Approximation Algorithms for Independent Sets in Map Graphs

Journal of Algorithms 41, 20–40 (2001)doi:10.1006/jagm.2001.1178, available online at http://www.idealibrary.com on

Approximation Algorithms for IndependentSets in Map Graphs1

Zhi-Zhong Chen

Department of Mathematical Sciences, Tokyo Denki University,Hatoyama, Saitama 350-0394, Japan

E-mail: [email protected]

Received September 25, 1999

This paper presents polynomial-time approximation algorithms for the problemof computing a maximum independent set in a given map graph G with or withoutweights on its vertices. If G is given together with a map, then a ratio of 1+ δ canbe achieved by a quadratic-time algorithm for any given constant δ > 0, no matterwhether each vertex of G is given a weight or not. In case G is given without a map,a ratio of 4 can be achieved by a low-degree polynomial-time algorithm if no vertexis given a weight, while a ratio of 5 can be achieved by a low-degree polynomial-timealgorithm otherwise. 2001 Academic Press

Key Words: independent set; planar graph; map graph; approximation algorithm;NP-hardness.

1. INTRODUCTION

An independent set in a graph G = �V�E� is a subset I of V such thatno two vertices of I are connected by an edge in E. If each vertex ofG is associated with a weight, the weight of an independent set I in Gis the total weight of vertices in I. Given a vertex-weighted graph G, themaximum independent set (MIS) problem requires the computation of anindependent set of maximum weight in G. The special case where eachvertex of G has weight 1, is called the unweighted maximum independentset (UMIS) problem. The MIS problem is a fundamental problem in manyareas; unfortunately, even the UMIS problem is NP-hard.

1 A preliminary version of this paper appeared, in “Proceedings of the Sixth Annual Interna-tional Computing and Combinatorics Conference, 2000,” Lecture Notes in Computer Science,Vol. 1858, pp. 105–114, Springer-Verlag, Berlin/New York.

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0196-6774/01 $35.00Copyright 2001 by Academic PressAll rights of reproduction in any form reserved.

independent sets in map graphs 21

Since the MIS problem is NP-hard, we are interested in designing approx-imation algorithms, i.e., algorithms that find an independent set of large butnot necessarily maximum weight in a given graph. An approximation algo-rithm � achieves a ratio of ρ if for every vertex-weighted graph G, theindependent set I found by � on input G has weight at least OPT �G�/ρ,where OPT �G� is the maximum weight of an independent set in G.

Hastad [10] gave strong evidence that no polynomial-time approximationalgorithm for the UMIS problem achieves a ratio of n1−ε for any ε > 0,where n is the number of vertices in the input graph. On the other hand,much work has been devoted to designing approximation algorithms forthe MIS problem restricted to certain classes of graphs (such as planargraphs, bounded-degree graphs, etc). This paper considers the MIS prob-lem restricted to map graphs, which were introduced by Chen et al. [9]recently.

Intuitively speaking, a map is a plane graph whose faces are classifiedinto nations and lakes. Traditional planarity says that two nations on a map� are adjacent if and only if they share at least a borderline. Motivatedby topological inference, Chen et al. [9] considered a modification of tradi-tional planarity, which says that two nations on � are adjacent if and onlyif they share at least a border-point. Consider the simple graph G whosevertices are the nations on � and whose edges are all �f1� f2� such thatf1 and f2 are nations sharing at least a border-point. G is called a mapgraph [9, 18]; if for some integer k, no point in � is shared by more thank nations, G is called a k-map graph [9]. The UMIS problem restricted tomap graphs is NP-hard, because planar graphs are exactly 3-map graphs [9]and the UMIS problem restricted to planar graphs is NP-hard.

It is known [1, 11] that the MIS problem restricted to planar graphs hasa polynomial-time approximation scheme (PTAS). Recall that a PTAS for amaximization problem � is a polynomial-time algorithm � such that givenan instance I of � and an error parameter ε > 0, � computes a solutionS of I in time polynomial in the size of I such that the value of S is atleast �1 − ε�·OPT �I�, where OPT �I� is the optimal value of a solution ofI. An interesting question is to ask whether the MIS problem restrictedto map graphs also has a PTAS. To answer this question, we prove thatif we require that the input graph be given together with a map, then theMIS problem restricted to map graphs has a PTAS. Combining this resultand Thorup’s polynomial-time algorithm [18] for constructing a map for agiven map graph, we obtain a PTAS for the MIS problem restricted to mapgraphs.

However, Thorup’s algorithm is very complicated and the exponent of itspolynomial time bound is about 120 [19]. So, it is natural to ask whetherthere is a PTAS for the MIS problem restricted to map graphs that does notconstruct a map for the given map graph. We partially answer this question

22 zhi-zhong chen

in the affirmative, by giving a PTAS for the MIS problem restricted tok-map graphs for any fixed integer k that does not construct a map for thegiven k-map graph. It is worth noting that the map constructed by Thorup’salgorithm for a given k-map graph may have a point shared by more thank nations [19].

We also present two nontrivial polynomial-time approximation algo-rithms for the MIS problem restricted to map graphs that do not constructa map for the given map graph. One of the algorithms achieves a ratioof 5. The other is for the UMIS problem and achieves a ratio of 4.

The rest of this paper is organized as follows. Section 2 gives precise defi-nitions and states known results that are needed later on. Section 3 presentsa PTAS for the MIS problem where the input graph is given together witha map. Section 4 describes two PTASs for the MIS problem restricted tok-map graphs. Section 5 presents nontrivial approximation algorithms forthe MIS and UMIS problems restricted to map graphs. Section 6 asks sev-eral open questions.

2. PRELIMINARIES

Throughout this paper, a graph may have multiple edges but no loops,while a simple graph has neither multiple edges nor loops. Let G = �V�E�be a graph. The size of G is the total number of vertices and edges in Gand is denoted by G. An induced subgraph of G is a subgraph H whoseedges are those of G between vertices of H. An induced subgraph of Gwith vertex-set U is denoted by G�U� and called the subgraph of G inducedby U . Let v ∈ V . The degree of v in G is the number of edges incident tov in G; v is an isolated vertex if its degree is 0. The neighborhood of v inG, denoted by NG�v�, is the set of vertices adjacent to v in G. Note thatNG�v� bounds the degree of v in G from below. For a subset U of V , letNG�U� = ⋃

v∈U NG�v�.An independent set of G is a subset U of V such that G�U� has no edge.

The independence number of G, denoted by α�G�, is the maximum numberof vertices in an independent set of G.

A simple cycle in G is a sequence of distinct vertices and edgesv1� e1� v2� e2� � � � � vk� ek with k ≥ 2 such that for all i ∈ �1� � � � � k�, eiis an edge between vi and vi+1, where vk+1 = v1. G is biconnected if (1) Ghas at least two edges and (2) for every pair of distinct edges e1 and e2 inG, G has a simple cycle on which both e1 and e2 appear. A biconnectedcomponent of a graph is a maximal biconnected subgraph.

