Average eccentricity, k -packing and k -domination in graphs

Discrete Mathematics 342 (2019) 1261–1274

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Average eccentricity, k-packing and k-domination in graphs P. Dankelmann

∗,1

, F.J. Osaye 2

University of Johannesburg, South Africa

article

info

a b s t r a c t

Article history: Received 26 September 2018 Received in revised form 29 December 2018 Accepted 3 January 2019 Available online xxxx Keywords: Eccentricity Average eccentricity Total eccentricity Packing number Independence number Domination number

Let G be a connected graph. The eccentricity e(v ) of a vertex v is the distance from v to a vertex farthest from v . The average eccentricity avec(G) of G is defined as the average ∑ of the eccentricities of the vertices of G, i.e., as |V1 | v∈V e(v ), where V is the vertex set of G. For k ∈ N, the k-packing number of G is the largest cardinality of a set of vertices of G whose pairwise distance is greater than k. A k-dominating set of G is a set S of vertices such that every vertex of G is within distance k from some vertex of S. The kdomination number (connected k-domination number) of G is the minimum cardinality of a k-dominating set (of a k-dominating set that induces a connected subgraph) of G. For k = 1, the k-packing number, the k-domination number and the connected k-domination number are the independence number, the domination number and the connected domination number, respectively. In this paper we present upper bounds on the average eccentricity of graphs in terms of order and either k-packing number, k-domination number or connected k-domination number. © 2019 Elsevier B.V. All rights reserved.

1. Introduction Let G be a connected graph with vertex set V (G) and edge set E(G). The distance dG (u, v ) between two vertices u and v is the length of a shortest (u, v )-path in G. The eccentricity eG (u) of a vertex u is the distance to a vertex farthest from u, i.e., eG (u) = maxv∈V (G) ∑dG (u, v ). The average eccentricity avec(G) of G is the average of the eccentricities of the vertices of G, 1 i.e., avec(G) = |V (G) u∈V (G) eG (u). The average eccentricity was introduced under the name eccentric mean by Buckley and | Harary [2], but it attracted major attention only after its first systematic study in [7], where, among several other results, a sharp upper bound on avec(G) in terms of order was given. Proposition 1 ([7]). Let G be a connected graph of order n. Then avec(G) ≤

1 3n2

⌊

n

− ⌋.

n 4 2 This bound is sharp and equality holds if and only if G is a path.

(1)

Several bounds on the average eccentricity in terms of other graph parameters or with given properties can be found in the literature. For example for graphs of given order and size [1] and [18], and for given order and minimum degree [7] ∗ Corresponding author. E-mail addresses: [email protected] (P. Dankelmann), [email protected], [email protected] (F.J. Osaye). 1 Financial Support by the South African National Research Foundation, Grant Number 103553, is gratefully acknowledged. 2 The results presented in this paper form part of the second author’s Ph.D. thesis. Financial support by the University of Johannesburg is gratefully acknowledged. https://doi.org/10.1016/j.disc.2019.01.004 0012-365X/© 2019 Elsevier B.V. All rights reserved.

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and [9]. The maximum average eccentricity of trees with given order and maximum degree was determined in [14], while the trees with given degree sequence that minimise the average eccentricity were determined in [17]. The maximum average eccentricity of maximal planar graphs was determined in [1]. In [3], Proposition 1 was generalised by showing that among all graphs of order n, the path maximises the average eccentricity of the strong product with any other connected graph. More specifically, it was shown in [3] that for every fixed connected graph H and every n ∈ N we have avec(G ⊠ H) ≤ avec(Pn ⊠ H) for every connected graph G of order n, where G ⊠ H denotes the strong product of G and H. The starting point for our investigations are sharp upper bounds on the average eccentricity of graphs in terms of independence number, domination number and connected domination number in [8]. A set S of vertices of G is an independent set if no two vertices in S are adjacent, and the independence number of G is the maximum cardinality of an independent set in G. A subset S ⊆ V (G) is a dominating set if every vertex not in S is adjacent to some vertex in S, and the minimum cardinality of a dominating set of G is the domination number, γ (G). The minimum cardinality of a dominating set that induces a connected graph is the connected domination number γc (G). Theorem 1 ([8]). Let G be a connected graph of order n and independence number β ≤ avec(G) ≤

1⌊ n

n . 2

Then

⌋ 2nβ − n − β 2 + β ,

and this bound is sharp. In the same paper [8], the authors also gave sharp upper bounds on the average eccentricity for graphs of given order and domination number, and of given order and connected domination number. We note that the relation between the average eccentricity and the domination number was also considered in [10,12], and that in [11] relationships between the average eccentricity and the independence number were found. The aim of this paper is to generalise the results on the maximum value of the average eccentricity of graphs with given order and domination number and independence number in [8]. A k-packing of G is a subset S ⊆ V (G) such that d(u, v ) > k for any u, v ∈ S, and the maximum cardinality of a k-packing of G is the k-packing number βk (G). A set S ⊆ V (G) is kdominating if every vertex not in S is within distance k from some vertex in S, and the minimum cardinality of a k-dominating set is the k-domination number γk (G). The minimum cardinality of a k-dominating set which induces a connected graph is the connected k-domination number γck (G). In this paper we give bounds on the average eccentricity of graphs of given order and k-packing number, k-domination number, or connected k-domination number. In addition, we construct graphs to show the sharpness of our bounds. Analogous results for the average distance of a graph, defined as the arithmetic mean of all distances were obtained in [6] and [19], where upper bounds on the average distance for graphs of given order and domination number (see [5]) and given order and independence number [4] were generalised to graphs of given order and k-packing and k-domination number. The approach used in this paper follows to some extent that in [6]. This paper is organised as follows. In Section 2 we present some tools which will be used in the proofs of our main results. Since by Proposition 2, the path Pn has maximum average eccentricity among all connected graphs of order n, it is unsurprising that bounds on the average eccentricity of a graph G of given k-packing number depend on whether βk (G) ≤ βk (Pn ) or βk (G) > βk (Pn ). In Section 3 we consider the former case, in Section 4 we consider the latter. We conclude the paper with Section 5, in which we make use of our results from the preceding sections to obtain bounds on the average eccentricity in terms of k-domination number and connected k-domination number. The notation we use is as follows. Let G be a connected graph. The diameter diam(G) of G is the largest of the distances between the vertices of G. If u is a vertex of G, then a vertex farthest from u, i.e., at distance eG (u) from u, is an eccentric vertex of u. The total eccentricity EX (G) of G is the sum of the eccentricities of all vertices of G. The smallest of the eccentricities of the vertices of G is the radius of G. The eccentric sequence of G is the nonincreasing sequence of the eccentricities of the vertices of G. If A and B are non-empty subsets of V , then d(A, B) is defined as mina∈A,b∈B d(a, b). If A = {u} then we write d(u, B) for dG ({u}, B). The neighbourhood of a vertex v , i.e., the set of vertices adjacent to v , is denoted by NG (v ). The closed neighbourhood NG [v] of v is defined as NG (v ) ∪ {v}. For k ∈ N we define NGk [v] to be the kth neighbourhood of v , i.e., the set of vertices at distance not more than k to v . The degree of a vertex v is defined as |NG (v )|. A vertex of degree one is an end vertex. The set of end vertices of G is denoted by Υ (G). If the graph G is clear from the context, then we often drop the subscript or argument G. For k ∈ N, the kth power Gk of G is the graph on the same vertex set as G, in which two vertices are adjacent if and only if their distance in G is not more than k. The path on n vertices is denoted by Pn . 2. Preliminary results The following results will be used for the proofs of our main results. Lemma 1 ([15]). Let T be a tree and let u, v be two vertices at distance diam(T). Then eG (x) = max{d(x, u), d(x, v )} for all x ∈ V (T ).

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Definition 1 ([7]). Let G be a connected graph and c : V (G) → R a nonnegative weight function on the vertices of G. Then the eccentricity of G with respect to c is defined by

∑

EXc (G) =

c(x)eG (x).

x∈V (G)

Let N =

∑

x∈V (G)

c(x) be the total weight of the vertices in G. If N > 0, then the average eccentricity of G with respect to c is

avecc (G) =

EXc (G)

.

