Abstract
The concept of 2-rainbow domination of a graph \(G\) coincides with the ordinary domination of the prism \(G\Box K_{2}\) (see Brešar et al., Taiwan J Math 12:213–225, 2008). Hence \(\gamma _{r2}(C_{m}\Box C_{n})\ge \frac{mn}{3}\). In this paper we give full characterization of graphs \(C_m\Box C_n\) with \(\gamma _{r2}(C_{m}\Box C_{n}) = \frac{mn}{3}\).
Similar content being viewed by others
Avoid common mistakes on your manuscript.
1 Introduction
For notation and graph theory terminology not given here, we follow Diestel (1997) and Haynes et al. (1998). Let \(G=(V(G),E(G))\) be a finite, simple and undirected graph with vertex set \(V(G)\) and edge set \(E(G).\) The open neighborhood of a vertex \(v\) is \(N(v)=\{u\in V(G):uv\in E(G)\}\). If \(A \subset V (G)\), then \(N(A)\) denotes the union of open neighborhoods of all vertices of \(A.\) For two subsets \(A,B\) of \(V(G),\) \(E(A,B)=\{ab\in E(G):a\in A,b\in B\}.\)
The Cartesian product \(G\Box H\) of graphs \(G\) and \(H\) is the graph with vertex set \(V(G)\times V(H),\) where two vertices are adjacent if and only if they are equal in one coordinate and adjacent in the other.
We restrict our attention to the Cartesian product of \(C_{n}\) and \(C_{m},\) \(n,m\ge 3.\) Let \(V(C_{n})=\{0,1,\ldots ,n-1\}\), \(E(C_{n})=\{i(i+1), (n-1)0:i=0,1,\ldots ,n-2\}\). Hence we will denote vertices of \(V(C_m\Box C_n)\) by \((i,j)\) for \(i=0,1,\ldots ,m -\! 1\) and \(j=0,1,\ldots ,n-\! 1\). For the arbitrary integers \(i\) and \(j\) we will use the following notation
A function \(f:V(G)\rightarrow {\fancyscript{P}}(\{1,\ldots ,k\})\) is called a \(k\)-rainbow dominating function of \(G\) (for short \(kRDF\) of \(G)\) if \( \bigcup \nolimits _{u\in N(v)}f(u)=\{1,\ldots ,k\},\) for each vertex \(v\in V(G)\) with \(f(v)=\varnothing .\) By \(w(f)\) we mean \(\sum _{v\in V(G)}\left| f(v)\right| \) and we call it the weight of a function \(f\) in \(G. \) The minimum weight of a \(kRDF\) of \(G\) is called the \(k\)-rainbow domination number of \(G\) and it is denoted by \(\gamma _{rk}(G).\) If \(f\) is a \(kRDF\) of \(G\) and \(w(f)=\gamma _{rk}(G),\) then \(f\) is called a \(\gamma _{rk}\)-function. For more information about rainbow domination we refer the reader to Brešar and Šumenjak (2007), Tong et al. (2009), Hartnell and Rall (1998), Wu and Rad (2013), Wu and Xing (2010), Šumenjak et al. (2013) and Xu (2009), where authors consider, in particular, connections between rainbow domination and Vizing conjecture.
Let \(f\) be any \(2RDF\) of \(C_{m}\Box C_{n}\). Define the following sets
We need the following technical lemma.
Lemma 1
(Stȩpień and Zwierzchowski 2012) Let \(f\) be any \(2RDF\) of \(C_{m}\Box C_{n}\). Then
where
Corollary 1
Let \(f\) be any \(2RDF\) of \(C_{m}\Box C_{n}\). Then \(w(f)\ge \frac{mn}{3}\) and equality holds if and only if \(\beta = 0.\)
In this paper we will use the following form of the Chinese Reminder Theorem.
Theorem 1
(Chinese Reminder Theorem) Two simultaneous congruences
are solvable if and only if \(a\equiv b \pmod {\gcd (m,n)}\). Moreover the solution is unique modulo \(\mathrm{lcm}(m, n)\).
2 Results
For any integer \(s\), let \(L_s=\left\{ [k, k-s]\in V(C_m\Box C_n):k=0,\pm 1, \pm 2, \ldots \right\} \). The following theorem is a consequence of the Chinese Reminder Theorem.
Theorem 2
We have
The sum is disjoint and \(|L_{s}|=\mathrm{lcm}(m,n).\)
Proof
By definition of \(L_s\) we have \(\bigcup _{s=0}^{\gcd (m,n)-1}\!\! L_s \subseteq V(C_m \Box C_n)\). Let \((i,j)\in V(C_m \Box C_n)\) and let \(s\in \{0,1,\ldots ,\gcd (m,n)-1\}\) be such that \(s\equiv i-j \pmod {\gcd (m,n)}\). By Theorem 1 there exists an integer \(k\) such that
Consequently, \((i,j) = [k, k-s] \in L_s\).
