

CASE 2:  Exactly one of and is an edge; by symmetry assume and . Since Rule 2 does not apply, . And then since Rule 2 does not apply, .

CASE 2.1:  is a double edge. We add to edge , and both copies of . Then and , so holds.

CASE 2.2:  is a simple edge. Since Rule 2 does not apply, and . We add to edges and , exactly one edge incident to which is not yet in and exactly one edge incident to which is not yet in . Then and , so holds. ∎

The following rule was shown to be correct by Abu-Khzam and Khuzam in [1].

Rule 2 Assume there is an induced path with endpoints and and with six internal vertices such that for some vertices , outside the path . If , then remove and decrease by one.

In [1] it was assumed that when Rule 2 described above is applied, does not contain an induced path such that for some vertex , we have . In our algorithm this is guaranteed by Rule 2 (slightly more general than their Rule 6). We are able to extend Rule 2 as follows.

Lemma 9.

Assume there is an induced path with endpoints and and with five internal vertices such that for some vertices , outside the path . Then there is an instance with such that is a yes-instance iff is a yes-instance and .

The proof of Lemma 9 involves five more rules and is quite technical; we defer it to Section 4. We stress here that Lemma 9 is not crucial for getting a substantial improvement of the kernel size. Indeed, if one uses Rule 2 instead of Lemma 9, the resulting kernel is of size at most (see Section 3). Let us also remark that by the analysis in Section 3, if someone manages to exclude paths described in Lemma 9 with only four internal vertices, the kernel size decreases further to .

To complete the algorithm we need a final rejecting rule which is applied when the resulting graph is too big. In Section 3 we prove that Rule 2 is correct.

Rule 2 If the graph has more than vertices, return a trivial no-instance (conclude that there is no feedback vertex set of size in ).

3 The size bound

In this section we prove the following theorem.

Theorem 10.

Let be a planar graph such that rules 1–2 do not apply and does not contain the configurations described in lemmas 7 and 8. Assume also that for every induced path with endpoints and and with internal vertices , …, the internal vertices have at least three neighbors outside the path, i.e., . If there is a feedback vertex set of size in , then .

Let be a feedback vertex set of size in (i.e., a “solution”), and let be the forest induced by . Denote the set of vertices of by . We call the vertices in solution vertices and the vertices in forest vertices.

A partition of Now we define some subsets of . Let denote the vertices whose degree in is two or at least three, respectively. The leaves of are further partitioned into two subsets. Let and be the leaves of that have two or at least three solution neighbors, respectively. By rules 2 and 2 all the vertices in have degree at least . Hence, if a leaf of has fewer than two solution neighbors, Rule 2 or Rule 2 applies. It follows that every leaf of belongs to . This proves claim of Lemma 11 below.

Lemma 11.

Graph satisfies the following properties.

  1. The sets , , , form a partition of .

  2. For every pair , of solution vertices there are at most two vertices such that