A graph is planar if it can be embedded in the plane so that any pair ofedges can only intersect at their endpoints; a plane graph is a planar onetogether with such an embedding. Let � be a plane graph. Consider the


set of all points in the plane that lie on no edge of �. This set consistsof a finite number of topologically connected regions; the closure of eachsuch region is called a face of �. Let f be a face of �. The boundaryvertices (respectively, edges) of f are those vertices (respectively, edges) of Gcontained in the closure f . The boundary of f is the graph �Vf � Ef �, whereVf and Ef are the set of boundary vertices and edges of f , respectively.The size of f , denoted by f , is Vf . We call f a cycle face if its boundaryis a simple cycle.

A map � is a pair �� ϕ�, where (1) � is a plane graph �� E�� and eachconnected component of � is biconnected, and (2) ϕ is a function that mapseach face f of � to 0 or 1 in such a way that whenever ϕ�f � = 1, f is a cycleface. A face f of � is called a nation on � if ϕ�f � = 1, while called a lakeon � otherwise. For an integer k, a k-point in � is a p ∈ � that appearson the boundaries of exactly k nations on �; a k+-point in � is a p ∈ �that appears on the boundaries of more than k nations on �. � is a k-mapif there is no k+-point in �. Two nations are adjacent if their boundariesshare a p ∈ � . The graph of � is the simple graph G = �V�EG�, where Vconsists of the nations on � and EG consists of all �f1� f2� such that f1 andf2 are adjacent. We call G a map graph, and we call G a k-map graph if �is a k-map. Whenever necessary for clarity, we will call the elements of �points and the elements of V nations. Note that both points and nations arecalled vertices. A nation f on � is isolated if f is an isolated vertex of G.

The half square of a bipartite graph H = �X�Y �EH� is the simple graphG = �X�E�, where E = ��x1� x2� there is a y ∈ Y with �x1� y� ∈ EH and�x2� y� ∈ EH�.

Lemma 2.1 [9]. Every map graph G = �V�E� is the half square of a bipar-tite planar graph H = �V�� EH� with � ≤ 3V − 6.

For a map � = �� ϕ� and a face f in �, we say that f touches a pointp if p appears on the boundary of f .

Lemma 2.2 [9]. Let G be the graph of a map � = �� ϕ�. Let C be aclique of G containing at least five nations. Then, at least one of the followingholds:

• C is a pizza; i.e., there is a point p in � such that all f ∈ C touch p.

• C is a pizza-with-crust; i.e., there are a point p in � and a nationf1 ∈ C such that (1) all nations of C − �f1� touch p and (2) for each nationf ∈ C − �f1�, some point q �= p in � is touched by both f1 and f .

• C is a hamantasch; i.e., there are three points in � such that eachnation f ∈ C touches at least two of the points and each point is touched byat least two nations.

24 zhi-zhong chen

3. AN EFFICIENT PTAS FOR THE CASE WITH A MAP

Let G = �V�E� be the input map graph given with a map� = �� E�� ϕ� and a nonnegative weight to each nation on �. Letε be the given error parameter.

Let H be the bipartite planar graph �V�� EH�, where EH = ��v�p� v ∈V�p ∈ � , and p is a point on the boundary of nation v�. Obviously, H canbe constructed in linear time from �. Moreover, G is the half square of H.Indeed, Lemma 2.1 is proved by further removing all redundant points fromH, where a point p ∈ � is redundant if for every pair �u� v� of neighborsof p in H, there is a point q ∈ � − �p� with �u� v� ⊆ NH�q�.

So, one naive method for computing an independent set of large weightin G is to emulate Lipton and Tarjan [12]. That is, we first remove redun-dant points from H to make H small, then find a small separator S ofH, and further compute an independent set IC in the half square of eachconnected component C of H − S. The output is

⋃C IC . However,

⋃C IC

may not be an independent set of G, implying that this naive method doesnot work.

Another approach to the design of PTASs for a certain class of NP-hardproblems restricted to planar graphs was suggested by Baker [1]. The essenceof this approach is the notion of k-outerplanar graphs defined inductively asfollows. 1-outerplanar graphs are outerplanar graphs, i.e., those planar graphshaving a plane embedding such that all vertices lie on the outer face. Forpositive integers k ≥ 2, a planar graph is k-outerplanar if it has a planeembedding such that the removal of all vertices on the outer face and alledges incident to them leaves a �k− 1�-outerplanar graph.

Our PTAS is a nontrivial application of the Baker approach. Let h = � 1ε�.

We may assume that G and hence H are connected. Starting at an arbitrarynation r ∈ V , we perform a breadth-first search (BFS) on H to obtain aBFS tree TH . For each vertex v in H, we define the level number &�v�of v to be the number of edges on the path from r to v in TH . Notethat only r has level number 0. For each i ∈ �0� � � � � h− 1�, let Hi be thesubgraph of H obtained from H by deleting all nations v with &�v� ≡ 2i(mod 2h). Obviously, each nation v ∈ V appears in exactly h − 1 of H0through Hh−1. Moreover, each connected component of Hi is clearly a�2h− 1�-outerplanar graph and so is Hi.

For each i ∈ �0� � � � � h − 1�, let Gi be the half square of Hi. The cruxis that each Gi is an induced subgraph of G, because all p ∈ � appear inHi. Our strategy is to compute an independent set Ii of maximum weightin Gi for each i ∈ �0� � � � � h− 1�. Since each nation of G appears in all butone of G0 through Gh−1, the independent set of maximum weight amongI0� � � � � Ih−1 has weight at least �1− 1/h� ·OPT �G� ≥ �1− ε� ·OPT �G�.Recall that OPT �G� is the maximum weight of an independent set in G.


So, it remains to show how to compute a maximum weight indepen-dent set Ii in Gi for each i ∈ �0� � � � � h − 1�. To do this, we first recallthe notion of treewidth, introduced by Robertson and Seymour [16]. Atree-decomposition of a graph K = �VK�EK� is a pair � = ��Xj j ∈ J�� ,where �Xj j ∈ J� is a family of subsets of VK and � is a tree with vertex-setJ such that the following hold:

•⋃

j∈J Xj = VK .

• For every edge �v�w� ∈ EK , there is a vertex j ∈ J with �v�w� ⊆ Xj .

• For all j1� j2� j3 ∈ J, if j2 lies on the path from j1 to j3 in � , thenXj1

∩Xj3⊆ Xj2

.

Each Xj is called the bag of vertex j. The treewidth of a tree-decomposition� is max�Xj − 1 j ∈ J�. The treewidth of K is the minimum treewidthof a tree-decomposition of K, taken over all possible tree-decompositionsof K.

It is widely known that the MIS problem restricted to graphs withbounded treewidth can be solved in linear time (see, e.g., [17]). However,the treewidths of our graphs G0� � � � �Gh−1 may not be bounded. Our strat-egy is to resort to H0� � � � �Hh−1 instead, and do a dynamic programmingon them as shown below.

