An Improved Bound for Upper Domination of Cartesian Products of Graphs
aa r X i v : . [ m a t h . C O ] M a r An Improved Bound for Upper Domination ofCartesian Products of Graphs
Yu-Yen ChienMathematics DivisionNational Center for Theoretical Science
Taiwan [email protected]
July 21, 2018
Abstract
In this paper, we prove a problem proposed by Breˇsar: for anygraphs G and H , Γ( G (cid:3) H ) ≥ Γ( G )Γ( H )+min {| V ( G ) |− Γ( G ) , | V ( H ) |− Γ( H ) } , where Γ( G ) denotes the upper domination number of G . For a simple graph G , a subset D of V ( G ) is a dominating set of G ifevery vertex in V ( G ) \ D has at least one neighbor in D . The dominationnumber γ ( G ) is the minimum cardinality of a dominating set of G . Forgraphs G and H , the Cartesian product G (cid:3) H is the graph with vertex set V ( G ) × V ( H ) and edge set { ( u , v )( u , v ) | u = u , v v ∈ E ( H ), or v = v , u u ∈ E ( G ) } . Vizing [11] suggested a conjecture regarding dominationin Cartesian products of graphs: Vizing Conjecture.
For any graphs G and H , γ ( G (cid:3) H ) ≥ γ ( G ) γ ( H ).This conjecture is one of the main problems in domination theory. See [3, 6, 8]for surveys and [1, 4, 10] for recent progress.For a simple graph G , a dominating set D of G is a minimal dominatingset if no proper subset of D is a dominating set of G . The upper dominationnumber Γ( G ) of G is the maximum cardinality of a minimal dominating setof G . The definition of the domination number γ ( G ) can be rephrased as theminimum cardinality of a minimal dominating set of G , and clearly we haveΓ( G ) ≥ γ ( G ) for any graph G . Nowakowski and Rall [9] conjectured thatfor any graphs G and H , Γ( G (cid:3) H ) ≥ Γ( G )Γ( H ). Breˇsar [2] proved a slightly1tronger bound: Γ( G (cid:3) H ) ≥ Γ( G )Γ( H ) + 1 for any nontrivial graphs G and H , where a nontrivial graph is a graph with at least one edge. Breˇsar alsoproposed the following question: does the inequalityΓ( G (cid:3) H ) ≥ Γ( G )Γ( H ) + min {| V ( G ) | − Γ( G ) , | V ( H ) | − Γ( H ) } hold for any graphs G and H ? We prove this inequality in this paper.We examine some basic properties about minimal dominating set first. G is a graph and D is a dominating set of G . We say a vertex in D is D - isolated if it is not adjacent to any other vertices in D . We say a vertex v ∈ D hasa private D - neighbor u if u / ∈ D is a neighbor of v and is no neighbor of anyother vertices in D . We have the following fundamental results. Lemma 1. [7] A dominating set D of a graph G is a minimal dominating setif and only if every vertex in D is D -isolated or has a private D -neighbor. Lemma 2.
Given a dominating set D of a graph G , we can always find asubset D ′ of D such that D ′ is a minimal dominating set of G . Moreover, if v ∈ D is D -isolated or has a private D -neighbor, then v must be in D ′ .For the graph G (cid:3) H , u ∈ V ( G ), v ∈ V ( H ) , we call { u } × V ( H ) row u and V ( G ) × { v } column v . In the following, we consider upper dominationin the Cartesian products of some basic graphs. Proposition 3.
For m, n ≥
2, Γ( K m (cid:3) K n ) = max { m, n } . Proof.
Assume n ≥ m . It is clear that a row of K m (cid:3) K n is a minimaldominating set, and Γ( K m (cid:3) K n ) ≥ n . Suppose K m (cid:3) K n has a minimal dom-inating set D with cardinality greater than n . If every column of K m (cid:3) K n has a vertex in D , the D = n . Therefore, there exists a column without avertex in D . To dominate vertices in this column, every row of K m (cid:3) K n musthave a vertex in D . Then, we have D = m , a contradiction.The following proposition is implicitly proved by Gutin and Zverovich [5]. Proposition 4.
For any graph G , Γ( K (cid:3) G ) = | G | . Proof.
