Wiener Index of Graphs with Radius Two

Abstract

The Wiener index of a graph is the sum of the distances between all pairs of vertices. It has been one of main descriptors that correlate a chemical compound's molecular graph with experimentally gathered data regarding the compound's characteristics. We characterize graphs with the maximum Wiener index among all graphs of order 𝑛. with radius two. In addition, we pose a conjecture concerning the minimum Wiener index of graphs with given radius. If this conjecture is true, it will be able to answer an open question by You and Liu (2011).

1. Introduction

Let 𝐺=(𝑉,𝐸) be a connected graph. For two vertices 𝑢,𝑣∈𝑉, the distance 𝑑_𝐺(𝑢,𝑣) between 𝑢 and 𝑣 in 𝐺 is the length of the shortest path connecting 𝑢 and 𝑣 in 𝐺. The eccentricity ecc(𝑣) of a vertex 𝑣 in 𝐺 is the largest distance from 𝑣 to another vertex of 𝐺; that is, max{𝑑(𝑣,𝑤)∣𝑤∈𝑉(𝐺)}. The diameter of 𝐺 is the maximum eccentricity in 𝐺, denoted by diam(𝐺). Similarly, the radius of 𝐺 is the minimum eccentricity in 𝐺, denoted by rad(𝐺). The Wiener index or total  distance of 𝐺, denoted by 𝑊=𝑊(𝐺), is the sum of all distances between unordered pairs of distinct vertices of 𝐺. In other words, rad(𝐺)=min_𝑣∈𝑉{(𝑣,(𝐺)ecc)}diam=max_𝑣∈𝑉{(𝑣,𝑊(𝐺)ecc)}=_{𝑢,𝑣∈𝑉}𝑑(𝑢,𝑣). (1)

The average distance 𝜇(𝐺) of 𝐺 is defined as 𝜇(𝐺)=𝑊(𝐺)/(^𝑛₂). A vertex 𝑣∈𝑉(𝐺) is called a center of 𝐺 if ecc(𝑣)=rad(𝐺). It is well known that every tree has either exactly one center or two adjacent centers. For the results on Wiener index of trees, we refer to Dobrynin et al. [1], and for some more results on Wiener index of graphs, one can see [2–12].

Plesník [13] addresses a problem on distance of graphs, which remains unresolved.

Problem 1. What is the maximum total or average distance among all graphs of order 𝑛 with diameter 𝑑?

To see how hard to solve Problem 1, one just consider the following conjecture, due to DeLaViña and Waller [14].

Conjecture 1 (see [14]). Let G be a graph with diameter 𝑑>2 and order 2d + 1. Then 𝑊(𝐺)≤𝑊(𝐶_2𝑑+1), where 𝐶_2𝑑+1 denotes the cycle of length 2𝑑+1.

For any connected graph 𝐺, rad(𝐺)≤diam(𝐺)≤2rad(𝐺). By considering the close relationship between the diameter and the radius of a graph, it is natural to consider the following problem.

Problem 2. What is the maximum total or average distance among all graphs of order 𝑛 with radius 𝑟?

By Lemma 3 in Section 2, Problem 2 is more tractable than Problem 1; the graph with maximum total distance among all graphs of order 𝑛 with radius 𝑟 must be a tree. But, Problem 2 still seems to be quite challenging. The main focus of this paper is to tackle Problem 2 when the case 𝑟=2. We will show that the graph with the maximum Wiener index among all connected graphs with radius two is a tree with somewhat fractal-like structure when the number of vertices of graphs goes to infinity. It provides some clues for the further investigation of Problem 2 for the case 𝑟≥3.

The following problem remains open, see You and Liu [10].

Problem 3. What is the minimum total or average distance among all graphs of order 𝑛 with radius 𝑟?

