Maximum Induced Linear Forests in Outerplanar Graphs

Michael J. Pelsmajer�

Department of Applied Mathematics, Illinois Institute of Technology, Chicago, IL 60616,USA. e-mail: pelsmajer@iit.edu

Abstract. Every simple n–vertex outerplanar graph has an induced subgraph with at least4nþ27

� �vertices that is a linear forest, and this bound is sharp.

1. Introduction

We consider only simple, non-empty graphs.Albertson and Berman [2] conjectured that every planar graph has an induced

subgraph with at least half of the vertices that is a forest. (This would directly implythat a planar graph contains an independent setwith at least a quarter of the vertices,without using the Four-Color Theorem.)Akiyama andWatanabe [4] gave examplesshowing that this would be best possible. Borodin [5] proved the best known lowerbound of 2=5 on this ratio by proving that the vertex set of a planar graph may bepartitioned into five independent sets, any two of which together induce a forest.Akiyama and Watanabe [4] conjectured a lower bound of 5=8 for every bipartiteplanar graph and gave examples to show that this would be best possible.

Hosono [8] proved that every outerplanar graph has an induced subgraph withat least 2=3 of the vertices that is a forest. This bound is best possible, as shown bythe square of a path with n vertices or by a disjoint union of triangles.

A linear forest is a graph in which every component is a path; equivalently, it isa graph with no cycles and no vertices of degree at least 3. Poh [10] showed thatevery planar graph has an induced linear forest using at least 1=3 of the vertices,by proving that the vertex set of a planar graph can be partitioned into three setsinducing linear forests (conjectured independently in [6] and [9]). Chappell con-jectured that every planar graph has an induced linear forest using more than 4=9of the vertices, and he found examples to show that this is best possible. Thiswould strengthen Albertson’s result [1] that a planar graph G has an independent

�Work done while at the Department of Mathematics, University of Illinois, Urbana, IL 61801,USA

Graphs and Combinatorics (2004) 20:121–129Digital Object Identifier (DOI) 10.1007/s00373-003-0528-x Graphs and

Combinatorics� Springer-Verlag 2004

set with at least 2=9 of the vertices (which was proved independently of the Four-Color Theorem).

These results suggest the analogous question for linear forests in outerplanargraphs. Three groups ([3], [6], [9]) independently showed that the vertex set of anouterplanar graph can be partitioned into two sets, each of which induces alinear forest. This proves a lower bound of n=2 for the largest induced linearforest. Chappell conjectured that every outerplanar graph has an induced linearforest with more than 4=7 of the vertices. In this paper, we prove this conjec-ture.

We first present Chappell’s examples showing that the bound is best possible.We use G½S� to denote the subgraph of G induced by the vertex set S, writeG� S ¼ G½V ðGÞ � S�, and let NðvÞ denote the set of neighbors of vertex v.

Example 1. Let Hi be the 10-vertex graph shown on the left in Fig. 1. Let Gn bethe subgraph of H1 [ � � � [ H n=7d e induced by fv1; . . . ; vng. The maximum order ofan induced subgraph of Gn that is a linear forest is 4nþ2

7

� �. Since these examples

show that our bound is sharp, we include a proof that they achieve the bound.

Suppose that U is a vertex set that induces a linear forest in Gn such that Ucontains x but not y, and NðyÞ � x � NðxÞ � y. Then U � xþ y also induces alinear forest, because it induces a subgraph of a linear forest. Hence, if Gn½U � is aninduced linear forest of maximum order and U does not contain all y such thatNðyÞ � x � NðxÞ � y, then there is an induced linear forest of maximum order thatavoids x.

A maximum induced linear forest in Gn must exclude at least one element fromeach set of the form fv7i�2; v7i�1; v7ig or fv7iþ1; v7iþ2; v7iþ3g. By the above argu-ment, Gn has an induced linear forest Gn½U � of maximum order that avoidsfv7i�1 : 7i � ng and fv7iþ3 : 7iþ 3 � ng. (See Fig. 2.) If 7iþ 1 � n or 7i� 1 ¼ n,then v7i�3 has three neighbors which may still be in U , and so U must avoid v7i�3or one of those neighbors. Avoiding fv7i�3 : 7iþ 1 � n or 7i� 1 ¼ ng yields alinear forest with d4nþ2

7 e vertices, and it is a largest induced linear forest.We prove Chappell’s conjecture by determining the extreme value for all n.