We prove the case that G is connected and | V ( G ) | ≥
2, and thenthe general case follows. Since a row of K (cid:3) G is a minimal dominating set,we have Γ( K (cid:3) G ) ≥ | G | . Suppose D is a minimal dominating set of K (cid:3) G .For v ∈ D having a private D -neighbor, assign one of its private D -neighboras v ′ . For v ∈ D having no private D -neighbor, assign the other vertex incolumn v as v ′ . For any v ∈ D , v ′ / ∈ D . For any different u , v in D , u ′ = v ′ .Therefore, | D | ≤ | K (cid:3) G | = | G | , and we have Γ( K (cid:3) G ) = | G | .2 roposition 5. For m ≥ n ≥
2, Γ( K m (cid:3) K ,n ) = m + n − Proof.
Let V ( K m ) = { u , ..., u m } and V ( K ,n ) = { v , v , ..., v n } , where v isthe vertex of degree n . { ( u i , v ) , ( u , v j ) | m ≥ i ≥ , n ≥ j ≥ } is a minimaldominating set of K m (cid:3) K ,n , and we have Γ( K m (cid:3) K ,n ) ≥ m + n −
2. Suppose K m (cid:3) K ,n has a minimal dominating set D with | D | > m + n −
2. If D has avertex in each column of K m (cid:3) K ,n , then | D | = n + 1 > m + n −
2, violatingthe assumption that m ≥
3. Therefore, there exists v j such that column v j contains no vertex in D . If j = 0, ( u i , v ) must be in D to dominate( u i , v j ), and we have D = m > m + n −
2, violating the assumption that n ≥
2. Therefore, column v contains no vertex in D . To dominate ( u i , v ),there exists some ( u i , v j ) ∈ D . Assign one of these vertices as ( u i , v ) ′ . Let D ′ = { ( u i , v ) ′ | m ≥ i ≥ } . Notice that D ′ can not be a column. Columnscontaining a vertex in D ′ do not contain any vertex in D \ D ′ . Except column v , columns containing no vertex in D ′ contains exact one vertex in D \ D ′ ,and there are at most n − | D | ≤ m + n −
2, acontradiction.
Proposition 6.
For l, m, n ≥
2, we haveΓ( K l (cid:3) K m,n ) = max { m + n, l, m + l − , n + l − } . Proof.
Suppose that V ( K l ) = { u , ..., u l } , and K m,n has partite sets V = { v , ..., v m } and V ′ = { v ′ , ..., v ′ n } . Row u , union of column v and column v ′ , { ( u i , v ) , ( u , v j ) | l ≥ i ≥ , m ≥ j ≥ } , and { ( u i , v ′ ) , ( u , v ′ j ) | l ≥ i ≥ , n ≥ j ≥ } are minimal dominating sets of K l (cid:3) K m,n , so we have Γ( K l (cid:3) K m,n ) ≥ max { m + n, l, m + l − , n + l − } . Suppose D is a minimal dominating set of K l (cid:3) K m,n . If each column of K l (cid:3) K m,n has a vertex in D , then | D | = m + n .If each of { u i } × V and { u i } × V ′ has a vertex in D , then | D | = 2 l . Suppose | D | ≥ max { m + n, l } , and then there exists a column v ∗ without vertex in D .If v ∗ ∈ V ′ , each of { u i }× V must have a vertex w i in D , and there exists some { u i } × V ′ without vertex in D . Without loss of generality, suppose { u } × V ′ has no vertex in D and w = ( u , v ). To dominate ( u , v j ) where column v j has no vertex in { w i } , column v j must have a vertex x j ∈ D . Noticethat { w i } ∪ { x j } is a dominating set of K l (cid:3) K m,n . If { w i } is a column, then { w i } ∪ { x j } can not be minimal. So { w i } is not a column, |{ x j }| ≤ m − | D | ≤ m + l −
2. If v ∗ ∈ V , similarly we have | D | ≤ n + l − X n and X ′ n here. X n has vertex set { u , u , ..., u n }∪{ v , v , ..., v n } and edge set { u i u j | i > j } ∪ { v i v j | i > j } ∪ { u i v i | i = 0 } . We callthe subgraph of X n induced by { u , u , ..., u n } an upper cell , the subgraphinduced by { v , v , ..., v n } a lower cell . We use X ′ n to denote the subgraphinduced by { u , ..., u n } ∪ { v , v , ..., v n } in X n . Consider Γ( X n ) and Γ( X ′ n ).3or X n , every cell of X n must have a vertex to dominate u and v , andthese two vertices dominate X n . Therefore, we have Γ( X n ) = 2. For X ′ n ,suppose D is a minimal dominating set of Γ( X ′ n ). If each of { u , ..., u n } and { v , v ..., v n } has a vertex in D , then | D | = 2. If D > { v , v ..., v n } must have a vertex in D to dominate v and therefore { u , ..., u n } has novertex in D . To dominate u i , v i must be in D . We have D = { v ..., v n } andΓ( X ′ n ) = n . X ′ n is the induced subgraph of X n and Γ( X ′ n ) is greater thanΓ( X n ). Similar situation also happens in Cartesian products of graphs. Thefollowing proposition gives an example. Proposition 7.
