the subtree center of a tree

The Subtree Center of a Tree

Juhani Nieminen, Matti Peltola

Department of Mathematics, Faculty of Technology, University of Oulu, PB 444,90571 Oulu, Finland

Received 5 November 1997; accepted 29 November 1998

Abstract: A method to determine the least central subtree of a tree is given. The structure of the treeshaving a single point as a least central subtree is described, and the relation of a least central subtree ofa tree to the centroid as well as to the center of that tree is given. © 1999 John Wiley & Sons, Inc. Networks34: 272–278, 1999

Keywords: central subtrees of a tree; center of a tree; centroid of a tree; joinsemilattice of subtrees ofa tree

1. INTRODUCTION

Various kinds of center problems have important applica-tions in transportation and business (facility planning andlocation problems) as well as in communication, social, andcomputing networks. Despite the importance of the prob-lem, the solution methods are limited to special types ofcenters like central paths or central points. In this paper, wepresent an idea for determining a central subtree of tree.Thus, we do not restrict the structure of the center. Theresult obtained can be a point or a path, if the point or pathis most central when compared with all subtrees of a tree. AsubtreeS of a treeT is a central subtree ofT if S has theminimum eccentricity in the joinsemilattice of all subtreesof T. Thus, the method of this paper picks up the bestconnected substructure among all connected substructuresof a tree. For example, if one likes to know what is the mostimportant/central suborganization of an organization of atree type, the method of this paper gives an answer. Thestructure of the trees, the central subtree of which is a singlepoint, is given. The relation of the central subtree to the

centroid as well as to the center of a tree is given. Variouskinds of centers in graphs were reported in the book [2] byBuckley and Harary and some recent results were givenin [3].

For every treeT, there is ajoinsemilattice L(T) ofsubtreesof T, where the meetS1 ` S2 equals the inter-sectionS1 ù S2 whenever the intersection is nonempty andS1 ~ S2 is the least subtree ofT containing the subtreesS1

andS2. Note that the empty graph is not a subtree ofT, and,thus, in general, there is no least element inL(T). Thesubtrees (convex subalgebras/ideals) of a tree constitute amedian algebra (see [1, Thm. 2.7] or [4, Thm. 4]) and, thus,the joinsemilatticeL(T) has a limited distributivity property(a median algebra property). This implies that, ifS1 andS2

are two subtrees ofT and if S1 ` S2 exists, the distancefrom S1 to S2 throughS1 ` S2 is the same as throughS1

~ S2 (see [1, Section 4]). Note that the distance in thejoinsemilattice L(T) is the same as the distance in the(undirected) Hasse diagram graphGL of L(T). We denotethe distance between two subtreesS1 andS2 in L(T) (aswell as inGL) by dL(S1, S2).

As is well known, theeccentricity e(v) of a point v ina connected graphG 5 (X, V) is the distance to a pointfarthest fromv: e(v) 5 max{dG(u, v)u u [ V}. TheCorrespondence to:J. Nieminen; e-mail [email protected]

© 1999 John Wiley & Sons, Inc. CCC 0028-3045/99/040272-07

272

center C of G consists of points with minimum eccentric-ity: C 5 { vue(v) 5 min{ e(u)uu [ V}}. A subtree S1 is acentral subtree of a treeT if it has the minimum eccen-tricity eL(S1) (briefly: minimum L-eccentricity) in thejoinsemilattice graphGL. A tree may contain several centralsubtrees. This can be seen by considering a path of sixpoints: eL(^{ v3, v4} &) 5 5 5 eL(^{ v2, v3, v4, v5} &),where^ A& means the subgraph induced by the point setA.A central subtree with the minimum number of points is aleast central subtreeof a treeT. The tree of two pathsv1v2v3v4v5v6v7 and v8v9v4v10v11 crossing at the pointv4

shows that there may be several least central subtrees in atree:eL(^{ v3, v4, v5, v10} &) 5 7 5 eL(^{ v3, v4, v5, v9} &).The least central subtrees of a treeT constitute the subtreecenter ofT. The goal of this paper was to describe theproperties of the least central subtrees of a tree.

If S1 andS2 are subtrees of a treeT, then thedistancedT(S1, S2) between S1 and S2 in T is the length of theshortest path joining two points ofS1 andS2 in T. Let A bea subtree of a treeT. The notationuAu means the number ofpoints inA.

At first, we prove a basic lemma on the relation betweenthe distancedL in the joinsemilattice graphGL of a treeTand the distancedT in the treeT.

Lemma 1. Let GL be the semilattice graph of all subtrees ofa tree T, and S1 and S2 be two subtrees of T. Then, thedistance dL(S1, S2) between S1 and S2 in GL is uS1u 1 uS2u1 2(dT(S1, S2) 2 1) if dT(S1, S2) $ 1, and uS1 ~ S2 u 2 uS1

` S2u if dT(S1, S2) 5 0.

Proof. The basic observation is that the existence of aline (S1, S2) in GL implies that the subtreeS1 is obtainedfrom the subtreeS2 by adding/removing a single point.Because of the median algebra property ofL(T), a shortestS1 2 S2 path goes through the pointS1 ~ S2 in GL. Now,the distancedL(S1, S1 ~ S2) gives the number of points onehas to add to the subtreeS1 in order to obtain the subtreeS1

~ S2. Thus,dL(S1, S1 ~ S2) 5 uS1 ~ S2u 2 uS1u. On theother hand,uS1 ~ S2u 5 uS1u 1 uS2u 1 dT(S1, S2) 2 1,and, hence,dL(S1, S1 ~ S2) 5 uS2u 1 dT(S1, S2) 2 1.Similarly, dL(S1 ~ S2, S2) 5 uS1u 1 dT(S1, S2) 2 1, and,accordingly,dL(S1, S2) 5 uS1u 1 uS2u 1 2(dT(S1, S2)2 1) whendT(S1, S2) $ 1.