It is easy to see that the complete graph 𝐾_𝑛 has the minimum Wiener index among all connected graphs of order 𝑛 (and so among all graphs of order 𝑛 with radius one). For the case of radius two, it is not hard to prove that the extremal graph 𝐺_𝑛 of Problem 3 should be the following: 𝐺_𝑛≅𝐾_𝑛⧵𝑀, if 𝑛 is even, where 𝑀 is a perfect matching; 𝐺_𝑛≅𝐾_𝑛⧵(𝐹∪{𝑒}), if 𝑛 is odd, where 𝐹 is a maximum matching of 𝐾_𝑛 and 𝑒 is an edge incident with the vertex of degree 𝑛−1 in 𝐾_𝑛⧵𝐹.

We denote by 𝐺_{𝑛,𝑟,𝑠} the graph obtained in the following way, where 𝑛 and 𝑟 are three positive integers with 𝑛≥2𝑟, 𝑟≥3, and 𝑛−2𝑟+1≥𝑠≥1: let 𝑢,𝑣,𝑥, and 𝑦 be four consecutive vertices on 𝐶_2𝑟; replace 𝑣 with a clique 𝐾_𝑠 of order 𝑠, replace 𝑥 with a clique of order 𝑛−2𝑟+2−𝑠, join each vertex of a clique to all vertices of the other clique, join 𝑢 to the all vertices of the 𝐾_𝑠, and join 𝑦 to all vertices of 𝐾_{𝑛−2𝑟+2−𝑠}. It is clear that 𝐺_{𝑛,𝑟,𝑠} is a graph of order 𝑛 with radius 𝑟, and 𝑊(𝐺_{𝑛,𝑟,𝑠})=𝑊(𝐺_{𝑛,𝑟,𝑠′}) for any 𝑠,𝑠^′∈{1,…,𝑟−1}.

Conjecture 2. Let 𝑛 and 𝑟 be two positive integers with 𝑛≥2𝑟 and 𝑟≥3. For any graph 𝐺 of order 𝑛 with radius 𝑟, 𝑊(𝐺)≥𝑊(𝐺_𝑛,𝑟,1), with equality if and only if 𝐺≅𝐺_{𝑛,𝑟,𝑠} for 𝑠∈{1,…,𝑟−1}.

2. Definitions and Lemmas

Lemma 3. For any connected graph 𝐺, there exists a spanning tree 𝑇 of 𝐺, such that 𝑟𝑎𝑑(𝑇)=𝑟𝑎𝑑(𝐺) and 𝑊(𝑇)≥𝑊(𝐺).

Proof. Let 𝑢 be a center of 𝐺 and 𝑇 be a BFS-tree of 𝐺 rooted at 𝑢. The tree 𝑇 is one, as we desired.

The path number 𝑝(𝐺) of a graph 𝐺 is defined to be the maximum cardinality of a subset of vertices that induces a path. Erdős et al. [15] showed that for any connected graph 𝐺, 𝑝(𝐺)≥2rad(𝐺)−1.

Lemma 4. For a tree 𝐺 with radius 𝑟, 2𝑟𝑎𝑑(𝐺)≤𝑝(𝐺)≤2𝑟𝑎𝑑(𝐺)+1.

Proof. Let 𝑃_𝑢𝑣 be the longest path of 𝐺. Since 𝐺 is a tree, 𝑝(𝐺)=|𝑉(𝑃)|. It is clear that a center 𝑤 of 𝐺 must be sitting on 𝑃_𝑢𝑣, min{𝑑_𝐺(𝑤𝑢),𝑑_𝐺(𝑤𝑣)}≥𝑟−1 and max{𝑑_𝐺(𝑤𝑢),𝑑_𝐺(𝑤𝑣)}≥𝑟. The result then follows.

Lemma 5. Let 𝑇 be a tree of order 𝑛 with the maximum Wiener index among all trees with radius two. If 𝑛=4, then 𝑇≅𝑃₄, and 𝑝(𝑇)=5, otherwise.

