Abstract

If 𝐺 is a simple connected graph with vertex 𝑉(𝐺), then the eccentric distance sum of 𝐺, denoted by 𝜉𝑑(𝐺), is defined as ∑𝑣∈𝑉(𝐺)ec𝐺(𝑣)𝐷𝐺(𝑣), where ec𝐺(𝑣) is the eccentricity of the vertex 𝑣 and 𝐷𝐺(𝑣) is the sum of all distances from the vertex 𝑣. Let 𝑛≥8. We determine the 𝑛-vertex trees with, respectively, the maximum, second-maximum, third-maximum, and fourth-maximum eccentric distance sums. We also characterize the extremal unicyclic graphs on 𝑛 vertices with respectively, the maximal, second maximal, and third maximal eccentric distance sums.

1. Introduction

Let 𝐺 be a simple connected graph with vertex set 𝑉(𝐺) and edge set 𝐸(𝐺). Let 𝑑𝐺(𝑢,𝑣) be the distance (or length of a shortest path) between vertices 𝑢 and 𝑣 in 𝐺. For a vertex 𝑣∈𝑉(𝐺), the eccentricity ec𝐺(𝑣) is the maximum distance from 𝑣 to any other vertex, and 𝐷𝐺∑(𝑣)=𝑢∈𝑉(𝐺)𝑑𝐺(𝑢,𝑣) is the sum of all distances from 𝑣. The eccentric distance sum of 𝐺 (EDS) is defined as 𝜉𝑑(𝐺)=𝑣∈𝑉(𝐺)ec𝐺(𝑣)𝐷𝐺(𝑣).(1.1)

This graph invariant was proposed by Gupta et al. in [1]. Like the Wiener index [2–4] and eccentric connectivity index [5–9], it turned to have high discriminating power and excellent predictability both with regard to biological and physical properties and to provide valuable leads for the development of safe and potent therapeutic agents of diverse nature. So, it is of interest to study the mathematical properties of this invariant.

Yu et al. [10] considered the 𝑛-vertex trees and unicyclic graphs with minimal eccentric distance sums, respectively. Ilić et al. [11] proved that path 𝑃𝑛 is the unique extremal trees with 𝑛 vertices having maximum eccentric distance sum, and provided various lower and upper bounds for the eccentric distance sum. In this paper, we determine the 𝑛-vertex trees with, respectively, the maximum, second-maximum, third-maximum, and fourth-maximum eccentric distance sums for 𝑛≥8. We also characterize the extremal unicyclic graphs with the maximal, second maximal, and third maximal eccentric distance sums.

2. The Trees with Maximal Eccentric Distance Sums

Lemma 2.1. Let 𝑢 be a vertex of a connected graph 𝐺 with at least two vertices. Let 𝐺𝑖 be the graph obtained by identifying u and a vertex 𝑣𝑖 of a path 𝑃𝑛+1=𝑣0𝑣1⋯𝑣𝑛−1𝑣𝑛, where 1≤𝑖≤⌊𝑛/2⌋. Then, 𝜉𝑑(𝐺𝑖−1)>𝜉𝑑(𝐺𝑖).