Next if the above system has any solution for a fixed \((i,j)\) and some \(s\), then again by Theorem 1 we have \(s\equiv i-j \pmod {\gcd (m,n)}.\) Hence \(L_{s_{1}}\cap L_{s_{2}}=\emptyset \) for \(s_{1}\ne s_{2}\) and \(s_{1},s_{2}\in \left\{ 0,1,\ldots ,\gcd (m,n)-1 \right\} .\) Finally, observe that cardinality of \(L_{s}\) is the same for each \(s\). Therefore \(|L_{s}|=\mathrm{lcm}(m,n).\) \(\square \)
For any integer \(s\), let us denote
Corollary 2
The following holds:
-
1.
for any integers \(i,j\) we have \([i,j] \in L_{[\![i-j ]\!]}\),
-
2.
if \(\gcd (m,n)>1\), then for any integer \(s\) we have
$$\begin{aligned} N\left( L_{[\![s ]\!]}\right) =L_{[\![s-1 ]\!]} \cup L_{[\![s+1 ]\!]}, \end{aligned}$$ -
3.
if \(\gcd (m,n)=1\), then we have
$$\begin{aligned} V(C_m\Box C_n)=L_0. \end{aligned}$$
Now we introduce some definitions. Let \(f\) be a \(2\)-rainbow dominating function of \(C_m\Box C_n\). We say that \(f\) is positive if for any \((i,j)\in V(C_m\Box C_n)\) the following implication holds:
Let \(L_s^- =\left\{ [k, -k+s]\in V(C_m\Box C_n):k=0, \pm 1, \pm 2, \ldots \right\} \). Note that \([i,j]\in L_{[\![i+j ]\!]}^-\). We say that \(f\) is negative if for any \((i,j)\in V(C_m\Box C_n)\) the following implication holds:
Lemma 2
Let \(f\) be a \(2\)-rainbow dominating function of \(C_{m}\Box C_{n}\) such that \(w(f)=\frac{mn}{3}\). Then \(f\) is either positive or negative.
Proof
Let \(f\) be a \(2\)-rainbow dominating function of \(C_{m}\Box C_{n}\) such that \(w(f)=\frac{mn}{3}\). By Corollary 1, we get \(\beta =0.\) Hence \(|V_{2}|=|E_{1}|=|V_{30}|=|V_{40}|=0.\)
Take any vertex \((i,j)\in C_m\Box C_n\) such that \((i,j) \in V_1\). Assume, without loss of generality, that \(f\left( (i,j)\right) =\left\{ 1\right\} .\) Since \(|E_{1}|= 0\), we have
We claim that also
To prove (2) suppose the contrary: assume, without loss of generality, that \([i,j+2]\in V_{1}.\) Then \([i+1,j+1],[i+1,j+2]\in V_{0}\) (otherwise \(|V_{30}\cup V_{40}\cup E_{1}|\ne 0).\) Since the vertex \([i+1,j+1]\) must be dominated in the sense of 2-rainbow domination, we get \(|V_{2}|\ne 0,\) a contradiction.
Observe that exactly one of \([i+1,j+1],\) \([i+1,j-1]\) belongs to \(V_{1}\). Indeed, on the one hand at most one of \([i+1,j+1],\) \([i+1,j-1]\) belongs to \(V_{1},\) since otherwise \(|V_{30}|\ne 0.\) On the other hand at least one of \([i+1,j+1],\) \([i+1,j-1]\) belongs to \(V_{1}\) and \(f\left( [i+1,j+1]\right) =\left\{ 2\right\} \) or \(f\left( [i+1,j-1]\right) =\left\{ 2\right\} \) (otherwise \([i+1,j]\) would not be dominated in the sense of 2-rainbow domination). Thus either \(f\left( [i+1,j+1]\right) =\left\{ 2\right\} \) or \(f\left( [i+1,j-1]\right) =\left\{ 2\right\} .\)
Assume that \(f\left( [i+1,j+1] \right) =\left\{ 2\right\} .\) This assumption combined with (1) and (2) imply that \([i+2,j+2] \in V_{1}.\) By induction we get \(L_{[[i-j]]}\subset V_{1}.\) Similarly, if \(f\left( [i+1,j-1] \right) =\left\{ 2\right\} \), then we get \(L^{-}_{[[i+j]]}\subset V_{1}.\) We have shown that if \((i,j)\in V_{1}\), then either \(L_{[[i-j]]}\subset V_{1}\) or \(L^{-}_{[[i+j]]}\subset V_{1}.\) Thus \(f\) is positive or negative.