Lemma 3.1. For every i ∈ �0� � � � � h − 1�, we can compute a maximumweight independent set Ii of Gi in O�Hi2 + 218hhHi� time.

Proof. We show how to compute I0. Since H0 is �2h− 1�-outerplanar, itstreewidth is at most 6h− 4 [4]. The proof of this fact given by Bodlaender[4] can be easily translated to an O�hH0�-time algorithm that computesa tree-decomposition � = ��Xj j ∈ J�� of H0 such that J = O�H0�and the treewidth of � is at most 6h− 4. We may further assume that thedegree of each vertex in � is at most 3 [5].

Root � at an arbitrary leaf vertex. Then, � is a binary tree. For eachvertex j of � , let H�j� be the subgraph of H0 induced by the union ofthe bags of j and its descendants; let H�j� = �V �j��j��E�j��, whereV �j� ⊆ V and ��j� ⊆ � . Let G�j� be the half square of H�j�. Note thatan independent set of G�j� may not be an independent set of G.

To compute I0, we process the vertices of � in postorder. Let j be a ver-tex of � . A target pair for j is �S�W �, where S ⊆ Xj ∩ V , S is an indepen-dent set of G0, W ⊆ Xj ∩ � , and NH0

�S� ∩Xj ⊆ W . We say that a subsetU of V �j� witnesses a target pair �S�W � for j, if the following conditions

26 zhi-zhong chen

(a) through (c) are satisfied:

(a) U ∩Xj = S.

(b) NH0�U� ∩Xj = W .

(c) U is an independent set of G0.

While processing j, for every target pair �S�W � for j, we want to computeand record a maximum weight subset I�j� S�W � of V �j� witnessing �S�W �.Of course, it is possible that no subset of V �j� witnesses �S�W �. In thiscase, we record I�j� S�W � = � while processing j.

The processing of each leaf j of � is straightforward, because H�j� hasat most 6h− 3 vertices. Next, consider the processing of a nonleaf vertex jof � . We assume that j has exactly two children in � , say j1� j2; the casewhere j has exactly one child in � is simpler. Let �S�W � be a target pairfor j. Let S1 = S ∩Xj1

and S2 = S ∩Xj2. We say that a target pair �S′

1�W1�for j1 and a target pair �S′

2�W2� for j2 are simultaneously compatible with�S�W �, if the following conditions (i) through (iv) are satisfied:

(i) S′1 ∩Xj = S1 and S′

2 ∩Xj = S2.

(ii) W1 ∩Xj ⊆ W −NH0�S − S1� and W2 ∩Xj ⊆ W −NH0

�S − S2�.(iii) W1 ∩W2 ⊆ NH0

�S1 ∩ S2�.(iv) W − ��W1 ∩ Xj� ∪ �W2 ∩ Xj� ∪ NH0

�S1� ∪ NH0�S2�� ⊆ NH0�S − �S1 ∪ S2��.

For a class �U1� � � � � U&� of subsets of V , let max�U1� � � � � U&� denote themaximum weight subset in the class. We claim that the following hold:

1. If no target pair for j1 and no target pair for j2 are simultaneouslycompatible with �S�W �, then I�j� S�W � = �.

2. Otherwise, I�j� S�W � = max�S ∪ I�j1� S′1�W1� ∪ I�j2� S′

2�W2� �S′

1� W1� is a target pair for j1, �S′2�W2� is a target pair for j2, and they are

simultaneously compatible with �S�W ��.To see Claim 1, suppose I�j� S�W � �= �. Since I�j� S�W � is an independentset of G0 and NH0

�I�j� S�W �� ∩ Xj = W , it holds that for each p ∈ W ,there is a unique nation a ∈ I�j� S�W � with p ∈ NH0

�a�; let -�p� denotethis unique nation a. For each i ∈ �1� 2�, let S′

i = I�j� S�W � ∩Xjiand let

Wi = NH0�I�j� S�W � ∩ V �ji�� ∩Xji

. Obviously, for each i ∈ �1� 2�, �S′i� Wi�

is a target pair for ji and I�j� S�W � ∩ V �ji� witnesses �S′i� Wi�. So, to prove

Claim 1, it suffices to prove that �S′1�W1� and �S′

2�W2� are simultaneouslycompatible with �S�W �.

Since I�j� S�W � ∩ Xj = S, �S′1�W1� and �S′

2�W2� satisfy Condition (i)above. To see Condition (ii) above, consider an arbitrary p ∈ W1 ∩ Xj .Since NH0

�I�j� S�W �� ∩ Xj = W , we have W1 ∩Xj ⊆ W ∩Xj1and


W2 ∩Xj ⊆ W ∩Xj2. In turn, by the construction of W1, -�p� ∈ I�j� S�W � ∩

V �j1�. So, -�p� �∈ I�j� S�W � − V �j1�. Recalling that S1 = S ∩ Xj1=

S ∩ V �j1� and S ⊆ I�j� S�W �, we now have that -�p� �∈ S − S1. Thistogether with the fact that I�j� S�W � is an independent set of G0 impliesthat p �∈ NH0

�S − S1�. Therefore, W1 ∩ Xj ⊆ W − NH0�S − S1�. Similarly,

W2 ∩ Xj ⊆ W − NH0�S − S2�. It follows that �S′

1�W1� and �S′2�W2� satisfy

Condition (ii) above.To see Condition (iii) above, consider an arbitrary p ∈ W1 ∩ W2. For

each i ∈ �1� 2�, there is exactly one nation ai ∈ I�j� S�W � ∩ V �ji� with p ∈NH0

�ai� because I�j� S�W � ∩ V �ji� witnesses �S′i� Wi�. If a1 were different

from a2, then I�j� S�W � would not have been an independent set of G0.So, a1 = a2. Since a1 ∈ V �j1� ∩ V �j2� and � is a tree decomposition of H0,a1 ∈ Xj ∩Xj1

∩Xj2and in turn a1 ∈ I�j� S�W � ∩Xj = S. Thus, a1 ∈ S1 ∩ S2.

It follows that �S′1�W1� and �S′

2�W2� satisfy Condition (iii) above.To see Condition (iv) above, consider an arbitrary p ∈ W − ��W1 ∩

Xj� ∪ �W2 ∩ Xj� ∪ NH0�S1� ∪ NH0

�S2��. Since W ⊆ Xj , p �∈ W1. Also,NH0

�I�j� S�W � ∩ V �j1�� = W1 because I�j� S�W � ∩ V �j1� witnesses�S′

1�W1�. So, -�p� �∈ I�j� S�W � ∩ V �j1�. Similarly, -�p� �∈ I�j� S�W � ∩V �j2�. Noting that I�j� S�W � ∩ Xj = S, we now have that -�p� ∈I�j� S�W � ∩ Xj = S. Further recall that p �∈ NH0

�S1� ∪ NH0�S2�. There-

fore, -�p� ∈ S − �S1 ∪ S2�. It follows that �S′1�W1� and �S′

2�W2� satisfyCondition (iv) above. This establishes Claim 1.

