AbstractLet γ(G) denote the domination number of a simple graph G and let G□H denote the Cartesian product of two simple graphs G and H. In this paper we prove that if γ(G)=3, then γ(G□H)≥γ(G)γ(H)

Sun, Liang

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Discrete Mathematics 275 (2004) 363–366
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A result on Vizing’s conjecture
Liang Sun
Department of Mathematics, Beijing Institute of Technology, Beijing 100081, China
Received 22 January 2003; received in revised form 5 September 2003; accepted 15 September 2003
Abstract
Let (G) denote the domination number of a simple graph G and let G H denote the Cartesian
product of two simple graphs G and H . In this paper we prove that if (G)=3, then (G H) ≥
(G)(H).
c© 2003 Elsevier B.V. All rights reserved.
MSC: 05035; 05070
Keywords: Graph; Domination number; Cartesian product
1. Introduction
For a simple graph G, a subset S of V (G) is a dominating set of G if for any
vertex v∈V (G)− S, there exists a vertex v′ ∈ S such that v and v′ are adjacent in G.
The domination number (G) of G is the minimum cardinality of a dominating set
of G. For a subset S of V (G), we de:ne NG(S) = {v∈V (G)− S : v is adjacent to a
vertex in S} and NG[S] = S ∪ NG(S). Then S is a dominating set of G if and only if
NG[S] = V (G).
Let G and H be two simple graphs. The Cartesian product G H of G and H has
V (G H)=V (G)×V (H) and vertices (u1; v1) and (u2; v2) in V (G H) are adjacent if
and only if either u1 = u2 and v1v2 ∈E(H) or v1 = v2 and u1u2 ∈E(G).
In 1963, Vizing [4] conjectured that for any graphs G and H , (G H) ≥ (G)(H):
For (G)=1 and 2, it is easy to verify the validity of Vizing’s conjecture [3]. Bre?sar
[2] proved that Vizing’s conjecture is true when (G) = (H) = 3. But for the general
graph H , Vizing’s conjecture for the case (G) = 3 is still open [3]. In this paper we
E-mail address: lsunn@bit.edu.cn (L. Sun).
0012-365X/$ - see front matter c© 2003 Elsevier B.V. All rights reserved.
doi:10.1016/j.disc.2003.09.003
364 L. Sun /Discrete Mathematics 275 (2004) 363–366
are going to settle this case and prove that Vizing’s conjecture is true for (G) = 3.
For the other progress on Vizing’s conjecture, the readers may refer to the survey paper
of Hartnell and Rall [3].
2. Main results
In this section we will prove that (G H) ≥ (G)(H), when (G) = 3. For this
purpose we need a classical theorem about bipartite graphs.
Theorem 1 (Bondy and Murty [1]). If G is a bipartite graph, then the minimum num-
ber of vertices in a covering of G is equal to the maximum number of edges in a
matching of G.
Next we give some notation. Denote V (G) = {1; : : : ; m}, V (H) = {1; : : : ; n}, and
V (G H) = {(i; j) : i = 1; : : : ; m; j = 1; : : : ; n}. Let Uj = {(i; j) : i = 1; : : : ; m}, and Gj
be the subgraph of G H induced by Uj, j = 1; : : : ; n. Let Vi = {(i; j) : j = 1; : : : ; n},
and Hi be the subgraph of G H induced by Vi, i = 1; : : : ; m. For any vertex (i; j) of
G H , the vertex j of H is the H -projection of (i; j), denoted j = H (i; j). For any
subset A= {(i1; j1); : : : ; (ik ; jk)} of V (G H), we denote H (A) = {j1; : : : ; jk} which is
a subset of V (H). For a subset A of V (G H), if H (A) = V (H), then A is H-full.
Obviously A is H -full if and only if A ∩ Uj 	= ∅ for j = 1; 2; : : : ; n. For simplicity, we
denote NG H (A) = N (A) and NG H [A] = N [A], where A is a subset of V (G H).
