EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 5, 2020, 1231-1240
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Special Issue Dedicated to
Professor Hari M. Srivastava

On the Occasion of his 80th Birthday

On the Wiener index of the dot product graph over
monogenic semigroups

Büşra Aydın1, Nihat Akgüneş1,∗, Ismail Naci Cangul2

1 Department of Mathematics-Computer Sciences, Necmettin Erbakan University, Konya,
Turkey
2 Department of Mathematics, Bursa Uludag University, Bursa, Turkey

Abstract. Algebraic study of graphs is a relatively recent subject which arose in two main streams:
One is named as the spectral graph theory and the second one deals with graphs over several
algebraic structures. Topological graph indices are widely-used tools in especially molecular graph
theory and mathematical chemistry due to their time and money saving applications. The Wiener
index is one of these indices which is equal to the sum of distances between all pairs of vertices in
a connected graph. The graph over the finite dot product of monogenic semigroups has recently
been defined and in this paper, some results on the Wiener index of the dot product graph over
monogenic semigroups are given.

2020 Mathematics Subject Classifications: 05C12, 05C25

Key Words and Phrases: Dot product, dot product graph, Wiener index, topological index,
monogenic semigroup

1. Introduction and preliminaries

The connections between graph theory and ring theory was first established in 1988 by
Beck [1]. Since then, different types of algebraic graphs have been introduced and stud-
ied. The most well-known example is the zero divisor graphs. Anderson and Livingston
studied the zero divisor graph of a commutative ring in [2]. Similarly, zero divisor graphs

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v13i5.3745

Email addresses: bsrcgn@gmail.com (B. Aydın),
nakgunes@erbakan.edu.tr (N. Akgünes), cangul@uludag.edu.tr (I. N. Cangul)

https://www.ejpam.com 1231 c© 2020 EJPAM All rights reserved.



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1232

of commutative semigroups have been studied in [3], [5] and [4]. The zero divisor graphs
have been studied intensively, see e.g. [6], [7], [9] and [8]. Further, Badawi has also studied
the dot product graphs of commutative rings in [10]. Akgüneş has defined the graph of
monogenic semigroups in [11] in a similar manner. The dot product graphs of monogenic
semigroups have been studied in [12].

Topological graph indices are very useful tools in graph theory and mathematical
chemistry. By modelling a chemical substance with a graph, we can apply these indices
and obtain physico-chemical properties of that substance solely by means of mathematical
calculations without any experiments in laboratory. The Wiener index is one of the sig-
nificant and surely the oldest topological indices used in mathematical chemistry. It was
introduced in 1947 by Wiener [13] and used for modelling the shape of organic molecules
and for calculating several of their physico-chemical properties, in particular the boiling
points of alkane isomers. Mathematicians are interested in the Wiener index as much as
chemists. Mathematical research on the Wiener index was started in 1976 in [14] and since
then, this distance-based quantity had been studied in [16], [17] and [15].

Now let us recall the definitions of the Wiener index and monogenic semigroups:

The vertex set of a graph G is denoted by V (G). The distance between two vertices u
and v in V (G) is denoted by dG(u, v). Then the Wiener index W (G) of G is defined by

W (G) =
∑

{u,v}⊆V (G)

dG(u, v).

Let Sn be a finite semigroup of order n. In [11], an undirected graph denoted by
d(Sn) = (V,E) was defined as follows: The vertex set consists of the non-zero elements
of Sn and two vertices x and y which are distinct non-zero elements in Sn are adjacent if
and only if xy = 0Sn . Hence

Sn =
{

0, x, x2, · · · , xn
}

where xi, xj ∈ V (d(Sn)) are adjacent ⇔ xi · xj = 0Sn ⇔ xi+j = 0Sn ⇔ i + j > n
(1 ≤ i, j ≤ n).

Some various types of monogenic semigroups including the tensor, lexicographic, strong
and disjunctive products were studied in [20], [21], [19] and [18]. Also some topological
indices of monogenic semigroups were calculated in [24], [22] and [23].

In [12], the dot product graph considered was d(S) where S is the cartesian product
of k times Sn:

S = Sn × Sn × . . .× Sn for k times, where 1 ≤ k <∞.