To prove Claim 2, consider an arbitrary target pair �S′1�W1� for j1 and an

arbitrary target pair �S′2�W2� for j2 that are simultaneously compatible with

�S�W �. It suffices to prove that S ∪ I�j1� S′1�W1� ∪ I�j2� S′

2�W2� witnesses�S�W �. For convenience, let U = S ∪ I�j1� S′

1�W1� ∪ I�j2� S′2�W2�.

To see that U satisfies Condition (a) above, first note that I�j1� S′1�W1� ∩

Xj ⊆ I�j1� S′1�W1� ∩ Xj1

because I�j1� S′1�W1� ⊆ V �j1� and � is a tree-

decomposition of H0. So, I�j1� S′1�W1� ∩Xj = �I�j1� S′

1�W1� ∩Xj1� ∩Xj =

S′1 ∩Xj = S1. Similarly, I�j1� S′

1�W1� ∩Xj = S2. From these facts, it followsthat U satisfies Condition (a) above.

To see that U satisfies Condition (b) above, we first prove thatNH0

�I�j1� S′1�W1� − S� ∩Xj ⊆ W . Consider an arbitrary p ∈ NH0

�I�j1� S′1�

W1� − S� ∩Xj and the unique nation b ∈ I�j1� S′1�W1� − S with p ∈ NH0

�b�.Since I�j1� S′

1�W1� ∩ Xj1= S′

1 and S′1 ∩ Xj = S, we have b �∈ Xj1

∩ Xj .Thus, b ∈ V �j1� − Xj . Let j3 be a vertex of � with �p� b� ⊆ Xj3

; j3 mustexist because �p� b� is an edge of H0 and � is a tree-decompositionof H0. Since b ∈ V �j1� − Xj and � is a tree-decomposition of H0,either j3 = j1 or j3 is a descendant of j1 in � . In either case, p ∈ Xj1or else � would not have been a tree-decomposition of H0. So,p ∈ NH0

�I�j1� S′1�W1� − S� ∩ Xj ∩ Xj1

⊆ W1 ∩ Xj ⊆ W . This finishesthe proof that NH0

�I�j1� S′1�W1� − S� ∩ Xj ⊆ W . Similarly, we can prove

that NH0�I�j2� S′

2�W2� − S� ∩ Xj ⊆ W . These two facts together with the

28 zhi-zhong chen

fact that NH0�S� ∩ Xj ⊆ W imply that NH0

�U� ∩ Xj ⊆ W . The reversedirection (namely, W ⊆ NH0

�U� ∩ Xj) follows from Condition (iv) aboveand the fact that for each i ∈ �1� 2�, Wi ∩ Xj ⊆ Wi = NH0

�I�ji� S′i� Wi��,

and W ∩NH0�Si� ⊆ NH0

�S�. Therefore, NH0�U� ∩Xj = W ; i.e., U satisfies

Condition (b) above.We next prove that U satisfies Condition (c) above. Toward a contradic-

tion, assume that two distinct nations a and b in U have a common neighborp in H0. Then, since �p� a� is an edge of H0 and � is a tree-decompositionof H0, there is a vertex j3 in � with �p� a� ⊆ Xj3

. Similarly, there is a vertexj4 in � with �p� b� ⊆ Xj4

. Moreover, by Condition (a) above and the factthat S, I�j1� S′

1�W1� and I�j2� S′2�W2� are independent sets of G0, exactly

one of the following three cases occurs:

• a ∈ I�j1� S′1�W1� −Xj and b ∈ I�j2� S′

2�W2� −Xj .

• a ∈ I�j1� S′1�W1� −Xj and b ∈ S −Xj1

.

• a ∈ I�j2� S′2�W2� −Xj and b ∈ S −Xj2

.

First, suppose that the first case occurs. Then, since� is a tree-decompositionof H0, either j3 = j1 or j3 is a descendant of j1 in � . Similarly, either j4 = j2or j4 is a descendant of j2 in � . Thus, p ∈ Xj ∩Xj1

∩Xj2or else � would

not have been a tree-decomposition of H0. Combining this, Condition (b)above and the fact that NH0

�I�ji� S′i� Wi�� ∩Xji

= Wi for each i ∈ �1� 2�, wehave that p ∈ W ∩W1 ∩W2. So, by Condition (iii) above, p ∈ NH0

�S1 ∩ S2�.Let c be a nation in S1 ∩ S2 with p ∈ NH0

�c�. Then, a �= c and both aand c belong to I�j1� S′

1�W1� and are adjacent to p in H0, contradicting thefact that I�j1� S′

1�W1� is an independent set of G0. Therefore, the first casecannot occur.

Now, suppose that the second case occurs. Then, since � is a tree-decomposition of H0, either j3 = j1 or j3 is a descendant of j1 in � .For the same reason, j4 cannot be j1 or a descendant of j1 in � . Thus,p ∈ Xj ∩Xj1

or else � would not have been a tree-decomposition of H0.In turn, p ∈ Xj ∩W1 because NH0

�I�j1� S′1�W1�� ∩Xj1

= W1. However, byCondition (ii) above, p �∈ NH0

�S − S1�, contradicting the assumption thatb ∈ S − Xj1

. So, the second case cannot occur. Similarly, the third casecannot occur, either. Since none of the three cases occurs, U must satisfyCondition (c) above.

Before processing the vertices of � , we compute and store the adjacencymatrix of H0 and that of its half square. Then, by Claims 1 and 2, processingeach vertex j of � takes O�218hh� time. Since � has J = O�H0� vertices,the whole processing takes O�218hhH0� time.

By Lemma 3.1, we have:

Theorem 3.2. There is a PTAS for the special case of the MIS problemrestricted to map graphs where the input graph is given together with a map


� = �� ϕ�. It runs in O�h�2 + 218hh2�� time, where h = � 1ε� and ε is the

given error parameter.

4. PTASs FOR k-MAP GRAPHS

We use an extension of the Baker approach, suggested in [8]. The onlydifference between the extension and the Baker approach is that instead ofdecomposing the given graph G into k′-outerplanar subgraphs, we decom-pose G into subgraphs of bounded treewidth such that each vertex of Gappears in all but one of these subgraphs. The next lemma shows that sucha decomposition is possible for a k-map graph.

Lemma 4.1. Given a k-map graph G and an integer h ≥ 2, it takesO�hG� time to compute a list � of h induced subgraphs of G such that(1) each subgraph in � is of treewidth ≤ k�6h− 4� and (2) each vertex of Gappears in exactly h− 1 of the subgraphs in � .

