The Italian Domination Numbers of Some Products of Directed Cycles

Abstract

An Italian dominating function on a digraph D with vertex set $V\left(D\right)$ is defined as a function $f:V\left(D\right)\to \{0,1,2\}$ such that every vertex $v\in V\left(D\right)$ with $f\left(v\right)=0$ has at least two in-neighbors assigned 1 under f or one in-neighbor w with $f\left(w\right)=2$. In this article, we determine the exact values of the Italian domination numbers of some products of directed cycles.

1. Introduction and Preliminaries

Let $D=(V,A)$ be a finite simple digraph with vertex set $V=V\left(D\right)$ and arc set $A=A\left(D\right)$. An arc joining v to w is denoted by $v\to w$. The maximum out-degree and maximum in-degree of D are denoted by ${\Delta }^{+}\left(D\right)$ and ${\Delta }^{-}\left(D\right)$, respectively.

Let $D_1=(V_1,A_1)$ and $D_2=(V_2,A_2)$ be two digraphs. The cartesian product of $D_1$ and $D_2$ is the digraph $D_1\square D_2$ with vertex set $V_1\times V_2$ and for two vertices $(u_1,u_2)$ and $(v_1,v_2)$,

$$(u_1,u_2)\to (v_1,v_2)$$

if one of the following holds: (i) $u_1=v_1$ and $u_2\to v_2$; (ii) $u_1\to v_1$ and $u_2=v_2$.

The strong product of $D_1$ and $D_2$ is the digraph $D_1\otimes D_2$ with vertex set $V_1\times V_2$ and for two vertices $(u_1,u_2)$ and $(v_1,v_2)$,

$$(u_1,u_2)\to (v_1,v_2)$$

if one of the following holds: (i) $u_1\to v_1$ and $u_2\to v_2$; (ii) $u_1=v_1$ and $u_2\to v_2$; (iii) $u_1\to v_1$ and $u_2=v_2$.

An Italian dominating function (IDF) on a digraph D is defined as a function $f:V\left(D\right)\to \{0,1,2\}$ such that every vertex $v\in V\left(D\right)$ with $f\left(v\right)=0$ has at least two in-neighbors assigned 1 under f or one in-neighbor w with $f\left(w\right)=2$. In other words, we say that a vertex v for which $f\left(v\right)\in \{1,2\}$ dominates itself, while a vertex v with $f\left(v\right)=0$ is dominated by f if it has at least two in-neighbors assigned 1 under f or one in-neighbor w with $f\left(w\right)=2$. An Italian dominating function $f:V\left(D\right)\to \{0,1,2\}$ gives a partition $\{V_0,V_1,V_2\}$ of $V\left(D\right)$, where $V_i:=\{x\in V\left(D\right)\mid f\left(x\right)=i\}$. The weight of an Italian dominating function f is the value $\omega \left(f\right)=f\left(V\left(D\right)\right)={\sum }_{u\in V\left(D\right)}f\left(u\right)$. The Italian domination number of a digraph D, denoted by ${\gamma }_I\left(D\right)$, is the minimum taken over the weights of all Italian dominating functions on D. A ${\gamma }_I\left(D\right)$-function is an Italian dominating function on D with weight ${\gamma }_I\left(D\right)$.

The Italian dominating functions in graphs and digraphs have been studied in [1,2,3,4,5,6,7]. The authors of [2] introduce the concept of Italian domination and give bounds, relating the Italian domination number to some other domination parameters. The authors of [3] characterize the trees T with $\gamma \left(T\right)+1={\gamma }_I\left(T\right)$ and also characterize the trees T with ${\gamma }_I\left(T\right)=2\gamma \left(T\right)$. After that, there are some studies on the cartesian products of undirected cycles or undirected paths in [4,8,9,10]. Recently, the author of [6] initiated the study of the Italian domination number in digraphs. In this article, we investigate the Italian domination numbers of $C_m\square C_n$ and $C_m\otimes C_n$.

The following results are useful to our study.

Proposition 1

([6]). Let D be a digraph of order n. Then ${\gamma }_I\left(D\right)\ge \lceil \frac{2n}{2+{\Delta }^{+}\left(D\right)}\rceil$.