Finally, suppose that \(f\) is positive and negative. This means that \(L_{s_{1}}\subset V_{1}\) and \(L^{-}_{s_{2}}\subset V_{1}\) for \(s_{1},s_{2}\in \left\{ 0, 1, \ldots , \gcd (m,n)-1 \right\} .\) To eliminate this possiblity, we will consider the following two cases.
-
(a)
There exists \(k\) such that \(s_{1}+s_{2} \equiv 2k \pmod {\gcd (m,n)}\). Consider the following system of simultaneous congruences
$$\begin{aligned}&l\equiv k \pmod m, \\&l\equiv -k+s_{1}+s_{2} \pmod n . \end{aligned}$$By Theorem 1, there exists a solution \(l\). Hence \([k,k-s_{1}]=[l,-l+s_{2}]\), which means that \(L_{s_{1}}\cap L^{-}_{s_{2}}\ne \emptyset \). It is easy to see that this contradicts the fact that \(|V_{30}|= 0.\)
-
(b)
For all \(k\) we have \(s_{1}+s_{2} \equiv 2k+1 \pmod {\gcd (m,n)}\). Consider the following system of simultaneous congruences for some fixed \(k\)
$$\begin{aligned}&l\equiv k \pmod m, \\&l\equiv -k+s_{1}+s_{2}-1 \pmod n . \end{aligned}$$By Theorem 1, there exists a solution \(l\). Hence \([l,-l+s_{2}-1]=[k,k-s_{1}]\in L_{s_1}\subset V_1\). Since vertices \([l,-l+s_{2}-1]\) and \([l,-l+s_{2}]\in L^{-}_{s_{2}}\subset V_1\) are adjacent, it contradicts the fact that \(|E_{1}| = 0.\)
Thus \(f\) is either positive or negative. \(\square \)
Lemma 3
Let \(f\) be a \(2\)-rainbow dominating function of \(C_{m}\Box C_{n}\) such that \(w(f)=\frac{mn}{3}\) then
-
1.
\(\mathrm{lcm}(m,n)\equiv 0\pmod 2,\)
-
2.
\(\gcd (m,n)\equiv 0\pmod 3.\)
Proof
Let \(f\) be a \(2\)-rainbow dominating function of \(C_{m}\Box C_{n}\) such that \(w(f)=\frac{mn}{3}\). By Lemma 2, \(f\) is either positive or negative. Assume, without loss of generality, that \(f\) is positive. Take any vertex \((i,j)\in C_m\Box C_n\) such that \((i,j)\in V_1.\) Hence \(L_{[\![i-j ]\!]} \subset V_1\).
The same argument as in the proof of Lemma 2 shows that
Hence by induction we have \(|L_{[[i-j]]}|\equiv 0 \pmod 2.\) This together with Theorem 2 proves (1).
Now we will prove (2). If \(\gcd (m,n)=1\), then by Corollary 2(3) we have \(L_0 = V(C_m\Box C_n)\). Since \(f\) is positive, we have \(L_0=V_1\). Therefore, \(w(f)=mn\). This contradicts our assumption that \(w(f)=\frac{mn}{3}\). Consequently \(\gcd (m,n)>1\). Suppose that \(\gcd (m,n)= 2.\) Then \((i,j)\), \([i+1,j+1]\), \([i+2,j]\in L_{[\![i-j ]\!]} \subset V_1\). This implies that \([i+1,j]\in V_{30}\). However, this contradicts the fact that \(|V_{30}|=0.\) Hence \(\gcd (m,n) \ge 3\). Assume now that \(\gcd (m,n)>3.\)
By Corollary 2 and inclusions (1), (2) we get
In particular \([i+1,j],\,[i+2,j+1],\,[i+2,j]\in V_{0}.\) To dominate \([i+2,j]\) we must have \([i+3,j]\in V_1\), and consequently \(L_{[\![i-j+3]\!]}\subset V_{1}\). Continuing in this way we get that for any \(l\ge 1\) we have
To prove (2) we must eliminate the following two possibilities.
-
(a)
Let \(\gcd (m,n)=3k+1\) for some \(k\ge 1\). Now \(L_{[\![i-j+3k ]\!]}\subset V_1\) and \(L_{[\![i-j+3k+1 ]\!]} = L_{[\![i-j ]\!]}\subset V_1\). This contradicts (3).