N In our proofs we make use of the following result from [7], which generalises Proposition 1. Proposition 2 ([7]). Let G be a connected graph, c a weight function on the vertices of G, and N = of the vertices of G. If c(v ) ≥ 1 for all v ∈ V (G), then

∑

v∈V (G)

c(v ) the total weight

avecc (G) ≤ avec(P⌈N ⌉ ). For the proofs of our main results we also require a bound on the average eccentricity of weighted trees in terms of total weight and diameter. It generalises a bound on the average eccentricity of trees of given order and diameter given in [8]. The proof of our bound makes use of a characterisation of eccentric sequences of trees due to Lesniak [15]. Proposition 3 ([15]). A nonincreasing sequence s1 , s2 . . . sn of positive integers is the eccentric sequence of some tree if and only if (i) for every integer k with sn ≤ k < s1 there exists i ∈ {1, 2, . . . , n − 1} such that si = si+1 = k, and s −1 s (ii) either sn = 21 and sn ̸ = sn−1 , or sn = sn−1 = 12 and sn−2 ̸ = sn−1 . Proposition 4. Let T be a tree of diameter d. Let c : V (T ) → R be a weight function on the vertices of T , and let N = c(V ). If c(v ) ≥ 1 for all v ∈ V (T ), then avecc (T ) ≤ d −

1 ⌈ d2 − 1 ⌉ N

4

.

Proof. Let s1 , s2 , . . . , sn be the eccentric sequence ∑ of T . For i ∈ N define Vi to be the set of vertices of eccentricity i. Denote the radius of T by r, so r = sn . Using c(Vi ) for v∈V c(v ), we have i

d

EXc (T ) =

∑

c(Vi )i.

i=r

We consider two cases, depending on the parity of d. Case 1: d is odd. 1 Then r = d+ . It follows from Proposition 3 that |Vi | ≥ 2 for i = r , r + 1, . . . , d − 1. Since by our hypothesis c(v ) ≥ 1 for all 2 ∑ v ∈ V , it follows that c(Vi ) ≥ 2 for all i ∈ {r , r + 1, . . . , d − 1}. Subject to this condition, di=r c(Vi )i is maximised if c(Vi ) = 2 for all i ∈ {r , r + 1, . . . , d − 1}, and c(Vd ) = N − 2(d − r) = N − d + 1. Hence EXc (T ) ≤

d−1 ∑

2i + (N − d + 1)d =

d−1 ∑

(

2i + (N − d + 1)d = d N −

i=(d+1)/2

i=r

d) 4

1

+ , 4

and division by N yields the proposition. Case 2: d is even. Then r = 2d . It follows from Proposition 3 that |Vr | = 1 and |Vi | ≥ 2 for i = r + 1, r + 2, . . . , d − 1. As in Case 1, we conclude ∑d that c(Vr ) ≥ 1 and c(Vi ) ≥ 2 for all i ∈ {r + 1, r + 2, . . . , d − 1}. Subject to this condition, i=r c(Vi )i is maximised if c(Vr ) = 1, c(Vi ) = 2 for all i ∈ {r + 1, r + 2, . . . , d − 1}, and c(Vd ) = N − 2(d − r) + 1 = N − d + 1. Hence d−1 ∑

EXc (T ) ≤ r +

2i + (N − d + 1)d =

i=r +1

d 2

+

d−1 ∑

(

2i + (N − d + 1)d = d N −

i=d/2+1

d) 4

,

and division by N yields the proposition. □ For unweighted trees, that is, tress in which every vertex has weight 1, we obtain the following corollary. Corollary 1 ([8]). Let T be a tree of order n and diameter d. Then avec(T ) ≤ d −

1 ⌈ d2 − 1 ⌉ n

4

.

(2)

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3. Bounds on the average eccentricity for small βk Definition 2. A k-star is a tree of radius at most k. −1 Theorem 2. Let G be a connected graph of order n ≥ 3 and k-packing number βk (G) ≤ 1 + ⌊ nk+ ⌋. 1 (a)If k is even, then

avec(G) ≤ (k + 1)(βk + 1) − 2 −

1 ⌈ [(k + 1)βk − 1]2 − 1 ⌉ n

4

.

(b) If k is odd, then avec(G) ≤ (k + 1)(βk + 1) −

1 ⌈ (k + 1)2 βk2 − 1 ⌉ n

.

4

These bounds are sharp apart from additive constants. Proof. (a) Let k be even. We find a maximum k-packing A = {v1 , v2 , . . . , vr } of G, using the following procedure. Choose an arbitrary vertex v1 of G and let A = {v1 }. If there exists a vertex v2 in G with dG (v2 , A) = k + 1, add v2 to A. Repeat adding vertices vi with dG (vi , A) = k + 1 to A until each of the vertices not in A is within distance k from A. Since A is a k-packing, r ≤ βk . k

For each vertex vi ∈ A, let Ti be a spanning tree of NG2 [vi ] such that dTi (vi , x) = dG (vi , x) for all x ∈ V (Ti ). By our construction of A, ⋃r there exist r − 1 edges in G, each joining an end vertex of some Ti to an end vertex of some Tkj , i ̸= j, whose addition to i=1 Ti yields a (not necessarily spanning) subtree H ≤ G. Since H is obtained from the union of 2 -stars T1 , T2 , . . . , Tr , each of diameter at most k, by adding r − 1 edges, we have diam(H) ≤ (k + 1)r − 1 ≤ (k + 1)βk − 1.

(3)

We now define a weight function c on V (H). For each vertex v of G choose a nearest vertex v of H. For each vertex u of G define ′

c(u) = |{v ∈ V (G) | v ′ = u}|. Note that the weight of each vertex not in H is zero. Let c1 be the constant weight function on V (G) which assigns the weight 1 to every vertex of G. Thus avec(G) = avecc1 (G). Since c can be thought of as being obtained from c1 by moving the weight of every vertex v to v ′ over a distance not exceeding 2k , we have avec(G) ≤ avecc (G) +

k 2

.

(4)

Since the weight of each vertex not in H is zero, we have v∈V (G) c(v )eG (v ) = v∈V (H) c(v )eG (v ). Since every vertex of G is within distance at most k/2 from some vertex of H, we have eG (v ) ≤ eH (v ) + k/2 for all v ∈ V (H). Thus,

∑

avecc (G) ≤ avecc (H) +

k 2

∑

.

Hence, avec(G) ≤ avecc (H) + k.

(5)

Now each vertex of H has weight at least 1, and the total weight of the vertices in H is n. Let d := diam(H). Then d ≤ (k + 1)βk − 1 by (3). By Proposition 4 we have avecc (H) ≤ d −

1 ⌈ d2 − 1 ⌉ n

4

≤ (k + 1)βk − 1 −

1 ⌈ [(k + 1)βk − 1]2 − 1 ⌉ n

4

.

This inequality, in conjunction with (5), yields part (a) of the theorem. (b) For k odd, we also find a maximal k-packing A = {v1 , v2 , . . . , vr } of G as in the proof of (a) above. For each i with 2 ≤ i ≤ r choose a vertex vj , j < i, such that dG (vi , vj ) = k + 1 and select a vertex wij with dG (vi , wij ) = dG (wij , vj ) = k+2 1 . k+1

Let W be the set of all vertices of the form wij . For i = 1, 2, . . . , r, let Ti′ be a spanning tree of N 2 [vi ] that preserves the distances from vi . The trees T1 , T2 , . . . , Tr are obtained from T1′ , T2′ , . . . , Tr′ , respectively, as follows. All vertices that are in more than one of the trees T1 , T2 , . . . , Tr , are deleted from all but one of the trees Ti , with exception of the vertices of the form wij which remain in both, Ti and Tj . Let H ′ be the union of the trees T1 , T2 , . . . , Tr . Equivalently, let H ′ be the tree obtained from the disjoint union of T1 , T2 , . . . , Tr by identifying the two vertices corresponding to the vertex wij for every contact vertex wij . Since each Ti is a k+2 1 -star, which has diameter at most k + 1, we have diam(H ′ ) ≤ (k + 1)r ≤ (k + 1)βk .