If dT(S1, S2) 5 0, thenS1 ù S2 Þ A, and the subtreeS1 ` S2 exists. Now, by the median algebra (limiteddistributivity) property,dL(S1 ~ S2, S2) 5 dL(S1, S1 ` S2),and, thus,dL(S1, S2) 5 uS1 ~ S2u 2 uS1u 1 uS1u 2 uS1 `S2u 5 uS1 ~ S2u 2 uS1 ` S2u, and the lemma follows.

2. ON POINTS OF LEAST L-ECCENTRICITY

The points of leastL-eccentricity eL are basic buildingstones when constructing least central subtrees of a tree,

and, hence, the points of leastL-eccentricity are consideredfirst. Throughout this paper, we use the brief notationv [ Tfor the expression “a pointv belongs to the (point set of the)treeT” instead of the long notationv [ V(T). Similarly, weuse the brief expressionT 2 v instead of the long expres-sion V(T) \ { v}.

Theorem 2. Let y be a point of the tree T such that there isat least one point x in T with eL(x) , eL(y). All the points{ xueL(x) , eL(y) and x [ T} constitute the point set of asubtree in the tree T. Moreover, eL(v) $ uTu 2 1 for everypoint v of T.

Proof. Consider an arbitrary pointv of T. SincedT(T, v)5 0, dL(v, T) 5 uTu 2 1 by Lemma 1, and, thus,eL(v)$ uTu 2 1 for any pointv of T.

If the setX 5 { xueL( x) , eL( y) andx [ T} is not thepoint set of a subtree ofT, there are three pointsx1, x2, andz such thatx1, x2 [ X, z¸ X, andz is on thex1 2 x2

geodesic. Sincez¸ X, eL( z) $ eL( y) . eL( x1), eL( x2)$ uTu 2 1. Then, by Lemma 1,eL( z) 5 dL( z, S) for somesubtreeS of T with z ¸ S. Since z is on thex1 2 x2

geodesic, then eitherdT( x1, S) . dT( y, S) or dT( x2, S). dT( y, S); thus, eithereL( x1) $ dL( x1, S) . dL( z, S)5 eL( z) or eL( x2) $ dL( x2, S) . dL( z, S) 5 eL( z),which is absurd, and the theorem follows.

Corollary 3. The pointsv of least L-eccentricity constitutethe point set of a subtree in a tree T.

We consider next the relation of central and centroidpoints ofT to points of leastL-eccentricity. A pointu of atree is an endpoint (a peripheral point) if its degree is one.

Lemma 4. If v is a point of the center of a tree T, then eL(v)# uTu, and if eL(v) 5 uTu, then T is a path with an evennumber of points.

Proof. The center of a tree consist of either a single pointor a pair of adjacent points. LetSbe a subtree ofT such thatdL(v, S) $ uTu. So,v is not inS. If the center ofT consistsof two adjacent points, assume then thatv is the point of thecenter farthest fromS. Then, by Lemma 1,dL(v, S) 5 uSu1 uvu 1 2(dT(v, S) 2 1). Two subcases arise: (i) dT(v, S)5 1, and (ii ) dT(v, S) . 1:

(i) If dT(v, S) 5 1, thenuSu 5 dL(v, S) 2 uvu 5 dL(v, S)2 1 $ uTu 2 1, which is possible only ifv is anendpoint ofT. Sincev is in the center ofT, S has to bea single point, and, thus,T is a path of two points.

(ii ) If dT(v, S) . 1, there is a pointv9 adjacent tov suchthat v9 is on thev 2 S geodesic andv9 ¸ S. Let y bea point ofT such thate(v9) 5 dT(v9, y), and sincev isin the center ofT, dT(v9, y) $ e(v). Clearly,v is on thev9 2 y geodesic. Now, consider the subtreeS* induced

SUBTREE CENTER OF A TREE 273

by the points of they 2 S geodesic andS. This subtreecontainsuSu 1 dT(v, S) 1 dT(v9, y) 2 1 points. SincedT(v9, y) $ dT(v, S), we haveuS* u $ uSu 1 2dT(v, S)2 1 5 dL(v, S) $ uTu. Moreover, sincedT(v9, y)5 e(v9), the equationdT(v9, y) 5 dT(v, S) holds ifand only if the subtreeSconsists of a single point. Now,if dT(v9, y) . dT(v, S), then uS* u . uTu, which isabsurd. But, then, they 2 S geodesic containsuTupoints. Hence,T is a path. Sincev andv9 are adjacentand dT(v, S) 5 e(v) 5 dT(v9, y) 5 e(v9), also thepoint v9 belongs to the center ofT. Thus,T is a pathwith an even number of points. This completes theproof.

A direct calculation even shows that for a path with aneven number of pointseL(v) 5 uTu for center pointsv andeL(w) 5 uTu 1 2k for pointsw of distance (inT) k from anearest center point, for values ofk satisfying 0# k # (uTu2 2)/ 2. In all other trees, by Lemma 4,eL(v) , uTu for apoint v in the center ofT. On the other hand, by Theorem2, eL(w) $ uTu 2 1 for every pointw of T, and, thus,eL(v)5 uTu 2 1 5 min{ eL( x)ux [ T} for all treesT, which arenot paths with an even number of points. Hence, we canwrite the following theorem:

Theorem 5. If the pointv belongs to the center of a tree T,then eL(v) 5 min{eL(x)ux [ T}. Moreover, min{eL(x)ux[ T} 5 uTu if T is path with an even number of points and,otherwise, min{eL(x)ux [ T} 5 uTu 2 1.