Proof. By the definition of EDS, we have 𝜉𝑑𝐺𝑙=𝑣∈𝑉(𝐺)⧵{𝑢}ec𝐺𝑙(𝑣)𝐷𝐺𝑙(+𝑣)𝑖−1𝑗=0ec𝐺𝑙𝑣𝑗𝐷𝐺𝑙𝑣𝑗+ec𝐺𝑙𝑣𝑛−𝑗𝐷𝐺𝑙𝑣𝑛−𝑗+𝑛−𝑖𝑘=𝑖ec𝐺𝑙𝑣𝑘𝐷𝐺𝑙𝑣𝑘.(2.1) Denoted by 𝐴𝑙, 𝐵𝑙, and 𝐶𝑙 respectively, the three sums of right equality above, we only need to prove the following three inequalities: 𝐴𝑖−1>𝐴𝑖, 𝐵𝑖−1>𝐵𝑖, 𝐶𝑖−1>𝐶𝑖.
Note that ec𝐺𝑖−1(𝑣)≥ec𝐺𝑖(𝑣) for any 𝑣∈𝑉(𝐺), and 𝐷𝐺𝑖−1(𝑣)−𝐷𝐺𝑖(𝑣)=𝜔∈𝑉(𝐺)𝑑(𝑣,𝜔)+𝑖−1𝑗=0(𝑑(𝑢,𝑣)+𝑗)×2+𝑛−𝑖+1𝑗=𝑖−(𝑑(𝑢,𝑣)+𝑗)𝜔∈𝑉(𝐺)𝑑(𝑣,𝜔)+𝑖𝑗=0(𝑑(𝑢,𝑣)+𝑗)×2+𝑛−𝑖𝑗=𝑖+1(𝑑(𝑢,𝑣)+𝑗)=𝑛−2𝑖+1>0,(2.2) and thus, the inequality 𝐴𝑖−1>𝐴𝑖 holds.
Note that for any 𝑖≤𝑘≤𝑛−𝑖, ec𝐺𝑖−1(𝑣𝑘)≥ec𝐺𝑖(𝑣𝑘), 𝐷𝐺𝑖−1(𝑣𝑘)>𝐷𝐺𝑖(𝑣𝑘), and thus 𝐶𝑖−1>𝐶𝑖. So, it suffices to prove 𝐵𝑖−1>𝐵𝑖.
Let ec𝐺(𝑢)=𝑠 and 𝑛1=|𝑉(𝐺)|, and we distinguish the following two cases:
Case 1 (𝑠≥𝑛−𝑖+1). By direct calculation, it follows that 𝐵𝑖−1=𝑖−1𝑗=0ec𝐺𝑖−1𝑣𝑗𝐷𝐺𝑖−1𝑣𝑗+ec𝐺𝑖−1𝑣𝑛−𝑗𝐷𝐺𝑖−1𝑣𝑛−𝑗=𝑖−1𝑗=0⎧⎪⎨⎪⎩ec𝐺𝑖𝑣𝑗⎡⎢⎢⎣−1𝑃𝑣∈𝑉𝑛+1𝑑𝑣𝑗+,𝑣𝑣∈𝑉(𝐺)⧵{𝑢}𝑑𝐺𝑖𝑣𝑗,𝑢−1+𝑑𝐺⎤⎥⎥⎦+(𝑢,𝑣)ec𝐺𝑖𝑣𝑛−𝑗⎡⎢⎢⎣+1𝑃𝑣∈𝑉𝑛+1𝑑𝑣𝑛−𝑗+,𝑣𝑣∈𝑉(𝐺)⧵{𝑢}𝑑𝐺𝑖𝑣𝑛−𝑗,𝑢+1+𝑑𝐺⎤⎥⎥⎦⎫⎪⎬⎪⎭(𝑢,𝑣)=𝐵𝑖+𝑖−1𝑗=0⎧⎪⎨⎪⎩𝑛1−1ec𝐺𝑖𝑣𝑛−𝑗−ec𝐺𝑖𝑣𝑗+𝑃𝑣∈𝑉𝑛+1𝑑𝑣𝑛−𝑗𝑣,𝑣−𝑑𝑗+𝑛,𝑣1𝑑𝑣−1𝑛−𝑗𝑣,𝑢−𝑑𝑗𝑛,𝑢+21⎫⎪⎬⎪⎭.−1(2.3) For 0≤𝑗≤𝑖−1, it is easily seen that ec𝐺𝑖(𝑣𝑛−𝑗)>ec𝐺𝑖(𝑣𝑗) and 𝑑(𝑣𝑛−𝑗,𝑢)≥𝑑(𝑣𝑗,𝑢). Furthermore, ∑𝑣∈𝑉(𝑃𝑛+1)[𝑑(𝑣𝑛−𝑗,𝑣)−𝑑(𝑣𝑗,𝑣)]=0, and thus, 𝐵𝑖−1>𝐵𝑖.
Case 2 (𝑠≤𝑛−𝑖). For 0≤𝑗≤𝑖−1, it is easily seen that ec𝐺𝑖−1(𝑣𝑗)=ec𝐺𝑖(𝑣𝑗), ec𝐺𝑖−1(𝑣𝑛−𝑗)≥ec𝐺𝑖(𝑣𝑛−𝑗). Then, 𝐵𝑖−1=𝑖−1𝑗=0ec𝐺𝑖−1𝑣𝑗𝐷𝐺𝑖−1𝑣𝑗+ec𝐺𝑖−1𝑣𝑛−𝑗𝐷𝐺𝑖−1𝑣𝑛−𝑗≥𝑖−1𝑗=0ec𝐺𝑖𝑣𝑗𝐷𝐺𝑖−1𝑣𝑗+ec𝐺𝑖𝑣𝑛−𝑗𝐷𝐺𝑖−1𝑣𝑛−𝑗=𝑖−1𝑗=0⎧⎪⎨⎪⎩ec𝐺𝑖𝑣𝑗⎡⎢⎢⎣𝑃𝑣∈𝑉𝑛+1𝑑𝑣𝑗+,𝑣𝑣∈𝑉(𝐺)⧵{𝑢}𝑑𝐺𝑖𝑣𝑗,𝑢−1+𝑑𝐺⎤⎥⎥⎦(𝑢,𝑣)+ec𝐺𝑖𝑣𝑛−𝑗⎡⎢⎢⎣𝑃𝑣∈𝑉𝑛+1𝑑𝑣𝑛−𝑗+,𝑣𝑣∈𝑉(𝐺)⧵{𝑢}𝑑𝐺𝑖𝑣𝑛−𝑗,𝑢+1+𝑑𝐺⎤⎥⎥⎦⎫⎪⎬⎪⎭(𝑢,𝑣)=𝐵𝑖+𝑛1−1𝑖−1𝑗=0ec𝐺𝑖𝑣𝑛−𝑗−ec𝐺𝑖𝑣𝑗>𝐵𝑖,(2.4) where the last inequality follows as ec𝐺𝑖(𝑣𝑛−𝑗)>ec𝐺𝑖(𝑣𝑗) for 0≤𝑗≤𝑖−1.