Now we prove the main result of this paper.
Theorem 2. If (G) = 3, then for any graph H , (G H) ≥ (G)(H).
Proof. Let G be a graph with domination number 3. Without loss of generality, we
assume that G is critical, that is, for any two non-adjacent vertices u and v of G, (G+
uv)=2. Let {1; p; q} be a dominating set of G such that NG[p]∪NG[q]=V (G)−{1}.
Then the vertex 1 of G can not be adjacent to p nor q. Let D be a dominating set of
G H . We are going to prove that |D| ≥ 3(H).
Denote A1 = V1 ∩D, A2 = V1 ∩N (A1), and A3 = V1 − A1 − A2. Let A4 be a maximal
independent set of the subgraph of G H induced by A3. Since A4 ∩ (A1 ∪ A2) = ∅,
for each vertex (1; j)∈A4, there exists a vertex (ij; j)∈D ∩ Uj such that (1; j) and
(ij; j) are adjacent in G H . Denote A5 ={(ij; j) : j∈H (A4)}, and D1 =A1∪A5. Since
A1 ∪ A4 dominates all vertices in H1, and |A5|= |A4|, we have that |D1| ≥ (H).
Since (G)=3, for any k ∈H (N (A5)∩Vp), there exists a vertex (tk ; k) in N [(p; k)]∩
Uk such that (tk ; k) is not adjacent to (ik ; k) nor (q; k). Otherwise, {(ik ; k), (q; k)} is a
dominating set of Gk , which contradicts the fact that (G) = 3. Denote C1 = {(tk ; k) :
k ∈H (N (A5) ∩ Vp)}, C2 = {(p; k) : k 	∈ H (N (A5) ∩ Vp)}, and F = C1 ∪ C2. Then
F∩N [D1]=∅, F ⊆ N [Vp], and F is H -full. Denote L1=Vp∩D. Let Y={(i; j)∈C1−D :
(i; j) can be dominated by D ∩ Vi and j 	∈ H (N [L1])}. Thus, for any vertex (i; j) in
Y , there exists a vertex (i; j′) in D ∩ Vi such that (i; j) and (i; j′) are adjacent. Denote
L2 = {(i; j′)∈D : (i; j)∈Y}.
L. Sun /Discrete Mathematics 275 (2004) 363–366 365
Now we discuss two cases according as p and q are adjacent or not in G.
Case 1: p and q are non-adjacent in G.
Let I be a maximal independent set of the subgraph of H induced by V (H) −
H (N [L1 ∪ L2]). Denote Q = {(i; j)∈F : j∈ I}. Then for any vertex (i; j) in Q,
there exists a vertex (i′; j) in Uj ∩ D such that (i; j) and (i′; j) are adjacent. Denote
L3 = {(i′; j)∈D : (i; j)∈Q}.
Denote D2 =L1 ∪L2 ∪L3. Obviously H (D2) is a dominating set of H . Thus |D2| ≥
(H). Since F ∩ N [D1] = ∅, we have that D1 ∩ D2 = ∅.
Now we show that for each j = 1; 2; : : : ; n, Uj − N [D1 ∪ D2] 	= ∅. By the de:nition
of D1 and D2, we have that 0 ≤ |Uj ∩ (D1 ∪ D2)| ≤ 2.
Case 1.1: |Uj ∩ (D1 ∪ D2)|= 0.
Since (D1 ∪ D2) ∩ Vq = ∅, we have that (q; j) 	∈ N [D1 ∪ D2].
Case 1.2: |Uj ∩ (D1 ∪ D2)|= 1.
Denote Uj ∩ (D1 ∪ D2) = {(i; j)}. If (i; j)∈A1 ∪ L1 ∪ L2, then (q; j) 	∈ N [D1 ∪ D2].
If (i; j)∈A5, then we have that Uj ∩ N [D1 ∪ D2] ⊆ N [(i; j); (p; j)], since H (A5) is
independent in H . Thus we have that Uj − N [D1 ∪ D2] 	= ∅.
Case 1.3: |Uj ∩ (D1 ∪ D2)|= 2.
Since H (A1∪A5)∩H (L2)=∅, we have j∈ (H (A1)∩H (L1))∪(H (A1)∩H (L3))∪
(H (A5) ∩ H (L1)) ∪ (H (A5) ∩ H (L3)). It is easy to verify that Uj ∩ N [D1 ∪ D2] ⊆
N [Uj ∩ (D1 ∪ D2)]. Thus we have that Uj − N [D1 ∪ D2] 	= ∅, since (G) = 3.
Thus, for each j = 1; 2; : : : ; n, there exists a vertex (xj; j) in Uj such that (xj; j) is
not dominated by D1 ∪ D2. Denote W1 = {(xj; j)∈Uj − N [D1 ∪ D2] : j = 1; 2; : : : ; n}.
Since W1 is H -full and W1 must be dominated by D3 = D − D1 − D2, we have that
|D3| ≥ (H). Therefore, |D|= |D1|+ |D2|+ |D3| ≥ 3(H).
Case 2: p and q are adjacent in G.
If there is only one vertex r in NG[q] − (NG[1] ∪ NG[p]), then 1, p and r consist
of a dominating set of G and the conclusion is true by the proof in Case 1. Thus we
