# Extremal results on stepwise transmission irregular graphs

Yaser Alizadeha Sandi Klavžarb,c,d

###### Abstract

The transmission ${\rm Tr}_{G}(v)$ of a vertex $v$ of a connected graph $G$ is
the sum of distances between $v$ and all other vertices in $G$. $G$ is a
stepwise transmission irregular (STI) graph if $|{\rm Tr}_{G}(u)-{\rm
Tr}_{G}(v)|=1$ holds for each edge $uv\in E(G)$. In this paper, extremal
results on STI graphs with respect to the size and different metric properties
are proved. Two extremal families appear in all the cases, balanced complete
bipartite graphs of odd order and the so called odd hatted cycles.

a Department of Mathematics, Hakim Sabzevari University, Sabzevar, Iran

e-mail<EMAIL_ADDRESS>
b Faculty of Mathematics and Physics, University of Ljubljana, Slovenia

e-mail<EMAIL_ADDRESS>
c Faculty of Natural Sciences and Mathematics, University of Maribor, Slovenia

d Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia

Key words: graph distance; transmission of vertex; stepwise transmission
irregular graph; eccentricity; Wiener index

AMS Subj. Class. 2020: 05C12

##  1 Introduction

All graphs throughout the paper are simple and connected. If $G=(V(G),E(G))$
is a graph and $u,v\in V(G)$, then the distance $d_{G}(u,v)$ is the number of
edges on a shortest $u,v$-path. The transmission ${\rm Tr}_{G}(v)$ of a vertex
$v\in V(G)$ is the sum of distances between $v$ and all other vertices. This
concept arose in different contexts, hence it is not surprising that it is
also known as the total distance, the farness, and the vertex Wiener value,
cf. [1, 22, 24, 25, 26]. The transmission also led to the Wiener complexity
(the number of different transmissions) [3], is closely related to other
topological indices [26], and characterizes the distance-balanced property and
the opportunity index [9, 13].

Quite recently, Dobrynin and Sharafdini [16] defined stepwise transmission
irregular graphs (STI graphs for short) as the graphs $G$ in which $|{\rm
Tr}_{G}(u)-{\rm Tr}_{G}(v)|=1$ holds for each edge $uv$ of $G$. The class was
proposed, among other things, as a new pebble in the investigation of
transmission dependent classes of graphs. A striking example of such classes
is the one of transmission irregular graphs which by definition contains the
graphs in which all its vertices have pairwise different transmissions. After
being introduced in [4], transmission irregular graphs received a lot of
attention [6, 7, 11, 14, 15, 28].

Just as transmission irregular graphs, STI graphs also turned out to be a very
interesting class of graphs. In the seminal paper [16], basic properties of
STI graphs were first established. In particular, STI graphs are bipartite,
$2$-connected (except $P_{3}$), and of odd order, so that no regular graph can
be STI. By computer search it was established that there are $1$, $1$, $3$,
$7$, $18$, $87$, and $1171$ STI graphs of respective orders $3$, $5$, $7$,
$9$, $11$, $13$, and $15$. Six interesting infinite families of STI graphs
were also presented and topological indices of STI graphs briefly discussed.
In the subsequent paper [8], Al-Yakoob and Stevanović confirmed a conjecture
from [16] by proving that all the graphs from an additional interesting
infinite family are STI. They also provided a further computational support
for a challenging conjecture from [16] that the girth of every STI graph is
$4$. As feasible candidates for a counterexample they checked all bipartite,
$2$-connected graphs on up to $23$ vertices with girth at least $6$.

In this paper we are interested in extremal properties of STI graphs. In the
second part of this section definitions and notation needed is given. In
Section 2 we determine the minimum and the maximum size of an STI graph and
characterize the extremal graphs. In Section 3 we determine extremal results
for different metric properties of STI graphs: the diameter, the eccentricity,
the Wiener index, and the transmission.

