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Abstract

A zero forcing set is a new concept in Graph Theory which was introduced in recent years. In this paper, we investigate the relationship between zero forcing sets and algebraic hyperstructures. To this end, we present some new definitions by considering a zero forcing process on a graph G. These definitions help us analyze the zero forcing process better and construct various hypergroups and join spaces on the vertex set of graph G. Finally, we give some examples to clarify these hyperstructures.
1
Some Algebraic Hyperstructures Related to Zero Forcing
Sets and Forcing Digraphs
M. Golmohamadian , M. M. Zahedi , N. Soltankhah
Department of Mathematics, Tarbiat Modares University, Tehran, Iran. E-mail address:
m.golmohamadian@modares.ac.ir
Department of Mathematics, Kerman Graduate University of Advanced Technology, Kerman,
Iran. E-mail address: zahedi_mm@kgut.ac.ir
Department of Mathematics, Alzahra University, Tehran, Iran. E-mail address:
soltan@alzahra.ac.ir
Abstract
A zero forcing set is a new concept in Graph Theory which was introduced in recent years. In this
paper, we investigate the relationship between zero forcing sets and algebraic hyperstructures.
To this end, we present some new definitions by considering a zero forcing process on a graph
𝐺. These definitions help us analyze the zero forcing process better and construct various
hypergroups and join spaces on the vertex set of graph 𝐺. Finally, we give some examples to
clarify these hyperstructures.
Keywords: Zero forcing set, Zero forcing process, Forcing digraph, Hypergroup, Join space.
Mathematics Subject Classification (2010): 05C50; 20N20.
1. Introduction and Preliminaries
PLEASE CITE AS: M. Golmohamadian, M.M. Zahedi and N. Soltankhah. Some Algebraic Hyper
structures Related to Zero Forcing Sets and Forcing Digraphs. J. of Algebra and its Appl,Vol:18,
No:10 (2019) 1950192 (19 pages), DOI: 10-1142/S0219498819501925
2
Let 𝐺 be a simple, undirected graph on the vertex set 𝑉 and each vertex of a graph 𝐺 be given
one of two colors, say, “black” and “white”. Let 𝑍 denote the (initial) set of black vertices of 𝐺.
The color-change rule converts the color of a vertex from white to black if the white vertex is the
only white neighbor of a black vertex. The set 𝑍 is said to be a zero forcing set of 𝐺 if all vertices
of 𝐺 will be turned black after finitely many applications of the color-change rule. This process
changes the color of all white vertices to black and is called the zero forcing process. We will often
use the adjective ‘forcing’ instead of ‘zero forcing’.
The forcing process is an instance of a propagation process on graphs. Such processes are a
common topic across mathematics and computer science (see [8, 13, 15]). In fact, diverse graph
processes are used to model technical or societal processes such as physics [3] and social network
analysis [10]. For an overview of the different models and applications, refer to the book [4].
The zero forcing process was introduced in [1] to bound the minimum rank, or equivalently, the
maximum nullity of a graph 𝐺. The purpose of the present paper is to construct new kinds of
commutative hypergroups and join spaces by considering the concept of zero forcing sets and
zero forcing process on a graph. The correlation between hyperstructures and Graph Theory also
has been investigated by several researches, such as Corsini [5], Davvaz [7], Golmohamadian [9],
Kalampakas [11-12], Leoreanu [14], Rosenberg [16], Spartalis [17], and so on.
In this paper, we want to continue the research on the zero forcing sets with new trend. To
achieve this goal, by taking idea from the set of white vertices that their color change to black in
each step of forcing process, we define two hyperoperations "𝑜 " and "𝑜" on 𝑉(𝐺). Then we
show that the hypergroupoids (𝑉,𝑜) and (𝑉,𝑜) are commutative hypergroups and join spaces.
Also, we introduce a forcing digraph 𝐺 and we define zero forcing roots and zero forcing leaves
of one vertex of 𝑉(𝐺). These new definitions made a way for us to present three hyperoperations
"", "" and "" on the vertex set of graph 𝐺. Then, we show that (𝑉,∗) is a commutative
hypergroup and a join space but (𝑉,∗) and (𝑉,∗) are just commutative hypergroups.
Now we briefly recall some of the basic concepts of the theories of graph and algebraic
hyperstructures and we refer to [2, 6] for more details.
let 𝐻 be a non-empty set and 𝑜:𝐻×𝐻𝒫(𝐻) be a hyperoperation, where 𝒫(𝐻) is the family
of non-empty subsets of 𝐻. The couple (𝐻,𝑜) is called a hypergroupoid.
Definition 1.1. [6] A hypergroupoid (𝐻,𝑜) is called a semihypergroup if for all 𝑎,𝑏,𝑐 of 𝐻 we
have (𝑎𝑜𝑏)𝑜𝑐=𝑎𝑜(𝑏𝑜𝑐), which means that
𝑢𝑜𝑐
∈ =𝑎𝑜𝑣
∈ .
