Wiener Complexity versus the Eccentric Complexity

Abstract

Let $w_G\left(u\right)$ be the sum of distances from u to all the other vertices of G. The Wiener complexity, $C_W\left(G\right)$, is the number of different values of $w_G\left(u\right)$ in G, and the eccentric complexity, $C_{\mathrm{ec}}\left(G\right)$, is the number of different eccentricities in G. In this paper, we prove that for every integer c there are infinitely many graphs G such that $C_W\left(G\right)-C_{\mathrm{ec}}\left(G\right)=c$. Moreover, we prove this statement using graphs with the smallest possible cyclomatic number. That is, if $c\ge 0$ we prove this statement using trees, and if $c<0$ we prove it using unicyclic graphs. Further, we prove that $C_{\mathrm{ec}}\left(G\right)\le 2C_W\left(G\right)-1$ if G is a unicyclic graph. In our proofs we use that the function $w_G\left(u\right)$ is convex on paths consisting of bridges. This property also promptly implies the already known bound for trees $C_{\mathrm{ec}}\left(G\right)\le C_W\left(G\right)$. Finally, we answer in positive an open question by finding infinitely many graphs G with diameter 3 such that $C_{\mathrm{ec}}\left(G\right)<C_W\left(G\right)$.

1. Introduction

Let G be a graph. Denote by $V\left(G\right)$ and $E\left(G\right)$ its vertex and edge sets, respectively. If $u\in V\left(G\right)$, then ${deg}_G\left(u\right)$ denotes the degree of u in G, and if $S\subseteq V\left(G\right)$ then $N\left(S\right)$ denotes the set S together with the vertices which have a neighbour in S. Obviously, $\left|N\left(u\right)\right|={deg}_G\left(u\right)+1$. If $R\subseteq E\left(G\right)$ then $G-R$ denotes a graph obtained when we remove all the edges of R from G. Similarly, if $S\subseteq V\left(G\right)$ then $G-S$ denotes a graph obtained when we remove all the vertices of S and all edges incident with a vertex of S from G. An edge $e\in E\left(G\right)$ is a bridge if $G-\left\{e\right\}$ has more components than G.

If $u,v\in V\left(G\right)$ then ${\mathrm{dist}}_G(u,v)$ is the length of a shortest path from u to v in G. The longest distance from a vertex u is its eccentricity $e_G\left(u\right)$. Hence, $e_G\left(u\right)=max\{{\mathrm{dist}}_G(u,v);v\in V\left(G\right)\}$. Using the eccentricity we define the radius $\mathrm{rad}\left(G\right)=min\{e_G\left(u\right);u\in V\left(G\right)\}$, and the diameter $\mathrm{diam}\left(G\right)=max\{e_G\left(u\right);u\in V\left(G\right)\}=max\{{\mathrm{dist}}_G(u,v);u,v\in V\left(G\right)\}$. The eccentric complexity of G is defined as

$$C_{\mathrm{ec}}\left(G\right)=\left|\{e_G\left(u\right);u\in V\left(G\right)\}\right|.$$

Observe that $C_{\mathrm{ec}}\left(G\right)=\mathrm{diam}\left(G\right)-\mathrm{rad}\left(G\right)+1$. The eccentric complexity has been introduced in [1]. Also see [2] for related connective eccentric complexity.

On the other hand the Wiener complexity of G is

$$C_W\left(G\right)=\left|\{w_G\left(u\right);u\in V\left(G\right)\}\right|,$$

where $w_G\left(u\right)={\sum }_{v\in V\left(G\right)}\mathrm{dist}(u,v)$ is the transmission of u in G. The parameter $\frac{1}{2}{\sum }_{u\in V\left(G\right)}{w}_G\left(u\right)$ is known as the Wiener index $W\left(G\right)$. Hence, $W\left(G\right)={\sum }_{u,v\in V\left(G\right)}\mathrm{dist}(u,v)$. The Wiener complexity $C_W\left(G\right)$ of a graph G was introduced in [3]. Further research on $C_W\left(G\right)$ can be found in [4,5,6]. For results on Wiener index see, e.g., [7].

In [8] the authors study the relation between $C_{\mathrm{ec}}\left(G\right)$ and $C_W\left(G\right)$. They prove the following statement.

Theorem 1.

If T is a tree then $C_{\mathrm{ec}}\left(T\right)\le C_W\left(T\right)$.

Next, using cartesian products they prove that for every $c\ge 0$ there are graphs G with $C_W\left(G\right)-C_{\mathrm{ec}}\left(G\right)=c$ and for every $k>0$ there are graphs G with $C_{\mathrm{ec}}\left(G\right)-C_W\left(G\right)=2^k$. Here we continue in their research. We prove that for every $c\ge 0$ there are infinitely many trees T such that $C_W\left(T\right)-C_{\mathrm{ec}}\left(T\right)=c$. By Theorem 1 to construct graphs G with $C_{\mathrm{ec}}\left(G\right)>C_W\left(G\right)$ we must abandon the class of trees. So we concentrate on graphs with cyclomatic number 1. We prove that for every $c>0$ there are infinitely many unicyclic graphs G such that $C_{\mathrm{ec}}\left(G\right)-C_W\left(G\right)=c$.

