An Upper Bound for the Independent Domination Number of a Connected Graph

1. Introduction

We consider only finite undirected graphs without loops or multiple edges. Notation and terminology not defined here follow those in [1]. For a graph $G=(V,E)$, we use n and e to denote its order $\left|V\right(G\left)\right|$ and size $\left|E\right(G\left)\right|$, respectively. The minimum degree of G is denoted by $\delta \left(G\right)$. The neighborhood of a vertex u in G is denoted by $N_G\left(u\right)$. For a subset S of V, we use $G\left[S\right]$ to denote the subgraph of G which is induced by S. We define $N\left(S\right)$ as ${\cup }_{u\in S}{N}_G\left(u\right)$. If a vertex $u\in V-S$, where S is a subset of the vertex set V, we define $d_{G\left[S\right]}\left(u\right)$ as $|N_G\left(u\right)\cap S|$. For disjoint vertex subsets $S_1$ and $S_2$ of V, We define $E(S_1,S_2)$ as $\{e:e=pq\in E,p\in S_1,q\in S_2\}$. A subset T of the vertex set V of G is called an independent set if no two vertices in T are adjacent in G. A subset D of the vertex set V of G is called a dominating set if every vertex v in G is either in the set D or is adjacent to a vertex in D. An independent dominating set in G is a subset of V such that it is both independent and dominating. The independent domination number of G, denoted $i\left(G\right)$, is the minimum size of an independent dominating set in G. In this note, we present an upper bounds for the independent domination number of a connected graph. We need to define the following families of graphs to state our results. The first family of the graphs is denoted by $\mathcal{F}$$(n,i(G\left)\right)$ and it is defined as $\{G=(V,E):|V|=n$ and there exists an independent dominating set X in G with $X=i\left(G\right)$ such that $\left|N\right(v,X\left)\right|\ge 1$ for each $v\in X$, where $N(v,X)=\{u\in V-X:N(u)\cap X=\{v\left\}\right\}$. The second family of the graphs is denoted by $\mathcal{H}$$(n,i(G\left)\right)$ and it is defined as $\{G=(V,E):V=X\cup A\cup B$, where $X,A,B$ are pairwise disjoint, $X=\{x_1,x_2,\dots ,x_{i\left(G\right)}\}$ is independent, $A=\{a_1,a_2,\dots ,a_{i\left(G\right)}\}$, $N\left(a_t\right)\cap X=\left\{x_t\right\}$ for each t with $1\le t\le i\left(G\right)$, $B=\{b_1,b_2,\dots ,b_{n-2i\left(G\right)}\}$, $x_p{b}_q\in E$ for each p with $1\le p\le i\left(G\right)$ and for each q with $1\le q\le n-2i\left(G\right)$, $G[V-X]$ is complete}. The main result of this note is as follows.

Theorem 1. 

Let G be a connected graph graph of order $n\ge 2$ and size e. If $G\in \mathcal{F}$$(n,i(G\left)\right)$, then

$$i\left(G\right)\le {\displaystyle \frac{3+\sqrt{9+12(n^2-n-2e)}}{6}}$$

with equality if and only if $G\in \mathcal{H}$$(n,i(G\left)\right)$.

2. Proofs

We use some ideas in the proof of Theorem 8 in [2] in our proofs.

Proof of Theorem 1. 

Since $G\in \mathcal{F}$$(n,i(G\left)\right)$, we can find an independent dominating set X in G with $X=i\left(G\right)$ such that $\left|N\right(v,X\left)\right|\ge 1$ for each $v\in X$, where $N(v,X)=\{u\in V-X:N(u)\cap X=\{v\left\}\right\}$. Thus we have that

$$e=\left|E\right(X,V-X\left)\right|+\left|E\right(G[V-X]\left)\right|$$

$$\le \left|E(X,V-X)\right|+{\displaystyle \frac{1}{2}}(n-i\left(G\right))(n-i\left(G\right)-1)$$

$$\le \sum _{v\in X}\left|N(v,X)\right|+\left|X\right|\left(n-\left|X\right|-\sum _{v\in X}\left|N(v,X)\right|\right)+{\displaystyle \frac{1}{2}}(n-i\left(G\right))(n-i\left(G\right)-1)$$

$$=i\left(G\right)(n-i\left(G\right))-(i\left(G\right)-1)\sum _{v\in X}\left|N(v,X)\right|+{\displaystyle \frac{1}{2}}(n-i\left(G\right))(n-i\left(G\right)-1)$$

$$\le i\left(G\right)(n-i\left(G\right))-i\left(G\right)(i\left(G\right)-1)+{\displaystyle \frac{1}{2}}(n-i\left(G\right))(n-i\left(G\right)-1).$$

Solving the inequality, we have that

$$i\left(G\right)\le {\displaystyle \frac{3+\sqrt{9+12(n^2-n-2e)}}{6}}.$$

If $i\left(G\right)={\displaystyle \frac{3+\sqrt{9+12(n^2-n-2e)}}{6}}$, we, from the above proofs, have that

$$\left|E\right(X,V-X\left)\right|$$

$$=\sum _{v\in X}\left|N(v,X)\right|+\left|X\right|\left(n-\left|X\right|-\sum _{v\in X}\left|N(v,X)\right|\right)$$

$$=i\left(G\right)(n-i\left(G\right))-(i\left(G\right)-1)\sum _{v\in X}\left|N(v,X)\right|$$

$$=i\left(G\right)(n-i(G\left)\right)-i\left(G\right)\left(i\right(G)-1).$$

Thus ${\sum }_{v\in X}\left|N(v,X)\right|=i\left(G\right)$. Since $\left|N\right(v,X\left)\right|\ge 1$ for each $v\in X$, we have that $\left|N\right(v,X\left)\right|=1$ for each $v\in X$. Set $A={\cup }_{x\in X}N(v,X)$. Then $\left|A\right|=\left|X\right|=i\left(G\right)$. Define $A:=\{a_1,a_2,...,a_{i\left(G\right)}\}$. Then $a_t$ has a unique neighbor in X for each t with $1\le t\le i\left(G\right)$. Without loss of generality, we can assume $N\left(a_t\right)\cap X=\left\{x_t\right\}$ for each t with $1\le t\le i\left(G\right)$. Set $B:=V-(X\cup A):=\{b_1,b_2,\dots ,b_{n-2i\left(G\right)}\}$. We also have that $\left|E\right(X,V-(X\cup A)|=\left|E\right(X,B|=\left|X\right|\left(n-\left|X\right|-{\sum }_{v\in X}\left|N(v,X)\right|\right)=\left|X\right|\left|B\right|$. Thus $x_p{b}_q\in E$ for each p with $1\le p\le i\left(G\right)$ and for each q with $1\le q\le n-2i\left(G\right)$. We, from the above proofs, also have that

$$\left|E\left(G[V-X]\right)\right|={\displaystyle \frac{1}{2}}(n-i\left(G\right))(n-i\left(G\right)-1).$$

Thus $G[V-X]$ is complete. Hence $G\in \mathcal{H}$$(n,i(G\left)\right)$.

If $G\in \mathcal{H}$$(n,i(G\left)\right)$, then $e=i\left(G\right)+i\left(G\right)(n-2i\left(G\right))+{\displaystyle \frac{1}{2}}(n-i\left(G\right))(n-i\left(G\right)-1)$. Thus $2e=n^2-n-3i^2\left(G\right)+3i\left(G\right)$. Therefore

$${\displaystyle \frac{3+\sqrt{9+12(n^2-n-2e)}}{6}}=i\left(G\right).$$

This completes the proof of Theorem 1. □