Proof. If 𝑛=4, the result is trivial. So, let 𝑛≥5. By Lemma 4, 4≤𝑝(𝑇)≤5. By contradiction, suppose that 𝑝(𝑇)=4, and let 𝑃=𝑣₁𝑣₂𝑣₃𝑣₄ be a path on four vertices in 𝑇. Every other vertex not on 𝑃 must be adjacent to 𝑣₂ or 𝑣₃. Without loss of generality, let 𝑑_𝑇(𝑣₂)≤𝑑_𝑇(𝑣₃), and let 𝑥 be a pendent vertex adjacent to 𝑣₃. Set 𝑇^′=𝑇−𝑣₂𝑣₃+𝑣₂𝑥. It is easy to check that 𝑊(𝑇^′)>𝑊(𝑇), contradicting the choice of 𝑇.

Lemma 6. Let 𝑇 be a tree. Then(1)the center of 𝑇 is a vertex or two adjacent vertices,(2)the center of 𝑇 is one vertex if and only if 𝑑𝑖𝑎𝑚(𝑇)=2𝑟𝑎𝑑(𝑇).

The above two lemmas assert that if 𝑇 is a tree of order 𝑛 with the maximum Wiener index among all trees of order at least five with radius two, then 𝑇 has the unique center. We denote by 𝑇_𝑛,𝑡(𝑛₁,…,𝑛_𝑡) the tree of order 𝑛 with radius two and unique center 𝑣_𝑐 such that 𝑑(𝑣_𝑐)=𝑡, 𝑁(𝑣_𝑐)={𝑣₁,…,𝑣_𝑡}, and 𝑑(𝑣_𝑖)=𝑛_𝑖+1 for each 𝑖. In particular, 𝑆(𝑛,𝑡)=𝑇_𝑛,𝑡(𝑛₁,…,𝑛_𝑡), with |𝑛_𝑖−𝑛_𝑗|≤1 for any 𝑖,𝑗.

Lemma 7. Let 𝑇=𝑇_𝑛,𝑡(𝑛₁,…,𝑛_𝑡) be the tree as described above. Then 𝑊(𝑇)=2𝑛²−5𝑛+3−(𝑛−2)𝑡−^𝑡_𝑖=1𝑛²_𝑖. (2)

Proof. Let 𝑥_𝑖=|{(𝑢,𝑣)∣𝑢,𝑣∈𝑉(𝑇),𝑑(𝑢,𝑣)=𝑖}|. Hence 𝑊(𝑇)=∑⁴_𝑖=1𝑖𝑥_𝑖, where 𝑥₁𝑥=𝑛−1,₂=()+𝑡2𝑛−𝑡−1+^𝑡_𝑖=1𝑛_𝑖2,𝑥₃=^𝑡_𝑖=1𝑛−𝑡−1−𝑛_𝑖=(𝑡−1)^𝑡_𝑖=1𝑛_𝑖,𝑥₄=^𝑡_𝑖=1𝑛_𝑖𝑛−𝑡−1−𝑛_𝑖2. (3) By simple calculation, 𝑊(𝑇)=(2𝑛²−5𝑛+3)−(𝑛−2)𝑡−∑^𝑡_𝑖=1𝑛²_𝑖.

The trees 𝑆(𝑛,𝑡) and 𝑆(𝑛,𝑡+1) when 𝑛=1+𝑡+𝑡² are illustrated in Figure 1. We will see that 𝑊(𝑆(𝑛,𝑡))=𝑊(𝑆(𝑛,𝑡+1)) if 𝑛=1+𝑡+𝑡² in the next corollary. Note that if 𝑛=1+𝑡+𝑡², each vertex in 𝑆(𝑛,𝑡), not a pendent vertex, has 𝑡 children. In this sense, 𝑆(𝑛,𝑡) has a fractal-like structure. Our main goal is to prove that if 𝑇 is a tree of order 𝑛 with the maximum Wiener index among all trees with radius two, then 𝑇≅𝑆(𝑛,𝑡₀), where 𝑡₀ can be determined. It will be shown in Theorem 9. For smaller value of 𝑛, one can see that those extremal graphs with radius two in Figure 2, and there exists two extremal graphs when 𝑛=1+𝑡+𝑡² for some 𝑡≥2; for instance, see the graphs for 𝑡=2 (𝑛=7) and 𝑡=3 (𝑛=13) in Figure 2.