All graphs G with $C_{\mathrm{ec}}\left(G\right)<C_W\left(G\right)$ found in [8] have diameter at least 4, and it was shown that there are no such graphs of diameter at most 2. So the authors posed in [8] the following problem.

Problem 1.

Does there exist a graph G with diameter 3 and $C_{\mathrm{ec}}\left(G\right)>C_W\left(G\right)$?

We answer Problem 1 affirmatively and we find infinitely many graphs satisfying its requirements.

The outline of the paper is as follows. In Section 2 we characterize all pairs $c_1$ and $c_2$ such that there is a tree T with $C_{\mathrm{ec}}\left(T\right)=c_1$ and $C_W\left(T\right)=c_2$. Analogously, in Section 3 we characterize all pairs $c_1$ and $c_2$ such that $c_1<c_2$ and there is a unicyclic graph G with $C_W\left(G\right)=c_1$ and $C_{\mathrm{ec}}\left(G\right)=c_2$. Finally, in Section 4 we deal with Problem 1.

2. Trees

In this section we characterize pairs $c_1$ and $c_2$ such that there are (infinitely many) trees T with $C_{\mathrm{ec}}\left(T\right)=c_1$ and $C_W\left(T\right)=c_2$. To do this, first we show that $w_T$ is a strictly convex function on paths consisting of bridges; observe that in a tree, every edge is a bridge. However, firstly we state the following easy lemma.

Lemma 1.

Let G be a connected graph with a bridge $u_1{u}_2$. Let $G_1$ and $G_2$ be the two components of $G-u_1{u}_2$, such that $u_i\in V\left(G_i\right)$, $1\le i\le 2$, and let $n_i$ be the number of vertices in $G_i$. Then $w_G\left(u_1\right)-w_G\left(u_2\right)=n_2-n_1$.

Proof. 

Let $w_i$ be the transmission of $u_i$ in $G_i$, $1\le i\le 2$. Then

$$w_G\left(u_1\right)=w_1+n_2+w_2\mathrm{and}{w}_G\left(u_2\right)=n_1+w_1+w_2.$$

Hence, $w_G\left(u_1\right)-w_G\left(u_2\right)=n_2-n_1$. □

Recall that a function $f\left(i\right)$ defined on $\{0,1,\dots ,t\}$ is strictly convex, if for every $i\in \{1,\dots ,t-1\}$, we have $2f\left(i\right)<f(i-1)+f(i+1)$, or equivalently $f\left(i\right)-f(i-1)<f(i+1)-f\left(i\right)$. We have the following statement.

Lemma 2.

Let G be a graph. Further, let $P=v_0{v}_1\dots v_t$ be a path in G such that every edge of P is a bridge. Then $f\left(i\right)=w_G\left(v_i\right)$ is a strictly convex function on $\{0,1,\dots ,t\}$.

Proof. 

Let $u_1{u}_2{u}_3$ be a subpath of P. Then $G-\{u_1{u}_2,u_2{u}_3\}$ has three components. Denote by $G_i$ the component of $G-\{u_1{u}_2,u_2{u}_3\}$ which contains $u_i$, for each $i\in \{1,2,3\}$. Moreover, denote $s_i=\left|V\left(G_i\right)\right|$. By Lemma 1, we have

$$w_G\left(u_1\right)+w_G\left(u_3\right)-2w_G\left(u_2\right)=(s_2+s_3-s_1)+(s_1+s_2-s_3)=2s_2>0.$$

Consequently, $f\left(i\right)=w_G\left(v_i\right)$ is strictly convex on $\{0,1,\dots ,t\}$. □

Observe that considering a diametric path, Lemma 2 directly implies Theorem 1. However, we use it in the following statement which characterizes all possible pairs $C_{\mathrm{ec}}\left(T\right)$, $C_W\left(T\right)$ for trees.

Theorem 2.

It holds:

(i) 

If $3\le c_1\le c_2$ then there are infinitely many trees T with $C_{\mathrm{ec}}\left(T\right)=c_1$ and $C_W\left(T\right)=c_2$.

(ii) 

If $c_1=2$ and $c_2\in \{2,4\}$ then there are infinitely many trees T with $C_{\mathrm{ec}}\left(T\right)=c_1$ and $C_W\left(T\right)=c_2$ and no trees with $C_{\mathrm{ec}}\left(T\right)=c_1$ and $C_W\left(T\right)\notin \{2,4\}$.

(iii) 

If $c_1=1$ then there are only two trees T with $C_{\mathrm{ec}}\left(T\right)=1$ and in this case $C_W\left(T\right)=1$ as well.

Proof. 