By Lemma 2.1 the inequality 𝜉𝑑(𝐺𝑖)<𝜉𝑑(𝐺0) follows easily. Note that 𝐺𝑖 has the number of pendent vertices greater than that of 𝐺0. Let 𝑇 be a tree with at least 𝑖 pendent vertices, where 3≤𝑖≤𝑛−1, by applying the above transformation to 𝑇 repeatedly, then we can obtain a new tree with exactly 𝑖−1 pendent vertices, which has larger eccentric distance sum. So it is easy to prove the following result, which is also obtained by Ilić et al. in [11].

Theorem 2.2. Among all trees with 𝑛 vertices, 𝑃𝑛 has the maximal eccentric distance sum.

Let 𝑇𝑛,𝑖 be the tree obtained from 𝑃𝑛−1=𝑣0𝑣1⋯𝑣𝑛−2 by attaching a pendent vertex 𝑣𝑛−1 to 𝑣𝑖, where 1≤𝑖≤⌊(𝑛−2)/2⌋.

Lemma 2.3. Let 𝑛≥8. Then, 𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛,2>2𝑛−4,𝜉𝑑𝑇𝑛,2−𝜉𝑑𝑇𝑛,3>4𝑛−10.(2.5)

Proof. Suppose that 𝑇𝑛,𝑖 has the same vertex labeling as the above definition. We have two cases based on the parity of 𝑛. If 𝑛≥8 is even, then 𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛,2=𝑛−1𝑖=0ec𝑇𝑛,1𝑣𝑖𝐷𝑇𝑛,1𝑣𝑖−ec𝑇𝑛,2𝑣𝑖𝐷𝑇𝑛,2𝑣𝑖𝑛=(𝑛−2)⋅(−1)+(𝑛−3)⋅(−1)+(𝑛−4)⋅1+⋯+2+𝑛⋅12𝑛−1⋅1+2+1⋅1+⋯+(𝑛−3)⋅1(𝑛−2)⋅1+23𝑛2=1−19𝑛+3649𝑛29−62𝑛+116=2(𝑛−2)8𝑛−448𝜉+7>2(𝑛−2),𝑑𝑇𝑛,2−𝜉𝑑𝑇𝑛,3=𝑛−1𝑖=0ec𝑇𝑛,2𝑣𝑖𝐷𝑇𝑛,2𝑣𝑖−ec𝑇𝑛,3𝑣𝑖𝐷𝑇𝑛,3𝑣𝑖=𝑛(𝑛−2)⋅(−1)+(𝑛−3)⋅(−1)+(𝑛−4)⋅(−1)+(𝑛−5)⋅1+⋯+2𝑛⋅1+2𝑛−1⋅1+2+1⋅1+⋯+(𝑛−3)⋅1+(𝑛−2)⋅123𝑛2=1−27𝑛+7249𝑛29−86𝑛+220=(4𝑛−10)16𝑛−127+3222716>4𝑛−10.(2.6)
If 𝑛≥9 is odd, then we make a similar calculation as above and obtain that 𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛,29=2(𝑛−2)8𝑛−448+294𝜉>2(𝑛−2),𝑑𝑇𝑛,2−𝜉𝑑𝑇𝑛,39=(4𝑛−10)16𝑛−143+3214916>4𝑛−10.(2.7) These complete the proof.

Lemma 2.4. Among all trees on 𝑛 vertices, where 𝑛≥8, 𝑇𝑛,3 has the maximal eccentric distance sum except 𝑃𝑛, 𝑇𝑛,1, and 𝑇𝑛,2.