(6)

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Similarly to the proof of part (a) of this theorem we define a weight function on the vertices of G. Let c(v ) be the number of vertices of G that have v as its nearest vertex in H ′ . As in the proof of part (a) we obtain avec(G) ≤ avecc (H ′ ) + k + 1.

(7)

Now each vertex of H ′ has weight at least 1 and the total weight of the vertices of H ′ is n. Hence avecc (H ′ ) ≤ (k + 1)βk −

1 ⌈ (k + 1)2 βk2 − 1 ⌉ n

.

4

This inequality in conjunction with (7) yields part (b) of the theorem. □ We show that for fixed k ≥ 2, the bounds in Theorem 2 are best possible apart from an additive constant. Example 1. For given k, βk , n ∈ N with k, βk ≥ 2 and n ≥ (k + 1)βk let Gk,βk ,n be the graph obtained from a path P(k+1)βk −1 by appending n + 1 − βk (k + 1) end vertices to one end of the path. It is easy to verify that avec(Gk,βk ,n ) = (k + 1)βk − 1 −

1 ⌈ [(k + 1)βk − 1]2 − 1 ⌉ n

4

.

Hence avec(Gk,βk ,n ) differs from the bound in Theorem 2 by not more than k + 1 + o(1) for large n. We will show below that the trees Gk,βk ,n have maximum average eccentricity among all trees of order n and k-packing number βk . Substituting k = 1 in Theorem 2 yields a bound on the average eccentricity in terms of order and independence number β which differs from the bound in Theorem 1 only by a small additive constant. The bound in Theorem 2 is asymptotically sharp for all graphs. For trees, however, we obtain a sharp bound as shown below in Theorem 3, provided that its k-packing number does not exceed that of the path of the same order. −1 ⌋, k > 2 and order n. Then Theorem 3. Let T be a tree with k-packing number βk ≤ 1 + ⌊ nk+ 1

avec(T ) ≤ (k + 1)βk − 1 −

1 ⌈ [(k + 1)βk − 1]2 − 1 ⌉ n

4

.

This bound is sharp. Proof. We first show that diam(T ) ≤ βk (k + 1) − 1. Suppose to the contrary that diam(T ) ≥ βk (k + 1). Let P = v0 , v1 , . . . , vd be a path in T of length d = diam(T ). Then v0 , vk+1 , v2(k+1) , . . . , v⌊ d ⌋ is a k-packing of P, and thus of T , with 1 + ⌊ k+d 1 ⌋ vertices. Thus, βk (T ) ≥ 1 + ⌊ k+d 1 ⌋. Since k+1

d ≥ βk (k + 1), βk (T ) ≥ 1 + ⌊ Now by Corollary 1, avec(T ) ≤ d −

βk (k+1) ⌋ k+1

1 ⌈ d2 − 1 ⌉ n

4

= 1 + βk , a contradiction. Hence, d ≤ βk (k + 1) − 1.

≤ (k + 1)βk − 1 −

1 ⌈ [(k + 1)βk − 1]2 − 1 ⌉ n

4

.

The trees Gk,βk ,n in Example 1 attain equality, hence the bound is sharp. □ 4. Bounds on the average eccentricity for large βk −1 We now consider graphs with k-packing number βk > βk (Pn ) = 1 + ⌊ nk+ ⌋. 1 The following construction will be used in this section. Let T be a tree with weight function c, u a vertex of T , and let ∅ ̸ = U ⊂ N(u) be a proper subset of the neighbourhood of u. Let U = {u1 , u2 , . . . , up } and let T0 , T1 , . . . , Tp be the components of T − {uu1 , uu2 , . . . , uup } containing u, u1 , . . . , up , respectively. If y ∈ V (T0 ), then we say that the tree T ′ = T − {uu1 , uu2 , . . . , uup } + {yu1 , yu2 , . . . , yup } is obtained from T by transferring U from u to y. This construction is related to the so-called α -transformation described by He, Li and Tu [13]. We now define the trees that maximise the average eccentricity among all graphs of order n and k-packing number βk , −1 where βk > 1 + ⌊ nk+ ⌋. These trees were shown in [6] to maximise the average distance among all graphs of order n and 1 −1 k-packing number βk , where βk > 1 + ⌊ nk+ ⌋. 1

Definition 3 ([6]). Given integers k > 0 and ℓ even with 0 ≤ ℓ ≤ 2k. We define Yk,ℓ (a, b, c) as the tree containing vertices a, b, c such that d(a, b) = d(a, c) = k and d(b, c) = ℓ, and either

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Fig. 1. The graph G0 (35, 4, 6).

(i) Υ (T ) = {a, b} and ℓ = 0, (ii) Υ (T ) = {b, c } and ℓ = 2k, or (iii) Υ (T ) = {a, b, c } and 0 < ℓ < 2k, where Υ (T ) denotes the set of end vertices of T . Note that a simple calculation shows that Yk,ℓ (a, b, c) has 21 ℓ + k + 1 vertices. Definition 4 ([6]). For given integers n, r , k > 0 with r(k + 1) ≤ n ≤ (r − 1)(2k + 1), let n − r(k + 1) = pk + q, where 0 ≤ q < k. Let ℓ1 , ℓ2 , . . . , ℓr be integers defined to be

{ r −p−1 } ℓi = 0 for i ∈ 1, 2, . . . , ⌊ ⌋ , 2

r −p+1

⌋, ℓi = 2q for i = ⌊ 2 { r −p+1 r −p+1 r −p−1 } ℓi = 2k for i ∈ ⌊ ⌋ + 1, ⌊ ⌋ + 2, . . . , r − ⌈ ⌉ , 2 2 2 { } r −p−1 r −p−1 ℓi = 0 for i ∈ r − ⌈ ⌉ + 1, r − ⌈ ⌉ + 2, . . . , r . 2

2

We define G0 (n, k, r) to be the tree of order n obtained from the union of vertex disjoint trees Ti = Yk,ℓi (vi , wi,1 , wi,2 ), i = 1, 2, . . . , r, by adding link edges wi,2 wi+1,1 for i = 1, 2, . . . , r − 1. Definition 5 ([6]). For given integers n, r , k > 0 with rk + 1 ≤ n ≤ 2k(r − 1), let n − rk − 1 = pk + q, where 0 ≤ q < k, and let ℓ1 , ℓ2 , . . . , ℓr be defined by the four terms in Definition 4. We define G1 (n, k, r) to be the tree of order n obtained from the union of vertex disjoint trees Ti = Yk,ℓi (vi , wi,1 , wi,2 ), i = 1, 2, . . . , r, by identifying wi,2 with wi+1,1 for i = 1, 2, . . . , r − 1. Example 2. The graphs G0 (35, 4, 6) and G1 (30, 4, 6) are given in Figs. 1 and 2. Both graphs are obtained from the trees T1 , . . . , T6 where T1 , T2 , T5 and T6 are isomorphic to Y4,0 , T3 is isomorphic to Y4,2 , and T4 is isomorphic to Y4,8 . Note that for each Ti , i = 1, 2, . . . , 6 we have wi,1 = wi,2 whenever ℓi = 0. In the graph G0 (35, 4, 6), the link edges wi,2 wi+1,1 are indicated with thin lines while the edges of each Ti are indicated with thick lines. In the graph G1 (30, 4, 6), for i = 1,2, w1,i = w2,i = w3,1 , w3,2 = w4,1 and w4,2 = w5,i = w6,i . For the proof of Lemma 2 we need some terminology and notation. We consider a tree G obtained from the disjoint union of r rooted trees T1 , T2 , . . . , Tr by adding edges e1 , e2 , . . . , er −1 , which we refer to as link edges, where each edge ei joins vertices in two different trees Ti . We denote the set of link edges by E ′ . We also consider a tree G′ obtained from the disjoint union of r trees T1 , T2 , . . . , Tr by successively identifying r − 1 pairs of vertices in different component. We denote the root of Ti by vi . We say a vertex of a tree Ti is a contact vertex if it is incident with a link edge or if it is identified with a vertex from some other component Tj . We denote the set of contact vertices of Ti by Wi and let Wi = {wi,1 , wi,2 , . . .}. In G, if Ti is joined to the remaining trees by only one link edge then we say Ti is an end tree. In G′ , if Ti has only one contact vertex, which is identified with a contact vertex of only one other tree Tj , then we say that Ti is an end tree. n−1 Lemma 2. Let G be a tree of order n and let r and k be positive integers such that r > 1 + ⌊ 2k ⌋. +1

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Fig. 2. The graph G1 (30, 4, 6).