Fig. 1. Chappell’s examples

122 M.J. Pelsmajer

Theorem 2. For every outerplanar graph G, there exists U � V ðGÞ such that G½U �is a linear forest and jU j � 4nðGÞþ2

7 .

In proving Theorem 2, we use the equivalent inequality jU j � 4jV ðGÞ�U jþ23 .

2. The Plan

We try to build up the desired inequality jU j � 4jV ðGÞ�U jþ23 inductively. Therefore,

in an outerplanar graph G, we seek a pair of disjoint vertex subsets U0; S0 suchthat G½U0� is a linear forest and jU0j � 4jS0j

3 . However, in applying the inductionhypothesis, we will not be able to ‘‘completely delete’’ all of U0 [ S0. We delete allof S0 but may need to retain some leaves of distinct components of the linearforest induced by U0 (see Fig. 3); we delete the rest of U0. The remaining vertex setwe call V 0, and we want to apply the induction hypothesis to G½V 0�.

The induction hypothesis will provide a set U 0 inducing a large linear forest inG½V 0�. We want U0 [ U 0 to induce a linear forest in G. In order to guarantee this,we will need to load the induction hypothesis with additional conditions to ensurethat vertices of V 0 \ U0 can only be used as leaves in G½U 0�. Also, we must avoidcounting vertices of V 0 \ U0 separately in both U 0 and U0.

The key idea in the proof is to account for the vertices of V 0 \ U0 partially asmembers of U0 and partially as members of V 0 using vertex weights, requiring thatthe vertices with ‘‘partial’’ weight occur (as leaves) in the induced linear forest.

Let G be the set of pairs ðG; PÞ, where G is an outerplanar graph and P is a setof vertices that have degree at most 1 in G. We call P the set of partial vertices ofG. The set P can be viewed as specifying a weight function w on V ðGÞ: partial

Fig. 2. This subgraph of G13 contains an induced linear forest of maximum order

Fig. 3. The Plan

Maximum Induced Linear Forests in Outerplanar Graphs 123

vertices have weight 2=3 and other vertices have weight 1. For a set X of vertices,we write wðX Þ for

P

x2XwðxÞ.

Definition 3. Let ðG; P Þ 2 G be a pair with associated weight function w. Supposethat V ðGÞ is partitioned into two sets U , S such that G½U � is a linear forest, Ucontains P , and wðUÞ � 4wðSÞþ2

3 . We say that ðU ; SÞ is a good partition for ðG; P Þ.

In order to prove Theorem 2, it suffices to show that every ðG; ;Þ 2 G has agood partition. To facilitate an inductive proof, we will show more generally thatevery ðG; P Þ 2 G has a good partition.

3. The Proof

We study the properties that can be assumed to hold for a pair ðG; P Þ 2 G that is acounterexample to Theorem 2 with minimal total weight. We call this a ‘‘minimalcounterexample’’. We may assume that G is not a linear forest, since otherwiseðG; PÞwould have good partition ðV ; ;Þ. Hence wemay also assume that P 6¼ V ðGÞ.

Definition 4. A pair ðG; P Þ 2 G is critical if it is a minimal counterexample withthe most edges.

Claim 5. If ðG; P Þ is critical, then G� P is a maximal outerplanar graph.

Proof. After adding edges to G� P to make it a maximal outerplanar graph, thepartial vertices may be restored while maintaining outerplanarity to form a graphG0. Because the edges of EðG0Þ � EðGÞ are not incident to P , the pair ðG0; P Þ is alsoin G. Any good partition for ðG0; P Þ is also a good partition for the original graph,so ðG0; P Þ is also a counterexample and has the same weight. (

Thus we may assume that G� P is connected and has no cut-vertices. If G isnot connected, then it has a component with vertex set VC � P . If ðG� VC; P � VCÞhas a good partition ðU ; SÞ, then ðU [ VC; SÞ is a good partition of ðG; P Þ. Hencewe may also assume that G is connected.