Consider $\left(i\right)$. Here $c_1\ge 3$. First suppose that $c_2>c_1$. Let $k=\lceil \frac{c_2-1}{c_1-1}\rceil$. Take k paths $P_0,P_1,\dots ,P_{k-1}$ of length $c_1-1$ and denote their vertices so that $P_i=v_{i,0}{v}_{i,1}\dots v_{i,c_1-1}$, where $0\le i\le k-1$. Denote

$$\ell =c_2-1-(⌈\frac{c_2-1}{c_1-1}⌉-1)(c_1-1).$$

Let $P_k$ be a path of length ℓ so that $P_k=v_{k,0}{v}_{k,1}\dots v_{k,\ell }$. Since $\lceil \frac{c_2-1}{c_1-1}\rceil (c_1-1)\ge (c_2-1)$, we have $\ell \le c_1-1$, and since $\lceil \frac{c_2-1}{c_1-1}\rceil (c_1-1)<(c_2-1)+(c_1-1)$, we have $\ell >0$. Thus, $1\le \ell \le c_1-1$. Now attach to $v_{i,c_1-2}$ exactly $i-1$ new pendant vertices, $2\le i\le k-1$, and attach to $v_{k,\ell -1}$ exactly $q-1$ new vertices. We expect that q is a big number. Finally, identify the vertices $v_{0,0},v_{1,0},\dots ,v_{k,0}$ into a single vertex, which we denote by c, and denote the resulting tree by T, see Figure 1. Observe that there is 1 pendant vertex attached to $v_{0,c_1-2}$ in T and exactly i pendant vertices are attached to $v_{i,c_1-2}$, $1\le i\le k-1$. Further, since $k\ge 2$ ($k\ge 1$ would suffice since there is also $P_0$) we have $\mathrm{rad}\left(T\right)=c_1-1$ and $\mathrm{diam}\left(T\right)=2(c_1-1)$, so that $C_{\mathrm{ec}}\left(T\right)=c_1$. Obviously, if u and v are pendant vertices attached to the same vertex in T then $w_T\left(u\right)=w_T\left(v\right)$. Also, $w_T\left(u\right)=w_T\left(v\right)$ if $u=v_{0,i}$ and $v=v_{1,i}$, $1\le i\le c_1-1$. In all other cases we show that $w_T\left(u\right)\ne w_T\left(v\right)$. Hence, we show that $c,v_{1,1},v_{1,2},\dots ,v_{k,\ell }$ have different transmissions.

Denote $r=c_1-1$. Let P be a path in T consisting of vertices of $P_a$ and $P_b$, $1\le a<b\le k-1$. Then $P=v_{a,r}\dots v_{a,1}c{v}_{b,1}\dots v_{b,r}$. Let $T^{\prime }$ be the nontrivial component of $T-\{v_{a,1},\dots ,v_{a,r},v_{b,1},\dots ,v_{b,r}\}$. Denote by $w^{\prime }$ the transmission of c in $T^{\prime }$ and denote by z the number of vertices of $T^{\prime }$. Then

$$\begin{array}{ccc}\hfill w_T\left(v_{a,r}\right)& =& 2(a-1)+\left(\genfrac{}{}{0pt}{}{2r+1}{2}\right)+2r(b-1)+w^{\prime }+(z-1)r;\hfill \\ \hfill w_T\left(v_{a,i}\right)& =& (r-i)(a-1)+\left(\genfrac{}{}{0pt}{}{r-i+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{r+i+1}{2}\right)+(r+i)(b-1)+w^{\prime }+(z-1)i,1\le i\le r-1;\hfill \\ \hfill w_T\left(c\right)& =& r(a-1)+\left(\genfrac{}{}{0pt}{}{r+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{r+1}{2}\right)+r(b-1)+w^{\prime };\hfill \\ \hfill w_T\left(v_{b,i}\right)& =& (r+i)(a-1)+\left(\genfrac{}{}{0pt}{}{r+i+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{r-i+1}{2}\right)+(r-i)(b-1)+w^{\prime }+(z-1)i,1\le i\le r-1;\hfill \\ \hfill w_T\left(v_{b,r}\right)& =& 2r(a-1)+\left(\genfrac{}{}{0pt}{}{2r+1}{2}\right)+2(b-1)+w^{\prime }+(z-1)r.\hfill \end{array}$$

Since $w_T\left(v_{a,i}\right)-w_T\left(v_{b,i}\right)=(r-i)(a-b)+(r+i)(b-a)=2i(b-a)>0$ if $1\le i\le r-1$ and $w_T\left(v_{a,r}\right)-w_T\left(v_{b,r}\right)=(2r-2)(b-a)>0$, we have $w_T\left(v_{a,j}\right)>w_T\left(v_{b,j}\right)$, $1\le j\le r$. And since q and consequently also z are big, the terms containing z in the above expressions are crucial. Therefore $w_T\left(v_{a,1}\right)<w_T\left(v_{b,2}\right)$ and in general $w_T\left(v_{a,i}\right)<w_T\left(v_{b,i+1}\right)$, where $1\le i<r$. So we conclude that

$$w_T\left(c\right)<w_T\left(v_{b,1}\right)<w_T\left(v_{a,1}\right)<w_T\left(v_{b,2}\right)<w_T\left(v_{a,2}\right)<w_T\left(v_{b,3}\right)<\dots <w_T\left(v_{a,r}\right).$$