For sufficiently large n , we have Γ( X ′ n (cid:3) K ) > Γ( X n (cid:3) K ). Proof.
Since { v , ..., v n } × V ( K ) is a minimal dominating set of X ′ n (cid:3) K , wehave Γ( X ′ n (cid:3) K ) ≥ n . Suppose D is a minimal dominating set of X n × K with | D | ≥ n . If a cell in X n × K has two or more vertices in D , thenthese vertices must have private D -neighbors, which must be in cells withoutvertex in D . If n is sufficiently large, it is easy to see that there are threecells without vertex in D and almost all vertices in the other three cells arein D . Cells without vertex in D are not all upper cells, or { u } × V ( K )can not be dominated. Similarly, cells without vertex in D are not all lowercells. Suppose two upper cells have no vertex in D . In the remaining uppercell, a vertex in D dominates at least two vertices in cells without vertexin D and prohibits these vertices from being private D -neighbors of othervertices in D . It follows that | D | < n , a contradiction. Therefore, we haveΓ( X ′ n (cid:3) K ) ≥ n > Γ( X n (cid:3) K ).Suppose x is a vertex having only one neighbor y in a graph G . If D is aminimal dominating set of G , D must contain exactly one vertex of x and y .To minimize D , it is a better choice to put y in D . However, if we want tomaximize D , putting x in D is not always better. For example, let G be thegraph with vertex set { x, y } ∪ V ( X n ) and edge set { xy, yu } ∪ E ( X n ), and D is a minimal dominating set of G . Using similar argument in determiningΓ( X n ) and Γ( X ′ n ), we know | D | = 3 if D contains x , and | D | = n + 1 if D contains y .Given a graph G , if we add some edges in G to get a graph G ′ , we alwayshave γ ( G ) ≥ γ ( G ′ ). However, upper domination in graphs does not have thisproperty. For example, let G ′ be the graph obtained from adding edge u v in G = X n . Then G ′ is isomorphic to K (cid:3) K n +1 and Γ( G ) = 2 < Γ( G ′ ) = n + 1for n >
1. The above examples show that upper domination in graphsprobably does not behave as well as one may expect, and the argumentof minimal counterexample probably does not work well in solving upperdomination problems. 4t is easy to see that a maximal independent set is a minimal dominatingset, and therefore we have Γ( G ) ≥ α ( G ) for any graph G . It is clear that α ( X n ) = α ( X ′ n ) = 2. Since Γ( X n ) = 2 and Γ( X ′ n ) = n , we know theinequality Γ( G ) ≥ α ( G ) is sharp, and the difference between Γ( G ) and α ( G )could be quite large. Lemma 8. G and H are two arbitrary graphs. I G and I H are the maximalindependent sets of G and H respectively. Let G ′ = G [ V ( G ) \ I G ] and H ′ = G [ V ( H ) \ I H ]. We have Γ( G (cid:3) H ) ≥ | I G || I H | + Γ( G ′ (cid:3) H ′ ). Proof.