(a) If G is obtained from vertex disjoint trees T1 , T2 , . . . , Tr , rooted at v1 , v2 , . . . , vr , respectively, by adding r − 1 link edges such that dG (vi , wi ) ≥ k for each contact vertex wi of Ti , i = 1, 2, . . . , r, then avec(G) ≤ avec(G0 (n, k, r)), with equality if and only if G = G0 (n, k, r). (b) If G is obtained from vertex disjoint trees T1 , T2 , . . . , Tr , rooted at v1 , v2 , . . . , vr , respectively, by identifying r − 1 pairs of contact vertices of different trees Ti such that dG (vi , wi ) ≥ k for each contact vertex wi of Ti , i = 1, 2, . . . , r, then avec(G) ≤ avec(G1 (n, k, r)), with equality if and only if G = G1 (n, k, r). Proof. Let d = diam(G) and P : p0 , p1 , . . . , pd a longest path of G. We may assume that G is chosen so that avec(G) is maximum among all trees of order n satisfying the hypothesis of the lemma. Also, let the decomposition of G into r rooted trees be such that the number of roots that are end vertices is as large as possible. We note that for much of the proof of Lemma 2, we use EX (G) instead of avec(G). (a) Let n and k be fixed. Our strategy is to first show that each Ti is incident with at most two link edges, so that the Ti ’s are linked in a path-like fashion. Then we show that each Ti is isomorphic to Yk,ℓi (a, b, c) for some li and that each ℓi satisfies the conditions in Definition 4, which eventually leads to identifying G0 and G1 as the extremal graphs. We first observe that G is not a path since the decomposition of a path of order n into r trees T1 , T2 , . . . , Tr , satisfying n−1 ⌋. Now each Ti has at least k + 1 vertices since the the hypothesis of the lemma, is only possible for r ≤ 1 + ⌊ 2k +1 distance between its roots and any of its contact vertices is at least k. Using the fact that n ≥ r(k + 1), it follows that n−1 ⌋ ≥ 1 + ⌊ r(k2k++1)1−1 ⌋, which only holds for r ≥ 3. r > 1 + ⌊ 2k +1 Claim 1: If Ti is an end tree of G then Ti is a path whose ends are vi and wi,1 . We first show that vi is an end vertex of Ti . Indeed, since Ti is an end tree, it contains only one contact vertex, wi,1 . If vi is not an end vertex of G, then we can replace vi by an end vertex of Ti farther from wi,1 , a contradiction to the fact that we choose as many vi as possible as end vertices of G. Hence vi is an end vertex of Ti . Now suppose to the contrary that Ti is not a path or that wi,1 is not an end vertex of Ti . Then Ti contains a vertex of degree at least 3 in G. Case 1: V (P) ∩ V (Ti ) = ∅. Let w be a vertex in Ti of degree at least 3 in G and let u be a neighbour of w in Ti such that the component Tu of G − uw containing u does not contain the root vertex vi of Ti . By Lemma 1, at least one of p0 and pd is an eccentric vertex for each vertex in G. Without loss of generality, we can assume that pd is an eccentric vertex of w and thus of every vertex in the component Tu . Consider the graph G′ = G − uw + up0 and make Tu part of the tree, Tj say, that contains p0 . Although Tu becomes part of Tj , this does not change the set of contact and root vertices. Hence G′ satisfies the hypothesis of the lemma. For each vertex ui in Tu , eG (ui ) = dG (ui , pd ) ≤ d and eG′ (ui ) = dG′ (ui , p0 ) + dG′ (p0 , pd ) ≥ d + 1. Hence each eccentricity has increased by at least 1 and so eG′ (ui ) > eG (ui ). On the other hand, for every vertex not in Tu , the distances to p0 and pd have remained unchanged, hence its eccentricity has not decreased. Therefore, EX (G′ ) ≥ EX (G) + |V (Tu )|, a contradiction to the choice of G. Case 2: V (P) ⊆ V (Ti ). By Lemma 1, at least one of p0 and pd of P is an eccentric vertex for each vertex in G. We can assume without loss of generality that for the root vertex vi , eG (vi ) = dG (vi , pd ), where dG (vi , pd ) ≥ dG (vi , wi,1 ) ≥ k. Let wj,1 be a contact vertex of the tree Tj adjacent to wi,1 of Ti in G. Let G′ = G − wj,1 wi,1 + wj,1 pd , so that pd becomes the new contact vertex of Ti in G′ . Since dG′ (vi , pd ) ≥ k, G′ satisfies the hypothesis of the lemma. For every vertex uj in V (G) − V (Ti ), we have eG (uj ) ≤ d and