Now let P be a path consisting of $P_a$ and $P_k$, $1\le a\le k-1$. Then $P=v_{a,r}\dots v_{a,1}c{v}_{k,1}\dots v_{k,\ell }$$=u_{a+\ell }{u}_{a+\ell -1}\dots u_0$. We remark that $u_j$ are just different labels for vertices of P which will be used later. Similarly as above, let $T^{\prime }$ be the nontrivial component of $T-\{v_{a,1},\dots .v_{a,r},v_{k,1},\dots ,v_{k,\ell }\}$. Denote by $w^{\prime }$ the transmission of c in $T^{\prime }$ and denote by z the number of vertices of $T^{\prime }$. Then

$$\begin{array}{ccc}\hfill w_T\left(v_{a,r}\right)& =& 2(a-1)+\left(\genfrac{}{}{0pt}{}{r+\ell +1}{2}\right)+(r+\ell )(q-1)+w^{\prime }+(z-1)r;\hfill \\ \hfill w_T\left(v_{a,i}\right)& =& (r-i)(a-1)+\left(\genfrac{}{}{0pt}{}{r-i+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{\ell +i+1}{2}\right)+(\ell +i)(q-1)+w^{\prime }+(z-1)i,1\le i\le r-1;\hfill \\ \hfill w_T\left(c\right)& =& r(a-1)+\left(\genfrac{}{}{0pt}{}{r+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{\ell +1}{2}\right)+\ell (q-1)+w^{\prime };\hfill \\ \hfill w_T\left(v_{k,i}\right)& =& (r+i)(a-1)+\left(\genfrac{}{}{0pt}{}{r+i+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{\ell -i+1}{2}\right)+(\ell -i)(q-1)+w^{\prime }+(z-1)i,1\le i\le \ell -1;\hfill \\ \hfill w_T\left(v_{k,\ell }\right)& =& (r+\ell )(a-1)+\left(\genfrac{}{}{0pt}{}{r+\ell +1}{2}\right)+2(q-1)+w^{\prime }+(z-1)\ell .\hfill \end{array}$$

Observe that $u_2=v_{k,\ell -2}$ if $\ell \ge 3$, $u_2=c$ if $\ell =2$ and $u_2=v_{a,1}$ if $\ell =1$. In any case, we have $w_T\left(v_{k,l}\right)-w_T\left(u_2\right)\ge \left(\genfrac{}{}{0pt}{}{r+\ell +1}{2}\right)-\left(\genfrac{}{}{0pt}{}{r+\ell -1}{2}\right)-\left(\genfrac{}{}{0pt}{}{2}{2}\right)>0$, and so $w_T\left(u_0\right)>w_T\left(u_2\right)$. And since q is big, analogously as above we conclude that

$$w_T\left(u_1\right)<w_T\left(u_2\right)<w_T\left(u_0\right)<w_T\left(u_3\right)<w_T\left(u_4\right)<\dots <w_T\left(u_{r+\ell }\right).$$

Let $S=\{c,v_{1,1},\dots ,v_{k,\ell }\}$. As shown above, vertices in S have pairwise different transmissions, while the vertices outside S have transmissions as some vertices in S. Since

$$\left|S\right|=1+(\lceil \frac{c_2-1}{c_1-1}\rceil -1)(c_1-1)+[c_2-1-(\lceil \frac{c_2-1}{c_1-1}\rceil -1)(c_1-1)]=c_2,$$

we have $C_W\left(T\right)=c_2$.

Now suppose $c_2=c_1$. Let $P=v_{0,c_1-1}\dots v_{0,1}c{v}_{1,1}\dots v_{1,c_1-1}$ be a path of length $2(c_1-1)$. We attach to both $v_{0,c_1-2}$ and $v_{1,c_1-2}$ exactly q pendant vertices and we denote by T the resulting tree, see Figure 2. Then T has $2c_1-1+2q$ vertices, $\mathrm{rad}\left(T\right)=c_1-1$ and $\mathrm{diam}\left(T\right)=2(c_1-1)$, so that $C_{\mathrm{ec}}\left(T\right)=c_1$. Denote $r=c_1-1$. By symmetry, we have $w_T\left(v_{0,i}\right)=w_T\left(v_{1,i}\right)$, $1\le i\le r$, and $w_T\left(u\right)=w_T\left(v\right)$ if u and v are pendant vertices of T. So it remains to show that the vertices $v_{0,r},\dots ,v_{0,1},c$ have different transmissions. However, since $w_T\left(v_{0,1}\right)=v_T\left(v_{1,1}\right)$, by Lemma 2 we get

$$w_T\left(c\right)<w_T\left(v_{0,1}\right)<\dots <v_T\left(v_{0,r}\right)$$

and so $C_W\left(T\right)=r+1=c_1$.