Figure 1 

The trees 𝑆(𝑛,𝑡) and 𝑆(𝑛,𝑡+1) when 𝑛=1+𝑡+𝑡2.

Figure 2 

The maximum wiener index with given radius two.

Corollary 8. Let 𝑛 and 𝑡 be two positive integers such that 𝑛≥𝑡+2 and 1+𝑡+𝑘𝑡≤𝑛<1+𝑡+(𝑘+1)𝑡 for some integer 𝑘≥0, and let 𝑝=𝑛−1−𝑡−𝑘𝑡. Then 𝑊(𝑆(𝑛,𝑡))=2𝑡²−𝑡𝑘²+3𝑡²+𝑘+𝑡4𝑝−1𝑡−2𝑝₂+3𝑡𝑝+2𝑝²−2𝑝, (4) and if 𝑘=𝑡 and 𝑝=0, then 𝑊(𝑆(𝑛,𝑡))=𝑊(𝑆(𝑛,𝑡+1))=2𝑡⁴+2𝑡³.

Proof. By the notation of the previous lemma, In 𝑆(𝑛,𝑡), the number of 𝑛^′_𝑖𝑠 equal to 𝑘+1 is 𝑝 and the number of 𝑛^′_𝑖𝑠 equal to 𝑘 is 𝑡−𝑝. So, by Lemma 7, 𝑊(𝑆(𝑛,𝑡))=2𝑛²−5𝑛+3−(𝑛−2)𝑡−^𝑡_𝑖=1𝑛²_𝑖=()−()𝑡−𝑝()2𝑛−3)(𝑛−1𝑛−2𝑘+1²+𝑡−𝑝𝑘²=−2𝑡+2𝑡𝑘+2𝑝−1𝑡+𝑡𝑘+𝑝𝑡+𝑡𝑘+𝑝−1𝑡−2𝑝𝑘+𝑝+𝑡𝑘²=2𝑡²−𝑡𝑘²+3𝑡²+𝑘+𝑡4𝑝−1𝑡−2𝑝₂+3𝑡𝑝+2𝑝²−2𝑝. (5) If 𝑘=𝑡 and 𝑝=0, by the above formula, 𝑊(𝑆(𝑛,𝑡))=2𝑡²−𝑡𝑡²+3𝑡²−𝑡𝑡+𝑡²=2𝑡⁴+2𝑡³. (6) Moreover, since 1+𝑡+𝑡²=1+(𝑡+1)+(𝑡+1)(𝑡−1) and 𝑊(𝑆(𝑛,𝑡+1))=2(𝑡+1)²−(𝑡+1)(𝑡−1)²+3(𝑡+1)²−(𝑡+1)(𝑡−1)+(𝑡+1)²=2𝑡⁴+2𝑡³; (7)
this proves 𝑊(𝑆(𝑛,𝑡))=𝑊(𝑆(𝑛,𝑡+1))=2𝑡⁴+2𝑡³.

3. Main Results

For a clear understanding of our main theorem as follows, we make a remark. If 𝑛=1+𝑡₀+𝑡²₀ for some positive integer 𝑡₀, then 𝑡₀=(√4𝑛−3−1)/2. By Corollary 8, 𝑊(𝑆(𝑛,𝑡₀))=𝑊(𝑆(𝑛,𝑡₀+1)).