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eG′ (uj ) = dG′ (uj , pd ) + d, and so eG′ (uj ) > eG (uj ). On the other hand, for every vertex in Ti , the distances to p0 and pd remain unchanged, hence its eccentricity has not decreased. Hence, EX (G′ ) ≥ EX (G) + |V (G) − V (Ti )|, contradicting the choice of G. Case 3: V (P) ∩ V (Ti ) ̸ = ∅ and V (P) ̸ ⊆ V (Ti ). Since V (P) ∩ V (Ti ) ̸ = ∅ and V (P) ̸ ⊆ V (Ti ), we have p0 ∈ V (Ti ) or pd ∈ V (Ti ). Without loss of generality we can assume that pd ∈ V (Ti ). Let pℓ ∈ V (P) ∩ V (Ti ) be a vertex of degree at least 3 in G. Let u ∈ N(pℓ ) − V (P) and let Tu be the component of G − upℓ containing u. Define G′ = G − upℓ + upd . If vi ∈ V (Tu ), then dG′ (vi , wi,1 ) = dG′ (vi , pd ) + dG′ (pd , wi,1 ) = dG′ (vi , pd ) + eTi (wi,1 ). Since P is a longest path and wi,1 is on P, eTi (wi,1 ) ≥ dTi (vi , wi,1 ) ≥ k and dG′ (vi , wi,1 ) ≥ dG′ (vi , pd ) + k. If v ̸ ∈ V (Ti ), then dG′ (vi , wi,1 ) = dG (vi , wi,1 ) ≥ k. Hence G′ satisfies the hypothesis of the lemma. Now for every vertex ui in Tu , eG (ui ) ≤ d and eG′ (ui ) ≥ d + 1. On the other hand, for every vertex not in Tu , the distances to p0 and pd remain unchanged, hence its eccentricity has not decreased. Therefore, EX (G′ ) ≥ EX (G) + |V (Tu )|, contradicting the choice of G. Hence Ti is a path with ends vi and wi,1 . Claim 2: All edges of E ′ lie on P. We first show that no contact vertex is within distance at most k − 1 of an end vertex of P. Let p0 be contained in a tree Tj and suppose to the contrary that there exists a contact vertex wj,1 of Tj that is within distance k − 1 of p0 . Let wi,1 wj,1 be a link edge in G, where wi,1 is a contact vertex of Ti . We consider the following two cases. Case 1: wi,1 is on P. If wi,1 is on P, then wj,1 is also on P. Since d(wj,1 , vj ) ≥ k by the hypothesis of the lemma, and d(wj,1 , p0 ) < k, replacing the (wj,1 , p0 )-section of P with the (wj,1 , vj )-path yields a path that is longer than P, a contradiction to the choice of P as a longest path in G. Case 2: wi,1 is not on P. Let ps be the last vertex of P that is on the (p0 , wi,1 )-path. Let Q be the (ps , vi )-path. Then the only vertex of Q that is also on P is ps since P contains no vertex of Ti and since ps is the last vertex of P that is on the (p0 , wi,1 )-path. Moreover, the length of Q is greater than k since Q strictly contains a (vi , wi,1 )-path, while the length of the (p0 , ps )-section of P is less than k since it contains a (p0 , wj,1 )-path. Hence replacing the (p0 , ps )-section of P with Q yields a path that is longer than P, a contradiction to the choice of P as a longest path. Hence every contact vertex is within distance at least k of p0 . The same holds for pd , and so every contact vertex is at distance at least k of both end vertices of P. Since p0 and pd are at distance at least k from every contact vertex, we may assume that p0 and pd are the roots of the trees in which they are. Since p0 and pd are end vertices, that does not change our assumption that the number of roots that are end vertices of G is maximal. Now suppose Claim 2 is false and G contains a link edge that is not on P. Among all such link edges let wi,1 wj,1 be a link edge that is farthest from V (P) and let wi,1 be farther from V (P) than wj,1 . Then Ti is necessarily an end tree. By Claim 1, Ti is a path whose end vertices are wi,1 and vi . Now eG (vi ) ≤ d and so eG (wi,1 ) = eG (vi ) − dG (vi , wi,1 ) ≤ d − k. Thus eG (wj,1 ) ≤ d − (k + 1). Now let pk be a vertex at distance k from p0 on P and let G′ be the graph obtained from G by transferring wi,1 from wj,1 to pk . Then pk is the only new contact vertex of G′ . Since p0 is the root of the tree containing pk and dG′ (p0 , pk ) = k, G′ satisfies the hypothesis of the lemma. Since eG (p0 ) = d, eG (pk ) = d − k and so for each vertex u in Ti , eG′ (u) ≥ eG (u) + 1. On the other hand, for every vertex not in Ti , the distances to p0 and pd remain unchanged, so its eccentricity has not decreased. It follows that EX (G′ ) ≥ EX (G) + |V (Ti )|, contradicting the choice of G as having maximum eccentricity. Hence all edges of E ′ lie on P. Claim 3: Each Ti is incident with either one or two link edges. By Claim 2, all link edges are on P. Since P enters Ti at most once and leaves Ti at most once each time using a link edge, it follows that at most two link edges are incident with vertices of Ti . This proves Claim 3. Claim 4: For all i, the set of all end vertices of Ti is in Wi ∪ {vi }. Suppose to the contrary that there exists an end vertex u in Ti which is not in Wi ∪ {vi }. By Claim 1, u ̸ = p0 and u ̸ = pd . Let v be the neighbour of u and let G′ = G − uv + up0 . Since u is neither a root nor a contact vertex, no distance between a contact vertex and a root has changed, so G′ satisfies the hypothesis of the lemma. Then eG′ (u) = d + 1 > eG (u), and for all other vertices the distances to p0 and pd have not changed, hence no eccentricity has decreased. It follows that EX (G′ ) > EX (G). This contradiction proves Claim 4. It follows from Claim 3 that each Ti is incident with either one or two link edges so that T1 , T2 , . . . , Tr are connected in a path-like pattern. We can assume that Ti is connected to Ti−1 and Ti+1 by link edges, for i = 2, 3, . . . , r − 1 and that T1 and Tr are the only end trees. Denote the contact vertex of Ti that is adjacent to a contact vertex of Ti−1 by wi,1 and that adjacent to Ti+1 by wi,2 . Claim 5: (i) T1 is a path of length at least k with end vertices v1 and w1,2 . (ii) Tr is a path of length at least k with end vertices vr and wr ,2 . T1 and Tr are the end trees of G. Hence both have only one contact vertex w1,2 and wr ,1 , respectively. By Claim 1, tree T1 (tree Tr ) is a path with end vertices v1 (vr ) and w1,i (wr ,i ). Since d(v1 , w1,2 ) ≥ k and d(vr , wr ,1 ) ≥ k, it follows that tree Ti (tree Tr ) has length at least k, as desired. Claim 6: Let vi ̸ ∈ Υ (Ti ). Then Ti is a path of length at least 2k and end vertices wi,1 and wi,2 . By Claim 4, Υ (Ti ) ⊆ Wi ∪ {vi }. If vi is not an end vertex of Ti , then Υ (Ti ) ⊆ Wi . By Claim 3, each Ti is incident with either one or two link edges. But Υ (Ti ) contains at least two vertices. Hence Υ (Ti ) = Wi and thus Ti is a path with end vertices wi,1 and