As is well known, thecentroid of a tree consists of asingle point or two adjacent points. According to the defi-nition, the centroid consists of the points of minimumbranch weight. A branch at a pointv is a subtree in whichthe pointv is an endpoint. The weight of a branch is thenumber of lines in the branch, and the branch weight of apoint v is the maximum weight of the branches at the pointv. On the other hand, the number of lines in a branch at apoint v containing the linevv1 equals the number of pointsin the subtree ofT 2 v containing the pointv1. Thefollowing theorem shows a connection between the leastL-eccentricity points and the points belonging to the cen-troid of a tree.

Theorem 6. At least one of the points y in the centroid ofa tree T has the property eL(y) 5 min{eL(v)uv [ T}.

Proof. Consider, first, the case where the centroid con-sists of two adjacent pointsy1 andy2. Remove the liney1y2

from T. If the subtreesT1 and T2 (whereTi contains thepointyi, i 5 1, 2) thus obtained contain an unequal numberof points, then the branch weights of the pointsy1 andy2 donot equal, which is absurd. Hence, the number of pointsuTuin T is 2n anduTi u 5 n, i 5 1, 2. If eL( yi) . min{ eL(v)uv[ T} for i 5 1, 2, then, by Theorem 2, the subtree of all

leastL-eccentricity points inT is contained either inT1 orT2; we assume that inT1 (the proof is the same in the caseof T2).

Let v0 be a point of leastL-eccentricity nearest toy1 andlet S be a subtree witheL( y1) 5 dL( y1, S). SinceeL( y1). eL(v0), theneL( y1) . uTu 2 1, and, thus,y1¸ S. If Sis in T1, then sincey1 is a point ofT1, we havedT( y1, S)# n 2 uSu anddL( y1, S) 5 1 1 uSu 1 2(dT( y1, S) 2 1)# 1 1 uSu 1 2(n 2 uSu 2 1) 5 2n 2 uSu 2 1 , uTu 2 1.So,S is in T2. Then,dL( y1, S) 5 1 1 uSu 1 2(dT( y1, S)2 1) # uSu 1 1 1 2(n 2 uSu 2 1) 5 2n 2 uSu 1 1,wheredT( y1, S) 5 n 2 uSu only if all points ofT2 2 S areon the path fromy1 to S. SinceeL( y) . uTu 2 1 5 2n 2 1,we haveuSu 5 1, and, thus,T2 must be a path startingfrom y2.

Similarly, the only case where the demandeL( y2) . uTu2 1 holds is that whereT1 is a path beginning from thepoint y1. But, then,T is a path of 2n points and the pointsy1 andy2 are the only two points of the leastL-eccentricityin T. This is absurd, and, hence, at least one of the twopoints of the centroid must satisfy the assertion of thetheorem.

Assume now that the centroid ofT consists of a singlepointy. By Theorem 2, min{eL(v)uv [ T} $ uTu 2 1, and,thus,eL( y) $ uTu 2 1. If uTu 5 2n 1 1, then, clearly, thebranch weightbw( y) of y is at mostn, and if uTu 5 2n, thenbw( y) # n 2 1. Consider the caseuTu 5 2n 1 1. If y [ S,thendL( y, S) 5 uSu 2 1 # uTu 2 1. If y¸ S, thenS iscontained in a branch aty, dT( y, S) # n 2 uSu 1 1 anddL( y, S) 5 1 1 uSu 1 2(dT( y, S) 2 1) # 1 1 uSu 1 2(n2 uSu 1 1 2 1) 5 2n 2 uSu 1 1 # 2n 5 uTu 2 1.Accordingly, uTu 2 1 $ eL( y), and, thus,y satisfies theassertion of the theorem. The proof is similar in the caseuTu5 2n and, hence, omitted. This completes the proof.

Consider a star of five pointsv1, v2, v3, v4, v5 with v4 asa middle point, join to this star a pathv5v6v7v8, and denotethe tree thus obtained byT. The treeT shows that one of thetwo points of a centroid need not be a leastL-eccentricitypoint of a tree: InT, the centroid is {v4, v5} but 8 5 eL(v4). eL(v5) 5 7 5 min{ eL(v)uv [ T}.

3. LEAST CENTRAL SUBTREES OF A TREE

In this section, we describe the properties and structure ofleast central subtrees of a tree.

The following theorem shows a connection between leastcentral subtrees and endpoints of a tree:

Theorem 7. If CL is a least central subtree of a tree T withat least three points, the subtree CL does not contain anyendpoint of T.

274 NIEMINEN AND PELTOLA

Proof. Let S be a subtree ofT, consider the numberdL(CL, S), and assume thatCL contains an endpointu of T.Let C*L 5 CL 2 u. We shall show thatC*L is a centralsubtree ofT, too, and, thus,CL cannot be a least centralsubtree ofT. It suffices to prove thatdL(C*L, S) # eL(CL).There are two main cases: (i) CL ù S 5 A and (ii ) CL ùS Þ A. The case (ii ) contains three subcases: (ii 1) u¸ S,(ii 2) u [ CL ù S with uSu $ 2, and (ii 3) u [ CL ù S withuSu 5 1.