Also (𝐻,𝑜) is called a quasihypergroup if for all 𝑎 of 𝐻 we have:
3
𝑎𝑜𝐻=𝐻𝑜𝑎=𝐻
Definition 1.2. [6] A hypergroupoid (𝐻,𝑜) which is both a semihypergroup and a
quasihypergroup is called a hypergroup. Furthermore, a hypergroup (𝐻,𝑜) is called commutative
if for all 𝑎, 𝑏𝐻, it holds 𝑎𝑜𝑏=𝑏𝑜𝑎.
In order to define a join space, we recall the following notation:
If 𝑎,𝑏 are elements of a hypergroup (𝐻, 𝑜), then we denote 𝑎𝑏
={𝑥𝐻|𝑎𝑥𝑜𝑏}.
Definition 1.3. [6] A commutative hypergroup (𝐻,𝑜) is called a join space if the following
condition holds for all elements of 𝑎,𝑏,𝑐,𝑑 of 𝐻:
𝑎𝑏
𝑐𝑑
𝑎𝑜𝑑𝑏𝑜𝑐
We also know that a graph 𝐺 is a triple consisting of a vertex set 𝑉(𝐺), an edge set 𝐸(𝐺), and a
relation that associates with each edge two vertices called its endpoints.
We draw a graph on paper by placing each vertex at a point and representing each edge by a
curve joining the locations of its endpoints.
Definition 1.4. [18] A cycle is a graph with an equal number of vertices and edges whose vertices
can be placed around a circle so that two vertices are adjacent if and only if they appear
consecutively along the circle.
Definition 1.5. [18] A graph with no cycle is acyclic. A jungle is an acyclic graph.
Definition 1.6. [18] A directed graph or digraph 𝐺 is a triple consisting of a vertex set 𝑉(𝐺), an
edge set 𝐸(𝐺), and a function assigning each edge an ordered pair of vertices. The first vertex of
the ordered pair is the tail of the edge, and the second is the head; together, they are the
endpoints. We say that an edge is an edge from its tail to its head. We write 𝑢𝑣 for an edge with
tail 𝑢 and head 𝑣. We also write 𝑢𝑣 for “there is an edge from 𝑢 to 𝑣”.
Definition 1.7. [18] The underlying graph of a digraph 𝐷 is the graph 𝐺 obtained by treating the
edges of 𝐷 as unordered pairs; the vertex set and edge set remain the same, and the endpoints
of an edge are the same in 𝐺 as in 𝐷, but in 𝐺 they become an unordered pair.
In what follows, we give more details to better understand the concepts of zero forcing set and
zero forcing process.
let 𝐺 be a graph in which each vertex is initially colored either black or white. If 𝑢 is a black vertex
of 𝐺 and 𝑢 has exactly one white neighbor, say 𝑣, then we change the color of 𝑣 to black; this
rule is called the color change rule. In this case we say “𝑢 𝑓𝑜𝑟𝑐𝑒𝑠 𝑣” which is denoted by 𝑢 𝑣
and the edge 𝑢𝑣 is called a forcing edge. Given an initial coloring of 𝐺, in which a set of the
vertices is black and all other vertices are white, the derived set is the set of all black vertices
4
resulting from repeatedly applying the color-change rule until no more changes are possible. If
the derived set for a given initial subset of black vertices is the entire vertex set of the graph, then
the set of initial black vertices is called a zero forcing set. The zero forcing number of a graph 𝐺 is
the size of the smallest zero forcing set of 𝐺; it is denoted by 𝑍(𝐺).
Example 1.8: let 𝐺 be the following graph
The following figures show that why the set {𝑣,𝑣} is a zero forcing set for 𝐺.
Let 𝑍 be a zero forcing set of a graph 𝐺. Construct the derived set, making a list of the forces in
the order in which they are performed. This list is called the chronological list of forces. A forcing
chain (for a particular chronological list of forces) is a sequence of vertices (𝑣,𝑣,,𝑣) such
that 𝑣𝑣, for 𝑖 = 1, …, 𝑘 − 1.
For the graph 𝐺 in above Example, we have the forcing chains (𝑣,𝑣,𝑣) and (𝑣,𝑣). A maximal
forcing chain is a forcing chain that is not a proper subsequence of another zero forcing chain
(the previous example has two maximal forcing chains). Note that a zero forcing chain can consist
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
5
of a single vertex (𝑣) and such a chain is maximal if 𝑣𝑍 and 𝑣 does not perform a force. In
each step of a forcing process, each vertex can force at most one other vertex. It is proved that
the derived set of as given set of black vertices is unique; however, a chronological list of forces
and the forcing chains of a particular zero forcing set usually is not. The number of chains in a
zero forcing process starting with a zero forcing set 𝑍 is equal to the size of 𝑍 and the elements
of 𝑍 are the initial vertices of the forcing chains.
2. Zero Forcing Sets and Hyperstructures
In this section, by considering the set of white vertices which are colored by black in 𝑖’𝑡ℎ step of
forcing process, we define two commutative hypergroups. Next, we show that these hypergroups
are also join spaces.