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1233

Let two nonzero elements of S be X = (xi1 , xi2 , · · · , xik) and Y = (xj1 , xj2 , · · · , xjk)
for {it}kt=1, {jt}kt=1 ∈ {0, 1, 2, · · · , n} where xit = 0Sn if and only if it = 0. A more explicit
definition of the dot product in terms of the elements of S is as follows:

X · Y = (xi1 , xi2 , · · · , xik) · (xj1 , xj2 , · · · , xjk)

= xi1 · xj1 + xi2 · xj2 + · · ·+ xik · xjk = xi1+j1 + xi2+j2 + · · ·+ xik+jk .

Then the dot product graph d(S) can be defined as an (undirected) graph with vertices
X, Y ∈ S∗ = S \ {0S} such that X and Y are adjacent iff X ·Y = 0Sn and this is denoted
by X ∼ Y .

The following result will be used in the rest of the paper:

Lemma 1 ([12]). d(S) is always connected and its diameter is equal to 2, that is

diam(d(S)) = 2.

2. Main results

Let n ∈ N and let Sn =
{

0, x, x2, · · · , xn
}

be a non-empty monogenic semigroup.
Let d(S) be the dot product graph where S = Sn × Sn is the cartesian product of two
Sn’s. Recall that its vertices are the non-zero elements in S. The following result is clear
as the diameter of d(S) is equal to 2 by Lemma 1:

Lemma 2 ([12]). The distance of any two vertices is less than or equal to 2. That is,

dd(S)(vi, vj) ≤ 2

where i, j ∈ N, vi, vj ∈ V (d(S)).

Theorem 1. The Wiener index of the dot product graph over a monogenic semigroup of
order n is that

W (S) =
1

2


(6n− 2)

(
n2 + 2n

)
+
(⌈

n
2

⌉
+ 1
)2 − 1−

∑
k,t∈{0,1,...,n−1}

[2n− k − t + (n− k)(n− t)]

−
∑

k∈{0,1,...,n−1}
t=n

[(n− k)(n + 1) + n]−
∑

t∈{0,1,...,n−1}
k=n

[(n− t)(n + 1) + n]

 .

Proof. Since the diameter of d(S) is equal to 2, we can write dd(S)(vi, vj) ≤ 2 for every
pair i, j ∈ N and vi, vj ∈ V (d(S)). Hence the distance of any two vertices is 1 or 2.
Consequently the Wiener index of d(S) is as follows:

W (d(S)) =
1

2

∑
{vi,vj}⊆V (d(S))

dd(S)(vi, vj).



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1234

Here, let
V1 = {vi : dd(S)(vi, vj) = 1, i, j ∈ N, vj ∈ V (d(S))}

and
V2 = {vi : dd(S)(vi, vj) = 2, i, j ∈ N, vj ∈ V (d(S))}.

Then the Wiener index formula becomes

W (d(S)) =
1

2

∑
v∈V (d(S))

[1 · |V1|+ 2 · |V2|] .

So if we find the cardinalities of the sets V1 and V2, the Wiener index of d(S) will be
calculated. Since the distance of any two vertices in d(S) is 1 or 2, it is clear that |V2| =
|V | − |V1|. Then

W (d(S)) = 1
2

∑
v∈V (d(S))

[1 · |V1|+ 2 · |V2|]

= 1
2

∑
v∈V (d(S))

[1 · |V1|+ 2 · (|V | − |V1|)]

= 1
2

∑
v∈V (d(S))

[1 · |V1|+ 2 · |V | − 2 · |V1|]

= 1
2

∑
v∈V (d(S))

[2 · |V | − |V1|]

and therefore finding |V1| will be enough.

In the next step, let us determine the distance of each vertex to other vertices. If we
can determine how many distances are 1, it means that we also determined the number
of elements of the set V1.

Any pair of two non-zero elements of S will be denoted by (xn−k, xn−t) where k, t ∈
{0, 1, 2, · · · , n}. As a special case, x0 = 0 is assumed. Let us now look at the three
cases where k, t ∈ {0, 1, 2, · · · , n− 1} or k = n or t = n, respectively.

First case: Let k, t ∈ {0, 1, 2, · · · , n − 1}. There are (n − k)(n − t) + 2n − k − t
vertices where the distance of the vertex (xn−k, xn−t) to another vertex (xa, xb) in V (d(S))
is 1.