(i) The relationCL ù S 5 A implies thatdL(CL, S) 5 uSu1 uCLu 1 2(dT(CL, S) 2 1). Now, dL(C*L, S) 5 uSu1 uC*Lu 1 2(dT(C*L, S) 2 1) 5 uSu 1 uCLu 2 11 2(dT(CL, S) 2 1) 5 dL(CL, S) 2 1, since anendpoint ofT cannot be on a shortestC*L 2 S path.

(ii1)SinceCL ù S Þ A, we havedL(CL, S) 5 uCL ~ Su2 uCL ` Su. The removal of the pointu from CL

implies thatuC*L ~ Su 5 uCL ~ Su 2 1 (sinceu¸ S)anduCL ` Su 5 uC*L ` Su. Thus,dL(C*L, S) 5 dL(CL,S) 2 1.

(ii2) If CL ù S Þ A, uSu $ 2 andu [ CL ù S, considerthe subtreeS* 5 S 2 u. Then,uCL ~ Su 5 uCL ~ S* uand uCL ` S*u 5 uCL ` Su 2 1; hence,dL(CL, S*)5 dL(CL, S) 1 1 anddL(C*L, S) 5 uC*L ~ Su 2 uC*L` Su 5 uCL ~ Su 2 (uCL ` Su 2 1) 5 dL(CL, S) 1 15 dL(CL, S*) # eL(CL).

(ii3)By a direct calculation, we see thateL(T) 5 uTu 2 1.On the other hand, by Theorem 5, min{eL(v)uv [ T}5 uTu 2 1, except whenT is a path with an evennumber of points, and, thus, usually,T cannot be a leastcentral subtree. IfT is a path with an even number ofpoints, we see by a direct calculation that the leastcentral subtreeCL consists of the center ofT with CL

Þ T except whenT consists of two points; the trees oftwo points are excluded in the theorem. Thus, in thecase of this theorem,T cannot be a least central subtree.If uSu 5 1, u [ CL ù S, there is an endpointx of Twith x¸ CL, sinceCL Þ T, anddL(CL, x) . dL(CL,S) 5 dL(C*L, S) 2 1. This implies thatdL(C*L, S)5 dL(CL, S) 1 1 # eL(CL).

The following theorem shows the connection betweenthe least central subtrees of a tree:

Theorem 8. Any two least central subtrees of a tree T havea nonempty intersection.

Proof. Let C1 andC2 be two least central subtrees ofTsuch thatC1 ù C2 5 A, and letu be a nearest point of theC1 2 C2 geodesic toC1 such thatu¸ C1 and letS be thesubtree determining the eccentricity ofC1 ø { u}. SinceC1

is a least central subtree,dL(C1 ø { u}, S) $ eL(C1). If S5 T, thendL(C1 ø { u}, T) 5 uTu 2 uC1u 2 1 , eL(CL),which is absurd, and, hence,S Þ T.

Assume thatS ù C1 Þ A. If u [ S, thenu(C1 ø { u})

ø Su 5 uC1 ø Su, u(C1 ø { u}) ù Su 5 uC1 ù Su 1 1 and,thus,dL((C1 ø { u}), S) 5 uC1 ø Su 2 (uC1 ù Su 1 1)5 dL(C1, S) 2 1 , eL(C1). Thus,u¸ S. SinceS ù C1

Þ A andu¸ S, we haveS ù C2 5 A. In the following,note thatuC1u 5 uC2u anddL(C2, S) # eL(C1), becauseC1

andC2 both are least central subtrees andu(C1 ø { u}) øSu 5 uC1 ø { u} u 1 uSu 2 u(C1 ø { u}) ù Su by theinclusion–exclusion principle. The relationS ù C2 5 A

implies thatdL(C2, S) 5 uC2u 1 uSu 1 2(dT(C2, S) 2 1)$ uC1u 1 uSu $ uC1u 1 uSu 1 2 2 2uC1 ù Su $ 1 1 uC1

ø { u} u 1 uSu 2 2u(C1 ø { u}) ù Su 5 1 1 u(C1 ø { u})ø Su 2 u(C1 ø { u}) ù Su 5 1 1 dL(C1 ø { u}, S). eL(C1), and, hence, the assumptionS ù C1 Þ A isabsurd.

Assume now thatS ù C1 5 A. If u [ S, thendL(C1 ø{ u}, S) 5 u(C1 ø { u}) ø Su 2 u(C1 ø { u}) ù Su 5 uC1u1 uSu 2 1 , uC1u 1 uSu # dL(C1, S), which is absurd. So,u¸ S. If u would now be on theC1 2 S geodesic; then,dT(C1 ø { u}, S) 5 dT(C1, S) 2 1 anddL(C1 ø { u}, S)5 uC1u 1 uSu 1 2(dT(C1, S) 2 1 2 1) 5 dL(C1, S) 2 2, eL(C1), which is absurd, and, hence,u is not on theC1

2 S geodesic. Now, sinceC1 ù C2 5 A andu is on theC1 2 C2 geodesic, we havedT(C2, S) $ dT(C1, S) 1 1anddT(C1, S) 5 dT(C1 ø { u}, S). Accordingly,dL(C2, S)5 uC2u 1 uSu 1 2(dT(C2, S) 2 1) $ uC1u 1 uSu 1 21 2(dT(C1, S) 2 1) 5 uC1 ø { u} u 1 uSu 1 2(dT(C1 ù{ u}, S) 2 1) 1 1 5 dL(C1 ø { u}, S) 1 1 . eL(C1), acontradiction. Thus, there cannot be two nonintersectingleast central subtrees in a tree.