Proof. suppose that 𝑇 is an 𝑛-vertex tree different from 𝑃𝑛, 𝑇𝑛,1, and 𝑇𝑛,2. Let 𝑝 be the number of pendent vertices of 𝑇, then 𝑝≥3.
If 𝑝=3, then 𝑇 is is a tree obtained by identifying three pendent vertices of three paths. Denote by 𝑟,𝑠, and 𝑡, respectively, the lengths of the three paths, where 𝑟≥𝑠≥𝑡≥1 and 𝑟+𝑠+𝑡+1=𝑛. Here, we denote it by 𝑇𝑛(𝑟,𝑠,𝑡). Clearly, 𝑇𝑛(𝑛−𝑖−2,𝑖,1)≅𝑇𝑛,𝑖(1≤𝑖≤⌊(𝑛−2)/2⌋). Suppose first that 𝑡=1. Then, 𝑇=𝑇𝑛,𝑖 with 𝑖≥3. For 3≤𝑖≤⌊(𝑛−4)/2⌋, by Lemma 2.1, we have 𝜉𝑑(𝑇𝑛,𝑖+1)<𝜉𝑑(𝑇𝑛,𝑖), and thus, 𝜉𝑑(𝑇)≤𝜉𝑑(𝑇𝑛,3) with equality if and only if 𝑇≅𝑇𝑛,3. Now suppose that 𝑡≥2, then 𝑟≥3 for 𝑛≥8, and by Lemma 2.1, we can obtain a new tree 𝑇𝑛(𝑟,𝑠+𝑡−1,1) or 𝑇𝑛(𝑠+𝑡−1,𝑟,1), which is not isomorphic to 𝑇𝑛,1, 𝑇𝑛,2, and 𝑇𝑛,3 and has larger eccentric distance sum.
If 𝑝≥5 by applying transformation of Lemma 2.1 to 𝑇 repeatedly, we can obtain a new tree with exactly four pendant vertices and larger eccentric distance sum. Thus, it suffices to consider the case 𝑝=4.
Now, suppose that 𝑝=4. In this case, 𝑇 has at most two vertices with degree more than 2.
Case 1. If 𝑇 has exactly two vertices with degree more than 2, say 𝑢 and 𝑣, then 𝑑(𝑢)=𝑑(𝑣)=3. Suppose that the length of path connecting 𝑢 and 𝑣 is 𝑎, the lengths of pendent paths at 𝑢 are 𝑏,𝑐, and the lengths of pendent paths at 𝑣 are 𝑑, 𝑓. We denote this tree by 𝑇𝑛(𝑎;𝑏,𝑐;𝑑,𝑓), where 𝑎≥1, 𝑏≥𝑐, 𝑑≥𝑓 and 𝑎+𝑏+𝑐+𝑑+𝑓+1=𝑛. If 𝑏≥3, then 𝑏,𝑑+𝑓+𝑎≥3, and by Lemma 2.1 and above proof, we have 𝜉𝑑(𝑇𝑛(𝑎;𝑏,𝑐;𝑑,𝑓))<𝜉𝑑(𝑇𝑛(𝑏,𝑑+𝑓+𝑎,𝑐))<𝜉𝑑(𝑇𝑛,3), where we suppose that 𝑏≥𝑑+𝑓+𝑎≥𝑐. If 𝑏=𝑐=2, then by Lemma 2.1 and the result above we have 𝜉𝑑(𝑇𝑛(𝑎;2,2;𝑑,𝑓))<𝜉𝑑(𝑇𝑛(𝑎;3,1;𝑑,𝑓))<𝜉𝑑(𝑇𝑛,3). If 