###  1.1 Preliminaries

The eccentricity $\varepsilon_{G}(v)$ of a vertex $v\in V(G)$ is the maximum
distance between $v$ and other vertices of $G$. The eccentricity ${\rm
ecc}(G)$ of a graph $G$ is the sum of the vertex eccentricities over all
vertices of $G$. A vertex $w$ is an eccentric vertex of $v$ if
$d_{G}(v,w)=\varepsilon_{G}(v)$. The diameter ${\rm diam}(G)$ of $G$ is the
largest eccentricity among its vertices.

The Wiener index $W(G)$ of a graph $G$ is the sum of distances between all
pairs vertices of $G$, that is, $W(G)=\sum_{\\{u,v\\}\subseteq
V(G)}d_{G}(u,v)=\frac{1}{2}\sum_{v\in V(G)}{\rm Tr}_{G}(v)$, see the survey
[23]. For an edge $vw\in E(G)$, the number of vertices lying closer to $v$
than to $w$ is denoted by $n_{v}(G)$. The value $n_{w}(G)$ is defined
analogously. The Mostar index ${\rm Mo}(G)$ of $G$ is defined as ${\rm
Mo}(G)=\sum_{uv\in E(G)}|n_{u}(G)-n_{v}(G)|$. It was introduced in [18] and
studied a lot afterwards, see the recent survey [2]. The first Zagreb index
${\rm Z}_{1}(G)$ of $G$ is defined as ${\rm Z}_{1}(G)=\sum_{u\in
V(G)}\deg(u)^{2}$, the second Zagreb index ${\rm Z}_{2}(G)$ of $G$ is ${\rm
Z}_{2}(G)=\sum_{uv\in E(G)}\deg(u)\deg(v)$. These two invariants were
respectively introduced in [21, 20], see also the survey [19]. We note that
the first Zagreb index can be equivalently expressed as ${\rm
Z}_{1}(G)=\sum_{uv\in E(G)}\deg(u)+\deg(v)$, see [17].

For further reference we recall the following properties of STI graphs.

###### Theorem 1.1

[16] If $G$ is an STI graph different from $P_{3}$, then $G$ is a bipartite,
$2$-connected graph of odd order.

To conclude the preliminaries we introduce a family of graphs that will be
ubiquitous throughout the rest of the paper. Let $n\geq 5$. The graph $G_{n}$
has the vertex set $\\{v_{1},\ldots,v_{n}\\}$, where the vertices
$v_{1},\ldots,v_{n-1}$ induce a cycle of length $n-1$, while $v_{n}$ is
adjacent to $v_{1}$ and $v_{3}$. Note that $G_{5}\cong K_{2,3}$, while for
$G_{11}$ see Fig. 1.

$v_{1}$$v_{2}$$v_{3}$$v_{4}$$v_{5}$$v_{6}$$v_{7}$$v_{8}$$v_{9}$$v_{10}$$v_{11}$
Figure 1: The graph $G_{11}$

We note in passing that in [12], the graphs $G_{n}$ were named hatted cycles
and investigated with respect to the Rall’s $1/2$-conjecture on the domination
game. From our current point of view, it was shown in [16] that $G_{n}$ is an
STI graph for every odd $n\geq 5$. Moreover, in [10] it was proved that hatted
cycles are the graphs that attain the third maximum Wiener index over
$2$-connected graphs.

##  2 Extremal STI graph with respect to size

In this section we bound the size of an STI graph and determine the extremal
graphs. As a consequence we do the same for the Mostar index and the two
Zagreb indices.

###### Theorem 2.1

If $G$ is an STI graph of order $n$ and size $m$, then

$n+1\leq m\leq\frac{n^{2}-1}{4}\,.$

Moreover, the left equality holds if and only if $G\cong G_{n}$, and the right
equality holds if and only if $G\cong K_{\frac{n-1}{2},\frac{n+1}{2}}$.

Proof. The right inequality was proved in [16, Proposition 4]. For the
equality, let $G$ be an arbitrary STI graph of order $n$ and size $m$. Then,
by Theorem 1.1, $G$ is bipartite and $n$ is odd. Since
$K_{\frac{n-1}{2},\frac{n+1}{2}}$ is the unique bipartite graph of order $n$
with the maximum size and $K_{\frac{n-1}{2},\frac{n+1}{2}}$ is an STI graph
(cf. [16]), it is thus the unique graph attaining the right equality.