Proposition 2

([6]). Let D be a digraph of order n. Then ${\gamma }_I\left(D\right)\le n$, and ${\gamma }_I\left(D\right)=n$ if and only if ${\Delta }^{+}\left(D\right),{\Delta }^{-}\left(D\right)\le 1$.

Proposition 3

([6]). If D is a directed path or a directed cycle of order n, then ${\gamma }_I\left(D\right)=n$.

2. The Italian Domination Numbers of Some Products of Directed Cycles

In this section, we determine the exact values of ${\gamma }_I(C_m\square C_n)$ and ${\gamma }_I(C_m\otimes C_n)$.

First, we consider $C_m\square C_n$. We denote the vertex set of a directed cycle $C_m$ by $\{1,2,\cdots ,m\}$, and assume that $i\to i+1$ is an arc of $C_m$. For every vertex $(i,j)\in V(C_m\square C_n)$, the first and second components are considered modulo m and n, respectively. For each $1\le k\le n$, we denote by $C_m^k$ the subdigraph of $C_m\square C_n$ induced by the set $\left\{\right(j,k)\mid 1\le j\le m\}$. Note that $C_m^k$ is isomorphic to $C_m$. Let f be a ${\gamma }_I(C_m\square C_n)$-function and set $a_k={\sum }_{x\in V\left(C_m^k\right)}f\left(x\right)$. Then ${\gamma }_I(C_m\square C_n)={\sum }_{k=1}^n{a}_k$. It is easy to see that $C_m\square C_n$ is isomorphic to $C_n\square C_m$. So, ${\gamma }_I(C_m\square C_n)={\gamma }_I(C_n\square C_m)$.

Theorem 1.

If $m=2r$ and $n=2s$ for some positive integers $r,s$, then ${\gamma }_I(C_m\square C_n)=\frac{mn}{2}$.

Proof. 

Define $f:V(C_m\square C_n)\to \{0,1,2\}$ by

$$f\left(\right(2i-1,2j-1\left)\right)=f\left(\right(2i,2j\left)\right)=1$$

for each $1\le i\le r$ and $1\le j\le s$, and

$$f\left(\right(x,y\left)\right)=0$$

otherwise. It is easy to see that f is an IDF of $C_m\square C_n$ with weight $\frac{mn}{2}$ and so ${\gamma }_I(C_m\square C_n)\le \frac{mn}{2}$. Since ${\Delta }^{+}\left(D\right)=2$, it follows from Proposition 1 that ${\gamma }_I(C_m\square C_n)\ge \frac{mn}{2}$. Thus, we have ${\gamma }_I(C_m\square C_n)=\frac{mn}{2}$. □

Theorem 2.

For an odd integer $n\ge 3$, ${\gamma }_I(C_2\square C_n)=n+1$.

Proof. 

Define $f:V(C_2\square C_n)\to \{0,1,2\}$ by

$$f\left(\right(1,2j-1\left)\right)=1$$

for each $1\le j\le \frac{n+1}{2}$,

$$f\left(\right(2,2j\left)\right)=1$$

for each $1\le j\le \frac{n-1}{2}$,

$$f\left(\right(2,n\left)\right)=1$$

and

$$f\left(\right(x,y\left)\right)=0$$

otherwise. It is easy to see that f is an IDF of $C_2\square C_n$ with weight $n+1$ and so ${\gamma }_I(C_2\square C_n)\le n+1$.

Now we claim that ${\gamma }_I(C_2\square C_n)\ge n+1$. Suppose to the contrary that ${\gamma }_I(C_2\square C_n)\le n$. Let f be a ${\gamma }_I(C_2\square C_n)$-function. If $a_k=0$ for some k, assume without loss of generality $k=3$, then $f\left(\right(1,3\left)\right)=f\left(\right(2,3\left)\right)=0$. To dominate the vertices $(1,3)$ and $(2,3)$, we must have $f\left(\right(1,2\left)\right)=f\left(\right(2,2\left)\right)=2$. Define $g:V(C_2\square C_n)\to \{0,1,2\}$ by