Definition 2.1. Let 𝐺=(𝑉(𝐺), 𝐸(𝐺)) be a graph and 𝑍 be a zero forcing set of 𝐺. Then we define
the class of 𝑇 as follows:
𝑇={𝑢𝑉(𝐺)| 𝑇ℎ𝑒 𝑐𝑜𝑙𝑜𝑟 𝑜𝑓 𝑢 𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑡𝑜 𝑏𝑙𝑎𝑐𝑘 𝑖𝑛 𝑡ℎ𝑒 𝑖𝑡ℎ 𝑠𝑡𝑒𝑝 𝑜𝑓 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑝𝑟𝑜𝑐𝑒𝑠𝑠}
In each step of forcing process, we consider all vertices that their color change to black by all
black vertices of previous steps. So, 𝑇 is unique and it is easy to see that 𝑍=𝑇.
Definition 2.2. Let 𝐺=(𝑉(𝐺), 𝐸(𝐺)) be a graph and 𝑍 be a zero forcing set of 𝐺. Then we define
the hyperoperation 𝑜𝑉×𝑉𝒫(𝑉) as follows:
For every 𝑢,𝑣𝑉 and 𝑖,𝑗{0}, we have
𝑢 𝑜 𝑣=𝑇
 if 𝑢𝑇, 𝑣𝑇 and 𝑙=max{𝑖,𝑗}
In fact, this hyperoperation shows the set of all vertices which have black color while 𝑣 and 𝑣
are colored by black at the first time.
Theorem 2.3. Let 𝐺 =(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then the hypergroupoid
(𝑉,𝑜) is a commutative hypergroup.
Proof. By definitions of the hyperoperation "𝑜" , it is easy to see that the hypergroupoid (𝑉,𝑜) is
commutative. First, we show that "𝑜 " is associative, i.e. (𝑢 𝑜 𝑤) 𝑜 𝑣= 𝑢 𝑜 (𝑤 𝑜 𝑣) for
all 𝑢, 𝑤, 𝑣 ∈𝑉(𝐺). Let 𝑢𝑇, 𝑤𝑇,𝑣𝑇 and 𝑙=max{𝑖,𝑗,𝑘}
According to the definition of hyperoperation "𝑜 " , we have
(𝑢 𝑜 𝑤) 𝑜 𝑣=𝑇
 =𝑢 𝑜 (𝑤 𝑜 𝑣)
Now, it is enough to prove that 𝑢 𝑜 𝑉=𝑉 𝑜 𝑢=𝑉 for all 𝑢𝑉(𝐺). We know that, for every
𝑢,𝑣𝑉(𝐺), {𝑢,𝑣}𝑢 𝑜 𝑣 . So, we have
6
𝑉 𝑜 𝑢= 𝑉=𝑢 𝑜 𝑉
and the hypergroupoid (𝑉,𝑜) is a commutative hypergroup.
Theorem 2.4. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then the hypergroup
(𝑉,𝑜) is a join space.
Proof. By Theorem 2.3, we found that (𝑉,𝑜) is a commutative hypergroup. According to the
definition of join space, we must prove
𝑣𝑣
 ∩𝑣𝑣
 ≠𝑣 𝑜 𝑣𝑣 𝑜 𝑣
for every 𝑣,𝑣,𝑣,𝑣𝑉(𝐺). Let 𝑣𝑇, 𝑣𝑇,𝑣𝑇,𝑣𝑇, 𝑘=max{𝑖,𝑗} and 𝑙=
max{𝑚,𝑛}. Then according to the definition of hyperoperation "𝑜", for every 𝑣,𝑣,𝑣,𝑣
𝑉(𝐺), in every situation, we have
𝑍=𝑇𝑣 𝑜 𝑣𝑣 𝑜 𝑣=𝑇
 𝑇

So, (𝑉,𝑜) is a join space.
Definition 2.5. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Also, assume that we
need 𝑡 step of forcing process to change the color of all white vertices to black. Then we define
the hyperoperation 𝑜 ∶ 𝑉 × 𝑉𝒫(𝑉) as follows:
For every 𝑢,𝑣𝑉 and 𝑖,𝑗{0}, we have
𝑢 𝑜𝑣=𝑇
 if 𝑢𝑇, 𝑣𝑇 and 𝑙=min {𝑖,𝑗}
Theorem 2.6. Let 𝐺=(𝑉, 𝐸) be a graph, 𝑍 be a zero forcing set of 𝐺 and we need 𝑡 step of
forcing process to change the color of all white vertices to black. Then the hypergroupoid (𝑉,𝑜)
is a commutative hypergroup.