can assume that there are two distinct vertices s and t in NG[q]− (NG[1] ∪ NG[p]).
Denote S1 = H (C2)− H (N [L1 ∪ L2]). Let I ′ be a maximal independent set of the
subgraph of H induced by S1. For every j∈ I ′, (p; j) must be dominated by a vertex
(p′; j) in Uj∩D. Denote L4 ={(p′; j) : j∈ I ′}, and S2 =H (C1)−H (N [L1∪L2∪L4]).
Since H (A5) is independent in H , S2 is independent in H also. Denote T={(i; j)∈C1 :
j∈ S2}. Then for any (i; j)∈T , (i; j) must be dominated by a vertex (i′; j) in Uj ∩D.
Denote T ′ = {(i′; j) : j∈ S2}, T1 = Vs ∩ T ′ and T2 = Vt ∩ T ′. Let B be a bipartite
graph such that V (B) = T1 ∪ T2, and a vertex (s; j1)∈T1 and a vertex (t; j2)∈T2 are
366 L. Sun /Discrete Mathematics 275 (2004) 363–366
adjacent in B if and only if there exists j∈V (H) such that j is adjacent to both j1
and j2 in H . Let B1 be the union of non-trivial components of B. By Theorem 1, we
assume that {$1; : : : ; $%} is a minimum vertex cover of B1 and {$1&1; : : : ; $%&%} is a
maximum matching of B1. Since there exists a vertex ki in H such that ki is adjacent
to both H ($i) and H (&i), i = 1; 2; : : : ; %, H (T ′) can be dominated by Z = H (T ′ −
{$1; : : : ; $%; &1; : : : ; &%})∪{k1; : : : ; k%} in H . Thus J=H (L1∪L2∪L4)∪Z is a dominating
set of H . Denote L5 = T ′ − {$1; : : : ; $%}, and D4 = L1 ∪ L2 ∪ L4 ∪ L5. Then D4 is a
subset of D − D1 and |D4| = |J | ≥ (H). Now we show that for each j = 1; 2; : : : ; n,
Uj − N [D1 ∪ D4] 	= ∅.
Case 2.1: |Uj ∩ (D1 ∪ D4)|= 0.
In this case, we have that either (s; j) or (t; j) can not be dominated by D1 ∪ D4.
Otherwise, we assume that both (s; j) and (t; j) are dominated by D1∪D4. Then, since
{s; t} ∩ (NG[1] ∪ NG[p]) = ∅ and L1 ∪ L2 ∪ L4 ⊂ N [Vp], there exist two vertices (s; j1)
and (t; j2) in L5 such that j is adjacent to both j1 and j2 in H , which contradicts the
fact that {$1; : : : ; $%} is a covering of B1.
Case 2.2: |Uj ∩ (D1 ∪ D4)|= 1.
Denote Uj∩(D1∪D4)={(i; j)}. If (i; j)∈L5, then j∈H (A5) and |Uj∩(D1∪D4)|=2, a
contradiction. Thus (i; j) 	∈ L5. If (i; j)∈A1∪A5, then Uj∩N [D1∪D4] ⊆ N [(i; j); (p; j)].
If (i; j)∈L1∪L4, then Uj∩N [D1∪D4] ⊆ N [(1; j); (i; j)]. If (i; j)∈L2, then (q; j)∈Uj−
N [D1 ∪ D4]. Thus we have that Uj − N [D1 ∪ D4] 	= ∅.
Case 2.3: |Uj ∩ (D1 ∪ D4)|= 2.
In this case, j∈ (H (A1) ∩ H (L1)) ∪ (H (A1) ∩ H (L4)) ∪ (H (A5) ∩ H (L1)) ∪
(H (A5) ∩ H (L5)). It is easy to verify that Uj ∩ N [D1 ∪ D4] ⊆ N [Uj ∩ (D1 ∪ D4)].
Thus we have that Uj − N [D1 ∪ D4] 	= ∅, since (G) = 3.
Thus, for each j = 1; 2; : : : ; n, there exists a vertex (yj; j) in Uj such that (yj; j) is
not dominated by D1 ∪ D4. Denote W2 = {(yj; j)∈Uj − N [D1 ∪ D4] : j = 1; 2; : : : ; n}.
Since W2 is H -full and W2 must be dominated by D5 = D − D1 − D4, we have that
|D5| ≥ (H). Therefore, |D|= |D1|+ |D4|+ |D5| ≥ 3(H):
References
[1] J.A. Bondy, U.S.R. Murty, Graph theory with applications, Macmillan Ltd. Press, New York, 1976.
[2] B. Bre?sar, On Vizing’s conjecture, Discuss. Math. Graph Theory 21 (2001) 5–11.
[3] B. Hartnell, D.F. Rall, Domination in Cartesian products: Vizing’s conjecture, in: T.W. Haynes, S.T.
Hedetniemi, P.J. Slater (Eds.), Domination in Graphs—Advanced Topics, Marcel Dekker, New York,
1998, pp. 163–189.
[4] V.G. Vizing, The cartesian product of graphs, Vy?cisl. Sistemy 9 (1963) 30–43.


A result on Vizing's conjecture 

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A result on Vizing's conjecture

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