Proof. We may assume that G is connected. Starting at an arbitraryvertex r, we perform a breadth-first search on G to obtain a BFS tree T ofG. For each vertex v in G, we define the level number &�v� of v to be thenumber of edges on the path from r to v in T . For each i ∈ �0� � � � � h− 1�,let Gi be the subgraph of G obtained from G by deleting all vertices v with&�v� ≡ i (mod h). Obviously, each vertex of G appears in exactly h− 1 ofthese subgraphs. It remains to show that each Gi has treewidth ≤ k�6h− 4�.

Let H = �V�� EH� be a bipartite planar graph whose half square is G.Such H exists, by Lemma 2.1. Starting at nation r, we perform a BFS onH to obtain a BFS tree T ′. For each vertex v in H, we define the levelnumber &′�v� of v to be the number of edges on the path from r to v in T ′.It can be verified that &′�v� = 2&�v� for all nations v of G. Thus, for eachi ∈ �0� � � � � h− 1�, Gi is the half square of the subgraph Hi of H obtainedfrom H by deleting all nations v with &′�v� ≡ 2i (mod 2h). Clearly, Hi is a�2h− 1�-outerplanar graph, and hence has treewidth ≤ 6h− 4 [4].

Let � = ��Xjj ∈ Ji��i� be a tree-decomposition of Hi withtreewidth ≤ 6h − 4. For each j ∈ Ji, let X ′

j be the set obtained fromXj by replacing each p ∈ Xj ∩� with the nations v ∈ V −Xj such that v isadjacent to p in Hi. Obviously, the treewidth of ��X ′

jj ∈ Ji��i� is at mostk�6h− 4�. It remains to show that ��X ′

jj ∈ Ji��i� is a tree-decompositionof Gi.

Since Gi is the half square of Hi,⋃

j∈Ji X′i is the set of all vertices in

Gi. Let �u� v� be an edge in Gi. Then, there is a p ∈ � such that bothu and v are adjacent to p in H. Since p is contained in Hi, there is aj ∈ Ji with p ∈ X ′

j . So, both u and v are in X ′j . Next, let j1� j2� j3 ∈ Ji and

suppose that j2 lies on the path P between j1 and j3 in � . We want to show

30 zhi-zhong chen

that X ′j1∩X ′

j3⊆ X ′

j2. To this end, let u ∈ X ′

j1∩X ′

j3. Consider the following

cases:Case 1: u ∈ Xj1

and u ∈ Xj3. Then, u ∈ Xj2

and certainly u ∈ X ′j2.

Case 2: u ∈ Xj1and u �∈ Xj3

. Then, there is a p ∈ Xj3∩ � such that u

is adjacent to p in Hi. Since �p�u� is an edge in Hi, there is a j′ ∈ Ji with�p�u� ⊆ Xj′ . Because � is a tree, j2 appears either on the path betweenj′ and j1 in � or on the path between j′ and j3 in � . In the former case,u ∈ Xj2

and certainly u ∈ X ′j2. In the latter case, p ∈ Xj2

and in turnu ∈ X ′

j2.

Case 3: u �∈ Xj1and u ∈ Xj3

—Similar to Case 2.Case 4: u �∈ Xj1

and u �∈ Xj3. Then, there are points p1 ∈ Xj1

∩ � andp3 ∈ Xj3

∩ � such that u is adjacent to both p1 and p3 in Hi. If p1 = p3,then p1 ∈ Xj2

and in turn u ∈ X ′j2. So, suppose p1 �= p3. Since �p1� u� is

an edge in Hi, there is a j′ ∈ Ji with �p1� u� ⊆ Xj′ . Because � is a tree, j2appears either on the path between j′ and j1 in � or on the path between j′

and j3 in � . In the former case, p1 ∈ Xj2and in turn u ∈ X ′

j2. So, suppose

it is the latter case. Since �p3� u� is an edge in Hi, there is a j′′ ∈ Ji with�p3� u� ⊆ Xj′′ . Because � is a tree, j2 appears either on the path betweenj′′ and j′ in � or on the path between j′′ and j3 in � . In the former case,u ∈ Xj2

and certainly u ∈ X ′j2. In the latter case, p3 ∈ Xj2

and in turnu ∈ X ′

j2.

Theorem 4.2. For any fixed integer k, there is a PTAS for the MIS problemrestricted to k-map graphs. It runs in O�232�6kh−4k�3hn� time, where n is thesize of the given k-map graph, h = � 1

ε�, and ε is the given error parameter.

Proof. Let G be the given vertex-weighted k-map graph and let ε bethe given error parameter. Let h = � 1

ε�. Using Lemma 4.1, we compute h

induced subgraphs of G in O�hG� time such that (1) each of the subgraphsis of treewidth ≤ k�6h − 4� and (2) each vertex of G appears in exactlyh − 1 of the subgraphs. For each such subgraph Gi, we first compute atree-decomposition of Gi in O�232�6kh−4k�3 Gi� time [6], and then use thetree-decomposition to find a maximum weight independent set Ii of Gi inO�26kh−4kk2h2Gi� time [17]. The Ii whose weight is maximized over allsuch subgraphs Gi has weight at least �1− ε� ·OPT �G�.

The PTAS in Theorem 4.2 is not practical. We next show a differentPTAS for the special case of the UMIS problem where the input graph isgiven together with a k-map. This PTAS is based on the Lipton and Tarjanapproach. Although this PTAS is not better than the one in Section 3, thetools used in its design seem to be fundamental. Indeed, the tools will beused in the next section, and we believe that they can be used in otherapplications.


A map � = �� ϕ� is well formed if (1) the degree of each point in � isat least 3, (2) no edge of � is shared by two lakes, (3) no lake touches apoint whose degree in � is 4 or more, and (4) the boundary of each lakecontains at least four edges.

Lemma 4.3. Given a map � = �� ϕ� with at least three nations andwithout isolated nations, it takes O�� time to obtain a well-formed map �′

such that (1) the graph of �′ is the same as that of � and (2) for every integerk ≥ 3, � is a k-map if and only if so is �′.

Proof. Let � = �� ϕ� be as in the lemma. We modify � by performingthe following steps in turn:

1. Delete all edges e of � such that there is no nation on � whoseboundary contains e.

2. Delete all isolated points in �. (Comment: Steps 1 and 2 have noeffect on the nations and their adjacencies on �.)

3. If a point p of degree 4 or more in � is touched by a lake, then foreach pair �e1� e2� of consecutive edges incident to p such that the boundaryof a lake L contains both e1 and e2, perform the following substeps:

(a) Split e2 into two edges by putting a new point x at the middleof e2.

(b) Split L into two smaller faces L and L′ by adding a new edgeex between x and the endpoint of e1 other than p, where the boundary ofL′ is a triangle formed by the edges e1, ex, and �x�p�.

(c) Find the nation f1 �= L′, whose boundary contains e1, anddelete e1 to merge L′ and f2 into a larger nation f1.

Repeat Step 3 until no such point as p exists in �. (Comment: After this,each connected component of � becomes biconnected and the adjacenciesof the nations remain the same as before.)