Proof. By definitions of the hyperoperation "𝑜" , it is obvious that the hypergroupoid (𝑉,𝑜) is
commutative. First, we show that "𝑜" is associative, i.e. (𝑢 𝑜 𝑤) 𝑜𝑣=𝑢 𝑜(𝑤 𝑜 𝑣) for
all 𝑢, 𝑤, 𝑣 ∈𝑉(𝐺). Let 𝑢𝑇, 𝑤𝑇,𝑣𝑇 and 𝑙=min{𝑖,𝑗,𝑘}
According to the definition of hyperoperation "𝑜" , we have
(𝑢 𝑜 𝑤) 𝑜𝑣=𝑇
 = 𝑢 𝑜(𝑤 𝑜 𝑣)
Now, it is enough to prove that 𝑢 𝑜𝑉=𝑉𝑜𝑢=𝑉 for all 𝑢𝑉(𝐺). We know that, for every
𝑢,𝑣𝑉(𝐺), {𝑢,𝑣}𝑢 𝑜𝑣 . So, we have
𝑉𝑜𝑢=𝑉=𝑢 𝑜𝑉
7
and the hypergroupoid (𝑉,𝑜) is a commutative hypergroup.
Theorem 2.7. Let 𝐺=(𝑉, 𝐸) be a graph, 𝑍 be a zero forcing set of 𝐺 and we need 𝑡 step of
forcing process to change the color of all white vertices to black. Then the hypergroup (𝑉,𝑜) is
a join space.
Proof. By Theorem 2.6, we found that (𝑉,𝑜) is a commutative hypergroup. Now, we must show
𝑣𝑣
 ∩𝑣𝑣
 ≠𝑣 𝑜𝑣𝑣 𝑜𝑣
for every 𝑣,𝑣,𝑣,𝑣𝑉(𝐺). Let 𝑣𝑇, 𝑣𝑇,𝑣𝑇,𝑣𝑇, 𝑘=min{𝑖,𝑗} and 𝑙=
min{𝑚,𝑛}. Then, for every 𝑣,𝑣,𝑣,𝑣𝑉(𝐺), in every situation, we have
𝑇𝑣 𝑜𝑣𝑣 𝑜𝑣=𝑇
 𝑇

So, (𝑉,𝑜) is a join space.
Example 2.8. let 𝐺 be the following graph and 𝑍={𝑣,𝑣} be a zero forcing set of it
The following figures show the steps of zero forcing process and shows the forcing edge
in each step:
Step 1
Step 2 Step 3
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
8
Step 4
So, according to above shapes we have
𝑇={𝑣,𝑣},𝑇={𝑣},𝑇={𝑣},𝑇={𝑣},𝑇={𝑣}
Now, following tables show the effect of hyperoperations "𝑜 " and "𝑜" on the vertices of graph
𝐺.
"
𝒐
"
𝒗
𝟏
𝒗
𝟐
𝒗
𝟑
𝒗
𝟒
𝒗
𝟓
𝒗
𝟔
𝒗
𝟏
{
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟐
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟑
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟒
{
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟓
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟔
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
"
𝒐
"
𝒗
𝟏
𝒗
𝟐
𝒗
𝟑
𝒗
𝟒
𝒗
𝟓
𝒗
𝟔
𝒗
𝟏
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟐
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟑
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟒
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
𝒗
𝟓
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
}
{
𝑣
,
𝑣
}
𝒗
𝟔
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
,
𝑣
}
{
𝑣
,
𝑣
}
{
𝑣
,
𝑣
}
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
9
3. Forcing Digraphs and Hyperstructures
In what follows, we expand the research on making relationships between zero forcing sets and
hyperstructures. First, by considering an arbitrary zero forcing set 𝑍 on a graph 𝐺 and it’s forcing
process, we construct a directed graph with forcing edges and all vertices of 𝐺 and we call it a
forcing digraph 𝐺. Then we define zero forcing roots and zero forcing leaves of one vertex. These
definitions provide the opportunity for us to define three hyperoperations on the vertex set of
𝐺. Then we show that all of these hypergroupoids are hypergroups but just one of them is a join
space.
Definition 3.1. let 𝐺=(𝑉(𝐺),𝐸(𝐺)) be a graph and 𝑍 be a zero forcing set of 𝐺. Then by
considering the forcing process of zero forcing set 𝑍, we construct a directed graph 𝐺=(𝑉,𝐸)
as follows:
- 𝑉=𝑉 (𝐺)
- 𝑢𝑣𝐸⟺ 𝑢 𝑓𝑜𝑟𝑐𝑒𝑠 𝑣
We call this directed graph a forcing digraph of 𝑍 on 𝐺 or briefly a forcing digraph 𝐺.
Theorem 3.2. let 𝐺=(𝑉(𝐺),𝐸(𝐺)) be a graph and 𝑍 be a zero forcing set of 𝐺. Then the
underlying graph of forcing digraph 𝐺=(𝑉,𝐸) is a jungle.
Proof. Let there is a cycle in the underlying graph of digraph 𝐺. Then we have two possible
shapes of this cycle in 𝐺:
The first shape cannot happen because every vertex in graph 𝐺 forces at most one vertex but in
this shape 𝑣 forces two vertices which is impossible.
Now, let 𝑣 be the first vertex in the second cycle which is colored by black. According to above
shape, 𝑣 is forced by 𝑢. So, 𝑢 was colored by black before 𝑣 and 𝑣 is not the first vertex which is
colored by black. Thus, it is a contradiction and the second shape also cannot happen.
By above discussion, we find that there is no cycle in the underlying graph of forcing digraph 𝐺
and it is a jungle.