-
(b)
Let \(\gcd (m,n)=3k+2\) for some \(k\ge 1\). Now \( L_{[\![i-j+3k]\!]}\subset V_1,\) \(L_{[\![i-j+3k+1]\!]}\subset V_0\) and \(L_{[\![i-j+3k+2]\!]}=L_{[\![i-j ]\!]}\subset V_1. \) This contradicts (3). \(\square \)
Lemma 4
For \(k,l\ge 1\)
Proof
By Lemma 1, we have \(\gamma _{r2}(C_{m}\Box C_{n}) \ge \frac{mn}{3}.\) Hence for the proof it suffices to find a 2RDF of \(C_{6k}\Box C_{3l}\) of weight \(\frac{6k3l}{3}.\) First we define \(f:V\left( C_{6}\Box C_{3}\right) \rightarrow {\fancyscript{P}}\left( \left\{ 1,2\right\} \right) \) as follows
It is easy to see that \(f\) is a 2RDF of \(C_{6}\Box C_{3}\) of weight \(6.\) The required function on \(C_{6k}\Box C_{3l}\) one can construct using this segment. Finally, the equality \(\gamma _{r2}(C_{6k}\Box C_{3l})=\gamma _{r2}(C_{3l}\Box C_{6k})\) follows by the symmetry. \(\square \)
We are ready to prove our main result.
Theorem 3
\(\gamma _{r2}(C_{m}\Box C_{n})=\frac{mn}{3}\) if and only if \(m=6k\) and \(n=3l\) or \(m=3k\) and \(n=6l\), \(k,l\ge 1\).
Proof
Let \(\gamma _{r2}(C_{m}\Box C_{n})=\frac{mn}{3}\) and \(f\) be a \(2RDF\) of \(C_{m}\Box C_{n}\) such that \(w(f)=\frac{mn}{3}\). By Lemma 3, we have \(\gcd (m,n)\equiv 0 \pmod 3\) and \(\mathrm{lcm}(m,n)\equiv 0 \pmod 2.\) Hence \(m\equiv 0 \pmod 3\), \(n\equiv 0 \pmod 3\) and at least one of \(m\) and \(n\) is even. This together with Lemma 4 proves our theorem. \(\square \)
The following theorem is the consequence of our considerations.
Theorem 4
Let \(m=6k\), \(n=3 l\). There are \( 6 \cdot 2^{\frac{\gcd (m,n)}{3}} \) \(\gamma _{r2}\)-functions of \(C_m\Box C_n\) and \( 2^{\frac{\gcd (m,n)}{3}} \) \(\gamma _{r2}\)-functions up to translations.
References
Brešar B, Šumenjak TK (2007) Note on the 2-rainbow domination in graphs. Discret Appl Math 155:2394–2400
Brešar B, Henning MA, Rall DF (2008) Rainbow domination in graphs. Taiwan J Math 12(1):213–225
Diestel R (1997) Graph theory. Springer-Verlag, New York
Hartnell B, Rall DF (1998) Domination in Cartesian products: Vizing’s conjecture. In: Haynes TW, Hedetniemi ST, Slater PJ (eds) Domination in graphs: advanced topics. Marcel Dekker, New York, pp 163–189
Haynes TW, Hedetniemi ST, Slater PJ (1998) Fundamentals of domination in graphs. Marcel Dekker, New York
Stȩpień Z, Zwierzchowski M (2012) 2-Rainbow domination number of Cartesian products: \(C_n\Box C_3\) and \(C_n\Box C_5\), J Comb Optim. doi:10.1007/s10878-012-9582-8
Šumenjak TK, Rall DF, Tepeh A (2013) Rainbow domination in the lexicographic product of graphs. Discret Appl Math 161:2133–2141. arXiv:1210.0514v2 [math.CO]
Tong C, Lin X, Yang Y, Luo M (2009) 2-Rainbow domination of generalized Petersen graphs \(P(n,2)\). Discret Appl Math 157:1932–1937
Wu Y, Rad NJ (2013) Bounds on the 2-rainbow domination number of graphs. Graphs Comb 29:1125–1133
Wu Y, Xing H (2010) Note on 2-rainbow domination and roman domination in graphs. Appl Math Lett 23:706–709
Xu G (2009) 2-Rainbow domination in generalized Petersen graphs \(P(n,3),\). Discret Appl Math 157:2570–2573
Author information
Authors and Affiliations
Corresponding author
Rights and permissions
Open Access This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.
About this article
Cite this article
Stȩpień, Z., Szymaszkiewicz, L. & Zwierzchowski, M. The Cartesian product of cycles with small 2-rainbow domination number. J Comb Optim 30, 668–674 (2015). https://doi.org/10.1007/s10878-013-9658-0
Published:
Issue Date:
DOI: https://doi.org/10.1007/s10878-013-9658-0