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wi,2 . By the hypothesis of the lemma, d(vi , wi,1 ) ≥ k and d(vi , wi,2 ) ≥ k. Hence Ti has length d(vi , wi,1 ) + d(vi , wi,2 ) ≥ 2k, as desired. This proves Claim 6. In what follows, denote the set V (Ti ) ∪ V (Ti+1 ) ∪ · · · ∪ V (Tj ) by Vi,j and let its induced subtree be denoted as Ti,j . Claim 7: If vi ∈ Υ (Ti ) for i ̸ = 1, r, then Ti = Yk,ℓ (vi , wi,1 , wi,2 ) for some integer 0 ≤ ℓ < 2k. Case 1: Ti is not a path. If Ti is not a path, then Ti has at least 3 end vertices, and so |Wi | ≥ 2 by Claim 4. By Claim 3, Wi contains at most 2 vertices. Hence, |Wi | = 2 where wi,1 ̸ = wi,2 . Thus, Υ (Ti ) = {vi , wi,1 , wi,2 }. Furthermore, Ti has exactly one vertex, say z, of degree 3 which lie on the (wi,1 , wi,2 )-path while all vertices in V (Ti ) − {vi , wi,1 , wi,2 , z } have degree 2. Let a, b, c be the neighbours of z closest to wi,1 , vi and wi,2 , respectively. Without loss of generality, we may assume that d(vi , wi,1 ) ≥ d(vi , wi,2 ). Let d(wi,1 , wi,2 ) = ℓ. To prove that Ti = Yk,ℓ (vi , wi,1 , wi,2 ), it suffices to show that dTi (vi , wi,1 ) ≤ k. Suppose to the contrary that dTi (vi , wi,1 ) ≥ k + 1. We consider two subcases: Subcase 1.1: dG (z , p0 ) ≤ dG (z , pd ). Let G1 = G − bz + ba. Then dG1 (vi , wi,1 ) = dG (vi , wi,1 ) − 1 ≥ k and dG1 (vi , wi,2 ) = dG (vi , wi,2 ) ≥ k. Hence G1 satisfies the hypothesis of the lemma. However, for all vertices in the component G1 − ab containing a the eccentricity remains unchanged since their distances to p0 and pd remain the same. On the other hand, for all other vertices of the component G1 − ab containing b the eccentricity goes up by 1. Hence EX (G1 ) > EX (G), a contradiction to the maximality of EX (G). Subcase 1.2: dG (z , p0 ) > dG (z , pd ). Let G2 = G − az + ab. Then G2 satisfies the hypothesis of the lemma as in Subcase 1.1 above. For all vertices on the (b, vi )-path the eccentricity drops by 1, but for all other vertices not on the (b, vi )-path the eccentricity goes up by 1. And there are more than k vertices for which the eccentricity goes up, while there are less than k vertices whose eccentricity drops. Hence the total eccentricity goes up. This contradicts the maximality of EX (G). Hence dTi (vi , wi,1 ) = dTi (vi , wi,2 ) = k and thus Ti = Yk,ℓ (vi , wi,1 .wi,2 ). Since vi is not on P(wi,1 , wi,2 ), we have that ℓ = d(wi,1 , wi,2 ) < dTi (vi , wi,1 ) + dTi (vi , wi,2 ) = 2k, as desired. Case 2: Ti is a path. If Ti is a path, then Ti has two end vertices vi and say, z. We claim that z = wi,1 = wi,2 . Suppose not. Then one of the contact vertices, say wi,1 , is an internal vertex of Ti . And so, Ti is joined to Ti−1 and Ti+1 by adding the link edges wi−1,2 , wi,1 and wi,2 , wi+1,1 , respectively. However, Ti is not of the form Ti = Yk,ℓ (vi , wi,1 , wi,2 ) since dTi (vi , wi,2 ) = dTi (vi , wi,1 ) + dTi (wi,1 , wi,2 ) ≥ k, implying that dTi (vi , wi,1 ) ≥ k − dTi (wi,1 , wi,2 ). Let a and b be the neighbours of wi,1 on the (wi,1 , vi )-path and on the (wi,1 , wi,2 )-path, respectively. Subcase 2.1: dG (wi,1 , p0 ) ≥ dG (wi,2 , pd ). Then let G′ = G − awi,1 + ab. As above, for every vertex in G − awi,1 containing a the eccentricity goes up by 1. For the remaining vertices in the component of G − awi,1 containing wi,1 the eccentricity has not decreased since their distances to p0 and pd remain unchanged. Hence EX (G′ ) > EX (G), a contradiction. Subcase 2.2: dG (wi,1 , p0 ) < dG (wi,2 , pd ). Let G′ = G − bwi,1 + ab. As above, EX (G′ ) > EX (G), a contradiction. Hence EX (G) is not maximal. This contradiction shows that z is the only contact vertex of Ti , and so z = wi,1 = wi,2 . We show that Ti is a path of length k. We know that if Ti is a path with vi ∈ Υ (Ti ), then Ti has end vertices vi and z, where z = wi,1 = wi,2 . By the hypothesis of the lemma, d(vi , z) ≥ k. Suppose to the contrary that d(vi , z) ≥ k + 1 and let a be the neighbour of z in Ti . We may assume without loss of generality that dG (z , p0 ) ≤ dG (z , pd ), so that all vertices in Ti have pd as an eccentric vertex. Let G′ = G − wi−1,2 z + wi−1,2 a. Then the eccentricity of all vertices in Ti remains unchanged. For vertices in V1,i−1 , the distance to pd has increased by one, so their eccentricity have increased. But for vertices in Vi+1,r , the distance to p0 and pd has not decreased, so no eccentricity has dropped. Hence the eccentricity has increased. This contradiction to the maximality of EX (G) shows that Ti is a path of length k, and so Ti = Yk,0. (vi , wi,1 , wi,2 ). Claim 8: For all i ∈ {1, 2, . . . , r }, there exist some ℓ ∈ {0, 1, . . . , 2k} with Ti = Yk,ℓ (vi , wi,1 , wi,2 ). Suppose to the contrary that there exists an i such that Ti is not of the form Ti = Yk,ℓ (vi , wi,1 , wi,2 ). It follows that either (i) from Claim 5 that i ∈ {1, r } and Ti is a path of length greater than k, or (ii) from Claims 6 and 7 that 1 < i < r and Ti is a path of length greater than 2k. Since G is not a path, there exists j ∈ {1, 2, . . . , r } such that Tj = Yk,ℓ (vj , wj,1 , wj,2 ) and 0 ≤ ℓ < 2k. Assume without loss of generality that i < j, and also that i and j are chosen such that j − i is minimum. Then the trees Ti+1 , Ti+2 , . . . , Tj−1 are paths of length 2k and their vertices induce the path P(wi,2 , wj,1 ). Now modify the decomposition of G into trees by removing the vertex wh,2 from Th and incorporating it into Th+1 for h = i, i + 1, i + 2, . . . , j − 1. For h = i + 1, i + 2, . . . , j − 1, we replace vh by its neighbours closer to wh,2 as the root of Th . It is easy to see that this decomposition into trees satisfies the hypothesis of the lemma and also has the maximum number of root vertices that are end vertices. However, the tree Tj′ is such that dT ′ (vj , wj,1 ) ≥ 2k + 1, and thus neither a path nor of j the form Ti = Yk,ℓ (vi , wi,1 , wi,2 ). This contradiction to Claim 7 proves Claim 8.

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We now show that the ℓi satisfy the conditions in Definition 4. Claim 9: Let 1 ≤ i < j ≤ r be such that ℓi+1 = ℓi+2 = · · · = ℓj−1 = 2k. (i) If dG (wi,1 , p0 ) < dG (wj,2 , pd ), then ℓi = 0 or ℓj = 2k. (ii) If dG (wi,1 , p0 ) > dG (wj,2 , pd ), then ℓi = 2k or ℓj = 0. (iii) If dG (wi,1 , p0 ) = dG (wj,2 , pd ), then ℓi ∈ {0, 2k} or ℓj ∈ {0, 2k}. To prove (i), suppose to the contrary that ℓi > 0 and ℓj < 2k. Consider the graph G′ obtained from G by replacing Ti by Ti′ := Yk,ℓi −2 (vi , wi,1 , wi,2 ) and Tj by Tj′ := Yk,ℓj +2 (vj , wj,1 , wj,2 ). So the sequence associated with G′ is ℓ1 , . . . , ℓi−1 , ℓi − 2, li+1 , . . . , ℓj−1 , ℓj + 2, ℓj+1 , . . . , ℓr . It is easy to see that dG (wi,1 , wj,2 ) = dG′ (wi,1 , wj,2 ) and that G′ has order n. Also, G′ satisfies the hypothesis of the lemma since dTi (vi , wi,1 ) = dT ′ (vi , wi,1 ) and dTj (vj , wj,1 ) = dT ′ (vj , wj,1 ). i

j

We now compare EX (G′ ) and EX (G). Replacing Ti by Ti′ and Tj by Tj′ in G′ only changes the eccentricity of vertices in Vi,j . Thus eccentricities of vertices in V1,i−1 ∪ Vj+1,r and on the path P(wi,1 , wj,2 ) remain unchanged since their distances to p0 and pd remain unchanged. Let Ti1 (Tj1 ) be the subtree of Ti (Tj ) that is not on P(wi,1 , wj,2 ) in G and denote their orders by ni1 = |V (Ti1 )| and nj1 = |V (Tj1 )|, respectively. Similarly, let Ti1′ (Tj1′ ) be the subtree of Ti′ (Tj′ ) that is not on P(wi,1 , wj,2 ) in G′ and denote their orders by n′i1 = |V (Ti1′ )| and n′j1 = |V (Tj1′ )|, respectively. Then, EX (G′ ) − EX (G) =

[ ∑

eG′ (x) −

x∈V (T ′ )

+

[ ∑i1

′ ) y∈V (Tj1

∑

eG (x)

]

x∈V (Ti1 )

eG′ (y) −

∑

eG (y)

]

y∈V (Tj1 )

However, by replacing Ti by Ti′ , and Tj by Tj′ , we have that n′i1 = ni1 + 1 and n′j1 = nj1 − 1, respectively. Let ei and ej be the least eccentricities in G among the vertices of Ti1 and Tj1 , respectively. It follows from dG (wi,1 , p0 ) < dG (wj,2 , pd ) that pd is an eccentric vertex of the vertices in Ti1 . Since the vertices of Ti1′ are attached to a vertex of P that is one step further away from vd than the vertex of P to which the vertices of Ti1 are attached, we have

∑

eG (x) = ei + (ei + 1) + · · · + (ei + ni1 − 1),

x∈V (Ti1 )

and

∑

eG′ (x) = (ei + 1) + (ei + 2) + · · · + (ei + ni1 − 1) + (ei + ni1 ) + (ei + ni1 + 1),

′ ) x∈V (Ti1

so that

∑

eG′ (x) −

′ ) x∈V (Ti1

∑

eG (x) = (ei + ni1 ) + (ei + ni1 + 1) − ei

x∈V (Ti1 )

= ei + 2ni1 + 1. In the same way,

∑

eG (y) = ej + (ej + 1) + · · · + (ej + nj1 − 1).

y∈V (Tj1 )

The vertices of Tj1 have either p0 or pd as an eccentric vertex. In the former case, the eccentricities of the vertices of Tj1′ in G′ are one less compared to the eccentricities of the vertices of Tj1 in G. In the latter case they are one more. Hence

∑

eG′ (y) ≥ (ej − 1) + ej + (ej + 1) + · · · + (ej + nj1 − 2) + (ej + nj1 − 3),

′ ) y∈V (Tj1

so that

∑

eG′ (y) −

′ ) y∈V (Tj1

∑

eG (y) ≥ (ej − 1) + (ej + nj1 − 2) + (ej + nj1 − 3)

y∈V (Tj1 )

= 3ej + 2nj1 − 6. Thus, we have EX (G′ ) − EX (G) ≥ (ei + 2ni1 + 1) + (3ej + 2nj1 − 6)

= (ei + 3ej ) + 2(ni1 + nj1 ) − 5. Now since G is not a star, we have ei , ej ≥ 2. Moreover, ni1 , nj1 ≥ 0. Hence, EX (G′ ) > EX (G), a contradiction to the maximality of G. This proves Claim 9(i).