$$g\left(\right(1,2\left)\right)=g\left(\right(2,1\left)\right)=g\left(\right(2,3\left)\right)=1,g\left(\right(2,2\left)\right)=0$$

and

$$g\left(\right(x,y\left)\right)=f\left(\right(x,y\left)\right)$$

otherwise. Then g is an IDF of $C_2\square C_n$ with a weight less than $\omega \left(f\right)$, which is a contradiction. Thus, $a_k\ge 1$ for each k. By assumption, $a_k=1$ for each k. Without loss of generality, we assume that $f\left(\right(1,2\left)\right)=1$. To dominate $(2,2)$, we must have $f\left(\right(2,1\left)\right)=1$. Since $a_3=1$ and $f\left(\right(2,2\left)\right)=0$, we have $f\left(\right(2,3\left)\right)=1$. By repeating this process, we obtain $f\left(\right(1,2i\left)\right)=1$ for each $1\le i\le \frac{n-1}{2}$, $f\left(\right(2,2i-1\left)\right)=1$ for $1\le i\le \frac{n+1}{2}$ and $f\left(\right(x,y\left)\right)=0$ otherwise. However, the vertex $(1,1)$ is not dominated, and this is a contradiction. Thus we have ${\gamma }_I(C_2\square C_n)\ge n+1$. This completes the proof. □

Theorem 3.

For an integer $n\ge 3$, ${\gamma }_I(C_3\square C_n)=2n$.

Proof. 

When $n=3r$ for some positive integer r, define $f_0:V(C_3\square C_n)\to \{0,1,2\}$ by

$$f_0\left((1,3j+1)\right)=f_0\left((2,3j+1)\right)=1$$

for each $0\le j\le n-1$,

$$f_0\left((2,3j+2)\right)=f_0\left((3,3j+2)\right)=1$$

for each $0\le j\le n-1$,

$$f_0\left((1,3j+3)\right)=f_0\left((3,3j+3)\right)=1$$

for each $0\le j\le n-1$ and

$$f_0\left((x,y)\right)=0$$

otherwise.

When $n=3r+1$ for some positive integer r, define $f_1:V(C_3\square C_n)\to \{0,1,2\}$ by

$$f_1\left((2,n)\right)=f_1\left((3,n)\right)=1$$

and

$$f_1\left((x,y)\right)=f_0\left((x,y)\right)$$

otherwise.

When $n=3r+2$ for some positive integer r, define $f_2:V(C_3\square C_n)\to \{0,1,2\}$ by

$$f_2\left((1,n-1)\right)=f_2\left((2,n-1)\right)=f_2\left((1,n)\right)=f_2\left((3,n)\right)=1$$

and

$$f_2\left((x,y)\right)=f_0\left((x,y)\right)$$

otherwise. It is easy to see that $f_i$$(i=0,1,2)$ is an IDF of $C_3\square C_n$ with weight $2n$ and so ${\gamma }_I(C_3\square C_n)\le 2n$.

Now we prove that ${\gamma }_I(C_3\square C_n)\ge 2n$. Let f be a ${\gamma }_I(C_3\square C_n)$-function.

Claim 1.

$a_k\ge 1$ for each $1\le k\le n$.

Suppose to the contrary that $a_k=0$ for some k, say $k=n$. To dominate $(1,n)$, $(2,n)$ and $(3,n)$, we must have $f\left(\right(1,n-1\left)\right)=f\left(\right(2,n-1\left)\right)=f\left(\right(3,n-1\left)\right)=2$. However, the function g defined by

$$g\left(\right(1,n-1\left)\right)=g\left(\right(2,n-1\left)\right)=g\left(\right(3,n-1\left)\right)=1,$$

$$g\left(\right(1,n\left)\right)=g\left(\right(2,n\left)\right)=1$$

and

$$g\left(\right(x,y\left)\right)=f\left(\right(x,y\left)\right)$$

otherwise, is an IDF of $C_3\square C_n$ with a weight less than $\omega \left(f\right)$. This is a contradiction. □

We choose a ${\gamma }_I(C_3\square C_n)$-function h so that the size of $M_h:=\{k\mid a_k=1\}$ is as small as possible.

