2.1 Introduction
Graph Theory is an important branch of discrete mathematics that studies
relationships between objects through structures known as graphs. Among
the fundamental concepts in this field are the independent set, which is a set
of vertices with no edges connecting any pair of them, and the dominating
set, which ensures that every vertex in the graph is either in the set or
adjacent to a vertex in it [19].
Combining these two ideas leads to the concept of the independent
domination number, which represents the minimum size of a set that is both
independent and dominating. This concept plays a significant role in
analyzing the structure of networks, including communication and computer
systems, as it helps achieve efficiency while minimizing interference [25].
This research aims to explore this concept in depth, discuss its main
properties, and highlight some of its practical applications in various
scientific fields.
2.2. Independent dominating set
Definition 2.2.1. [25] A set π· is an independent dominating set if it is both
independent and dominating set.
Definition 2.2.2.[25] A dominating set π· is called a minimal independent
dominating set if no proper subset π ⊂ π· is independent dominating set.
Definition 2.2.3. [25] The independent domination number πΎπ (πΊ) of a
graph πΊ equals the minimum cardinality of a set of minimal independent
dominating set of πΊ. Such a set is called the πΎπ −set of πΊ.
Proposition 2.2.4.
(i)
πΎπ (πΎπ ) = 1
(ii)
πΎπ (ππ ) = πΎπ (πΆπ ) = ⌈ ⌉
(iii)
πΎπ (ππ ) = 1
π
3
1, ππ π ππ π = 1
(iv) πΎπ (πΎπ,π ) = {
, πΎ (π ) = 1, where
πππ{π, π}, ππ‘βπππ€ππ π π π
ππ ≡ πΎ1,π−1
(v) πΎπ (ππ ) = π.
Proof.
(i) and (iii) Let πΊ be graph of order π, then πΎπ (πΎπ ) = πΎπ (ππ ) = 1.
Proof:
Let πΊ be graph of order π, then πΎπ (πΎπ ) = πΎπ (ππ ) = 1. Let π· = {π£}, where
v is any vertex in the graph πΎπ and is the center in the wheel graph. Thus,
the vertex v dominates all other vertices and it is obvious that the set d has
minimum cardinality. Therefore, πΎπ (πΎπ ) = πΎπ (ππ ) = 1.
(ii) Proof. Let {π£1 , π£2 , … , π£π }be the vertices of a path graph ( as an example,
see Figure 1).
π£1
π£2
π£3
π£4
π£5
π£6
π£7
π£8
π£9
Figure 1: Path of order 9 (π9 )
Now, to create the minimum dominating set we follow the following steps:
The vertex π£1 dominates to π£1 and π£2 , while the vertex π£2 dominates to
π£1 , π£2 and π£3 , so the vertex π£2 will be the chosen one in the dominating set.
As same manner, the vertex π£5 dominates to π£4 , π£5 and π£6 , so the vertex π£5
will be the chosen one in the dominating set and so on… let π· =
π
{π£2+3π , π = 0,1,2, … , ⌈ 3 ⌉ − 1}, then there are three two cases depend on n as
follows.
Case 1. If π ≡ 0 (πππ 3), then the set D is a independent dominating set
and has minimum cardinality, since each vertex in it has the maximum
closed neighborhood ( as an example, see Figure 2).
π£1
π£3
π£2
π£5
π£4
π£8
π£7
π£6
π£9
Figure 2: The independent dominating set of the path of order 9 (π9 )
π
Thus, πΎπ (ππ ) = .
3
Case 2. If π ≡ 2 (πππ 3), then again the set π· is an independent
dominating set and has minimum cardinality, since each vertex in it has the
maximum closed neighborhood except the vertex π£π dominates to only two
vertices π£π and π£π−1 ( as an example, see Figure 3)
π£1
π£2
π£3
π£4
π£5
π£6
π£7
π£8
π£10
π£9
π£11
Figure 3: The independent dominating set of the path of order 11
(π11 ).
π
Thus, πΎπ (ππ ) = ⌈ ⌉.