𝑏=2, 𝑐=1 or 𝑏=1, 𝑐=1. Applying similar proof of Lemma 2.3, it is easily proven that 𝜉𝑑(𝑇𝑛(𝑛−6;1,1;2,1))−𝜉𝑑(𝑇𝑛(𝑛−7;2,1;2,1))>0, 𝜉𝑑(𝑇𝑛(𝑛−5;1,1;1,1))−𝜉𝑑(𝑇𝑛(𝑛−6;1,1;2,1))>0, and thus, we only need to prove that 𝜉𝑑(𝑇𝑛(𝑛−5;1,1;1,1))<𝜉𝑑(𝑇𝑛,3).
Here, we write 𝑇𝑛 instead of 𝑇𝑛(𝑛−5;1,1;1,1). Then, 𝑇𝑛 is a tree obtained from the path 𝑣0𝑣1⋯𝑣𝑛−4𝑣𝑛−3 by joining an isolated vertex 𝑣𝑛−2 to 𝑣𝑛−4 and an isolated vertex 𝑣𝑛−1 to 𝑣1. Let 𝑇𝑛,1 be a tree obtained from 𝑇𝑛 by deleting the edge 𝑣1𝑣𝑛−1 and adding the edge 𝑣0𝑣𝑛−1. If 𝑛≥8 is odd, then ec𝑇𝑛,1𝑣0𝐷𝑇𝑛,1𝑣0−ec𝑇𝑛𝑣0𝐷𝑇𝑛𝑣0=−(𝑛−3),ec𝑇𝑛,1𝑣1𝐷𝑇𝑛,1𝑣1−ec𝑇𝑛𝑣1𝐷𝑇𝑛𝑣1=𝑛−4,ec𝑇𝑛,1𝑣𝑛−1𝐷𝑇𝑛,1𝑣𝑛−1−ec𝑇𝑛𝑣𝑛−1𝐷𝑇𝑛𝑣𝑛−1=𝐷(𝑛−2)𝑇𝑛𝑣𝑛−1−+𝑛−3(𝑛−3)𝐷𝑇𝑛𝑣𝑛−1=3𝑛2−13𝑛+162,ec𝑇𝑛,1𝑣𝑛−3𝐷𝑇𝑛,1𝑣𝑛−3−ec𝑇𝑛𝑣𝑛−3𝐷𝑇𝑛𝑣𝑛−3=ec𝑇𝑛,1𝑣𝑛−2𝐷𝑇𝑛,1𝑣𝑛−2−ec𝑇𝑛𝑣𝑛−2𝐷𝑇𝑛𝑣𝑛−2𝐷=(𝑛−2)𝑇𝑛𝑣𝑛−2+1−(𝑛−3)𝐷𝑇𝑛𝑣𝑛−2=𝑛2−𝑛2,ec𝑇𝑛,1𝑣(𝑛−3)/2𝐷𝑇𝑛,1𝑣(𝑛−3)/2−ec𝑇𝑛𝑣(𝑛−3)/2𝐷𝑇𝑛𝑣(𝑛−3)/2=𝑛−32𝐷+1𝑇𝑛𝑣(𝑛−3)/2−+1𝑛−32𝐷𝑇𝑛𝑣(𝑛−3)/2=𝑛2+2𝑛−114,(2.8) for 2≤𝑖≤(𝑛−5)/2, ec𝑇𝑛,1(𝑣𝑖)𝐷𝑇𝑛,1(𝑣𝑖)−ec𝑇𝑛(𝑣𝑖)𝐷𝑇𝑛(𝑣𝑖)=𝑛−3−𝑖; for (𝑛−1)/2≤𝑖≤𝑛−4, ec𝑇𝑛,1𝑣𝑖𝐷𝑇𝑛,1𝑣𝑖−ec𝑇𝑛𝑣𝑖𝐷𝑇𝑛𝑣𝑖=𝐷(𝑖+1)𝑇𝑛𝑣𝑖+1−𝑖𝐷𝑇𝑛𝑣𝑖=𝑖2𝑛−(𝑛−4)𝑖+2−3𝑛+22.(2.9)
It follows that 𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛=(𝑛−5;1,1;1,1)5𝑛2−15𝑛+142+(𝑛−5)/2𝑖=2𝑛(𝑛−3−𝑖)+2+2𝑛−114+𝑛−4𝑖=(𝑛−1)/2𝑖2𝑛−(𝑛−4)𝑖+2−3𝑛+22=25𝑛2−48𝑛+1118+(𝑛−5)(𝑛+3)(2𝑛−5)=124𝑛3+57𝑛2−304𝑛+337>2418𝑛2−148𝑛+3374=𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛,3.(2.10) Thus, 𝜉𝑑(𝑇𝑛,3)>𝜉𝑑(𝑇𝑛(𝑛−5;1,1;1,1)).
Similarly, for even 𝑛, we have 𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛=(𝑛−5;1,1;1,1)5𝑛2−15𝑛+142+𝑛/2−2𝑖=2(𝑛−3−𝑖)+𝑛−4𝑖=𝑛/2−1𝑖2𝑛−(𝑛−4)𝑖+2−3𝑛+22>18𝑛2−148𝑛+2564=𝜉𝑑𝑇𝑛,1−𝜉𝑑𝑇𝑛,3,(2.11) and the inequality 𝜉𝑑(𝑇𝑛,3)>𝜉𝑑(𝑇𝑛(𝑛−5;1,1;1,1)) also holds.