4. If a point p is incident to exactly two edges e1 = �p� q1� ande2 = �p� q2� in �, then replace edges e1 and e2 by a single edge �q1� q2�.(Comment: q1 �= q2 because each connected component of � is bicon-nected and no nation is isolated.) Repeat Step 4 until no such point as pexists in �.

5. For each lake L on � whose boundary contains exactly two edgese1 and e2, find the nation f1, whose boundary contains e1, and delete e1 tomerge f1 and L into a larger nation f1. (Comment: After Step 5, if a pointp is incident to exactly two edges e1 and e2, then neither e1 nor e2 appearson the boundary of a lake.)

6. Same as Step 4. (Comment: After Step 6, no lake on � can containexactly two edges.)

32 zhi-zhong chen

7. For each lake L whose boundary contains exactly three edges e1 =�p1� p2�, e2 = �p2� p3�, and e3 = �p3� p1�, perform the following substeps:

(a) Put a new point q inside L.

(b) Split L into three smaller faces L1, L2 and L3 by adding threenew edges �q�p1�, �q�p2� and �q�p3�, where the boundary of Li containsei for each i ∈ �1� 2� 3�.

(c) For each i ∈ �1� 2� 3�, find the nation fi �= Li whose boundarycontains ei, and delete ei to merge Li and fi into a larger nation fi.

After the above steps are performed, � becomes well formed and itsgraph remains unchanged.

Let � = �� ϕ� be a well-formed map. Then, each point on the boundaryof a lake on � is a 2-point in � and has degree 3 in �; moreover, for allintegers i ≥ 3, each i-point in � has degree i in �. Suppose that a nation Lon � touches a 3+-point p. Then, the two edges e1 and e2 on the boundaryof L and incident to p must appear around p consecutively. Detaching Lfrom p is the operation of modifying � by performing Substeps 3a through3c in the proof of Lemma 4.3.

Lemma 4.4. Let k ≥ 3 and let G be a k-map graph with n ≥ 3 verticesand m edges. Then, m ≤ kn− 2k.

Proof. We may assume that G is connected. Let � = �� ϕ� be a well-formed k-map whose graph is G. For each integer j ≥ 4, let hj be thenumber of j-points in �. Let H be the simple and bipartite plane graphconstructed from � by (1) putting a new point pf inside each nation f ,(2) connecting pf to each 3+-point on the boundary of f , and (3) deletingall original edges. H has at most 2�n + ∑k

j=4 hj� − 4 edges and exactly∑k

j=4 jhj edges. Thus,∑k

j=4�j − 2�hj ≤ 2n− 4.After detaching a nation from a j-point p with j ≥ 4 in �, we lose

at most j − 3 edges from G; moreover, p becomes a �j − 1�-pointand no new 3+-point is created. Thus, after repeatedly detaching anation from a 3+-point until no 3+-point exists in �, we lose at most∑k

j=4��j − 3��j − 2�/2�hj edges from G; moreover, G becomes planar.Hence, m ≤ 3n− 6+∑k

j=4��j − 3��j − 2�/2�hj .Combining the two inequalities above, we obtain m ≤ kn− 2k.

It is known that k-map graphs are 2k-colorable [15], but the followingnew proof has the advantage of being algorithmic.

Corollary 4.5. Given an integer k and a k-map graph G, it takes O�G�time to color G with at most 2k colors. Consequently, given a k-map � =�� ϕ�, it takes O�k�� time to color the graph of � with at most 2k colors.


Proof. The corollary trivially holds when k ≤ 2. So, assume k ≥ 3. ByLemma 4.4, the vertex v with minimum degree in G has degree 2k− 1 orless. We delete v from G, recursively color the resulting graph H whichmust be a k-map graph, and extend the returned 2k-coloring of H to a2k-coloring of the original graph G.

It is known that k-map graphs with k ≥ 8 can be colored with at most2k− 3 colors [7], but the proof does not yield an efficient algorithm.

Let G be a connected graph with nonnegative weights on its vertices. Theweight of a subgraph of G is the total weight of vertices in the subgraph.A separator of G is a subgraph H of G such that deleting the verticesof H from G leaves a graph whose connected components each haveweight ≤ 2W /3, where W is the weight of G. The size of a separator H isthe number of vertices in H.

Lemma 4.6. Given a k-map � = �� ϕ� on which each nation has anassociated nonnegative weight and at least two nations exist, it takes O��time to find a separator H of the graph G of � such that the size of H is atmost 2

√2k�n− 1�.

Proof. Obviously, a separator of the heaviest connected component G′

of G is also a separator of G, and the lemma trivially holds when G′ hasat most two vertices. So, we may assume that G is connected and n ≥ 3.Moreover, if k ≤ 3, then G is planar and the lemma holds [11]. So, assumek ≥ 4. Then, by Lemma 4.3, we may assume that � is a well formed k-map.In turn, � is biconnected and the dual plane graph �∗ of � is biconnected.Since � is a well-formed k-map, the boundary of each face in �∗ containsat most k edges. Each vertex of �∗ that is a nation on � has its associatedweight. To each vertex of �∗ that is a lake on �, we assign a weight 0. Then,we find a separator C of �∗ that is a simple cycle and is of size at most2√k�n+ nL� [13], where nL is the number of lakes on �. Note that the

vertices of C that are nations on � form a separator of G. So, it suffices toshow that nL ≤ n− 2.

Let np be the number of points in �. Then, np ≥ 4nL because � iswell formed. Let H be the simple and bipartite plane graph constructedfrom � by (1) putting a new point qf inside each face f of �, (2) connect-ing qf to each point on the boundary of f , and (3) deleting all originaledges. H has at most 2�n + nL + np� − 4 edges and at least 3np edges.Thus, nL ≤ n− 2.

Using Lemma 4.6, we can emulate Lipton and Tarjan [12] to prove thefollowing theorem:

Theorem 4.7. For any fixed integer k, there is a PTAS for the special caseof the UMIS problem restricted to k-map graphs where the input graph is giventogether with a k-map � = �� ϕ�. It runs in O�� log � + 2288h2k3

n� time,

34 zhi-zhong chen

where n is the number of nations on �, h = � 1ε�, and ε is the given error

parameter.

Proof. By Lemma 4.6 and the discussions in [12], we can prove the fol-lowing: Given a k-map � = �� ϕ� and a positive integer x, we can computea set S of nations on � in O�� log �� time such that S ≤ 6n

√2k/x and

the removal of S from the map graph G of � leaves a graph G′ whose con-nected components each contain at most x vertices. For each connectedcomponent C of G′, we compute an independent set IC of maximum sizein O�C2x� time. The output is I = ⋃

C IC . By Lemma 4.4, I/α�G� ≥1− 12k

√2k/x. Thus, setting x = 288h2k3 yields I/α�G� ≥ 1− 1

h.

5. APPROXIMATION ALGORITHMS FOR MAP GRAPHS

This section presents four approximation algorithms. The first three arefor the UMIS problem restricted to map graphs. The fourth is for the MISproblem restricted to map graphs.