Now, consider $\left(ii\right)$. So, let $c_1=2$. If T is a tree with $\mathrm{rad}\left(T\right)\ge 3$, then $\mathrm{diam}\left(T\right)\ge 5$ and consequently $C_{\mathrm{ec}}\left(T\right)\ge 3$, a contradiction. Hence, either $\mathrm{rad}\left(T\right)=1$ and $\mathrm{diam}\left(T\right)=2$, in which case T is a star $K_{1,t}$, where $t\ge 2$, or $\mathrm{rad}\left(T\right)=2$ and $\mathrm{diam}\left(T\right)=3$, in which case T is a double star $D_{a,b}$, i.e., a graph on $a+b+2$ vertices obtained by attaching a pendant vertices to one vertex of $K_2$ and b pendant vertices to the other vertex of $K_2$, where $1\le a\le b$. If T is a star $K_{1,t}$, $t\ge 2$, then $C_W\left(T\right)=2$ since the central vertex has transmission smaller than is the transmission of pendant vertices. This establishes the case $c_1=c_2=2$. On the other hand if T is a double star then since pendant vertices adjacent to a common vertex have the same transmission, we have $C_W\left(T\right)\le 4$. In the next we consider $T=D_{a,b}$, where $a<b$, since $C_W\left(D_{a,a}\right)=2$, a case already solved by stars. Let $v_0,v_1,v_2,v_3$ be a path in $D_{a,b}$ such that ${deg}_T\left(v_1\right)=a+1$ and ${deg}_T\left(v_2\right)=b+1$. Then

$$\begin{array}{ccc}\hfill w_T\left(v_0\right)& =& 2(a-1)+3+3b;\hfill \\ \hfill w_T\left(v_1\right)& =& a+1+2b;\hfill \\ \hfill w_T\left(v_2\right)& =& 2a+1+b;\hfill \\ \hfill w_T\left(v_3\right)& =& 3a+3+2(b-1).\hfill \end{array}$$

Since $0<a<b$, it is obvious that $2a+1+b<a+1+2b<3a+1+2b<2a+1+3b$. Thus. $w_T\left(v_2\right)<w_T\left(v_1\right)<w_T\left(v_3\right)<w_T\left(v_0\right)$, and so $C_W\left(T\right)=4$.

Finally, consider $\left(iii\right)$. Since there are only two trees T such that $\mathrm{rad}\left(T\right)=\mathrm{diam}\left(T\right)$, namely the complete graphs $K_1$ and $K_2$, this part of Theorem 2 is trivial. □

Theorem 2 has the following consequence.

Corollary 1.

For every $c\ge 0$ there are infinitely many trees T such that $C_W\left(T\right)-C_{\mathrm{ec}}\left(T\right)=c$.

3. Unicyclic Graphs

In this section, we give counterparts of the previous results for unicyclic graphs. We also bound the eccentric complexity in term of Wiener complexity and characterize the pairs $c_1,c_2$ such that $c_1<c_2$ and there are (infinitely many) unicyclic graphs G with $C_W\left(G\right)=c_1$ and $C_{\mathrm{ec}}\left(G\right)=c_2$. We start with the following lemma.

Lemma 3.

Let G be a unicyclic graph with a cycle C. Further, let $u_2,v\in V\left(C\right)$ and let $u_1$ be a neighbour of $u_2$ which is not in C. If $w_G\left(u_1\right)\le w_G\left(u_2\right)$ then $w_G\left(u_2\right)<w_G\left(v\right)$.

Proof. 

Observe that $u_1{u}_2$ is a bridge in G. Hence, $G-u_1{u}_2$ has two components, say $G_1$ and $G_2$. Assume that $u_i\in V\left(G_i\right)$ and $n_i=\left|V\left(G_i\right)\right|$, $1\le i\le 2$. By the assumptions and by Lemma 1, $w_G\left(u_1\right)-w_G\left(u_2\right)=n_2-n_1\le 0$.

Let T be a tree obtained from G by removing an edge of C which is opposite (i.e., antipodal) to v. Observe that if C has odd length, then there is a unique edge opposite to v, while if C has even length, then there are two edges opposite to v. Obviously, $w_G\left(v\right)=w_T\left(v\right)$ and $w_G\left(u_2\right)\le w_T\left(u_2\right)$. Observe also that

$$0\ge w_G\left(u_1\right)-w_G\left(u_2\right)=w_T\left(u_1\right)-w_T\left(u_2\right).$$

Now, consider a path from $u_1$ to v in T. Assume that the length of this path is $k-1$ and denote their vertices by $u_1{u}_2{u}_3\dots u_k(=v)$. Since $u_1{u}_2$ is a bridge in T, we have $w_T\left(u_1\right)-w_T\left(u_2\right)=n_2-n_1$ again. And by Lemma 2 we get $w_T\left(u_1\right)+w_T\left(u_3\right)>2w_T\left(u_2\right)$ or equivalently $w_T\left(u_1\right)-w_T\left(u_2\right)>w_T\left(u_2\right)-w_T\left(u_3\right)$. Applying Lemma 2 several times we get

$$0\ge w_T\left(u_1\right)-w_T\left(u_2\right)>w_T\left(u_2\right)-w_T\left(u_3\right)>\dots >w_T\left(u_{k-1}\right)-w_T\left(v\right)$$

which implies $w_T\left(u_1\right)\le w_T\left(u_2\right)<w_T\left(u_3\right)<\dots <w_T\left(v\right)$ and consequently $w_G\left(u_2\right)\le w_T\left(u_2\right)<w_T\left(v\right)=w_G\left(v\right)$. □

The following statement characterizes all possible pairs $C_W\left(G\right)$, $C_{\mathrm{ec}}\left(G\right)$ for unicyclic graphs, provided that $C_W\left(G\right)<C_{\mathrm{ec}}\left(G\right)$.

Theorem 3.