Recall that a graph $G$ is $2$-connected if and only if $G$ admits an ear
decomposition (see [27] for the definition of the ear decomposition). Hence
the only $2$-connected graph $G$ with $m=n$ is $C_{n}$. As cycles are not STI
graphs, we conclude that if $G$ is an STI graph, then $m\geq n+1$. Let now $G$
be an arbitrary STI graph with $m=n+1$. Then it is $2$-connected and the ear
decomposition implies that $G$ consists of two vertices of degree $3$, say $x$
and $y$, which are connected by three internally disjoint paths. Let these
paths be of lengths $r,s,t$, where $1\leq r\leq s\leq t$.

Since $G$ is bipartite, $r$, $s$, and $t$ are either all odd or all even.
Suppose first that $r=1$, that is, $x$ and $y$ are connected by an edge. Then
clearly ${\rm Tr}_{G}(x)={\rm Tr}_{G}(y)$ and so $G$ is not an STI graph.
Assume in the rest that $r\geq 2$. If $r=s=2$, then $G\cong G_{n}$ which was
shown to be an STI graph in [16, Proposition 7]. Hence let $s\geq 3$. If
$P_{1}$, $P_{2}$, and $P_{3}$ are the $x,y$-paths of respective lengths $r$,
$s$, and $t$, and $w$ is the neighbor of $x$ on $P_{3}$, then we can estimate
as follows:

$\displaystyle{\rm Tr}_{G}(w)-{\rm Tr}_{G}(x)$ $\displaystyle=$
$\displaystyle\sum_{z\in{P_{1}}}(d_{G}(w,z)-d_{G}(x,z))+\sum_{z\in{P_{2}\cup
P_{3}}\setminus\\{x,y\\}}(d_{G}(w,z)-d_{G}(x,z))$ $\displaystyle\geq$
$\displaystyle\sum_{z\in{P_{1}}}(d_{G}(w,z)-d_{G}(x,z))\geq 2\,.$

We conclude that $G_{n}$ is the only STI graph with $m=n+1$. $\square$

###### Corollary 2.2

If $G$ is an STI graph of order $n\geq 5$, then the following holds.

  1. (i)

$\displaystyle{n+1\leq{\rm Mo}(G)\leq\frac{n^{2}-1}{4}\,.}$

  2. (ii)

$\displaystyle{4n+1\leq{\rm Z}_{1}(G)\leq\frac{n(n^{2}-1)}{4}\,.}$

  3. (iii)

$\displaystyle{4n+16\leq{\rm
Z}_{2}(G)\leq\left(\frac{n^{2}-1}{4}\right)^{2}\,.}$

Moreover, in each of the cases the left equality holds if and only if
$G=G_{n}$ and the right equality holds if and only if
$G=K_{\frac{n-1}{2},\frac{n+1}{2}}$.

Proof. (i) This follows from the fact that if $G$ is an arbitrary graph, then
${\rm Mo}(G)$ can be equivalently written as $\sum_{uv\in E(G)}|{\rm
Tr}_{G}(u)-{\rm Tr}_{G}(v)|$, see [5, Corollary 2.2].

(ii) By Theorem 2.1, and since $G$ is $2$-connected, $G$ contains at least two
vertices of degree $3$, while the other vertices are of degree at least $2$.
Thus

${\rm Z}_{1}(G)=\sum_{v\in V(G)}\deg(v)^{2}\geq 4(n-2)+9=4n+1\,.$

Equality holds here if and only if $G$ has exactly two vertices of degree $3$
and the other vertices are of degree $2$ which means (having the ear
decomposition in mind) that $G\cong G_{n}$. For the right inequality observe
first that if $uv$ is an edge of (a bipartite graph) $G$, then
$\deg(u)+\deg(v)\leq n$. Therefore,

${\rm Z}_{1}(G)=\sum_{uv\in E(G)}\deg(u)+\deg(v)\leq
n|E(G)|\leq\frac{n(n^{2}-1)}{4}.$

Using Theorem 2.1 again we infer that the right equality holds if and only if
$G\cong K_{\frac{n-1}{2},\frac{n+1}{2}}$.