Indeed, any vertex X = (xa, xb) in V (d(S)) is adjacent to Y = (xn−k, xn−t) for
k, t ∈ {0, 1, 2, · · · , n − 1} if X · Y = 0, so xn−k+a + xn−t+b = 0. This equality
holds if n− k + a > n or a = 0Sn and n− t + b > n or a = 0Sn .

Alternatively, all elements of S can be thought to form a matrix [C](n+1)×(n+1) for
n ∈ N, in [12]:



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1235

[C](n+1)×(n+1) =



. . . c0(t+1) c0(t+2) . . . c0(n−1) c0n
... . . .

...
... . . .

...
...

c(k+1)0 . . . c(k+1)(t+1) c(k+1)(t+2) . . . c(k+1)(n−1) c(k+1)n

c(k+2)0 . . . c(k+2)(t+1) c(k+2)(t+2) . . . c(k+2)(n−1) c(k+2)n
... . . .

...
... . . .

...
...

c(n−1)0 . . . c(n−1)(t+1) c(n−1)(t+2) . . . c(n−1)(n−1) c(n−1)n

cn0 . . . cn(t+1) cn(t+2) . . . cn(n−1) cnn


.

For xa, the values of a may be placed in the zero-row and also in the rows (k + 1) to
n.

For xb, the values of b may be placed in the zero-column and also in the columns (t+1)
to n.

Let us find out the number of vertices at their intersection: (n− (k + 1) + 1)+(n− (t + 1) + 1)+
(n− (k + 1) + 1) (n− (t + 1) + 1) = 2n−k−t+(n− k) (n− t) = 2n−k−t+(n− k) (n− t)
vertex pairs have distance 1.

Second case: Let k = n. Here d((x0, xn−t), (xa, xb)) = d((0, xn−t), (xa, xb)) = 1
so there are (n− t)(n + 1) + n vertices.

Indeed, if k = n and the vertices X = (0, xn−t) and Y = (xa, xb) are adjacent, then
0 +xn−t+b = 0. So n− t+ b > n or b = 0Sn . Hence similarly to the previous case, we have

[C](n+1)×(n+1) =



. . . c0(t+1) c0(t+2) . . . c0(n−1) c0n
... . . .

...
... . . .

...
...

c(k+1)0 . . . c(k+1)(t+1) c(k+1)(t+2) . . . c(k+1)(n−1) c(k+1)n

c(k+2)0 . . . c(k+2)(t+1) c(k+2)(t+2) . . . c(k+2)(n−1) c(k+2)n
... . . .

...
... . . .

...
...

c(n−1)0 . . . c(n−1)(t+1) c(n−1)(t+2) . . . c(n−1)(n−1) c(n−1)n

cn0 . . . cn(t+1) cn(t+2) . . . cn(n−1) cnn


.

For xb, the values of b may be placed in the zero-column and also in the columns (t+1)
to n. So (n− (t + 1) + 1) (n + 1) + n = (n− t) (n + 1) + n = (n− t) (n + 1) + n vertex
pairs have distance 1.

Third case: Let t = n. Now d((xn−k, x0), (xa, xb)) = d((xn−k, 0), (xa, xb)) = 1,
so there are (n− k)(n + 1) + n vertices.

In this case, if t = n and X = (xn−k, 0) and Y = (xa, xb) vertices are adjacent then
xn−k+b + 0 = 0. So n− k + b > n or a = 0Sn . Similarly to the first case, we obtain



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1236

[C](n+1)×(n+1) =



. . . c0(t+1) c0(t+2) . . . c0(n−1) c0n
... . . .

...
... . . .

...
...

c(k+1)0 . . . c(k+1)(t+1) c(k+1)(t+2) . . . c(k+1)(n−1) c(k+1)n

c(k+2)0 . . . c(k+2)(t+1) c(k+2)(t+2) . . . c(k+2)(n−1) c(k+2)n
... . . .

...
... . . .

...
...

c(n−1)0 . . . c(n−1)(t+1) c(n−1)(t+2) . . . c(n−1)(n−1) c(n−1)n

cn0 . . . cn(t+1) cn(t+2) . . . cn(n−1) cnn


.

For xa, the values of a are placed in zero-row and also in the rows (k + 1) to n. So
(n− (k + 1) + 1) (n + 1) + n = (n− k − 1 + 1) (n + 1) + n = (n− k) (n + 1) + n vertex
pairs have distance 1.