Definition 3.3. let 𝐺=(𝑉(𝐺),𝐸(𝐺)) be a graph and 𝑍 be a zero forcing set of 𝐺 . The set of zero
forcing
roots of vertex 𝑣𝑉(𝐺) is denoted by (𝑣) and is defined as follows:
𝒖
𝒗
𝟐
𝒖
𝒗
1
10
(𝑣)={𝑢𝑉(𝐺)| 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑐ℎ𝑎𝑖𝑛 𝑖𝑛 𝐺 𝑓𝑟𝑜𝑚 𝑢 𝑡𝑜 𝑣}∪ {𝑣}
We also define ℒ(𝑣) as the set of zero forcing leaves of vertex 𝑣 by
(𝑣)={𝑢𝑉(𝐺)| 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡𝑠 𝑎 𝑓𝑜𝑟𝑐𝑖𝑛𝑔 𝑐ℎ𝑎𝑖𝑛 𝑖𝑛 𝐺 𝑓𝑟𝑜𝑚 𝑣 𝑡𝑜 𝑢}∪ {𝑣}
Example 3.4. let 𝐺 be the following graph and 𝑍={𝑣,𝑣,𝑣,𝑣} be a zero forcing set of 𝐺.
The following figures show the steps of zero forcing process and shows the forcing edge
in each step:
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
11
Now, we could draw the forcing digraph 𝐺 as follows:
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣
𝑣
𝑣
𝑣
𝑣
𝑣
𝑣

𝑣

𝑣

12
According to above shape, we have
(𝑣)={𝑣,𝑣,𝑣,𝑣,𝑣,𝑣}, (𝑣)={𝑣,𝑣,𝑣}
(𝑣)={𝑣,𝑣,𝑣},(𝑣)={𝑣,𝑣,𝑣,𝑣}
Theorem 3.5. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. If 𝑢(𝑣), then
(𝑢)(𝑣).
Proof. According to the coloring rule, each vertex could force at most one vertex. So, in 𝐺, all
leaves of each vertex are on the unique path. So, if 𝑢(𝑣), then (𝑢)(𝑣).
Definition 3.6. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then we define the
hyperoperation :𝑉×𝑉𝒫(𝑉) as follows:
For every 𝑢, 𝑣𝑉(𝐺) 𝑢𝑣=(𝑢)(𝑣)
This hyperoperation shows a set of vertices that change the color of these vertices to black,
depending on the blackness of the colors of 𝑢 and 𝑣.
Theorem 3.7. Let 𝐺=(𝑉,𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then the hypergroupoid
(𝑉,∗) is a commutative hypergroup.
Proof. By definitions of the hyperoperation "" , it is easy to see that the hypergroupoid (𝑉,∗)
is commutative. First, we show that "" is associative, i.e. (𝑢𝑤)𝑣=𝑢(𝑤𝑣) For
all 𝑢, 𝑤, 𝑣 ∈𝑉(𝐺). According to the definition of hyperoperation "" , we have
(𝑢𝑤)𝑣=[(𝑢)(𝑤)]𝑣=(𝑢)
∈ℒ()∪ℒ() ℒ(𝑣)
For every 𝑢ℒ(𝑢)ℒ(𝑤), we have
𝑒𝑖𝑡ℎ𝑒𝑟 𝑢(𝑢) 𝑜𝑟 𝑢(𝑤)
So, By Theorem 3.5, we get that
𝑒𝑖𝑡ℎ𝑒𝑟 ℒ(𝑢)(𝑢) 𝑜𝑟 ℒ(𝑢)ℒ(𝑤)
We also know that 𝑢,𝑤(𝑢)(𝑤). So, by above discussion we conclude that
(𝑢𝑤)𝑣=(𝑢)(𝑤)(𝑣). Similarly, we could show that 𝑢(𝑤𝑣)=(𝑢)
(𝑤)(𝑣) . Thus "" is associative.
Now, it is enough to prove that 𝑢𝑉=𝑉𝑢=𝑉 for all 𝑢 ∈𝑉(𝐺). For every 𝑣𝑉(𝐺), we
have 𝑣(𝑣) . Thus, we have
𝑉𝑢=(𝑣)
∈() (𝑢)=𝑢𝑉𝑢𝑉=𝑉=𝑉𝑢
13
Now, we conclude that the hypergroupoid (𝑉,∗) is a commutative hypergroup.
Theorem 3.8. Let 𝐺=(𝑉,𝐸) be a graph. Then (𝑉,∗) is a join space.
Proof. By Theorem 3.7, we found that (𝑉,∗) is a commutative hypergroup. According to the
definition of a join space, we must prove
𝑣𝑣
 ∩𝑣𝑣
 ≠𝑣𝑣𝑣𝑣
for every 𝑣,𝑣,𝑣,𝑣𝑉(𝐺).