(iii) This is proved using a similar argument as (ii). $\square$

##  3 Metric properties

In this section we consider extremal behavior of STI graphs with respect to
metric invariants. In the first subsection we begin with their diameter and
determine the extremal graphs. Based on this result, the extremal STI graphs
are determined with respect to the eccentricity. In the second subsection we
prove parallel results for the Wiener index and the transmission.

###  3.1 Eccentricity

###### Lemma 3.1

If $G$ is an STI graph of order $n$, then

$2\leq{\rm diam}(G)\leq\frac{n-1}{2}\,.$

Moreover, the left equality holds if and only if $G\cong
K_{\frac{n-1}{2},\frac{n+1}{2}}$, and the right equality holds for $G_{n}$.

Proof. The left inequality follows from the fact that the only bipartite graph
of diameter $1$ is $K_{2}$ which is not an STI graph. If the left equality
holds, that is, if ${\rm diam}(G)=2$, then $G$ is a complete bipartite graph.
And since $G$ is an STI graph, we conclude that $G\cong
K_{\frac{n-1}{2},\frac{n+1}{2}}$.

By Theorem 1.1, $G$ is a 2-connected graph of odd order which in turn implies
the right inequality. It is clear that ${\rm diam}(G_{n})=(n-1)/2$. $\square$

###### Theorem 3.2

If $G$ is an STI graph of order $n\geq 4$, then

$2n\leq{\rm ecc}(G)\leq\frac{n(n-1)}{2}\,.$

Moreover, the left equality holds if and only if $G\cong
K_{\frac{n-1}{2},\frac{n+1}{2}}$ and the right equality holds if and only if
$G\cong G_{n}$.

Proof. By Lemma 3.1, $2\leq{\rm diam}(G)\leq(n-1)/2$.

If ${\rm diam}(G)=2$, then Theorem 1.1 implies that $G$ is a complete
bipartite graph. Among them only $K_{\frac{n-1}{2},\frac{n+1}{2}}$ is an STI
graph.

As already said, ${\rm diam}(G)\leq(n-1)/2$ which in turn implies that ${\rm
ecc}(G)\leq n\frac{(n-1)}{2}$. Suppose now that the right equality holds for
an STI graph $G$. Then each vertex $v$ has $\varepsilon_{G}(v)=(n-1)/2$. Let
$v$ and $w$ be two vertices with $d_{G}(v,w)=(n-1)/2$. As $G$ is 2-connected,
there exists internally disjoint $v,w$-paths $P_{1}$ and $P_{2}$. Then each of
these paths is of length at least $(n-1)/2$, which means that $|V(P_{1})\cup
V(P_{2})|\geq n-1$. We now distinguish two cases.

Suppose first that $|V(P_{1})|=(n-1)/2$ and $|V(P_{2})|=(n+1)/2$. In this
case, $V(G)=V(P_{1})\cup V(P_{2})$. Since $G$ is an STI graph and cycles are
not, there exists an edge $pq\notin E(P_{1}\cup P_{2})$. As $G$ is bipartite,
$d_{P_{1}\cup P_{2}}(p,q)\geq 3$. But this implies that
$\varepsilon_{G}(p)<(n-1)/2$ which is not possible. In this case we thus have
no equality situation.

Suppose second that $|V(P_{1})|=|V(P_{2})|=(n-1)/2$. Then there exists exactly
one vertex $x\notin V(P_{1}\cup P_{2})$. As $G$ is $2$-connected, $x$ has at
least two neighbors in $V(P_{1})\cup V(P_{2})$, let $y$ and $z$ be its
arbitrary neighbors. If $d_{P_{1}\cup P_{2}}(y,z)\geq 3$, then
$\varepsilon_{G}(y)<(n-1)/2$. Hence we must have $d_{P_{1}\cup P_{2}}(y,z)=2$.
If $\deg(x)=3$, then $\varepsilon_{G}(y)<(n-1)/2$ or
$\varepsilon_{G}(z)<(n-1)/2$. Also, if there exists an edge not in
$E(P_{1})\cup E(P_{2})\cup\\{xy,xz\\}$, say $st$, then
$\varepsilon_{G}(s)<(n-1)/2$. We conclude that $G\cong G_{n}$. $\square$