Let us be careful about the fact that we have not yet excluded the vertices which are
adjacent to themselves and the distances between such pairs of vertices do not contribute
to Wiener index. So these values should be taken away from the sum we have just found.

Now, let us find how many vertices are adjacent to itself: There are
(
n−

⌊
n
2

⌋)2
+

2
(
n−

⌊
n
2

⌋)
vertices with distance 1 to itself. Hence there are |V |−

[(
n−

⌊
n
2

⌋)2
+ 2

(
n−

⌊
n
2

⌋)]
vertices having distance 2 to itself. When k, t ∈

{
0, 1, 2, · · · ,

⌈
n
2

⌉
− 1,

⌈
n
2

⌉
,
⌈
n
2

⌉
+ 1, · · · , n− 1

}
,

if any vertex X = (xn−k, xn−t) is adjacent to itself, then X · X = 0 so that x2(n−k) +
x2(n−t) = 0. This equality holds when 2(n − k) > n or k = n and 2(n − t)n or t = n. It
means that 2n − 2k > n or k = n and 2n − 2t > n or t = n and equivalently n > 2k or
k = n and n > 2t or t = n. This is equivalent to k < n

2 or k = n and t < n
2 or t = n.

Therefore, we get

d((xn−k, xn−t), (xn−k, xn−t)) = 1⇔ k ∈
{

0, 1, 2, · · · ,
⌈n

2

⌉
− 1, n

}
and

t ∈
{

0, 1, 2, · · · ,
⌈n

2

⌉
− 1, n

}
.

So (n− k) ∈
{
n, n− 1, · · · ,

⌊
n
2

⌋
+ 1, 0

}
and (n− t) ∈

{
n, n− 1, · · · ,

⌊
n
2

⌋
+ 1, 0

}
since

n−
⌈
n
2

⌉
=
⌊
n
2

⌋
. In this case, we obtain the following vertices

[C](n+1)×(n+1) =



. . . c0(bn2 c+1) c0(bn2 c+2) . . . c0(n−1) c0n

... . . .
...

... . . .
...

...
c(bn2 c+1)0 . . . c(bn2 c+1)(bn2 c+1) c(bn2 c+1)(bn2 c+2) · · · c(bn2 c+1)(n−1) c(bn2 c+1)n

c(bn2 c+2)0 . . . c(bn2 c+2)(bn2 c+1) c(bn2 c+2)(bn2 c+2) · · · c(bn2 c+2)(n−1) c(bn2 c+2)n

... · · ·
...

... · · ·
...

...
c(n−1)0 · · · c(n−1)(bn2 c+1) c(n−1)(bn2 c+2) · · · c(n−1)(n−1) c(n−1)n

cn0 · · · cn(bn2 c+1) cn(bn2 c+2) · · · cn(n−1) cnn


.



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1237

Thus there are (n −
⌊
n
2

⌋
)2 + 2(n −

⌊
n
2

⌋
) vertices with a distance of 1 to itself. Hence

there are |V | − [(n−
⌊
n
2

⌋
)2 + 2(n−

⌊
n
2

⌋
)] vertices with a distance of 2 to itself.

Considering the above claims, we obtain the following result:

Since V1 = {vi : vi = (xa, xb), (xa, xb) ∼ (xn−k, xn−t), k, t ∈ {0, 1, 2, . . . , n}},
we find

|V1| =
∑

k,t∈{0,1,...,n−1}

[2n− k − t + (n− k)(n− t)]

+
∑

k∈{0,1,...,n−1}
t=n

[(n− k)(n + 1) + n]

+
∑

t∈{0,1,...,n−1}
k=n

[(n− t)(n + 1) + n] .