Theorem 9. Let 𝑇 be a tree with the maximum Wiener index among all trees of order 𝑛 with radius two. If 𝑛=1+𝑡₀+𝑡²₀ for some positive integer 𝑡₀, then 𝑇∈{𝑆(𝑛,(√4𝑛−3−1)/2),𝑆(𝑛,(√4𝑛−3+1)/2)}, and 𝑇≅𝑆(𝑛,𝑡₀) otherwise, where 𝑡₀=min{𝑖∣𝑛−1−𝑖−𝑖²<0}.

Proof. The result is trivial for 𝑛≤4, and assume that 𝑛≥5. By Lemma 5, let 𝑣_𝑐 be the unique center of 𝑇, and let 𝑑(𝑣_𝑐)=𝑡. Set 𝑁(𝑣_𝑐)={𝑣₁,…,𝑣_𝑡} and |𝑁(𝑣_𝑖)⧵{𝑣_𝑐}|=𝑛_𝑖 for 1≤𝑖≤𝑡. Lemma 7 implies that |𝑛_𝑖−𝑛_𝑗|≤1 for any 𝑖,𝑗, and thus 𝑇≅𝑆(𝑛,𝑡) for some 𝑡. To show 𝑡=𝑡₀, as specified in the theorem, it suffices to show that 𝑊(𝑆(𝑛,𝑡))<𝑊(𝑆(𝑛,𝑡₀)) for any 𝑡≠𝑡₀.
By Corollary 8, 𝑊(𝑆(𝑛,𝑡))=(2𝑛²−5𝑛+3)−(𝑛−2)𝑡−[𝑞(𝑘+1)²+(𝑡−𝑞)𝑘²], where 𝑞=𝑛−1−𝑡−𝑡⌊(𝑛−𝑡−1)/𝑡⌋, 0≤𝑞≤𝑡−1, we have 𝑊(𝑆(𝑛,𝑡))≤2𝑛²−()𝑡−()−5𝑛+3𝑛−2𝑛−𝑡−1²𝑡. (8)
Define a function 𝑓(𝑡)=(2𝑛²−5𝑛+3)−(𝑛−2)𝑡−(𝑛−𝑡−1)²/𝑡, which is an upper bound of 𝑊(𝑆(𝑛,𝑡)). Its derivative 𝑓^′(𝑡)=(𝑛−1)²/𝑡²−(𝑛−1) has the unique root 𝑡=√𝑛−1. Observe that √4𝑛−3−12<√√𝑛−1≤4𝑛−3+12, (9) and since 𝑡₀=min{𝑖∣𝑛−1−𝑖−𝑖²<0}, √4𝑛−3−12<𝑡₀≤√4𝑛−3+12. (10)
By the definition of 𝑡₀, 𝑛=1+𝑡₀+𝑡₀𝑘+𝑝, where 𝑡₀−2≤𝑘≤𝑡₀ and 0≤𝑝≤𝑡₀−1, and moreover, if 𝑘=𝑡₀, then 𝑝=0; if 𝑘=𝑡₀−2, then 𝑝>0.
Case  1 (𝑡∈[2,𝑡₀−1]). Since 𝑓^′(𝑡)>0for𝑡∈2,𝑡₀,𝑊(𝑆((𝑡)𝑡−1𝑛,𝑡))≤𝑓≤𝑓₀−1. (11)