###  3.2 Wiener index and transmission

###### Theorem 3.3

If $G$ is an STI graph of order $n$, then

$\frac{3n^{2}-4n+1}{4}\leq W(G)\leq\frac{n^{3}-n^{2}-n+17}{8}\,.$

Moreover, the left equality holds if and only if $G\cong
K_{\frac{n-1}{2},\frac{n+1}{2}}$, and the right equality holds if and only if
$G\cong G_{n}$.

Proof. By Theorem 1.1, $G$ is bipartite. Let $V(G)=V_{1}\cup V_{2}$ be its
bipartition. Since adding an edge between two nonadjacent vertices decrease
the Wiener index of the graph, we have $W(G)\geq W(K_{|V_{1}|,|V_{2}|})$.
Moreover, $K_{\frac{n-1}{2},\frac{n+1}{2}}$ is the only complete bipartite of
order $n$ which is also an STI graph, hence the left inequality holds, as well
as its equality part.

For the right inequality, let $H_{n,p,q}$ be a graph of order $n$ comprised of
three internally disjoint paths with the same end-vertices and of respective
lengths $p$, $q$, and $n-p-q+1$. In [10, Theorem 1] it was proved that if
$n\geq 11$, and $G$ is a $2$-connected graph of order $n$ different from
$C_{n}$, $H_{n,1,2}$, and $H_{n,2,2}$, then
$W(G)<W(H_{n,2,2})<W(H_{n,1,2})<W(C_{n})$. Note now that $G_{n}\cong
H_{n,2,2}$ and that $H_{n,1,2}$ is not an STI graph. Hence, as for odd $n$ we
have $W(G_{n})=(n^{3}-n^{2}-n+17)/8$, cf. [10, Table 2], the right inequality
follows together with the equality part. For the latter, one needs to consider
the small cases separately, that is, [10, Theorem 7] for the case $n=9$ and
seven sporadic STI graphs with $n=7$ from [16, Figure 1]. $\square$

###### Theorem 3.4

If $G$ is an STI graph of order $n$, then

$\max\\{{\rm Tr}_{G}(v):\ v\in V(G)\\}\leq\frac{n^{2}-1}{4}\,.$

Moreover, the equality holds if and only if $G\cong G_{n}$.

Proof. Let $v$ be a vertex of $G$ and consider the BFS-tree rooted at $v$.
Since $G$ is 2-connected, there must be at least two vertices in each of the
levels of the BFS-tree, that is, there exists at least two vertices at each
possible distance from $v$. Since $n$ is odd, this implies that

${\rm Tr}_{G}(v)\leq 2(1+\cdots+\frac{n-1}{2})=\frac{n^{2}-1}{4}\,,$

which proves the upper bound.

The equality holds if and only if for each positive integer $k$, $1\leq
k\leq(n-1)/2$, there are exactly two vertices at distance $k$ from $v$ and
$\varepsilon_{G}(v)=(n-1)/2$. Let $G$ be an STI graph for which the equality
holds, and let $v\in V(G)$ be a vertex with ${\rm Tr}_{G}(v)=(n^{2}-1)/4$. Let
$w$ be an eccentric vertex of $v$, so that $d_{G}(v,w)=(n-1)/2$. As $G$ is
$2$-connected, there exist two internally vertex disjoint $v,w$-paths $P_{1}$
and $P_{2}$. Then $V(G)=V(P_{1})\cup V(P_{2})\cup\\{w^{\prime}\\}$, where
$w^{\prime}$ is the second eccentric vertex of $v$. Using the 2-connectivity
of $G$ again, we infer that $w^{\prime}$ is adjacent to the two vertices at
distance $(n-3)/2$ from $v$. Now, if these two edges, together with the edges
of $P_{1}$ and $P_{2}$ are all the edges of $G$, then $G\cong G_{n}$. Suppose
next that $G$ contains some other edge $f$. Since $P_{1}$ and $P_{2}$ are
shortest paths, such an edge connects a vertex $x$ of $P_{1}$ with a vertex
$y$ of $P_{2}$. Suppose that $x$ is selected such that it is closest to $v$
among all veretices that are endpoints of such additional edges $f$. As $G$ is
bipartite this implies that $d_{G}(v,y)=d_{G}(v,x)+1$. Let $x^{\prime}$ be the
predecessor of $x$ on $P_{1}$ (so that $d_{G}(v,x)=d_{G}(v,x^{\prime})+1$). It
is possible that $x^{\prime}=v$.