Theorem 9 determines the structure of the trees, the leastcentral subtree of which is a single point. Its result strength-ens the widely accepted opinion that the most centrallymanaged organizations of tree shape are a star and a pathwith an odd number of points.

Theorem 9. The least central subtree of a tree T is a singlepoint if and only if T is either a path with an odd number ofpoints or a star.

Proof. By a direct calculation, one sees that the leastcentral subtree of a path with an odd number of points or astar is a single point. Conversely, assume that the leastcentral subtree ofT is a single pointy. Suppose thatT is notan odd path. Then, since the least central subtree of a pathwith an even number of points is the centroid of the pathconsisting of two points, we may assume thatT is not apath. Since the least central subtree here consists of a singlepoint y, it is unique by Theorem 8, and, thus,y is the onlypoint having the minimumL-eccentricity inT. By Theorem5, we may assume thateL( y) 5 uTu 2 1. Consider now thepoints v1, v2, . . . , vs adjacent toy and branchesY1,Y2, . . . , Ys at y such thatvi [ Yi for i 5 1, . . . , s. ByTheorem 7, we haves $ 2. Sincey is the only point havingthe leastL-eccentricity, we haveeL(vi) . eL( y) for i5 1, . . . , s. If eachv i is an endpoint, thendL(v i, T 2 v i)


5 uTu andeL(v i) . eL( y) andT is a star. If somevi is notan endpoint, the relationeL(vi) . eL( y) 5 uTu 2 1 ispossible only if there exists a subtreeSi such thatdL(vi, Si)$ uTu. Clearly,v i ¸ Si. If y is not on thev i 2 Si geodesic,thendL( y, Si) . dL(v i, Si), which is absurd. Ify [ Si, thensinceSi ù Yi # { y}, we haveuSi u # uTu 2 uYi u 1 1, andsincev i ¸ Si, we havedL(v i, Si) 5 1 1 uSi u 1 2(dT(v i,Si) 2 1) 5 1 1 uSi u # uTu 2 uYi u # uTu 2 1, which isabsurd. Thus,y¸ Si. Sincey¸ Si, there exists a branchYk (k Þ i ) containing Si. If Si consists of an endpointadjacent toy, thendL(v i, Si) 5 1 1 1 1 2(2 2 1) 5 3# uTu 2 1, which is absurd. Thus, eitheruSi u $ 2 or dT(vi,Si) $ 3, which imply uYku . 2. Sincey, v i ¸ Si, v i is notan endpoint ofT, andT is not a path with four points, wehaveuTu . uSi u 1 3. Now,dT(vi, Si) . 2, since, otherwise,dL(vi, Si) # 1 1 uSi u 1 2(2 2 1) 5 uSi u 1 3 , uTu and,consequently,dL(v i, Si) # uTu 2 1, which is absurd. Thus,there must exist a pointvj Þ v i adjacent toy such thatvj isnot an endpoint ofT. As above, we see that there is asubtreeSj such thatdL(v j, Sj) 5 eL(v j) $ eL( y) 1 15 uTu, y is on thevj 2 Sj geodesic, anddT(vj, Sj) . 2. Ifa branchYt does not contain the subtreeSi, thendL(vt, Si)5 dL(vi, Si), and, similarly, if Sj ÷ Yt, then dL(vt, Sj)5 dL(v j, Sj). Thus, we may deduce that there exist twobranches, sayY1, Y2, and two subtrees, sayS1 # Y2, S1

# Y2, such thatdL(v1, S1) $ uTu anddL(v2, S2) $ uTu, andy is on thevm 2 Sm geodesic withy¸ Sm for m 5 1, 2.Since dT(v1, Y2) 5 dT(v2, Y1) 5 1, we have uSmu1 dT(vm, Sm) # uYl u 1 1 whenm Þ l andm, l 5 1, 2.Then, 2uTu # dL(v1, S1) 1 dL(v2, S2) 5 1 1 uS1u1 2(dT(v1, S1) 2 1) 1 1 1 uS2u 1 2(dT(v2, S2) 2 1)# uY1u 1 dT(v1, S1) 1 uY2u 1 dT(v2, S2). Since uY1u1 uY2u # uTu 1 1, v1, y, v2 are on theS1 2 S2 geodesicanddT(v1, v2) 5 2, the inequality above implies that 2uTu# uTu 1 1 1 dT(S1, S2) 2 dT(v1, v2) 5 uTu 1 dT(S1, S2)2 1, and, thus,dT(S1, S2) $ uTu 1 1. But this implies thatT is a path, which is a contradiction. This completes theproof.

The next theorem shows a connection between the leastcentral subtrees and centroids of a tree.

Theorem 10. There is at least one least central subtree ofa tree T containing at least one point of the centroid of T.

Proof. If uTu # 3, then the assertion holds obviously.According to Theorem 6, one of the points of the centroid,say y, has the propertyeL( y) 5 min{ eL(v)uv [ T}. IfeL( y) 5 uTu, then, by Theorem 5,T is a path with an evennumber of points and the centroid, the least central subtreeas well as the center ofT coincide. Thus, ify does notbelong to a least central subtreeCL of T, theneL( y) 5 uTu2 1 andCL must contain at least two points: If not, thenthere are two nonintersecting least central subtreesCL and

y, which is absurd by Theorem 8. Assume that from allpoints of the centroidy is the nearest toCL.