Claim 2.

$|M_h|=0$.

Suppose to the contrary that $|M_h|\ge 1$. Without loss of generality, assume that $a_n=1$ and $h\left(\right(1,n\left)\right)=1$. To dominate $(2,n)$ and $(3,n)$, we must have $h\left(\right(2,n-1\left)\right)=1$ and $h\left(\right(3,n-1\left)\right)=2$. If $n=3$, then clearly $a_1\ge 2$ and so ${\gamma }_I(C_3\square C_3)\ge 6$. However, when $n=3$, the previously defined function $f_0$ is a ${\gamma }_I(C_3\square C_3)$-function such that $\omega \left(f_0\right)=6$ and $|M_{f_0}|=0$. This contradicts the choice of h. From now on, assume $n\ge 4$. We divide our consideration into the following two cases.

Case 1. $a_{n-2}=1$.

By the same argument as above, we have $a_{n-3}\ge 3$. So $a_{n-3}+a_{n-2}+a_{n-1}+a_n\ge 8$. If $n=4$, then the previously defined function $f_1$ induces a contradiction. Suppose $n\ge 5$. Since $a_{n-4}\ge 1$ by Claim 1, $h\left(\right(i,n-4\left)\right)=1$ or 2 for some $i\in \{1,2,3\}$. Without loss of generality, we may assume $h\left(\right(1,n-4\left)\right)=1$ or 2. Define $t:V(C_3\square C_n)\to \{0,1,2\}$ by

$$t\left(\right(1,n-3\left)\right)=t\left(\right(2,n-2\left)\right)=t\left(\right(1,n-1\left)\right)=t\left(\right(2,n\left)\right)=0,$$

$$t\left(\right(2,n-3\left)\right)=t\left(\right(1,n-2\left)\right)=t\left(\right(2,n-1\left)\right)=t\left(\right(1,n\left)\right)=1,$$

$$t\left(\right(3,n-3\left)\right)=t\left(\right(3,n-2\left)\right)=t\left(\right(3,n-1\left)\right)=t\left(\right(3,n\left)\right)=1$$

and

$$t\left(\right(x,y\left)\right)=h\left(\right(x,y\left)\right)$$

otherwise. Then it is easy to see that t is an IDF of $C_3\square C_n$ such that $|M_t|<|M_h|$. This contradicts the choice of h.

Case 2. $a_{n-2}\ge 2$.

Now $a_{n-2}+a_{n-1}+a_n\ge 6$. Since $a_{n-3}\ge 1$ by Claim 1, $h\left(\right(i,n-3\left)\right)=1$ or 2 for some $i\in \{1,2,3\}$. Without a loss of generality, we may assume $h\left(\right(1,n-3\left)\right)=1$ or 2. Define $t:V(C_3\square C_n)\to \{0,1,2\}$ by

$$t\left(\right(1,n-2\left)\right)=t\left(\right(2,n-1\left)\right)=t\left(\right(3,n\left)\right)=0,$$

$$t\left(\right(2,n-2\left)\right)=t\left(\right(1,n-1\left)\right)=t\left(\right(1,n\left)\right)=1,$$

$$t\left(\right(3,n-2\left)\right)=t\left(\right(3,n-1\left)\right)=t\left(\right(2,n\left)\right)=1$$

and

$$t\left(\right(x,y\left)\right)=h\left(\right(x,y\left)\right)$$

otherwise. Then it is easy to see that t is an IDF of $C_3\square C_n$ such that $|M_t|<|M_h|$. This contradicts the choice of h.

By Claims 1 and 2, we have ${\gamma }_I(C_3\square C_n)\ge 2n$. This completes the proof. □

Next, we consider $C_m\otimes C_n$. We denote the vertex set of a directed cycle $C_m$ by $\{1,2,\cdots ,m\}$, and assume that $i\to i+1$ is an arc of $C_m$. For every vertex $(i,j)\in V(C_m\otimes C_n)$, the first and second components are considered modulo m and n, respectively. For each $1\le k\le n$, we denote by $C_m^k$ the subdigraph of $C_m\otimes C_n$ induced by the set $\left\{\right(j,k)\mid 1\le j\le m\}$. Note that $C_m^k$ is isomorphic to $C_m$. Let f be a ${\gamma }_I(C_m\otimes C_n)$-function and set $a_k={\sum }_{x\in V\left(C_m^k\right)}f\left(x\right)$. Then ${\gamma }_I(C_m\otimes C_n)={\sum }_{k=1}^n{a}_k$.