Every unicyclic graph G satisfies

$$C_{\mathrm{ec}}\left(G\right)\le 2C_W\left(G\right)-1.$$

Moreover, for any positive integers $c_1$ and $c_2$ with $c_1<c_2\le 2c_1-1$ there are infinitely many unicyclic graphs G such that $C_W\left(G\right)=c_1$ and $C_{\mathrm{ec}}\left(G\right)=c_2$.

Proof. 

Let G be a unicyclic graph with a cycle C of length k. Further, let $P_1$ and $P_2$ be two longest paths starting in different vertices of C and which contain only edges which are not in C. Observe that if the length of $P_1$ is positive, then the path terminates in a pendant vertex of G. Similar statement holds for $P_2$. Let ${\ell }_i$ be the length of $P_i$, $1\le i\le 2$, and let $P_i=v_{i,0}{v}_{i,1}\dots v_{i,{\ell }_i}$, where $v_{i,0}\in V\left(C\right)$. Observe that in each of $P_1$ and $P_2$, there are at most two vertices with the same transmission, by Lemma 2. If there are three vertices, say $u_1$, $u_2$ and $u_3$, in $V\left(P_1\right)\cup V\left(P_2\right)$ which have the same transmission in G, then two of them are in one of the paths $P_1$ and $P_2$ while the third one is in the other. Without loss of generality we may assume that $u_1,u_2\in V\left(P_1\right)$ and $u_3\in V\left(P_3\right)$. Then $w_G\left(v_{1,1}\right)\le w_G\left(v_{1,0}\right)$ by Lemma 2, and so $w_G\left(v_{1,0}\right)<w_G\left(v_{2,0}\right)$ by Lemma 3. If $w_G\left(v_{2,1}\right)\le w_G\left(v_{2,0}\right)$ then $w_G\left(v_{2,0}\right)<w_G\left(v_{1,0}\right)$ by Lemma 3, a contradiction. Hence $w_G\left(v_{2,0}\right)<w_G\left(v_{2,1}\right)$, and by Lemma 2 $w_G\left(v_{2,0}\right)<w_G\left(v_{2,i}\right)$ for every i, $1\le i\le {\ell }_2$. Consequently $w_G\left(u_1\right)=w_G\left(u_2\right)\le w_G\left(v_{1,0}\right)<w_G\left(v_{2,0}\right)\le w_G\left(u_3\right)$. Hence, there are not three vertices in $V\left(P_1\right)\cup V\left(P_2\right)$ which have the same transmission in G. Therefore $C_W\left(G\right)\ge \lceil \frac{{\ell }_1+{\ell }_2}{2}\rceil +1$.

On the other hand $\mathrm{diam}\left(G\right)\le {\ell }_1+{\ell }_2+\lfloor k/2\rfloor$ and $\mathrm{rad}\left(G\right)\ge \lfloor k/2\rfloor$. So $C_{\mathrm{ec}}\left(G\right)=\mathrm{diam}\left(G\right)-\mathrm{rad}\left(G\right)+1\le {\ell }_1+{\ell }_2+1$, and hence

$$2C_W\left(G\right)-C_{\mathrm{ec}}\left(G\right)\ge 2⌈\frac{{\ell }_1+{\ell }_2}{2}⌉+2-l_1-l_2-1\ge 1.$$

Now we prove the second result. Let $c_1$ and $c_2$ satisfy $c_1<c_2\le 2c_1-1$. Denote $\mathsf{\Delta }=c_2-c_1$. Let C be a cycle of length $4\mathsf{\Delta }$ and let $u_1$ and $v_1$ be opposite vertices on C. Attach to $u_1$ (resp. $v_1$) a path of length $c_1-1$$u_1{u}_2\dots u_{c_1}$ (resp. $v_1{v}_2\dots v_{c_1}$). Finally, attach to both $u_{c_1-1}$ and $v_{c_1-1}$ exactly $q\ge 0$ pendant vertices, and denote the resulting graph by G, see Figure 3.

Obviously, $\mathrm{diam}\left(G\right)=2(c_1-1)+2\mathsf{\Delta }=2c_2-2$. Since $c_2\le 2c_1-1$, we have $c_2-c_1\le c_1-1$, and so $2\mathsf{\Delta }\le \mathsf{\Delta }+(c_1-1)$. Thus, $\mathrm{rad}\left(G\right)=max\{\mathrm{rad}\left(C\right),\lceil \mathrm{diam}\left(G\right)/2\rceil \}=\mathsf{\Delta }+(c_1-1)$, which means that $C_{\mathrm{ec}}\left(G\right)=\mathrm{diam}\left(G\right)-\mathrm{rad}\left(G\right)+1=c_2$.

On the other hand, denote by $w^T$ the transmission of $u_1$ in the tree attached to C and denote by $w^C$ the transmission of $u_1$ in C. Then $w_G\left(u_1\right)=w^T+w^C+2\mathsf{\Delta }(c_1-1+q)+w^T$, and similarly for every vertex v of C we have $w_G\left(v\right)=2w^T+w^C+2\mathsf{\Delta }(c_1-1+q)$ as well. By Lemmas 2 and 3 it holds $w_G\left(u_1\right)<w_G\left(u_2\right)<\dots <w_G\left(u_{c_1}\right)$ and by symmetry $w_G\left(u_i\right)=w_G\left(v_i\right)$, $1\le i\le c_1$. Thus $C_W\left(G\right)=c_1$, and so G satisfies the assumptions of the theorem. □

Theorem 3 has the following consequence.