Since 𝑣𝑣
 ∩𝑣𝑣
 ≠, we assume that 𝑢𝑣𝑣
 ∩𝑣𝑣
and 𝑢𝑉(𝐺). We get that
𝑣∈ 𝑢 ∗𝑣=ℒ(𝑢)ℒ(𝑣)
𝑣𝑢𝑣=(𝑢)ℒ(𝑣)
Now, we have the following possible situations:
- 𝑣(𝑢) 𝑎𝑛𝑑 𝑣ℒ(𝑢)
By Theorem 3.5, we know that all leaves of each vertex are on the unique path. So
𝑒𝑖𝑡ℎ𝑒𝑟 ℒ(𝑣)(𝑣) 𝑜𝑟 ℒ(𝑣)(𝑣)(𝑣)(𝑣)
𝑣𝑣𝑣𝑣=[(𝑣)(𝑣)][ℒ(𝑣)(𝑣)]
- 𝑣(𝑣)
We know that for every 𝑤𝑉(𝐺). We have 𝑤(𝑤). Thus, we get that
𝑣𝑣𝑣𝑣𝑣=[(𝑣)(𝑣)][ℒ(𝑣)(𝑣)]
- 𝑣∈ ℒ (𝑣)
The proof is similar to the previous situation and we have
𝑣𝑣𝑣𝑣𝑣=[(𝑣)(𝑣)][ℒ(𝑣)(𝑣)]
We conclude that (𝑉,∗) is a join space.
Theorem 3.9. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. If 𝑢ℛ(𝑣), then
(𝑢)(𝑣).
Proof. Let 𝑤(𝑢). Then there is a forcing chain from 𝑤 to 𝑢. We know that there is also a
forcing chain from 𝑢 to 𝑣. So, there exists a forcing chain from 𝑤 to 𝑣 and 𝑤ℛ(𝑣). Thus,
(𝑢)(𝑣).
Definition 3.10. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then we define the
hyperoperation :𝑉×𝑉𝒫(𝑉) as follows:
14
For every 𝑢, 𝑣𝑉(𝐺) 𝑢𝑤=(𝑢)(𝑣)
This hyperoperation shows the set of vertices that the blackness of 𝑢 and 𝑣 is dependent on the
blackness of these vertices.
The proof of the next Theorem is similar to Theorem 3.7.
Theorem 3.11. Let 𝐺=(𝑉,𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then the
hypergroupoid (𝑉,∗) is a commutative hypergroup.
From the following example, we obtain that (𝑉,∗) is not a join space.
Example 3.12. In the following graph 𝐺=𝐺 and 𝑍={𝑢,𝑢,𝑢,𝑢}.
Since 𝑣𝑢𝑣=(𝑢)(𝑣) and 𝑣𝑢𝑣=(𝑢)(𝑣) , then we have
𝑢𝑣𝑣
 ∩𝑣𝑣
 
But according to the above shape we get that
𝑣𝑣𝑣𝑣=[(𝑣)(𝑣)][(𝑣)(𝑣)]=
This shows that in this graph, (𝑉,∗) is not a join space.
Definition 3.13. Let 𝐺=(𝑉, 𝐸) be a graph, 𝑣,𝑣𝑉 and 𝑍 be a zero forcing set of 𝐺. Then we
say that 𝑣 and 𝑣 are affected by each other if there exists a forcing chain between them (from
𝑣 to 𝑣 or from 𝑣 to 𝑣). Since each vertex could force at most one vertex so the forcing chain
between two distinct vertices is unique. Thus, the set of vertices which appear in this chain is
unique too and we denote this set by "𝐶ℎ𝑎𝑖𝑛,".
Definition 3.14. Let 𝐺=(𝑉, 𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then we define the
hyperoperation *𝑉×𝑉𝒫(𝑉) as follows:
For every 𝑢, 𝑣𝑉(𝐺)
𝑢
𝑣
𝑢
𝑢
𝑣
𝑢
𝑣
𝑢
𝑣
15
𝑢𝑣=𝐶ℎ𝑎𝑖𝑛. 𝑖𝑓 𝑢𝑣 𝑎𝑛𝑑 𝑢, 𝑣 𝑎𝑟𝑒 𝑎𝑓𝑓𝑒𝑐𝑡𝑒𝑑 𝑏𝑦 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟
{𝑢,𝑣} 𝑖𝑓 𝑢≠ 𝑣 𝑎𝑛𝑑 𝑢, 𝑣 𝑎𝑟𝑒 𝑛𝑜𝑡 𝑎𝑓𝑓𝑒𝑐𝑡𝑒𝑑 𝑏𝑦 𝑒𝑎𝑐ℎ 𝑜𝑡ℎ𝑒𝑟
{𝑢} 𝑖𝑓 𝑢=𝑣
Theorem 3.15. Let 𝐺=(𝑉,𝐸) be a graph and 𝑍 be a zero forcing set of 𝐺. Then the
hypergroupoid (𝑉,∗) is a commutative hypergroup.