We make a further reduction. If V ðGÞ � P contains a vertex x such thatdegGðxÞ ¼ 1, then G� P has at most two vertices, since it has no cut-vertices. Let vbe the neighbor of x. If G is not a path, then G is a star centered at v, andðV ðGÞ � v; fvgÞ is a good partition for ðG; P Þ. Accordingly, we may assumehenceforth that P is the set of all vertices with degree 1 in G.

We will obtain a good partition for ðG; P Þ by considering peripheral ‘‘pieces’’of the graph. Our terminology for such pieces appears in [11, p.221].

Definition 6. For X � V ðGÞ, an X-lobe is a subgraph of G induced by the union ofX and the vertex set of one of the components of G� X .

Claim 7. Let H0; . . . ;H5 be the graphs in Fig. 4. If ðG; P Þ is critical, then H0 is not theunion of two fxg-lobes of G, and each Hi with 1 � i � 5 is not an fx1; x2g-lobe of G.

124 M.J. Pelsmajer

Proof. For each case i, suppose that Hi arises in the undesired way. The defini-tions and computations for case i appear in the table below. Theproof for each case is the same. Recall that P is the set of vertices of degree 1 in G.

If G ¼ Hi, then ðV ðGÞ � T ; T Þ is a good partition for ðG; PÞ. Otherwise, thepair ðG0; P 0Þ as defined belongs to G. By the minimality of G, there is a goodpartition ðU 0; S0Þ of ðG0; P 0Þ. With ðU ; SÞ as defined, we have P � U . Also, G½U �is obtained from G0½U 0� by lengthening one path component or adding one ortwo path components. Thus G½U � is a linear forest. Finally, sincewðUÞ � w0ðUÞ � 4

3 jS � S0j ¼ 43 ðwðSÞ � wðS0ÞÞ, where w and w0 are the weight

functions associated with ðG; P Þ and ðG0; P 0Þ, we have wðUÞ � 4wðSÞþ23 . (Note that

in Cases 2 and 4 vertices dropped from P 0 to obtain P acquire 13 additional

weight.) Thus ðU ; SÞ is a good partition for ðG; P Þ. (

From Case 0 we immediately conclude that when ðG; PÞ is critical, each vertexin G is adjacent to at most one partial vertex.

Definition 8. Given a vertex v0 in a graph G, let Vk ¼ fu 2 V ðGÞ : dðv0; uÞ ¼ kg.

Lemma 9. If v0 is a vertex in an outerplanar graph G and k � 0, then G½Vk� is alinear forest.

Fig. 4. The graphs of Claim 7

i T G0 P 0 U S wðUÞ � w0ðU 0Þ

0 ; G� V ðH0Þ P � fy1; y2g U 0 [ fy1; y2g S0 [ fxg 4=31 fy2g G� V ðH1Þ P U 0 [ fy1; y2; y3g S0 [ fx1; x2g 32 fy1g G� fx2; y2g P [ fy1g U 0 [ fy2g S0 [ fx2g 1þ 1=33 fy1g G� V ðH3Þ P � z U 0 [ fy1; y2; zg S0 [ fx1; x2g 2þ 2=34 fy2g G� fy2; zg ðP � zÞ [ fy1; y3g U 0 [ fzg S0 [ fy2g 2=3þ 2ð1=3Þ5 fyg G� z P � z U 0 [ fzg S0 2=3

Maximum Induced Linear Forests in Outerplanar Graphs 125

Proof. (This is essentially the proof of the result in [3], [6], [9] mentioned inSection 1.)

For a vertex ui in Vk, let Pi be a uiv0 path of length k that only intersects Vk atendpoint ui.

Suppose that u1, u2, u3 are three vertices on a cycle C in G½Vk�. The union of P1,P2, P3 and C contains a subdivision of K4, which is forbidden in outerplanargraphs (see [11], p240 or 256).