Consequently, we conclude that

W (Γ(S)) = 1
2

∑
i,j∈N

dΓ(S)(vi, vj)

= 1
2


∑

vi∈V (d(S))

[2 · |V | − |V1|]− 1 ·
[(
n−

⌊
n
2

⌋)2
+ 2

(
n−

⌊
n
2

⌋)]
−2 ·

[
|V | −

((
n−

⌊
n
2

⌋)2
+ 2

(
n−

⌊
n
2

⌋))]


= 1
2



∑
k,t∈{0,1,...,n−1}

[2 |V | − (n− k)− (n− t)− (n− k)(n− t)]

+
∑

k∈{0,1,...,n−1}
t=n

[2 |V | − (n− k)(n + 1)− n]

+
∑

t∈{0,1,...,n−1}
k=n

[2 |V | − (n− t)(n + 1)− n]− 2 |V |+ 2
[(
n−

⌊
n
2

⌋)2
+ 2

(
n−

⌊
n
2

⌋)]



= 1
2


(6n− 2)

(
(n + 1)2 − 1

)
+
(⌈

n
2

⌉
+ 1
)2 − 1−∑

k,t∈{0,1,...,n−1}

[2n− k − t + (n− k)(n− t)]

−
∑

k∈{0,1,...,n−1}
t=n

[(n− k)(n + 1) + n]−
∑

t∈{0,1,...,n−1}
k=n

[(n− t)(n + 1) + n]


since |V | = (n + 1)2 − 1 and n−

⌊
n
2

⌋
=
⌈
n
2

⌉
.

3. Examples

Let us give the following examples to strengthen the theory.



B. Aydın, N. Akgüneş, I. N. Cangul / Eur. J. Pure Appl. Math, 13 (5) (2020), 1231-1240 1238

Example 1. Let n = 5. Then S = S5 × S5 where S5 =
{

0, x, x2, . . . , x5
}

. Then we
can list the vertices of S as follows:

V (S) =



(0, x) (0, x2) (0, x3) (0, x4) (0, x5)
(x, 0) (x, x) (x, x2) (x, x3) (x, x4) (x, x5)
(x2, 0) (x2, x) (x2, x2) (x2, x3) (x2, x4) (x2, x5)
(x3, 0) (x3, x) (x3, x2) (x3, x3) (x3, x4) (x3, x5)
(x4, 0) (x4, x) (x4, x2) (x4, x3) (x4, x4) (x4, x5)
(x5, 0) (x5, x) (x5, x2) (x5, x3) (x5, x4) (x5, x5)


.

Let now k = 4 and t = 4. The values (xa, xb) satisfying the condition

dΓ(S)((x
5−4, x5−4), (xa, xb)) = dΓ(S)((x, x), (xa, xb)) = 1

are 

(0, x5)

(x5, 0) (x5, x5)


.

Hence there are

(n− k)(n− t) + 2n− k − t = (5− 4)(5− 4) + 2 · 5− 4− 4 = 1 · 1 + 1 + 1 = 3

vertices.

Let us see the same example for another values of k and t: Let k = 1 and t = 2. Then
the pairs of vertices (xa, xb) which satisfy the condition

dΓ(S)((x
5−1, x5−2), (xa, xb)) = dΓ(S)((x

4, x3), (xa, xb)) = 1

are listed as 

(0, x3) (0, x4) (0, x5)

(x2, 0) (x2, x3) (x2, x4) (x2, x5)
(x3, 0) (x3, x3) (x3, x4) (x3, x5)
(x4, 0) (x4, x3) (x4, x4) (x4, x5)
(x5, 0) (x5, x3) (x5, x4) (x5, x5)


.

Since, in the above example, n = 5, k = 1 and t = 2, we obtain the number of vertices
satisfying this condition as

(n− k)(n− t) + 2n− k − t = (5− 1)(5− 2) + 2 · 5− 1− 2 = 4 · 3 + 4 + 3 = 19.



REFERENCES 1239

So for n = 5, the vertices adjacent to itself, in other words, the vertices whose distance to
itself is 1 can be listed as follows:

(0, x3) (0, x4) (0, x5)

(x3, 0) (x3, x3) (x3, x4) (x3, x5)
(x4, 0) (x4, x3) (x4, x4) (x4, x5)
(x5, 0) (x5, x3) (x5, x4) (x5, x5)


.

The total number of these vertex pairs is 15 which can also be found by the formula

(n−
⌊n

2

⌋
)2+2(n−

⌊n
2

⌋
) = (5−

⌊
5

2

⌋
)2+2(5−

⌊
5

2

⌋
) = (5−2)2+2(5−2) = 32+2·3 = 9+6 = 15

vertices.

Consequently, the Wiener index of the Γ(S5 × S5) which is equal to 895, can also be
found by the above formula.

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