Let y, xm, xm21, . . . , x2, x1, c1 be they 2 CL geodesicand remove the lineyxm from T. We obtain two subtrees:One of them isY containing the pointy and anotherYc

containingCL. Consider the eccentricityeL(CL). Note that,sincey belongs to the centroid ofT, Yc can contain at mostn points if uTu is 2n or 2n 1 1. Clearly,eL(CL) $ dL(CL,T) 5 uTu 2 uCLu. Let eL(CL) 5 dL(CL, S), and assume,first, thatYc contains all the points ofS. Then, if S ù CL

5 A, we havedL(CL, S) 5 uSu 1 uCLu 1 2(dT(CL, S)2 1) # uSu 1 uCLu 1 2(uYcu 2 (uSu 1 uCLu) 1 1 2 1)5 2uYcu 2 uCLu 2 uSu # 2n 2 uCLu 2 uSu # uTu 2 uCLu2 uSu , uTu 2 uCLu, a contradiction. On the other hand, ifCL ù S Þ A (andYc contains the points ofS), thendL(CL,S) 5 uCL ~ Su 2 uCL ` Su 5 uCLu 1 uSu 2 2uCL ù Su# uYcu # n , 2n 2 uCLu # uTu 2 uCLu, a contradiction,too. Hence, the subtreeS either contains the pointy or thepoint y is on theCL 2 S geodesic.

The subtreeCL contains at least two endpoints, one ofwhich, saya1, is not on the pathy, xm, xm21, . . . , x1.Thus, we obtain a new subtreeCL

1 by joining the pointx1 tothe subtreeCL and removing the pointa1, and, moreover,uCLu 5 uCL

1u. Now, by repeating the proof of the precedingparagraph, whereCL is replaced byCL

1, we see that eithereL(CL

1) 5 uTu 2 uCL1u or the subtreeS1 determining the

eccentricityeL(CL1) either contains the pointy or the CL

1

2 S1 geodesic contains the pointy. In any case,eL(CL1)

# eL(CL). The subtreeCL1 contains an endpoint, saya2,

which is not on the pathy, xm, . . . , x2, and, thus, we obtaina new subtreeCL

2 by joining the pointx2 to CL1 and remov-

ing the pointa2. As above, we see thateL(CL2) # eL(CL

1).By repeating this process, we obtain a sequence of subtreesCL, CL

1, CL2, . . . , CL

m21, CLm, CL

y, whereCLy is obtained

from CLm by joining the pointy and removing an endpoint

am11 Þ xm, and whereeL(CLy) # eL(CL

m) # . . .# eL(CL).This completes the proof.

The following theorem shows a connection between theleast central subtrees and the center of a tree:

Theorem 11. If C is a center of a tree T, then Cù CL Þ A

for at least one least central subtree CL of T.

Proof. Assume thatC ù CL 5 A for every least centralsubtreeCL of T. Then, we can choose the subtreeCL suchthat the distancedT(CL, C) is the least possible. By Theo-rem 9, we may assume thatuCLu . 1, or, otherwise, thetheorem follows immediately. Denote byv the point nearestto CL on theCL 2 C geodesic (v ¸ CL) and bySv thebranch atv containingC. Letn be the minimum eccentricityof the treeT [i.e., the radiusr (T) 5 n]. SinceC # Sv, thesubtreeSv contains at least one path of lengthn, and, thus,uSvu $ n 1 1.


The subsequent proof is based on the results of theClaims (1)–(6) given and proved below.

Consider, first, the case whereCL is a path andu1 is theonly endpoint ofCL not adjacent tov. Let u1 be the pointand S1 the subtree of Claim (3). Then,eL(CL) # uCLu1 uS1u 1 2(dT(CL, S1) 2 1) 1 1 and the number of pointsof T satisfies the following equality:uTu $ uSvu 1 uCLu1 dT(CL, S1) 2 1 1 uS1u. SinceCL is a path,r (T) 5 n,and theC 2 S1 geodesic goes throughCL, we haveuCLu1 dT(CL, S1) 2 1 , n. Then, uTu 2 uCLu $ uSvu 1 uS1u1 dT(CL, S1) 2 1 $ n 1 1 1 uS1u 1 dT(CL, S1) 2 1. uCLu 1 dT(CL, S1) 2 1 1 uS1u 1 dT(CL, S1) 5 dL(CL,S1) 1 1 $ eL(CL), and, hence,uTu 2 uCLu . eL(CL). NotethateL(CL) $ dL(CL, T) 5 uTu 2 uCLu, and, thus, the resultuTu 2 uCLu . eL(CL) implies a contradiction.

Assume now thatCL containsm $ 2 endpointsu1,u2, . . . , um not adjacent tov. By Claim (3), there existmsubtreesS1, S2, . . . , Sm such thatSi ù CL 5 A, dL(CL,Si) $ eL(CL) 2 1, andui is on theCL 2 Si geodesic forall i 5 1, 2, . . . ,m. Clearly,Si ù Sj 5 A for all i Þ j ,i , j 5 1, 2, . . . ,m. The subscripts can always be chosenso thatdT(S2, CL) # dT(S1, CL). It suffices now to provethat the contradictionuTu 2 uCLu 2 eL(CL) . 0 follows.This is done by using Claims (4), (5), and (6), the proofs ofwhich are given below: (4)uS1u # 1 1 dT(u1, u2), (5) uCLu# mn 2 ¥i51