Suppose that u1; u2; u3; v 2 Vk, with u1; u2; u3 2 NðvÞ. The union of P1, P2, P3

with fu1v; u2v; u3vg contains a subdivision of K2;3, which is also forbidden inouterplanar graphs (see [11], p240 or 256). (

Definition 10. If v0 is a vertex in an outerplanar graph G, and u 2 Vk has aneighbor u0 2 Vk�1, then u0 is a parent of u, and u is a child of u0.

In an outerplanar graph, every vertex has at most two parents, since otherwisewe obtain a subdivision of K2;3.

Claim 11. Suppose that ðG; P Þ is critical, with v0 2 V ðGÞ � P . If Y is the vertex setof a component of G½Vk� with k > 1, then altogether Y has at most two parents. Ifalso jY j > 1, then Y has exactly two parents. If Y has two parents x1 and x2, thenthey are adjacent, and Y decomposes into two paths with a common vertex, inducedby the children of x1 and the children of x2, respectively.

Proof. Since k > 1, Y has a parent. If jY j > 1 and Y has only one parent, then theparent is a cut-vertex of G� P . By Claim 5, this is impossible.

If Y has at least three parents, then consider the subgraph consisting of G½Y �,an edge from each parent to Y , and a path of length k � 1 from v0 to each parent.This contains a subdivision of K2;3, which contradicts outerplanarity.

Now suppose that Y has parents x1 and x2. Let Q1 be a shortest v0x1 path, and letQ2 be a shortest v0x2 path. Consider the cycle formed by an x1x2–path withinQ1 [ Q2, some edges x1y and x2y0 with y; y0 2 Y , and the path from y to y0 in G½Y �. Inan outerplanar embedding of G� P , every inner face is a triangle, so the regioninside this cycle is triangulated. As there are no edges from ðQ1 � x1Þ [ ðQ2 � x2Þ toY , no edge of the triangulated region separates x1 from x2. Thus x1 and x2 areadjacent.

The final claim is trivial if jY j ¼ 1, so assume that jY j > 1. Let fy1; . . . ; yjY jg bethe vertices of Y in order along the path, indexed so that y1 is a child of x1. If thereexist yi; yj; yk with i < j < k such that yi; yk 2 NðxrÞ and yj 2 Nðx3�rÞ, then these

Fig. 5. A component of Vk with its parents, as in Claim 11

126 M.J. Pelsmajer

five vertices are the branch vertices in a subdivision of K2;3. Also, if x1 and x2 havetwo common neighbors in Y , then we have a subdivision of K4. Furthermore,every element of Y has a parent in fx1; x2g, so yi 2 Nðx1Þ for all i � j andyi 2 Nðx2Þ for all i > j for some 1 � j < jY j. The last neighbor of x1 andfirst neighbor of x2 in Y must be the same, since G� P is a maximal outerplanargraph. (

Definition 12. Let m ¼ maxfi : Vi 6¼ ;g; m is the eccentricity of vertex v0.

Claim 13. If ðG; P Þ is critical and v0 2 V ðGÞ � P , then Vm is an independent set andm � 2.

Proof. If Vm is not independent, then let Y be the vertex set of a non-trivialcomponent of G½Vm�. The set X of parents of Y is nonempty, because otherwisem ¼ 0 and jV j ¼ 1.

If jX j ¼ 1, then X can have no parents, for otherwise the element of X is a cut-vertex of G� P . Thus X ¼ fv0g and m ¼ 1, but then ðV � X ;X Þ is a good par-tition for G, since jY j � 2 ¼ 4�1þ2

3 .Therefore, m � 2, and Y has two parents. Let y1; . . . ; yt be the path induced by

Y , with y1 a child of x1 and yl the common child of x1 and x2. If l � 3, thenG½fx1; y1; y2; y3g� is an fx1; y3g-lobe forbidden by Case 2 of Claim 7. If l � t � 2,then G½fx2; yt�2; yt�1; ytg� is an fx2; yt�2g-lobe similarly forbidden. Therefore,2 � l � t � 1, which yields jY j � 3. This is forbidden by Case 1 of Claim 7 whenjY j ¼ 3 and by Case 2 when jY j ¼ 2. (

Suppose that ðG; P Þ 2 G is critical with v0 2 V ðGÞ � P . Let Y be the vertex setof a component of G½Vm�1� with at least one child. Let X be the set of parents of Y ,let Z be the set of children of Y , and let H ¼ G½X [ Y [ Z�.