Moreover, since 𝑊(𝑆(𝑛,𝑡₀))=(2𝑛²−5𝑛+3)−(𝑛−2)𝑡₀−(2𝑝𝑘+𝑝+𝑡₀𝑘²) and 𝑓(𝑡₀−1)=(2𝑛²−5𝑛+3)−(𝑛−2)𝑡₀+(𝑛−2)−(𝑛−𝑡₀)²/(𝑡₀−1), we have 𝑡₀−1𝑊𝑆𝑛,𝑡₀−𝑓𝑡₀=−1𝑛−𝑡₀²−2𝑝𝑘+𝑝+𝑡₀𝑘²𝑡₀−()𝑡−1𝑛−2₀=−11+𝑡₀𝑘+𝑝²−2𝑝𝑘+𝑝+𝑡₀𝑘²𝑡₀−𝑡−1₀+𝑡₀𝑘+𝑝−1𝑡₀−1=𝑝²+4+2𝑘−2𝑡₀𝑝+𝑡₀𝑘²+3𝑡₀𝑘+2𝑡₀−(𝑘+1)𝑡²₀≥𝑡₀𝑘²+3𝑡₀𝑘+2𝑡₀−(𝑘+1)𝑡²₀=𝑡₀(𝑘+1)𝑘+2−𝑡₀≥0. (12)
Thus, 𝑊(𝑆(𝑛,𝑡₀))≥𝑓(𝑡₀−1), and together with (11), we have 𝑊(𝑆(𝑛,𝑡₀))≥𝑊(𝑆(𝑛,𝑡)) for any 𝑡∈[2,𝑡₀−1]. Observe from the above equation that 𝑊(𝑆(𝑛,𝑡₀))=𝑊(𝑆(𝑛,𝑡)) for some 𝑡∈[2,𝑡₀−1] if and only if 𝑘=𝑡₀−2,𝑝=0. But, this contradicts the fact that if 𝑘=𝑡₀−2, then 𝑝=0. It follows that 𝑊(𝑆(𝑛,𝑡₀))>𝑊(𝑆(𝑛,𝑡)) for any 𝑡∈[2,𝑡₀−1].
Case  2 (𝑡∈[𝑡₀+1,𝑛−1)). Since 𝑓^′(𝑡)<0 for 𝑡∈[𝑡₀+1,𝑛), 𝑊(𝑆(𝑛,𝑡))≤𝑓(𝑡)≤𝑓(𝑡₀+1). Since 𝑓(𝑡₀+1)=(2𝑛²−5𝑛+3)−(𝑛−2)𝑡₀−(𝑛−2)−(𝑛−𝑡₀−2)²/(𝑡₀+1), we have 𝑡₀+1𝑊𝑆𝑛,𝑡₀−𝑓𝑡₀=+1𝑛−𝑡₀−2²+(𝑛−2)𝑡₀−+12𝑝𝑘+𝑝+𝑡₀𝑘²𝑡₀=𝑡+1₀𝑘+𝑝−1²+𝑡₀+𝑡₀𝑘+𝑝−1𝑡₀−+12𝑝𝑘+𝑝+𝑡₀𝑘²𝑡₀+1=𝑝²−(2+2𝑘)𝑝+𝑡₀(𝑘+1)𝑡₀=−𝑘𝑝−𝑘−1²+(𝑘+1)𝑡²₀−𝑡₀𝑘−𝑘−1. (13) Recall that 𝑡₀−2≤𝑘≤𝑡₀ and 0≤𝑝≤𝑡₀−1. Since 𝑡²₀−𝑡₀𝑡𝑘−𝑘−1=⁰+1,if𝑘=𝑡₀−2,0,if𝑘=𝑡₀−1,−𝑡₀−1,if𝑘=𝑡₀, (14) and the fact that if 𝑘=𝑡₀, then 𝑝=0, we have 𝑡₀+1𝑊𝑆𝑛,𝑡₀−𝑓𝑡₀=+1𝑝−𝑘−1²+(𝑘+1)𝑡²₀−𝑡₀𝑘−𝑘−1≥0, (15) with equality only if 𝑘=𝑡₀. It follows that 𝑊(𝑆(𝑛,𝑡))≤𝑓(𝑡)≤𝑓(𝑡₀+1)≤𝑊(𝑆(𝑛,𝑡₀)), with 𝑊(𝑆(𝑛,𝑡))=𝑊(𝑆(𝑛,𝑡₀)) if and only if 𝑡=𝑡₀+1 and 𝑛=1+𝑡₀+𝑡²₀. The proof is completed.

Acknowledgments

The research was supported by KPCME (no. 210243), NSFC (no. 11161046), and the Tianyuan Special Fund of NSFC (no. 11126113).