For a well formed map � = �� ϕ�, let �� be the plane graph obtainedfrom the dual plane graph �∗ of � by deleting all vertices that are lakeson �. Note that corresponding to each 2+-point p in �, there is a uniqueface p in �� such that the vertices on the boundary of p are exactlythe nations on � that touch p; the boundary of p is a simple cycle. Wecall p a point-face of ��. Those faces of �� that are not point-facesare called lake-faces.

Consider the following algorithm, called MapIS1:

Input: A map � = �� ϕ�.1. If at most two nations are on �, then output a maximum indepen-

dent set of the graph of � and halt.

2. If there is an isolated nation on �, then find such a nation f ;otherwise, perform the following substeps:

2.1. Modify � to a well-formed map, without altering the graphof �.

2.2. Construct ��.2.3. Find a minimum-degree vertex f in ��.

3. Construct a new map �′ = �� ϕ′�, where for all faces f1 of �, iff1 = f or f1 is a neighbor of f in the graph of �, then ϕ′�f1� = 0, whileϕ′�f1� = ϕ�f1� otherwise.


4. Recursively call MapIS1 on input �′, to compute an independentset J in the graph of �′.

5. Output �f� ∪ J and halt.

Theorem 5.1. MapIS1 achieves a ratio of 5 and runs in O��2� time.

Proof. The proof is by induction on n, the number of nations on theinput map �. The case where n ≤ 2 is trivial. Consider the case wheren ≥ 3. Let G be the graph of �. Let f be the nation found in Step 2. Iff is an isolated nation, then we are done. So, assume that f is found inSubstep 2.3. Since � is modified to a well-formed map in Substep 2.1, ��is a plane graph in which the boundary of each face contains at least threeedges. Thus, the number of edges in �� is at most 3n− 6. So, the degreeof f in �� is at most 5.

We claim that the neighborhood of f in G can be divided into at most fivecliques. To see this, let 1� � � � �& be the faces in �� whose boundarieseach contain vertex f . Then, & ≤ 5. For each i ∈ �1� � � � � &�, if i is apoint-face, then the vertices on i form a clique Ci of G. For each neighborfj of f in G that is in no Ci, �f� fj� must be an edge in ��, and fjforms a singleton clique of G. The total number of such cliques as Ci

and fj is bounded from above by &. This establishes the claim. By theclaim, each independent set of G contains at most five vertices among fand its neighbors in G. This fact and the inductive hypothesis imply thetheorem.


Input: A map graph G = �V�E�.1. If G has at most two vertices, then output a maximum independent

set of G and halt.

2. Find a vertex f in G such that α�Gf � ≤ 5, where Gf = G�NG�f ��.(Comment: α�Gf � is the independence number of graph Gf . See Section 2for its definition.)

3. Remove f and its neighbors from G.

4. Recursively call MapIS2 on input G, to compute an independentset J.

5. Output �f� ∪ J and halt.

Theorem 5.2. MapIS2 achieves a ratio of 5, and runs in O�V ·∑f∈V NG�f �6� time.

Proof. By the proof of Theorem 5.1, Step 2 of MapIS2 can always bedone. Moreover, it can be done in O�∑f∈V NG�f �6� time.

36 zhi-zhong chen


Input: A map graph G = �V�E�, an integer k ≥ 4, and a positive realnumber ε less than 1.

1. If G has at most two vertices, then output a maximum independentset of G and halt.

2. If G has a vertex f with α�Gf � ≤ 4, where Gf = G�NG�f ��, thenperform the following substeps:

2.1. Remove f and its neighbors from G.2.2. Recursively call MapIS3 on input G to compute an indepen-

dent set J.2.3. Output �f� ∪ J and halt.

3. Repeat the following substeps until G has no clique of size ≥ k+ 1(and hence is a k-map graph):

3.1. Find a maximal clique C in G such that C contains at leastk+ 1 vertices.

3.2. Remove all vertices of C from G.

4. Color G with at most 2k colors (cf. Corollary 4.5); let J1 be themaximum set of vertices with the same color.

5. Compute an independent set J2 in G with J2 ≥ �1− ε�α�G�.6. Output the larger one among J1 and J2, and halt.

Theorem 5.3. If we fix k = 28 and ε = 0�25, then MapIS3 achieves aratio of 4 and runs in O�V ·∑f∈V NG�f �5 + V 3 · E� time.

Proof. Let G0 be the input graph. We may assume that G0 is connected.By the proof of Theorem 5.2, it suffices to prove that if α�Gf � ≥ 5 for allvertices f of G, then the J1 found in Step 4 or the J2 found in Step 5 hassize ≥ α�G0�/4. So, suppose that α�Gf � ≥ 5.

Let n be the number of vertices in G0. Fix a well-formed map � of G0.For each integer i > k, let hi be the number of cliques C in G0 such that(1) C contains exactly i vertices and (2) C is found in Substep 3.1 by thealgorithm. We claim that the number x of edges in �� is bounded fromabove by 3n− 6−∑∞

i=k+1 hi�i− 4�. To see this claim, let 1� � � � �t be thepoint-faces in �� whose boundaries each contain at least three vertices.For each j ∈ �1� � � � � t�, fix a vertex fj on the boundary of j . We cantriangulate each face j by adding exactly j − 3 new edges incident tofj; call these new edges imaginary edges incident to fj . There are exactlyj − 3 imaginary edges incident to fj . Thus, x ≤ 3n− 6−∑t

j=1�j − 3�.Let C be a clique of G0 found in Substep 3.1. We say that an imaginary


edge �fj� f ′� with j ∈ �1� � � � � t� is contributed by C, if f ′ is contained in C.Let i be the number of vertices in C. To prove the claim, it remains to provethat the number of imaginary edges contributed by C is at least i − 4. ByLemma 2.2, only the following three cases can occur:

• Case 1: C is a pizza. Then, there is a point p in � such that allnations f ′ in C touch p on �. Let j be the point-face in �� that corre-sponds to p. No matter whether fj is in C or not, at least i − 3 imaginaryedges incident to fj are contributed by C.

• Case 2: C is a pizza-with-crust. Then, there is a point p in � suchthat C − 1 nations in C touch p on �. Similar to Case 1, at least �i− 1�− 3imaginary edges are contributed by C.

• Case 3: C is a hamantasch. Then, there are three points p1 throughp3 in � such that each nation f ′ ∈ C touches at least two of the threepoints on � and each point is touched by at least two nations in C. Sincei ≥ k+ 1 ≥ 5, at most one of p1 through p3 is a 2-point in �. Moreover,if one of p1 through p3 is a 2-point, then C is actually a pizza-with-crustand Case 2 applies. So, we may assume that p1 through p3 are 2+-pointsin �. Let j1

be the point-face in �� that corresponds to p1, and leta1 be the number of nations in C that touch p1 on �. Define j2

, a2,j3

, and a3 similarly. Then, the number of imaginary edges incident to fj1(respectively, fj2 , or fj3� contributed by C is at least a1 − 3 (respectively,a2 − 3, or a3 − 3). Thus, the number of imaginary edges contributed by Cis at least �a1 + a2 + a3� − 9 ≥ 2i− 9 ≥ i− 4.

