The Italian domination numbers of some products of directed cycles

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

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1. Introduction and preliminaries

Let be a finite simple digraph with vertex set and arc set . An arc to join to is denoted by . The maximum out-degree and maximum in-degree of are denoted by and , respectively.

Let and be two digraphs. The cartesian product of and is the digraph with vertex set and for two vertices and ,

 (x1,x2)→(y1,y2)

if one of the following holds:

1. and ;

2. and .

The strong product of and is the digraph with vertex set and for two vertices and ,

 (x1,x2)→(y1,y2)

if one of the following holds:

1. and ;

2. and ;

3. and .

An Italian dominating function (IDF) on a digraph is defined as a function such that every vertex with has at least two in-neighbors assigned under or one in-neighbor with . An Italian dominating function gives a partition of , where . The weight of an Italian dominating function is the value . The Italian domination number of a digraph , denoted by , is the minimum taken over the weights of all Italian dominating functions on . A -function is an Italian dominating function on with weight .

The study of Italian dominating functions in graphs and digraphs have done in [1, 2, 3, 4, 8, 9]. In particular, there are many studies on the cartesian products of undirected cycles or undirected paths in [5, 6, 7]. Recently, the author of [8] initiated the study of the Italian domination number in digraphs. In this paper, we investigate the Italian domination numbers of cartesian products and strong products of directed cycles.

The following results are useful to our study.

Proposition 1.1 ([8]).

Let be a digraph of order . Then .

Proposition 1.2 ([8]).

Let be a digraph of order . Then and if and only if .

Proposition 1.3 ([8]).

If is a directed path or a directed cycle of order , then .

2. The Italian domination numbers of some products of digraphs

In this section, we determine the exact values of the Italian domination numbers of some products of digraphs.

First, we consider the cartesian product of directed cycles. We denote the vertex set of a directed cycle by , and assume that is an arc of . For every vertex , the first and second components are considered modulo and , respectively. For each , we denote by the subdigraph of induced by the set . Note that is isomorphic to . Let be a -function and set . Then . It is easy to see that is isomorphic to . So, .

Theorem 2.1.

If and for some positive integers , then .

Proof.

Define by

 f((2i−1,2j−1))=f((2i,2j))=1

for each and , and

 f((x1,x2))=0

otherwise. It is easy to see that is an IDF of with weight and so . Since , it follows from Proposition 1.1 that . Thus, we have . ∎

Theorem 2.2.

For an odd integer

, .

Proof.

Define by

 f((1,2j−1))=1

for each ,

 f((2,2j))=1

for each ,

 f((2,n))=1

and

 f((x1,x2))=0

otherwise. It is easy to see that is an IDF of with weight and so .

Now we claim that . Suppose to the contrary that . Let be a -function. If for some , say , then . To dominate the vertices and , we must have . Define by

 g((1,2))=g((2,1))=g((2,3))=1,g((2,2))=0

and

 g((x1,x2))=f((x1,x2))

otherwise. Then is an IDF of with weight less than , which is a contradiction. Thus, for each . By assumption, for each . Without loss of generality, we assume that . To dominate , we must have . Since and , we have . By repeating this process, we obtain for each , for and otherwise. But, the vertex is not dominated, a contradiction. Thus we have . This completes the proof. ∎

Theorem 2.3.

For an integer , .

Proof.

When for some positive integer , define by

 f0((1,3j+1))=f0((2,3j+1))=1

for each ,

 f0((2,3j+2))=f0((3,3j+2))=1

for each ,

 f0((1,3j+3))=f0((3,3j+3))=1

for each and

 f0((x1,x2))=0

otherwise.

When for some positive integer , define by

 f1((2,n))=f1((3,n))=1

and

 f1((x1,x2))=f0((x1,x2))

otherwise.

When for some positive integer , define by

 f2((1,n−1))=f2((2,n−1))=f2((1,n))=f2((3,n))=1

and

 f2((x1,x2))=f0((x1,x2))

otherwise. It is easy to see that is an IDF of with weight and so .