We claim that ${\rm Tr}(x^{\prime})-{\rm Tr}(x)\geq 2$. By the way the edge
$xy$ is selected, the vertices that are at distance at most $d_{G}(v,x)$ from
$v$ together with the vertex $y$ induce a cycle $C$. Moreover, again by the
way the edge $xy$ is selected, $C$ is an isometric cycle, that is, if
$z,z^{\prime}\in V(C)$, then $d_{C}(z,z^{\prime})=d_{G}(z,z^{\prime})$. It
follows that the vertices of $C$ contribute the same value to ${\rm
Tr}_{G}(x)$ and to ${\rm Tr}_{G}(x^{\prime})$. Since for every vertex
$z^{\prime\prime}$ from $V(G)\setminus V(C)$ we have
$d_{G}(x,z^{\prime\prime})<d_{G}(x^{\prime},z^{\prime\prime})$, the claim
follows.

From the above claim we conclude that there is no such edge $f$ in $G$ and
hence $G\cong G_{n}$. To complete the argument observe that $G_{n}$ contains
exactly one vertex, say $v$, with $\varepsilon_{G}(v)=(n-1)/2$, and for $v$ we
have ${\rm Tr}_{G_{n}}(v)=(n^{2}-1)/4$. $\square$

###### Theorem 3.5

If $G$ is an STI graph of order $n$, then

$\min\\{{\rm Tr}_{G}(v):\ v\in V(G)\\}\geq\frac{3n-5}{2}\,.$

Moreover, equality holds if and only if $G\cong
K_{\frac{n-1}{2},\frac{n+1}{2}}$.

Proof. Let $(X,Y)$ be the bipartition of $G$ and assume without loss of
generality that $|X|\leq|Y|$. Since the order $n$ of an STI graph $G$ is odd,
we actually have $|X|<|Y|$. Let $v\in X$ has the minimum transmission among
vertices of $G$. Then ${\rm Tr}_{G}(v)\geq|Y|+2(|X|-1)$, and equality holds if
and only if $v$ is adjacent to all vertices in $Y$ and is at distance $2$ from
each vertex of $X$. Let $w\in Y$ be adjacent to $v$. Since $G$ is an STI graph
and $|X|<|Y|$, we have ${\rm Tr}_{G}(w)={\rm Tr}_{G}(v)+1=|Y|+2|X|-1$. If
there would be a vertex $z\in X$ non adjacent to $w$, or a vertex $z\in Y$
with $d_{G}(w,z)\geq 4$, then

${\rm Tr}_{G}(w)\geq(|X|-1)+2(|Y|-1)+3=|X|+2|Y|>{\rm Tr}_{G}(v)+1\,,$

which is not possible. Hence $w$ must be adjacent to all vertices of $X$ and
at distance $2$ from each vertex in $Y$. As $w$ was an arbitrary neighbor of
$v$ we conclude that $v$ has the minimum transmission ${\rm
Tr}_{G}(v)=|Y|+2(|X|-1)$ if and only if $G$ is a complete bipartite graph. We
conclude that $G\cong K_{\frac{n-1}{2},\frac{n+1}{2}}$ and ${\rm
Tr}_{G}(v)=(3n-5)/2$. $\square$

## Acknowledgements

Sandi Klavžar acknowledges the financial support from the Slovenian Research
Agency (research core funding P1-0297 and projects N1-0095, J1-1693, J1-2452).

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