Proof: By definition of hyperoperation "" , it is easy to see that the hypergroupoid (𝑉,∗) is
commutative. Also, for every 𝑢,𝑣 𝑉, we have
{𝑢,𝑣}𝑢𝑣𝑢𝑉=𝑉=𝑉𝑢
Now, it is enough to check the associativity of " ∗ ", i.e. (𝑢𝑣)𝑤=𝑢(𝑣𝑤) for all 𝑢, 𝑣, 𝑤 ∈
𝑉 . To achieve this aim, we check the following situations for every 𝑢, 𝑣, 𝑤𝑉:
- 𝑢=𝑣=𝑤
(𝑢𝑣)𝑤=(𝑢𝑢)𝑢=({𝑢})𝑢={𝑢}=𝑢({𝑢})=𝑢(𝑢𝑢)=𝑢(𝑣𝑤)
- 𝑢=𝑣𝑤, 𝑢 and 𝑤 are affected by each other
(𝑢𝑣)𝑤=(𝑢𝑢)𝑤=𝑢𝑤=𝐶ℎ𝑎𝑖𝑛, =𝑢𝐶ℎ𝑎𝑖𝑛,=𝑢(𝑣𝑤)
- 𝑢=𝑣𝑤, 𝑢 and 𝑤 are not affected by each other
(𝑢𝑣)𝑤=(𝑢𝑢)𝑤={𝑢}𝑤={𝑢,𝑤}=𝑢({𝑢,𝑤})=𝑢(𝑣𝑤)
The proofs of the following situations are the same as above situations:
- 𝑢=𝑤𝑣, 𝑢 and 𝑣 are affected by each other
- 𝑢=𝑤𝑣, 𝑢 and 𝑣 are not affected by each other
- 𝑣=𝑤𝑢, 𝑢 and 𝑣 are affected by each other
- 𝑣=𝑤𝑢, 𝑢 and 𝑣 are not affected by each other
To check the next situations, we consider that 𝑢𝑣𝑤𝑢.
- None of the vertices are affected by each other
(𝑢𝑣)𝑤=({𝑢,𝑣})𝑤={𝑢,𝑣,𝑤}=𝑢({𝑣,𝑤})=𝑢(𝑣𝑤)
- 𝑢 and 𝑣 are affected by each other but 𝑤 is not affected by 𝑢 and 𝑣
In this case, let there is a chain between 𝑤 and one of the vertices in 𝐶ℎ𝑎𝑖𝑛,. Then according
to the following shape (the direction of the arrows can change), we find that 𝑤 is affected by
𝑢 or 𝑣 which is a contradiction. So 𝑤 is not affected by any vertices in the 𝐶ℎ𝑎𝑖𝑛,.
16
Thus, we have
(𝑢𝑣)𝑤=𝐶ℎ𝑎𝑖𝑛,𝑤=𝐶ℎ𝑎𝑖𝑛,{𝑤}=𝑢({𝑣,𝑤})=𝑢(𝑣𝑤)
- 𝑣 and 𝑤 are affected by each other but 𝑢 is not affected by 𝑣 and 𝑤
Similar to the previous situation, we get that
(𝑢𝑣)𝑤=({𝑢,𝑣})𝑤=𝐶ℎ𝑎𝑖𝑛,{𝑢}=𝑢𝐶ℎ𝑎𝑖𝑛,=𝑢(𝑣𝑤)
- 𝑢 and 𝑤 are affected by each other but 𝑣 is not affected by 𝑢 and 𝑤
Similarly, in this case, we have
(𝑢𝑣)𝑤=({𝑢,𝑣})𝑤=𝐶ℎ𝑎𝑖𝑛,{𝑣}=𝑢({𝑣,𝑤})=𝑢(𝑣𝑤)
- 𝑢 and 𝑣 are affected by each other, 𝑢 and 𝑤 are affected by each other but 𝑣 is not
affected by 𝑤
In this situation, there is a chain between 𝑢 and 𝑤. There is also a chain between 𝑢 and 𝑣
and there is not a chain between 𝑣 and 𝑤. So, we have
If the chain between 𝑢 and 𝑣 is from 𝑢 to 𝑣 and the chain between 𝑢 and 𝑤 is from 𝑢
to 𝑤. Then since each vertex can force at most one vertex, we have
𝑒𝑎𝑡ℎ𝑒𝑟 𝐶ℎ𝑎𝑖𝑛, 𝐶ℎ𝑎𝑖𝑛, 𝑜𝑟 𝐶ℎ𝑎𝑖𝑛,𝐶ℎ𝑎𝑖𝑛,
Thus, there is a chain between 𝑣 and 𝑤, which is a contradiction.
If the direction of the chain between 𝑢 and 𝑣 is from 𝑣 to 𝑢 and the direction of the
chain between 𝑢 and 𝑤 is from 𝑢 to 𝑤. Then, it is obvious that there is a chain between
𝑣 and 𝑤, which is a contradiction.
If the chain between 𝑢 and 𝑣 is from 𝑢 to 𝑣 and the chain between 𝑢 and 𝑤 is from 𝑤
to 𝑢. Then, it is easy to see that there is a chain between 𝑣 and 𝑤, which is a
contradiction.
From above discussion, we find that the chain between 𝑢 and 𝑣 is from 𝑣 to 𝑢 and the chain
between 𝑢 and 𝑤 is from 𝑤 to 𝑢. These two chains are disjoint and there is not any chain
𝑢
𝑣
𝑤
17
between the vertices of these two chains. Because in the following shapes there exists one
vertex that forces two vertices which is impossible.