Lemma 1.

For positive integers $m,n\ge 2$, ${\gamma }_I(C_m\otimes C_n)\ge \lceil \frac{mn}{2}\rceil$.

Proof. 

Note that the vertices of $C_m^k$ are dominated by vertices of $C_m^{k-1}$ or $C_m^k$. It suffices to verify that ${\sum }_{k=1}^n{a}_k\ge \lceil \frac{mn}{2}\rceil$. In order to do so, we claim $a_k+a_{k+1}\ge m$ for each k. First of all, we assume that $a_{k+1}=0$. Then, to dominate $(i,k+1)$ for each $1\le i\le m$, we must have

$$f\left(\right(i-1,k\left)\right)+f\left(\right(i,k\left)\right)\ge 2.$$

Then, $2a_k={\sum }_{i=1}^m(f\left((i-1,k)\right)+f\left((i,k)\right))\ge 2m$ and hence $a_k+a_{k+1}\ge m$. If $a_{k+1}=t>0$, then there will be at least $m-t$ vertices in $V_0$ that will be dominated only by vertices of $C_m^k$. This fact induces $a_k\ge m-t$ and so $a_k+a_{k+1}\ge m$. Therefore, we have

$$2{\gamma }_I(C_m\otimes C_n)=2\sum _{k=1}^n{a}_k=\sum _{k=1}^n(a_k+a_{k+1})\ge nm.$$

This completes the proof. □

Theorem 4.

For positive integers $m,n\ge 2$, ${\gamma }_I(C_m\otimes C_n)=\lceil \frac{mn}{2}\rceil$.

Proof. 

We divide our consideration into the following two cases.

Case 1. m or n is even.

Since $C_m\otimes C_n$ is isomorphic to $C_n\otimes C_m$, we may assume that $n=2s$ for some positive integers s.

Define $f:V(C_m\otimes C_n)\to \{0,1,2\}$ by

$$f\left(\right(i,2j-1\left)\right)=1$$

for each $1\le i\le m$ and $1\le j\le s$, and

$$f\left(\right(x,y\left)\right)=0$$

otherwise. It is easy to see that f is an IDF of $C_m\otimes C_n$ with weight $\frac{mn}{2}$ and so ${\gamma }_I(C_m\otimes C_n)\le \lceil \frac{mn}{2}\rceil$. Thus, it follows from Lemma 1 that ${\gamma }_I(C_m\otimes C_n)=\lceil \frac{mn}{2}\rceil$.

Case 2. $m=2r+1,n=2s+1$ for some positive integers $r,s$.

Define $f:V(C_m\otimes C_n)\to \{0,1,2\}$ by

$$f\left(\right(2i+1,2j+1\left)\right)=1$$

for each $0\le i\le r$ and $0\le j\le s$,

$$f\left(\right(2i,2j\left)\right)=1$$

for each $1\le i\le r$ and $1\le j\le s$ and

$$f\left(\right(x,y\left)\right)=0$$

otherwise. It is easy to see that f is an IDF of $C_m\otimes C_n$ with weight $(r+1)(s+1)+rs$ and so ${\gamma }_I(C_m\otimes C_n)\le \lceil \frac{mn}{2}\rceil$. Thus, it follows from Lemma 1 that ${\gamma }_I(C_m\otimes C_n)=\lceil \frac{mn}{2}\rceil$. □

3. Conclusions

In this article, we determined the exact values of ${\gamma }_I(C_2\square C_l)$, ${\gamma }_I(C_3\square C_l)$ and ${\gamma }_I(C_m\square C_n)$ for an integer l and even integers $m,n$. The other cases are still open. We conclude by giving a conjecture.

Conjecture 1.

For an odd integer n, ${\gamma }_I(C_4\square C_n)=2n+2$.