Now we prove that . Let be a -function.

Claim 1. for each .

Proof. Suppose to the contrary that for some , say . To dominate , and , we must have . But, the function defined by

 g((1,n−1))=g((2,n−1))=g((3,n−1))=1,
 g((1,n))=g((2,n))=1

and

 g((x1,x2))=f((x1,x2))

otherwise, is an IDF of with weight less than . This is an contradiction. ∎

We choose a -function so that the size of is as small as possible. Claim 2. .

Proof. Suppose to the contrary that . Without loss of generality, assume that and . To dominate and , we must have and . If , then clearly and so . From now on, assume . We divide our consideration into the following two cases.

Case 1. .

By argument as above, we have . So . If , then we are done. Suppose . Since by Claim 1, or for some . Without loss of generality, we may assume or . Define by

 t((1,n−3))=t((2,n−2))=t((1,n−1))=t((2,n))=0,
 t((2,n−3))=t((1,n−2))=t((2,n−1))=t((1,n))=1,
 t((3,n−3))=t((3,n−2))=t((3,n−1))=t((3,n))=1

and

 t((x1,x2))=h((x1,x2))

otherwise. Then it is easy to see that is an IDF of such that . This contradicts the choice of .

Case 2. .

Now . Since by Claim 1, or for some . Without loss of generality, we may assume or . Define by

 t((1,n−2))=t((2,n−1))=t((3,n))=0,
 t((2,n−2))=t((1,n−1))=t((1,n))=1,
 t((3,n−2))=t((3,n−1))=t((2,n))=1

and

 t((x1,x2))=h((x1,x2))

otherwise. Then it is easy to see that is an IDF of such that . This contradicts the choice of . ∎

By Claims 1 and 2, we have . This completes the proof. ∎

Next, we consider the strong product of directed cycles. We denote the vertex set of a directed cycle by , and assume that is an arc of . For every vertex , the first and second components are considered modulo and , respectively. For each , we denote by the subdigraph of induced by the set . Note that is isomorphic to . Let be a -function and set . Then .

Lemma 2.4.

For positive integers , .

Proof.

Note that the vertices of are dominated by vertices of or . It suffices to verify that . In order to do, we claim for each . First of all, we assume that . Then to dominate for each , we must have

 f((i−1,k))+f((i,k))≥2.

Then and hence . If , then there exist at least vertices in which is only dominated by vertices of . This fact induces and so . Therefore, we have

 2γI(Cm⊗Cn)=2n∑k=1ak=n∑k=1(ak+ak+1)≥nm.

This completes the proof. ∎

Theorem 2.5.

For positive integers , .

Proof.

We divide our consideration into the following four cases.

Case 1. for some positive integers .

Define by

 f((2i−1,2j−1))=2

for each and , and

 f((x1,x2))=0

otherwise. It is easy to see that is an IDF of with weight and so . Thus, it follows from Lemma 2.4 that .

Case 2. for some positive integers .

Define by

 f((i,2j−1))=1

for each and , and

 f((x1,x2))=0

otherwise. It is easy to see that is an IDF of with weight and so . Thus, it follows from Lemma 2.4 that .

Case 3. for some positive integers .

Define by

 f((2i−1,j))=1

for each and , and

 f((x1,x2))=0

otherwise. It is easy to see that is an IDF of with weight and so . Thus, it follows from Lemma 2.4 that .

Case 4. for some positive integers .

Define by

 f((2i+1,2j+1))=1

for each and ,

 f((2i,2j))=1

for each and and

 f((x1,x2))=0

otherwise. It is easy to see that is an IDF of with weight and so . Thus, it follows from Lemma 2.4 that . ∎

3. Conclusions

In this paper, we determined the exact values of , and for an integer and even integers . The other cases are still open. We conclude by giving a conjecture.

Conjecture 3.1.

For an odd integer , .

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