By the claim, x ≤ 3n − 6 −∑∞i=k+1 hi�i − 4�. The degree of f in ��

is at most � 2xn�. On the other hand, since α�Gf � ≥ 5, the degree of f

in �� is at least 5. Thus, 2∑∞

i=k+1 hi�i − 4� ≤ n. In turn,∑∞

i=k+1 hii ≤n�k+ 1�/2�k− 3� and

∑∞i=k+1 hi ≤ n/2�k− 3�.

Let J be the larger one among J1 and J2. By Step 5, J ≥ �1 − ε��α�G0� −

∑∞i=k+1 hi�. By Step 4 and Corollary 4.5, J ≥ 1/2k · �n −∑∞

i=k+1 ihi�. Thus, J/α�G0� ≥ �1 − ε��1 − 1/2�k− 3� · n/α�G0�� andJ/α�G0� ≥ k− 7/4k�k− 3� · n/α�G0�. As the value of n/α�G0�decreases, the first lower bound on J/α�G0� increases while the sec-ond lower bound on J/α�G0� decreases. The worst case happenswhen n/α�G0� = 4k�k− 3��1− ε�/3k− 2kε− 7. Therefore, J/α�G0� ≥�1 − ε� · k− 7/3k− 2kε− 7. By fixing ε = 1/4 and k = 28, we haveJ/α�G0� ≥ 1/4.

By Theorem 5.2, the total time needed by Steps 1 and 2 is O�V ·∑f∈V NG�f �5�. G has only O�V � maximal cliques [9], and hence the max-

imal cliques can be enumerated in O�V 2 · E� time [14]. By Corollary 4.5and Theorem 4.2, Steps 5 and 6 take O�G� time. So, the total time neededis as claimed.

38 zhi-zhong chen

Remark. MapIS3 cannot give a ratio better than 4 no matter how wechoose k and ε.

Finally, we show that the proof of Theorem 5.1 together with thelocal-ratio technique developed by Bar-Yehuda and Even [3] leads to anapproximation algorithm for the MIS problem restricted to map graphsthat achieve a ratio of 5. The local-ratio technique in its most general formis stated in Theorem 2.1 in [2]. Theorem 2.1 in [2] applied to the MISproblem can be stated as follows.

Theorem 5.4 [2]. Let G = �V�E� be a graph. Let w1 and w2 be twofunctions assigning a nonnegative weight to each v ∈ V . Let w3 = w1 + w2;i.e., w3�v� = w1�v� + w2�v� for every v ∈ V . For each i ∈ �1� 2� 3�, let OPTi

be the maximum weight of an independent set of G when each v ∈ V hasweight wi�v�. Suppose that I is an independent set of G such that for somereal number r > 0,

∑v∈I w1�v� ≥ r ·OPT1 and

∑v∈I w2�v� ≥ r ·OPT2. Then,∑

v∈I w3�v� ≥ r ·OPT3.

Consider the following modification of MapIS2, called MapIS4:

Input: A map graph G whose vertices each have a positive weight.

1. If G has at most two vertices, then output a maximum weightindependent set of G and halt.

2. Find a vertex f in G such that α�Gf � ≤ 5, where Gf = G�NG�f ��.3. Let 4 = �f� ∪NG�f �. Find the minimum weight c of a vertex of

4 in G.4. For every v ∈ 4, subtract c from the weight of v.5. Compute the set U of those v ∈ 4 whose weight is 0.6. Remove the vertices in U from G.7. Recursively call MapIS4 on input G, to compute an independent

set J.8. Find a subset J ′ of U such that J ∪ J ′ is an independent set of

the original input graph but J ∪ J ′ ∪ �v� is not an independent set of theoriginal input graph for every v ∈ U − J ′.

9. Output J ∪ J ′ and halt.

Theorem 5.5. MapIS4 achieves a ratio of 5 and runs in O�G ·∑f∈V NG�f �6� time.

Proof. The proof is by induction on n, the number of vertices in theinput map graph. The case where n ≤ 2 is trivial. Consider the case wheren ≥ 3. Let V be the vertex-set of the input map graph. Let w3 be thefunction with domain V such that w3�v� is the weight of vertex v in the


input map graph. Moreover, let w1 be the function with domain V suchthat w1�v� = c for every v ∈ 4, while w1�v� = 0 for every v ∈ V − 4. Letw2 = w3 −w1; i.e., w2�v� = w3�v� −w1�v� for every v ∈ V .

For i ∈ �1� 2� 3�, let OPTi be the maximum weight of an independentset of the input map graph when each vertex v ∈ V has weight wi�v�.By Step 2, OPT1 ≤ 5c. On the other hand, 4 ∩ �J ∪ J ′� �= � by Step 8, and∑

v∈J∪J ′ w1�v� ≥ c in turn. So,∑

v∈J∪J ′ w1�v� ≥ OPT1/5. Moreover, by theinductive hypothesis,

∑v∈J∪J ′ w2�v� =

∑v∈J w2�v� ≥ OPT2/5. Therefore, by

Theorem 5.4,∑

v∈J∪J ′ w3�v� ≥ OPT3/5. Consequently, MapIS4 achieves aratio of 5. Obviously, each recursion of MapIS4 removes at least one vertexfrom G and takes O�∑f∈V NG�f �6� time. Hence, MapIS4 runs in O�G ·∑

f∈V NG�f �6� time.

6. CONCLUDING REMARKS

The impracticality of the PTAS in Theorem 4.2 and hence that of MaxIS3mainly result from the impracticality of the Bodlaender algorithm [6] fortree-decomposition. A more practical algorithm for tree-decomposition canmake our algorithms more practical.

It would be nice if one could design approximation algorithms for theproblems that achieve better ratios than ours. Can MapIS1 and MapIS2 beimplemented to run faster than as claimed?

It also seems interesting to design approximation algorithms for otherNP-hard optimization problems than the MIS problem. A particularlyinteresting problem is the minimum coloring problem where we want tocolor a given map graph with as few colors as possible. As a consequenceof Corollary 4.5, we can easily obtain a polynomial-time approximationalgorithm for this problem that achieves a ratio of 2. Can this ratio beimproved?

ACKNOWLEDGMENTS

The author thanks Professors Magnus M. Halldorsson, Xin He, and Seinosuke Toda forhelpful discussions. He also thanks the referees for their helpful comments. Especially, onereferee pointed out an error in an earlier proof of Lemma 3.1 and another referee suggestedthe improvement of an original logarithmic ratio in Theorem 5.5 to 5.

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Approximation Algorithms for Independent Sets in Map Graphs