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The proof of (ii) is analogous to that of (i). To prove (iii), we may assume, without loss of generality, that ℓi ≤ ℓj . If ℓi = 0 or ℓj = 2k, then the statement holds. Now suppose to the contrary that ℓi > 0 and ℓj < 2k. We obtain a graph G2 similar to G′ in proof of Claim 9(i), so that by comparison, EX (G2 ) > EX (G). A contradiction which proves Claim 9(iii). If ℓ1 = · · · = ℓr = 0, then each Ti has exactly k + 1 vertices and so n = r(k + 1). By Definition 4, n − r(k + 1) = pk + q. Thus, p = q = 0 and so sequence ℓ1 , ℓ2 . . . , ℓr satisfies Definition 4. Hence we can assume that not all ℓi equal 0. Claim 10: There exists a, b ∈ {1, 2, . . . , r }, a ≤ b, such that (i) ℓi = 0 for i ∈ {1, 2, . . . , a − 1} ∪ {b, b + 1, . . . , r }, (ii) ℓi = 2k for all i ∈ {a + 1, a + 2, . . . , b − 1}, (iii) ℓa = 2q, where q is as defined in the statement of the lemma, (iv) r − b + 1 ≤ a ≤ r − b + 2. We first prove (i). For given i, dG (wi,1 , p0 ) − dG (wi+1,2 , pd ) increases as i increases. Thus there exists a smallest integer s ∈ {1, 2, . . . , r } such that dG (wi,1 , p0 ) − dG (wi+1,2 , pd ) ≥ 0. If i < s, then dG (wi,1 , p0 ) − dG (wi+1,2 , pd ) < 0 so that by Claim 9(i) with j = i + 1, ℓi = 0 or ℓi+1 = 2k. This yields that there exists an a ∈ {1, 2, . . . , s} such that ℓi = 0 for i ∈ {1, 2, . . . , a − 1} and ℓi = 2k for i ∈ {a + 1, a + 2, . . . , s}. If i > s, then dG (wi,1 , p0 ) − dG (wi+1,2 , pd ) > 0. By Claim 9(ii) with j = i + 1, ℓi = 2k or ℓi+1 = 0. Hence there exists b ∈ {s + 1, s + 2, . . . , r } such that ℓi = 2k for i ∈ {s + 1, s + 2, . . . , b − 1} and ℓi = 0 for i ∈ {b + 1, b + 2, . . . , r }. Therefore ℓi = 0 for i ∈ {1, 2, . . . , a − 1} ∪ {b + 1, b + 2, . . . , r } and ℓi = 2k for i ∈ {a + 1, a + 2, . . . , b − 1}. This proves part (ii) of Claim 10, and also part (i), except for the value of ℓb . In order to complete the proof of (ii) we now consider the values of ℓa and ℓb . We have shown that ℓa+1 = ℓa+2 = · · · = ℓb−1 = 2k. Hence it follows from Claim 9 (i)-(iii) that ℓa ∈ {0, 2k} or ℓb ∈ {0, 2k}. We may assume that ℓb ∈ {0, 2k}. If ℓb = 0 then (i) holds and if ℓb = 2k then let b := b + 1. In both cases (i) holds. To prove (iii) note that 0 ≤ ℓa ≤ 2k by Claim 8. Since Ti has order k + 1 for i ∈ {1, 2, . . . , a − 1} ∪ {b, b + 1, . . . , r } and order 2k + 1 for i ∈ {a + 1, a + 1, . . . , b − 1}, we have

∑

|V (Ti )| = (k + 1)(a − 1) + (k + 1)(r − (b − 1)) + (2k + 1)(b − 1 − a)

i̸ =a

= r(k + 1) + k(b − a) − 2k − 1, so that

|V (Ta )| = n −

∑

|V (Ti )|

i̸ =a

= n − r(k + 1) + k(a − b + 2) + 1. By Definition 4, n − r(k + 1) = pk + q. Thus,

|V (Ta )| = k(p + a − b + 2) + (q + 1) ≡ q + 1(mod k). Since G satisfies the hypothesis of the lemma, it follows that Ta = Yk,ℓa (va , wa,1 , wa,2 ) and thus |V (Ta )| = k + 1 + 21 la . Hence k + 1 + 21 ℓa = k(p + a − b + 2) + (q + 1), so that 21 ℓa = k(p + a − b + 1) + q ≡ q (mod k). Since q ∈ {1, 2, . . . , k − 1} and 0 ≤ ℓa < 2k, ℓ2a = q. Thus, ℓa = 2q. This proves (iii). We now prove part (iv) of Claim 10. We first show that a ≥ r − b + 1. If q = 0, then ℓa = 0 so that ℓi = 0 for i ∈ {1, 2, . . . , a} ∪ {b, b + 1, . . . , r }. In this case, we can assume, without loss of generality, that a ≥ r − b + 1, otherwise we reverse the order of the ℓi and still have the same G. Hence we may assume that q ̸ = 0, that is 0 < q < k. Using Claim 9, let i = a and j = b so that ℓa+1 = ℓa+2 = · · · = ℓb−1 = 2k. Since ℓa = 2q and ℓb = 0, it follows from Claim 9(ii) and (iii) that dG (wa,1 , p0 ) ≥ dG (wb,2 , pd ). This yields k + a − 1 ≥ k + r − b, which implies that a ≥ r − b + 1, as desired. To derive the second part of the inequality a ≤ r − b + 2, let i = a and j = b − 1 so that ℓa+1 = ℓa+2 = · · · = ℓb = 2k. Since ℓa < 2k and ℓb−1 = 2k, it follows from Claim 9(i) and (iii) that dG (wa,1 , p0 ) ≤ dG (wb−1,2 , pd ). This yields k+a−1 ≤ r −b+1+k, which implies that a ≤ r − b + 2, as desired. By Claims 1 to 10, G is obtained from the union of vertex disjoint trees Ti = Yk,ℓi (vi , wi,1 , wi,2 ) by adding suitable edges joining Ti to Ti+1 for i = 1, 2, . . . , r − 1, and the ℓi satisfy the properties in Definition 4. Hence G = G0 (n, k, r), and so part (a) of the lemma follows. (b) The proof of (b) is similar to that of (a) but some of the claims need to be modified. Here G is obtained from the vertex disjoint trees T1 , T2 , . . . , Tr rooted at v1 , v2 , . . . , vr by identifying r − 1 pairs of contact vertices of different trees Ti . As in (a), we assume that G is chosen so that EX (G) is maximum among all trees satisfying the hypothesis of the lemma and the decomposition of G into r rooted trees is such that the number of roots that are end vertices is as large as possible. Claim 2 is rephrased as follows: Let P be a longest path of G. Then all contact vertices Wi lie on P, and all contact vertices except the two contact vertices closest to the two ends of P belong to exactly two of the trees T1 , . . . , Tr . The proof is similar to the proof in (a). Here we suppose that there is a contact vertex wi in a tree Ti which is not on P. Let wk be a contact vertex in tree Tk identified with wi and let wj be a contact vertex of an end tree Tj . By Claim 1, Tj is a path whose ends are, say wj and vj . By transferring wi from wk and identifying with wj in an end tree Tj yields a graph G′ with EX (G′ ) > EX (G), a contradiction to the maximality of G.