Corollary 2.

For every integer $c>0$ there are infinitely many unicyclic graphs G such that

$$C_{\mathrm{ec}}\left(G\right)-C_W\left(G\right)=c.$$

We remark that the attachment vertices $u_1$ and $u_2$ do not need to be opposite on C if $c_1$ is big enough (compared to $\mathsf{\Delta }=c_2-c_1$). We can use also even cycles of length $\not\equiv 0(mod4)$ and odd cycles, but again $c_1$ must be big enough. Though for small order graphs, one with even cycle are quite abundant, the smallest unicyclic graph G with a cycle of odd length satisfying $C_W\left(G\right)<C_{\mathrm{ec}}\left(G\right)$ has 13 vertices, and its cycle has length 9.

4. Graphs with Diameter 3

In this section we solve Problem 1. Observe that if $\mathrm{diam}\left(G\right)=3$ and $C_{\mathrm{ec}}\left(G\right)>C_W\left(G\right)$ then $\mathrm{rad}\left(G\right)=2$, $C_{\mathrm{ec}}\left(G\right)=2$ and $C_W\left(G\right)=1$. Hence, there is no unicyclic graph G satisfying the requirements of Problem 1, by Theorem 3.

Let G be a graph with diameter 3. For every vertex $u\in V\left(G\right)$, by $d_G^i\left(u\right)$ we denote the number of vertices of G which are at distance i from u. Denote ${\sigma }_G\left(u\right)=d_G^1\left(u\right)-d_G^3\left(u\right)$. We have the following statement.

Lemma 4.

Let G be a graph with diameter 3. Then $C_W\left(G\right)=1$ if and only if all vertices of G have the same value of σ.

Proof. 

Let $u\in V\left(G\right)$. Then $w_G\left(u\right)=d_G^1\left(u\right)+2d_G^2\left(u\right)+3d_G^3\left(u\right)$. Since $d_G^1\left(u\right)+d_G^2\left(u\right)+d_G^3\left(u\right)=n-1$, where $n=\left|V\right(G\left)\right|$, we have $d_G^2\left(u\right)=n-1-d_G^1\left(u\right)-d_G^3\left(u\right)$, and consequently $w_G\left(u\right)=2n-2-d_G^1\left(u\right)+d_G^3\left(u\right)$. Hence, if $v\in V\left(G\right)$ with $v\ne u$, then $w_G\left(v\right)=w_G\left(u\right)$ is equivalent with ${\sigma }_G\left(u\right)={\sigma }_G\left(v\right)$. □

By Lemma 4, in graphs G of diameter 3 with $C_{\mathrm{ec}}\left(G\right)=2$ and $C_W\left(G\right)=1$, the vertices of eccentricity 3 must have degree greater than is the degree of vertices of eccentricity 2. This looks surprising, nevertheless, there exist such graphs.

Let G be a graph on $2r$ vertices and let $S\subseteq V\left(G\right)$ such that $\left|S\right|=r$. By $A(G,S)$ we denote the graph obtained from G by adding two vertices, v and $v^{\prime }$, where v is connected to all vertices of S and $v^{\prime }$ is connected to all vertices of $V\left(G\right)\backslash S$. We have:

Proposition 1.

Let G be a k-regular graph of diameter 2 on $2(k+2)$ vertices. Moreover, let $S\subseteq V\left(G\right)$, $\left|S\right|=k+2$, such that every vertex of S has a neighbour in $V\left(G\right)\backslash S$ and every vertex of $V\left(G\right)\backslash S$ has a neighbour in S. Then $\mathrm{diam}\left(A\right(G,S\left)\right)=3$, $C_{\mathrm{ec}}\left(A(G,S)\right)=2$ and $C_W\left(A(G,S)\right)=1$.

Proof. 

First observe that $N_G\left(S\right)=V\left(G\right)=N_G(V\left(G\right)\backslash S)$. Since every vertex of S has a neighbour in $V\left(G\right)\backslash S$ and every vertex of $V\left(G\right)\backslash S$ has a neighbour in S, we have $e_{A(G,S)}\left(u\right)=2$ for every $u\in V\left(G\right)$. Since ${\mathrm{dist}}_{A(G,S)}(v,v^{\prime })=3$, we have $\mathrm{diam}\left(A\right(G,S\left)\right)=3$ and $C_{\mathrm{ec}}\left(A(G,S)\right)=2$.

If $u\in V\left(G\right)$ then ${deg}_{A(G,S)}\left(u\right)={\sigma }_{A(G,S)}\left(u\right)=k+1$. On the other hand ${deg}_{A(G,S)}\left(v\right)={deg}_{A(G,S)}\left(v^{\prime }\right)=k+2$. Moreover, since in G holds $N\left(S\right)=V\left(G\right)=N\left(V\right(G)\backslash S)$, we have $d_{A(G,S)}^3\left(v\right)=d_{A(G,S)}^3\left(v^{\prime }\right)=1$. Thus ${\sigma }_{A(G,S)}\left(v\right)={\sigma }_{A(G,S)}\left(v^{\prime }\right)=k+1$ and $C_W\left(A(G,S)\right)=1$, by Lemma 4. □

Since there is no 2-regular graph on 8 vertices with diameter 2, the smallest graph G satisfying assumptions of Proposition 1 is the Petersen graph in which S is the set of vertices of one of its 5-cycles. If G is the Petersen graph and S is the set of vertices of one of its 5-cycles, then $A(G,S)$ has 12 vertices.