m dT(CL, Si), and (6) uTu $ uCLu 1 n(2m1 1) 1 m 1 1 2 ¥ i51

m dT(CL, Si).By Claim (3),eL(CL) # dL(CL, S1) 1 1 5 uCLu 1 uS1u

1 2(dT(CL, S1) 2 1) 1 1 5 uCLu 1 uS1u 1 2dT(CL, S1)2 1. This and Claims (4)–(6) imply now thatuTu 2 uCLu2 eL(CL) $ uTu 2 uCLu 2 uCLu 2 uS1u 2 2dT(CL, S1) 1 1$ uCLu 1 n(2m 1 1) 1 m 1 1 2 ¥i51

m dT(CL, Si)2 2uCLu 2 uS1u 2 2(dT(CL, S1) 1 1) $ n(2m 1 1) 1 m1 1 2 ¥i51

m dT(CL, Si) 2 nm 1 ¥i51m dT(CL, Si) 2 uS1u

2 2dT(CL, S1) 1 1 $ nm 1 n 1 m 1 1 2 1 2 dT(u1,u2) 2 2dT(CL, S1) 1 1 5 n(m 1 1) 1 m 1 1 2 dT(u1,u2) 2 2dT(CL, S1). Since dT(u1, u2) # dT(u1, v)1 dT(u2, v), dT(CL, S1) 5 dT(u1, S1) # n anddt(v, u1)1 dT(u1, S1) 5 dT(v, S1) # n, we further obtainuTu2 uCLu 2 eL(CL) $ n(m 1 1) 1 m 1 1 2 dT(u1, v)2 dT(u2, v) 2 dT(u1, S1) 2 dT(u1, S1) $ n(m 1 1) 1 m1 1 2 n 2 n 2 n 5 n(m 1 1 2 3) 1 m 1 1. Sincem $ 2, we haveuTu 2 uCLu 2 eL(CL) . 0, which isabsurd, and the assertion of the theorem follows.

Claim (1). There is a subtreeS Þ T such thatdL(CL, S)$ eL(CL) 2 1.

Proof of Claim (1).Assume on the contrary thatdL(CL,S) 1 1 , eL(CL) for every subtreeS Þ T. Then,dL(CL

ø { v}, S) # dL(CL, S) 1 1 , eL(CL) for every subtreeS Þ T. Thus,eL(CL ø { v}) 5 dL(CL ø { v}, T) 5 uTu2 (uCLu 1 1) , uTu 2 uCLu # eL(CL), which is acontradiction, sinceCL is a least central subtree ofT.

Claim (2). If dL(CL, S) $ eL(CL) 2 1 for a subtreeSÞ T,thenS ù CL 5 A or v [ S.

Proof of Claim (2).Assume on the contrary that there isa subtreeS Þ T such thateL(CL) # dL(CL, S) 1 1, Sù CL Þ A andv¸ S. SinceS ù CL Þ A andv¸ S, thepoint v is on theC 2 S geodesic. Then, sincer (T) 5 n, wehavedT(v, S) # n. The subtreeSv is the branch (at thepoint v) containing the centerC. SinceSv ù CL 5 A, wehavedL(CL, ^S ø Sv&) 5 u^S ø Sv& ø CLu 2 u^S ø Sv&ù CLu $ uSvu 1 uS ø CLu 2 (uS ù CLu 1 (dT(v, S) 2 1))$ n 1 1 1 dL(CL, S) 2 dT(v, S) 1 1 . dL(CL, S) 1 1$ eL(CL), a contradiction. Thus,dL(CL, S) # eL(CL) 2 2for every subtreeS Þ T with S ù CL Þ A and v ¸ S.

Claim (3). If u is an endpoint ofCL not adjacent tov, thenthere exists a subtreeSu such thatdL(CL, Su) 1 1 $ eL(CL),CL ù Su 5 A andu is on theCL 2 Su geodesic.

Proof of Claim (3).Let C*L 5 (CL 2 u) ø { v}). SincedT(C*L, C) , dT(CL, C), then C*L is not a least centralsubtree, and thus there exists a subtreeS Þ T, such thateL(C*L) 5 dL(C*L, S) . eL(CL).

Assume, first, thatS ù CL Þ A. If C*L ù S Þ A andv [ S, then uC*L ø Su # uCL ø Su and uC*L ù Su $ uCL

ù Su; hence,eL(C*L) 5 dL(C*L, S) 5 uC*L ø Su 2 uC*Lù Su # uCL ø Su 2 uCL ù Su # eL(CL), which is absurd.If C*L ù S Þ A and v ¸ S, then uC*L ø Su # uCL ø Su1 1, uC*L ù Su $ uCL ù Su 2 1 anddL(C*L, S) 5 uC*Lø Su 2 uC*L ù Su # uCL ø Su 1 1 2 (uCL ù Su 2 1)5 dL(CL, S) 1 2. SinceS ù CL Þ A andv¸ S, then bythe proof of Claim (2), we havedL(CL, S) # eL(CL) 2 2,and, thus,dL(C*L, S) # eL(CL), which is absurd. IfC*L ù S5 A, thenCL ù S 5 { u}, and sinceu is not adjacent tov, then v ¸ S. Moreover,dT(C*L, S) 5 1, uC*Lu 1 uSu5 uCL ø Su 1 1 5 uCL ø Su 2 uCL ù Su 1 2. Then,dL(C*L, S) 5 uC*Lu 1 uSu 1 2(dT(C*L, S) 2 1) 5 uCL ø Su1 1 5 uCL ø Su 2 uCL ù Su 1 2 5 dL(CL, S) 1 2. SinceCL ù S Þ A andv ¸ S, then by the proof of Claim (2),dL(CL, S) # eL(CL) 2 2, which implies a contradiction.