3
Case 3. If π ≡ 1 (πππ 3), then the last vertex in the set π· will come out of
π
the vertices of a path, then let π·1 = {π£2+3π , π = 0,1,2, … , ⌈ ⌉ − 2}. As same
3
way in case 1, the set dominates all the vertices of the path except the vertex
π£π . Thus, the set π·1 ∪ {π£π } is an independent dominating set and has
minimum cardinality(as an example, see Figure 4)
π£1
π£2
π£3
π£4
π£5
π£6
π£7
π£8
π£9
π£10
Figure 4: The independent dominating set of the path of order 10
(π10 ).
π
Thus, πΎπ (ππ ) = ⌈ ⌉.
3
From all cases above, the result is obtained.
(iv) Proof: πΎπ (πΎπ,π ) = πππ{π, π}; π, π ≥ 2, since the two vertices which
are chosen to determine the domination number are adjacent. Thus, to
determine an independent dominating set, there are two choice the set X or
Y. Thus, πΎπ (πΎπ,π ) = πππ{π, π}; π, π ≥ 2.
Proposition 2.25: [25] Let πΊ be a graph of order n, then
1) Every isolated vertex belong to each dominating set.
2) If there is a vertex of degree n-1, then πΎ(πΊ) = 1.
(v) Proof. According to Proposition 2.25, each isolated vertex belongs to a
dominating set. Thus, πΎπ (ππ ) = π
2.3 A Study of Operations on the Independent Domination Number of
Graphs
Proposition 2.3.1: Let πΊ be a graph such that πΊ ≡ ππ − π then πΎπ (πΊ) =
π
π
3
3
⌈ 1⌉ + ⌈ 2⌉
Proof: Let {π£1 , π£2 , … , π£π }be the vertices of a path graph ππ , since a path
graph is connected graph has π vertices and π − 1 edges then removing an
edge splits the graph into exactly two disjoint path graphs ππ1 and ππ2 total
number of vertices π = π1 + π2 . Therefore, πΊ ≡ ππ1 ∪ ππ2 then
independent domination number of its two components, thus:
πΎπ (πΊ) = πΎπ (ππ1 ∪ ππ2 ) = πΎπ (ππ1 ) + πΎπ (ππ2 )
π
π
3
3
We get πΎπ (πΊ) = ⌈ 1⌉ + ⌈ 2⌉
Illustrative Example:
1- If π ≡ 0 (πππ3) then π = 3,6,9,12,15, … ..
Let π = 6
π£1
π£3
π£2
π£1
π6 − π1
π£4
π£3
π£2
π£5
π£6
π£4
π£5
π£6
Then πΎπ (π6 − π1 ) = 1 + 2 = 3
2- If π ≡ 1 (πππ3) then π = 1,4,7,10,13, … ..
Let π = 7
π£1
π£2
π£3
π£1
π7 − π2
π£5
π£4
π£2
π£3
π£6
π£7
π£4
π£5
π£6
π£7
π£8
π£9
π£10
Then πΎπ (π7 − π2 ) = 1 + 2 = 3
3- If π ≡ 2 (πππ3) then π = 2,5,8,11,14, … ..
If π = 11
π£1
π£2
π£3
π£4
π£5
π£6
π£7
π£11
π11 − π3
π£1
π£2
π£3
π£1
π£2
π£3
π£4
π£5
π£6
π£7
π£8
Then πΎπ (π11 − π3 ) = 1 + 3 = 4
Proposition 2.3.2: Let πΊ be a graph such that πΊ ≡ πΆπ − π then πΎπ (πΊ) ≡
π
πΎπ (ππ ) = ⌈ ⌉
3
Proof: when we delete an edge π from cycle graph then πΆπ − π ≡ ππ and
π
π
3
3
since πΎπ (ππ ) = ⌈ ⌉, we get πΎπ (πΆπ − π) ≡ πΎπ (ππ ) = ⌈ ⌉.
Proposition 2.3.3: Let πΊ be a graph such that πΊ ≡ ππ + π then πΎπ (πΊ) ≡
π
πΎπ (πΆπ ) = ⌈ ⌉
3
Proof: Let π = π’π£ such that π’ and π£ are leaf vertices such that size of path
graph π = π − 1 after add an edge π then π = π then πΎπ (πΊ) ≡ πΎπ (πΆπ ) =
π
⌈ ⌉
3
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