However, there are also other graphs satisfying the assumptions of Proposition 1.

Lemma 5.

Let $k\ge 6$ be an even number, and let $D=\left(\right\{1,4,7,\dots \}\cap \{1,2,\dots ,k+1\left\}\right)\cup \{k+1,k,k-1,\dots ,i\}$ with $\left|D\right|=k/2$. Let G be the Cayley graph with $V\left(G\right)=Z_{2k+4}$ and $E\left(G\right)=\{ij;i-j\in D\cup -D\}$. Finally, let $S=\{0,2,\dots ,2k+2\}$. Then G and S satisfy the assumptions of Proposition 1.

Proof. 

Obviously, G is k-regular. Since $1\in D$ and $S=\{0,2,\dots ,2k+2\}$, S satisfies the assumptions of Proposition 1. Hence, it remains to prove that $\mathrm{diam}\left(G\right)=2$.

We only show that $e_G\left(0\right)=2$, and since G is vertex-transitive, we conclude that $\mathrm{diam}\left(G\right)=2$. So it is enough to show that if $1\le r\le k+2$, then either $0r\in E\left(G\right)$, or $0(r-1)\in E\left(G\right)$ or $0(r+1)\in E\left(G\right)$, because $\alpha :u\to 2k+4-u$ is an isomorphism of G. Let $t=k/2$ and let $D^{\prime }$ be a set of t numbers starting with 1 and continuing with difference 3. Then $D^{\prime }=\{1,4,7,\dots ,3t-2\}$. Since $k\ge 6$, we have $t\ge 3$ and $3t-2\ge k+1$. Hence, it follows that $k+1\in D$ which means that ${\mathrm{dist}}_G(0,k+2)=2$. And since $D\supseteq D^{\prime }\cap \{1,2,\dots ,k+1\}$, we have $0r\in E\left(G\right)$ or $0(r-1)\in E\left(G\right)$ or $0(r+1)\in E\left(G\right)$ for every r with $1\le r\le k+1$. Thus, $e_G\left(0\right)=2$. □

Let G be the Petersen graph or a graph from Lemma 5 and let S be as described. Then $A(G,S)$ has diameter 3 and $C_{\mathrm{ec}}\left(A(G,S)\right)>C_W\left(A(G,S)\right)$. However, all these graphs have exactly 2 vertices with eccentricity 3. Next statement shows that there are required graphs with $2t$ vertices with eccentricity 3 for arbitrary $t\ge 1$.

Let H be a graph. By $B_t\left(H\right)$ we denote a graph on $t\left|V\right(H\left)\right|$ vertices obtained from H by replacing every vertex by $K_t$. Moreover, vertices from different copies of $K_t$ are adjacent in $B_t\left(H\right)$ if and only if these copies of $K_t$ are obtained from adjacent vertices in H.

Theorem 4.

Let G be a graph and $S\subseteq V\left(G\right)$ such that G and S satisfy the assumptions of Proposition 1. Moreover, let $t\ge 1$. Then $\mathrm{diam}\left(B_t\left(A(G,S)\right)\right)=3$, $C_{\mathrm{ec}}\left(B_t\left(A(G,S)\right)\right)=2$ and $C_W\left(B_t\left(A(G,S)\right)\right)=1$.

Proof. 

For $t=1$ the statement reduces to Proposition 1. Therefore, in the following we assume $t\ge 2$. Denote $H=B_t\left(A(G,S)\right)$. Let u be a vertex of H obtained from a vertex of G. Then $e_H\left(u\right)=2$ and ${deg}_H\left(u\right)=d_H^1\left(u\right)=(t-1)+kt+t$, and so ${\sigma }_H\left(u\right)=kt+2t-1$.

Now let $u^{\prime }$ be a vertex of H obtained from v or $v^{\prime }$ (i.e., from the vertices of $A(G,S)$ which are not in G). Then $e_H\left(u^{\prime }\right)=3$, ${deg}_H\left(u^{\prime }\right)=d_H^1\left(u^{\prime }\right)=(t-1)+(k+2)t$ and $d_H^3\left(u^{\prime }\right)=t$. Hence $\sigma \left(u^{\prime }\right)=kt+2t-1$ as well. Thus, $\mathrm{diam}\left(H\right)=3$, $C_{\mathrm{ec}}\left(H\right)=2$ and by Lemma 4 we have $C_W\left(H\right)=1$. □

In [8] the authors checked all graphs on at most 10 vertices and none of them had $C_W<C_{\mathrm{ec}}$ and diameter 3. We checked the same for graphs on 11 vertices. Thus, the smallest graph with the above properties has 12 vertices and it is obtained using Proposition 1.