Case 2. If 𝑇 has an unique vertex of degree greater than 2, say 𝑤, then 𝑑(𝑤)=4, and 𝑇 is a tree obtained by identifying four pendent vertices of four paths. Denote by 𝑇𝑛(𝑟,𝑠,𝑡,𝑙) this tree, where 𝑟,𝑠,𝑡,𝑙 are the lengths of the four pendent paths respectively and 𝑟≥𝑠≥𝑡≥𝑙≥1, 𝑟+𝑠+𝑡+𝑙+1=𝑛. If 𝑡=2, by Lemma 2.1, we have 𝜉𝑑(𝑇𝑛(𝑟,𝑠,𝑡,𝑙))<𝜉𝑑(𝑇𝑛(𝑟+1,𝑠+1,𝑙))<𝜉𝑑(𝑇𝑛,3). Similarly, we can show the result for 𝑡=𝑙=1 and 𝑟≥2. In the following, we will prove that 𝜉𝑑(𝑇𝑛(𝑛−4,1,1,1))<𝜉𝑑(𝑇𝑛,3). We label the vertices of 𝑇𝑛(𝑛−4,1,1,1) and 𝑇𝑛,3 such that 𝑇𝑛(𝑛−4,1,1,1) can be viewed as obtained from the path 𝑣0𝑣1⋯𝑣𝑛−4 by joining three isolated vertices 𝑣𝑛−3, 𝑣𝑛−2 and 𝑣𝑛−1 to 𝑣𝑛−4, and 𝑇𝑛,3 can be viewed as obtained from the path 𝑣0𝑣1⋯𝑣𝑛−2 by joining an isolated vertex 𝑣𝑛−1 to 𝑣𝑛−5. Clearly, ec𝑇𝑛,3(𝑣𝑖)≥ec𝑇𝑛(𝑛−4,1,1,1)(𝑣𝑖) for 0≤𝑖≤𝑛−3, 𝐷𝑇𝑛,3(𝑣𝑖)=𝐷𝑇𝑛(𝑛−4,1,1,1)(𝑣𝑖) for 0≤𝑖≤𝑛−5 and 𝑖=𝑛−3, and 𝐷𝑇𝑛,3(𝑣𝑛−4)=𝐷𝑇𝑛(𝑛−4,1,1,1)(𝑣𝑛−4)+2. Note that ec𝑇𝑛,3𝑣𝑛−2𝐷𝑇𝑛,3𝑣𝑛−2−ec𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−2𝐷𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−2𝐷=(𝑛−2)𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−2+𝑛−2−(𝑛−3)𝐷𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−21=(𝑛−2)(𝑛−2)+2=(𝑛−2)(𝑛−3)+43𝑛2−13𝑛+222,ec𝑇𝑛,3𝑣𝑛−1𝐷𝑇𝑛,3𝑣𝑛−1−ec𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−1𝐷𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−1=𝐷(𝑛−4)𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−2−−𝑛+8(𝑛−3)𝐷𝑇𝑛(𝑛−4,1,1,1)𝑣𝑛−21=−(𝑛−2)(𝑛−8)−2(𝑛−2)(𝑛−3)−4=−3𝑛2−25𝑛+462,(2.12) then 𝜉𝑑𝑇𝑛,3−𝜉𝑑𝑇𝑛>(𝑛−4,1,1,1)3𝑛2−13𝑛+222−3𝑛2−25𝑛+462>0,(2.13) and this completes the proof.