So, the only shape of this situation is as follows:
Therefore, we have
(𝑢𝑣)𝑤=𝐶ℎ𝑎𝑖𝑛,𝑤=𝐶ℎ𝑎𝑖𝑛,𝐶ℎ𝑎𝑖𝑛, =𝑢({𝑣,𝑤})=𝑢(𝑣𝑤)
- 𝑢 and 𝑣 are affected by each other, 𝑣 and 𝑤 are affected by each other but 𝑢 is not
affected by 𝑤
By similar discussion of the previous situation, the only shape of this situation is the same as
following shape:
So, we have
(𝑢𝑣)𝑤=𝐶ℎ𝑎𝑖𝑛,𝑤=𝐶ℎ𝑎𝑖𝑛,𝐶ℎ𝑎𝑖𝑛, =𝑢𝐶ℎ𝑎𝑖𝑛,=𝑢(𝑣𝑤)
- 𝑢 and 𝑤 are affected by each other, 𝑣 and 𝑤 are affected by each other but 𝑢 is not
affected by 𝑣
Similar to the previous situation, we have the following shape:
𝑢
𝑣
𝑤
𝑢
𝑣
𝑤
𝑢
𝑣
𝑤
𝑣
𝑢
𝑤
18
According to the above shape, we get that
(𝑢𝑣)𝑤=({𝑢,𝑣})𝑤=𝐶ℎ𝑎𝑖𝑛,𝐶ℎ𝑎𝑖𝑛, =𝑢𝐶ℎ𝑎𝑖𝑛,=𝑢(𝑣𝑤)
- 𝑢 and 𝑣 are affected by each other, 𝑢 and 𝑤 are affected by each other and 𝑣 and 𝑤 are
affected by each other
The only possible shape for this situation is a chain between 𝑢,𝑣 and 𝑤. It means that for 𝑖, 𝑗, 𝑘 ∈
{𝑢,𝑣,𝑤}and 𝑖𝑗𝑘𝑖, we have the below shape
Thus, we have
(𝑢𝑣)𝑤=𝐶ℎ𝑎𝑖𝑛,𝑤=𝐶ℎ𝑎𝑖𝑛, =𝑢𝐶ℎ𝑎𝑖𝑛,=𝑢(𝑣𝑤)
The proof is complete and we conclude that (𝑉,∗) is a commutative hypergroup.
The following example shows that (𝑉,∗) is not a join space.
Example 3.16. Let 𝐺=(𝑉,𝐸) be the following graph and 𝑍={𝑢,𝑢}.
Then the shape of 𝐺=(𝑉,𝐸) is the following shape
𝑤
𝑣
𝑢
𝑗
𝑘
𝑖
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
𝑢
19
According to above shape and the definition of hyperoperation "", we have the following table
"
"
𝒖
𝟏
𝒖
𝟐
𝒖
𝟑
𝒖
𝟒
𝒖
𝟓
𝒖
𝟔
𝒖
𝟏
{
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
,
𝑢
}
{
𝑢
,
𝑢
,
𝑢
,
𝑢
}
𝒖
𝟐
{
𝑢
,
𝑢
}
{
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
,
𝑢
}
𝒖
𝟑
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
𝒖
𝟒
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
,
𝑢
}
𝒖
𝟓
{
𝑢
,
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
}
{
𝑢
,
𝑢
}
𝒖
𝟔
{
𝑢
,
𝑢
,
𝑢
,
𝑢
}
{
𝑢
,
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
,
𝑢
,
𝑢
}
{
𝑢
,
𝑢
}
{
𝑢
}
Above table shows that
𝑢𝑢𝑢 and 𝑢𝑢𝑢𝑢𝑢𝑢
 ∩𝑢𝑢
 
But according to the definition of hyperoperation "", we have
𝑢𝑢𝑢𝑢={𝑢,𝑢}{𝑢,𝑢}=
So, (𝑉,∗) is not a join space.
Acknowledgement
This work was partially supported by Center of Excellence of Algebraic Hyperstructures and its
Applications of Tarbiat Modares University (CEAHA).
References
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20
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Constanta, Seria Matematica. 22 (2014) 141-153.
[13] D. Kempe, J. Kleinberg and E. Tardos, Maximizing the spread of influence in a social network, In ACM SIGKDD
International Conference on Knowledge Discovery and Data Mining (KDD) (2003) 137 – 146.
[14] V. Leoreanu, L. Leoreanu, Hypergroups associated with hypergraphs, Italian Journal of Pure and Applied
Mathematcs. 4 (1998) 119-126.
[15] Z. Montazeri, N. Soltankhah, Zero forcing number for cartesian product of some graphs, submitted.
[16] I. Rosenberg, Hypergroups induced by pathes of a directed graph, Italian Journal of Pure and Applied
Mathematics. 4 (1998).
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[18] D. B. West, Introduction to Graph Theory, 2/E, Pearson Education, Delhi, India, 2002.
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