Assume now thatS ù CL 5 A. If v [ S, theneL(CL)$ dL(CL, S) 5 uCLu 1 uSu . uCLu 1 uSu 2 1 2 1 5 uC*Lø Su 2 uC*L ù Su 5 dL(C*L, S) $ eL(C*L), which isabsurd. Ifu is not on theC*L 2 S geodesic, thendT(C*L, S)# dT(CL, S), and sinceuC*Lu 5 uCLu, we haveeL(CL), dL(C*L, S) 5 uC*Lu 1 uSu 1 2(dT(C*L, S) 2 1) # uCLu1 uSu 1 2(dT(CL, S) 2 1) # eL(CL), which is absurd.Thus, u must be on theC*L 2 S geodesic, and, conse-quently, on theCL 2 S geodesic, too. Then,dT(CL, S)5 dT(C*L, S) 2 1 and dL(CL, S) 1 1 5 uCLu 1 uSu1 2(dT(CL, S) 2 1) 1 1 5 uC*Lu 1 uSu 1 2(dT(C*L, S)2 1) 2 1 5 eL(C*L) 2 1 $ eL(CL), which proves theassertion of Claim (3).


Claim (4). uS1u # 1 1 dT(u1, u2).

Proof of Claim (4).Let S 5 ^S1 ø S2& be the leastsubtree containingS1 and S2. Then, the subtreeS ù CL

equals theu1 2 u2 geodesic, and, thus,uS ù CLu 5 dT(u1,u2) 1 1. SincedT(CL, Si) 5 dT(ui, Si) for i 5 1, 2, wehave uS ø CLu 5 uS1u 1 (dT(CL, S1) 2 1) 1 uCLu1 (dT(CL, S2) 2 1) 1 uS2u, and sinceS ù CL Þ A andv¸ S, we have by Claim (2)eL(CL) 2 1 $ dL(CL, S). Bythe facts dL(CL, S2) $ eL(CL) 2 1 and dT(CL, S2)$ dT(CL, S1), we now haveeL(CL) 2 1 $ dL(CL, S) 5 uSø CLu 2 uS ù CLu $ uS1u 1 (uCLu 1 uS2u 1 2(dT(CL, S2)2 1)) 2 dT(u1, u2) 2 1 $ uS1u 1 eL(CL) 2 1 2 dT(u1,u2) 2 1 and Claim (4) follows.

Claim (5). uCLu # mn 2 ¥i51m dT(CL, Si).

Proof of Claim (5).Since no point ofCL is an endpointof T, Si Þ A for every i , and thus any point ofCL is onsomev 2 Si geodesic. Sinceu1, u2 . . . um are all end-points ofCL not adjacent tov and sinceui is on the geodesicv 2 Si for eachi 5 1, 2, . . . ,m, we haveuCLu # ¥ i51

m

(dT(v, Si) 2 dT(ui, Si)), wheredT(ui, Si) 5 dT(CL, Si) foreachi . Sincer (T) 5 n andv is on theCL 2 C geodesic,dT(v, Si) # n for all i 5 1, 2, . . . , m and Claim (5)follows.

Claim (6). uTu $ uCLu 1 n(2m1 1) 1 m1 1 2 ¥i51m dT(CL,

Si).

Proof of Claim (6).Let v9 be a point farthest fromv.According to the choice of the pointv, there is at least onepoint v9 such that no point ofCL is on thev 2 v9 geodesic.Since r (T) 5 n, we havedT(v, v9) $ n and dT(CL, v9)

$ n 1 1. Then,dL(CL, v9) 5 uCLu 1 1 1 2(dT(CL, v9)2 1) $ uCLu 1 2n 1 1. SincedL(CL, Si) 1 1 5 uCLu1 uSi u 1 2(dT(CL, Si) 2 1) 1 1 $ eL(CL) # dL(CL, v9)for all i 5 1, 2, . . . ,m, we haveuCLu 1 uSi u 1 2(dT(CL,Si) 2 1) 1 1 $ uCLu 1 2n 1 1, and, thus,uSi u $ 2(n 1 12 dT(CL, Si)) for all i 5 1, 2, . . . ,m. Recall thatSv ù CL

5 CL ù Si 5 Sv ù Si 5 A for all i 5 1, 2, . . . ,m andSi ù Sj 5 A for all i , j 5 1, 2, . . . ,m, i Þ j . Since theCL 2 Si geodesic containsdT(CL, Si) 2 1 points containedneither inSi nor in CL, we haveuTu $ uSvu 1 uCLu 1 ¥i51

m

(dT(CL, Si) 2 1) 1 ¥i51m uSi u $ n 1 1 1 uCLu 1 ¥i51

m

(dT(CL, Si) 2 1) 1 ¥i51m (2(n 1 1 2 dT(CL, Si)) and

Claim (6) follows. This completes the proof of Theorem 11.By the observations, the points in every least central

subtree appear to belong to the set of leastL-eccentricitypoints; unfortunately, we have not been able to prove this.This and the extension of the results to the case of allmedian graphs is the goal of further research.

The authors like to thank the anonymous referee for her/hisnumerous valuable comments for improving this paper.

REFERENCES

[1] H.-J. Bandelt and J. Hedlikova, Median algebras, DiscrMath 45 (1983), 1–30.

[2] F. Buckley and F. Harary, Distance in graphs., Addison-Wesley, Reading MA, 1990.

[3] J. Nieminen, The structure of the centroid in a Ptolemaicgraph, Appl Math Lett 8 (1995), 35–38.

[4] J. Nieminen, The ideal structure of simple ternary algebras,Colloq Math 40 (1978), 23–29.


the subtree center of a tree

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