For the remaining discussion, we adopt a canonical labeling of X [ Y . Label Yby y1; . . . ; yjY j, with x1 a parent of y1. Let l ¼ jY j if m ¼ 2 (and jX j ¼ 1); otherwiselet yl be the common child of x1 and x2. By Claim 11 with k ¼ m, the set of parentsof each z 2 Z has the form fyig or fyi; yiþ1g. Also, since the neighborhood of zcontains just its parents, z has a single parent only if z 2 P .

Definition 14. We say that H with these labels and structure is a critical config-uration. For x 2 X in a critical configuration, the subgraph of H induced by x andall its descendants is a side of H .

Fig. 6. A side of H

Maximum Induced Linear Forests in Outerplanar Graphs 127

First we consider the side whose intersection with Y is fyi : i � lg, with sharedparent x.

Claim 15. If l � 2 for a critical configuration in a critical ðG; P Þ, then y1 has nochildren.

Proof. Suppose that y1 does have children.Consider the subgraph inducedby y1 andall its neighbors; this is an fx; y2g-lobe of G (see Fig. 7a). If y1 has two children, thenCase 0 of Claim 7 implies that one of them is also a child of y2, and now Case 3 ofClaim 7 applies. If the only child of y1 is a partial vertex, then Case 5 of Claim 7applies. If the only child of y1 is also a child of y2, then Case 2 of Claim 7 applies. (

Claim 16. If H is a critical configuration in a critical ðG; PÞ, then l < 3.

Proof. Suppose that l � 3. Consider the fx; y3g-lobe of the critical configurationthat is induced by y2 and all of its neighbors. By Claim 15, y1 has no children. If y2also has no children, then Case 2 of Claim 7 applies. Now by Claim 11 and Case 0of Claim 7, y2 may have one partial vertex child, one child shared with parent y3,and no other children. If y2 has two children, then Case 4 of Claim 7 applies. If theonly child of y2 is a partial vertex, then Case 3 of Claim 7 applies. If the only childof y2 is also a child of y3, then Case 1 of Claim 7 applies. (

Of course a critical configuration may label Y in either direction along G½Y �, soClaim 15 and Claim 16 may be applied to both sides. We use this to reduce criticalconfigurations to a few remaining special cases.

Claim 17. There is no critical ðG; P Þ.

Proof. Let H be a critical configuration for ðG; P Þ. If m ¼ 2, then Claim 16 impliesthat jY j � 2. Now Claim 15, applied to both ends of G½Y �, shows that Y has nochildren; this contradicts m ¼ 2. Hence we may assume that m � 3, which requiresjX j ¼ 2. Since Y has exactly one common child of x1 and x2, applying Claim 16from both ends of G½Y � yields jY j � 2þ 2� 1 ¼ 3.

If jY j ¼ 3, then each side has two elements of Y , and Claim 15 implies that allchildren of Y are partial vertices adjacent to y2. Since y2 has at most one partialneighbor, it has exactly one, and this is Case 4 of Claim 7.

y1 y1

? ?

y2

x

? ?

y2 3y

x

Fig. 7. For Claim 15 and Claim 16

128 M.J. Pelsmajer

If jY j ¼ 2, then we may let l ¼ 2 by symmetry. Now Claim 15 implies that y1has no children, and so y2 must have one partial child. This is Case 1 of Claim 7.

If jY j ¼ 1, then Y has one partial child, and this is Case 5 of Claim 7. (

Since we have eliminated all possibilities for critical ðG; P Þ, we have completedthe proof of Theorem 2.

Acknowledgments. The author wishes to thank Douglas B. West for vastly improving theexposition of this paper. The author would also like to thank Radhika Ramamurthi foruseful discussions.

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Received: April 25, 2001Final version received: June 5, 2003

Maximum Induced Linear Forests in Outerplanar Graphs 129