From Lemmas 2.1 and 2.4, We have the following

Theorem 2.5. If 𝑛≥8, then 𝑇𝑛,1, 𝑇𝑛,2, and 𝑇𝑛,3 are the unique trees with the second-maximal, third-maximal, and fourth-maximal eccentric distance sums among the trees on 𝑛 vertices.

3. The Unicyclic Graphs with Maximal Eccentric Distance Sums

Let 𝑈𝑛 be the graph obtained from a path 𝑃𝑛−1=𝑣0𝑣1⋯𝑣𝑛−3𝑣𝑛−2 by joining the vertex 𝑣𝑛−1 to 𝑣𝑛−3 and 𝑣𝑛−2. Let 𝑄𝑛 be the graph obtained from a path 𝑃𝑛−1=𝑣0𝑣1⋯𝑣𝑛−3𝑣𝑛−2 by joining the vertex 𝑣𝑛−1 to 𝑣𝑛−4 and 𝑣𝑛−2, and 𝐵𝑛 be obtained by joining 𝑣𝑛−1 to 𝑣𝑛−4 and 𝑣𝑛−3.

Theorem 3.1. Let 𝐺 be a graph with 𝑛 vertices and 𝑛 edges; that is, 𝐺 is an unicyclic graph, where 𝑛≥8. Then, 𝜉𝑑(𝐺)≤𝜉𝑑(𝑈𝑛), with equality if and only if 𝐺≅𝑈𝑛.

Proof. If 𝐺≇𝐶𝑛 and 𝐺≇𝑈𝑛, then we can always find an edge 𝑒 of 𝐺 such that 𝐺−𝑒 is a tree with at least three pendent vertices and 𝐺−𝑒≇𝑇𝑛,1, 𝑃𝑛. It follows from Lemma 2.3 that 𝜉𝑑(𝐺)<𝜉𝑑(𝐺−𝑒)≤𝜉𝑑𝑇𝑛,2<𝜉𝑑𝑇𝑛,1−2(𝑛−2)=𝜉𝑑𝑈𝑛.(3.1)
Suppose that 𝐺≅𝐶𝑛 and 𝑣 is any vertex of 𝐶𝑛. If 𝑛≥8 is even, then 𝑛ec(𝑣)𝐷(𝑣)=2𝑛1+2+⋯+2𝑛−1×2+2=𝑛38.(3.2) By the definition of EDS, we have 𝜉𝑑𝑈𝑛=𝑣ec0𝐷𝑣0𝑣+ec𝑛/2−1𝐷𝑣𝑛/2−1+𝑣ec𝑛/2−2𝐷𝑣𝑛/2−2𝑣+ec𝑛/2𝐷𝑣𝑛/2+𝑖≠0,𝑛/2−1,𝑛/2−2,𝑛/2𝑣ec𝑖𝐷𝑣𝑖.(3.3) It is easily checked that 𝑣ec0𝐷𝑣0𝑣+ec𝑛/2−1𝐷𝑣𝑛/2−1>𝑛34,𝑣ec𝑛/2−2𝐷𝑣𝑛/2−2𝑣+ec𝑛/2𝐷𝑣𝑛/2=𝑛34,(3.4) and for 𝑖≠0, 𝑛/2−1, 𝑛/2−2, 𝑛/2, ec(𝑣𝑖)≥𝑛/2+1, 𝐷(𝑣𝑖)>𝑛2/4. Then, 𝜉𝑑𝑈𝑛>𝑛34+𝑛34𝑛+(𝑛−4)2𝑛+124=𝑛48+38𝑛2𝑛(𝑛−2)>48=𝜉𝑑𝐶𝑛.(3.5) We can prove the result for odd number 𝑛 similarly, and thus complete the proof.

Theorem 3.2. Let 𝑛≥8. Then, 𝐵𝑛, 𝑄𝑛 are the graph with, respectively, the second-maximal and third-maximal eccentric distance sums among all unicyclic graphs on 𝑛 vertices.

Proof. If 𝐺 has a spanning tree 𝑇 such that 𝑇≇𝑃𝑛,𝑇𝑛,1,𝑇𝑛,2, then 𝜉𝑑(𝐺)<𝜉𝑑(𝑇)≤𝜉𝑑𝑇𝑛,3<𝜉𝑑𝑇𝑛,2−4𝑛+10=𝜉𝑑𝑄𝑛.(3.6) Now suppose that any spanning tree of 𝐺 is one of {𝑃𝑛,𝑇𝑛,1,𝑇𝑛,2}. Then, 𝐺 must be isomorphic to 𝐶𝑛, 𝑄𝑛 or 𝐵𝑛. It can be proven easily that 𝜉𝑑(𝐶𝑛)<𝜉𝑑(𝑄𝑛) by similar proof of Theorem 3.1. And by directed computation, we obtain that 𝜉𝑑(𝐵𝑛)=𝜉𝑑(𝑄𝑛)+1. Thus, the result follows.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (Grant no. 11001089), and by the Foundation for Distinguished Young Talents in Higher Education of Guangdong, China (Grant no. (2009) 400). The authors thank the referees for valuable comments and suggestions.