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Claim 3 is rephrased as follows: Each Ti has either one or two contact vertices. Since G is formed by identifying r − 1 trees, there exists a contact vertex w of three or more trees. Then at most one of the trees Ti containing w has more than one contact vertex. This implies that the trees Ti can be numbered such that there exists s, t ∈ {1, 2, . . . , r } where T1 , T2 , . . . , Ts−1 are end trees with a common contact vertex which they share with Ts . Also, Tt +1 , . . . , Tr −1 , Tr are end trees with a common contact vertex which they share with Tt . The remaining trees Ts , Ts+1 , . . . , Tt have two contact vertices each so that Ti and Ti+1 share a contact vertex, for i ∈ {s, s + 1, . . . , t − 1}. The remaining Claims 1, 4, 5, 6, 7, 8, 9 and 10 remain the same as in (a). Thus by Claims 1 − 10 in (b), G is the union of vertex disjoint trees Ti = Yk,li (vi , wi,1 , wi,2 ). Hence G = G1 (n, k, r), and so part (b) of the lemma follows. □ −1 ⌋. Theorem 4. Let G be a connected graph of order n and k-packing number βk > 1 + ⌊ nk+ 1 (a) If k is even, then

avec(G) ≤ avec(G0 (n, k/2, βk )), with equality if and only if G = G0 (n, k/2, βk ). (b) If k is odd, then avec(G) ≤ avec(G1 (n, (k + 1)/2, βk )), with equality if and only if G = G1 (n, (k + 1)/2, βk ). Proof. (a) We first consider the case that k is even. Let S = {v1 , v2 , . . . , vr } ⊆ V (G) be a maximum k-packing of G so βk = r. Let H be an arbitrary spanning tree of G. Then S is also a k-packing in H. For i = 1, 2, . . . , r let Ti′ be the subtree of H induced ⋃r k/2 by NH (vi ) ∪ {vi }. Clearly, we can extend the trees Ti′ to subtrees Ti of H such that i=1 V (Ti ) = V (H). It is easy to see that H is obtained from the union of the Ti by adding r − 1 edges, and that each contact vertex of Ti is at distance at least k/2 from vi . Hence H satisfies the hypothesis of Lemma 2(a), with k replaced by k/2, and so EX (G) ≤ EX (H) ≤ EX (G0 (n, k/2, r)). Equality implies by Lemma 2(a) that H = G0 (n, k/2, r). Since this holds for every spanning tree H of G, we conclude that equality implies that G = G0 (n, k/2, r). (k+1)/2 ⋃ (b) The proof of (b) is similar to that of (a). Here we consider the case where k is odd so that for each vi ∈ S, NG {vi } is disjoint. As in (a), we obtain a spanning tree H ′ with EX (G) ≤ EX (H ′ ) ≤ EX (G1 (n, (k + 1)/2, βk )) so that if equality holds, then G = H ′ = G1 (n, (k + 1)/2, βk ). This completes the proof. □ 5. Bounds on average eccentricity in terms of γk and γck In this section we give sharp upper bounds on the average eccentricity for graphs of given order and k-domination number

γk or connected k-domination number γck . Lemma 3. Let G be a connected graph of order n and S a minimum connected k-dominating set of G. Let s = |S | and T = G[S ]. If s ≥ 2, then EX (G) ≤ EX (T ) + k(2n − k − s + 1) + (n − s) diam(T ). Proof. Every vertex not in S is within distance k from S. For i = 1, 2, . . . , k define Si as the set of vertices at distance i from S, and let si = |Si |. We first show that si ≥ 2

for i = 1, 2 . . . , k − 1.

(8)

Suppose to the contrary that there exists i ∈ {1, 2, . . . , k − 1} such that si ≤ 1. We may assume that i is minimum with this property. Let Si = {vi }, and let u ∈ S be a vertex at distance i from vi . If w is a vertex of G at distance greater than i from S, then every path from w to S goes through vi and so u is a vertex at distance at most k from w . Now let x ∈ S be a non-cut vertex of T distinct from u. Then T − x is connected and non-empty. Furthermore, S − {x} is a k-dominating set. Indeed, if there was a vertex y that is not within distance k of S − {x}, then x ∈ Sk since otherwise a neighbour of x in T would be within distance k of y. But dG (u, y) ≤ k by the above, and so y is within distance k of S − {x}. We conclude that S − {x} is a connected k-dominating set. This contradiction to S being a minimum connected k-dominating set proves (8). Since every vertex not in S is within distance k from S, we have eG (v ) ≤ eT (v ) + k for all v ∈ S.

(9)

If v ∈ Si , then dG (v, S) = i, so v is within distance diam(T ) + i of every vertex of S. Since S is k-dominating, we have eG (v ) ≤ i + diam(T ) + k for all v ∈ Si .

(10)

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Summing (9) over all v ∈ S and (10) over all v ∈ Si for i = 1, 2, . . . , k we get EX (G) =

∑

eG (v ) +

e G (v )

i=1 v∈Si

v∈S

≤

k ∑ ∑

∑

k ∑ (

v∈S

i=1

(eT (v ) + k) +

)

si i + diam(T ) + k

= EX (T ) + sk + (n − s)(diam(T ) + k) +

k ∑

i si .

i=1

∑k

∑k

Now i=1 si = n − s and si ≥ 2 for i = 1, 2, . . . , k − 1. Subject to these constraints, the sum i=1 isi is clearly maximised if s1 = s2 = · · · = sk−1 = 2 and sk = n − s − 2k + 2. Substituting these values for s1 , s2 , . . . , sk yields EX (G) ≤ EX (T ) + sk + (n − s)(diam(T ) + k) + k(k − 1) + k(n − s − 2k + 2)

= EX (T ) + k(2n − k − s + 1) + (n − s)diam(T ). □ We use Lemma 3 to obtain a bound on the average eccentricity of a graph of given order and connected k-domination number. This generalises a bound in terms of order and connected domination number in [8], which is the special case k = 1 of the following theorem. Theorem 5. Let G be a connected graph of order n and connected k-domination number γck . Then, avec(G) ≤ 2k + γck − 1 −

k(k + γck − 1) n

+

1⌊1 n 2

1 ⌋ γck − γck2 , 4

and this bound is sharp. Proof. Let S be a minimum connected k-dominating set of G and let T be the subgraph of G induced by S. Now |V (T )| = γck , so that diam(T ) ≤ γck − 1. By Proposition 1, EX (T ) ≤

⌊ 3γ 2

ck

−

γck ⌋ ,

4 2 and by Lemma 3 we have

EX (G) ≤ EX (T ) + k(2n − k − γck + 1) + (n − γck )(γck − 1). It follows by Lemma 3 in conjunction with Proposition 2 that

⌊ 3γ 2

γck ⌋

+ k(2n − k − γck + 1) + (n − γck )(γck − 1), 4 2 and division by n yields, after simplification, the bound of the corollary. □ EX (G) ≤

ck

−

We conclude the paper with a sharp upper bound on the average eccentricity of a graph of given order and k-domination number. Our result generalises a bound on the average eccentricity of graphs of given order and domination number in [8], which is the special case k = 1. For the proof of our bound we require the following result by Meir and Moon [16]. Theorem 6 ([16]). Let T be a tree of order n and k a positive integer. Then γk (T ) = β2k (T ). Corollary 2. Let G be a connected graph of order n and k-domination number γk . (a) If γk ≤ γk (Pn ), then, avec(G) ≤ (2k + 1)γk − 1 −

1 ⌈ [(2k + 1)γk − 1]2 − 1 ⌉ n

4

,

and this bound is sharp. (b)(i) If γk > γk (Pn ) and k is even, then avec(G) ≤ avec(G0 (n, k, γk )). (b)(ii) If γk > γk (Pn ) and k is odd, then avec(G) ≤ avec(G1 (n, k + 1, γk ))s. Proof. We make use of the well-known fact that every connected graph has a spanning tree with the same k-domination number. Let T be such a tree. By Theorem 6 we have β2k (T ) = γk . Applying Theorem 6 now yields the bound and sharpness in (a), and applying Theorem 4 yields the bound in (b). □

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