Suppose D = I G × I H and D is the maximum minimal dominatingset of G ′ (cid:3) H ′ . We claim that D ∪ D is a minimal dominating set of G (cid:3) H and then we have the inequality. Since I G and I H dominate V ( G ) and V ( H )respectively, D dominates I ( G ) × V ( H ) and V ( G ) × I ( H ), and D ∪ D dominates V ( G ) × V ( H ). Notice that there is no edge between D and D . Vertices in D are ( D ∪ D )-isolated, and D -isolated vertices are still( D ∪ D )-isolated. If a vertex u in D has a private D -neighbor v in G ′ (cid:3) H ′ , v is not adjacent to any vertex in D and is still a private ( D ∪ D )-neighborof u in G (cid:3) H . By Lemma 1, we know that D ∪ D is a minimal dominatingset of G (cid:3) H .In G (cid:3) H , we can choose vertices from different rows and columns toform an independent set. Therefore, we have Γ( G (cid:3) H ) ≥ α ( G (cid:3) H ) ≥ min {| V ( G ) | , | V ( H ) |} . The following lemma improves this result. Lemma 9.
For any nontrivial graph G and arbitrary graph H , we haveΓ( G (cid:3) H ) ≥ | V ( H ) | , and the equality holds only if G is a complete graph or K , . Proof.
Since G is nontrivial, we can always find a component G of G with | V ( G ) | ≥
2. Choose a vertex u in V ( G ). The set N ( u ) of neighbors of u in G is not empty. In case row u is a dominating set of G (cid:3) H , it is a minimaldominating set of G (cid:3) H and we have Γ( G (cid:3) H ) ≥ Γ( G (cid:3) H ) ≥ | V ( H ) | .Suppose row u is not a dominating set of G (cid:3) H . Let D = ( V ( G ) \ N ( u )) × V ( H ). Clearly, D is a dominating set of G (cid:3) H , and we can find a subset D ′ of D such that D ′ is a minimal dominating set of G (cid:3) H . For every vertex v ∈ V ( H ), D ′ must have a vertex in column v of G (cid:3) H to dominate verticesin N ( u ) × { v } , and we have Γ( G (cid:3) H ) ≥ Γ( G (cid:3) H ) ≥ | V ( H ) | .Suppose Γ( G (cid:3) H ) = | V ( H ) | . G must be connected. Otherwise, we haveΓ( G (cid:3) H ) > Γ( G (cid:3) H ) ≥ | V ( H ) | . If G is not bipartite, an induced subgraphof G from removing a maximal independent set of G is nontrivial. If G is bipartite but not complete bipartite, choose two vertices u and u ′ from5ifferent partite sets of G such that u and u ′ are not adjacent. Suppose I is amaximal independent set of G containing u and u ′ . Then G [ V ( G ) \ I ] must benontrivial. Otherwise, G is not connected. Therefore, if G is not a completebipartite graph, we can always find a maximal independent set I G of G suchthat G ′ = G [ V ( G ) \ I G ] is nontrivial. Suppose I H is a maximal independentset of H and H ′ = H [ V ( H ) \ I H ]. By Lemma 8 and the inequality provenabove, we have Γ( G (cid:3) H ) = | V ( H ) | ≥ | I G || I H | + Γ( G ′ (cid:3) H ′ ) ≥ | I G || I H | + | V ( H ′ ) | . It implies | I G | = 1, which means G is a complete graph. Proposition3 shows that the equality does hold while H is also a complete graph with V ( H ) ≥ V ( G ).Consider the case that G is complete bipartite. If G = K , , which isalso a complete graph, Proposition 4 shows that the equality always holds.If G = K , , Proposition 5 shows that the equality does hold while H is acomplete graph with V ( H ) ≥ G = K ,m , m ≥
3, let V ( G ) = { u , u , ..., u m } , where u is the vertexof degree m . If H has a vertex of degree 0, it is easy to see that Γ( G (cid:3) H ) > | V ( H ) | . Suppose every vertex of H has degree at least 1, and D H is a minimaldominating set of H . Let R H = V ( H ) \ D H . Notice that R H is not empty,and every vertex in D H has a neighbor in R H . In other words, R H is adominating set of H . By Lemma 1, it is easy to check that ( { u m } × R H ) ∪ ( { u , ..., u m − } × D H ) is a minimal dominating set of G (cid:3) H , and we haveΓ( G (cid:3) H ) > | V ( H ) | . If G = K m,n , m, n ≥
2, let u and u ′ be two vertices indifferent partite sets of G . Then { u, u ′ } × V ( H ) is a minimal dominating setof G (cid:3) H , and we have Γ( G (cid:3) H ) > | V ( H ) | .Let D be the minimal dominating set of a graph G . We define D P asthe set of vertices in D having a private D -neighbor, and D I = D \ D P . ByLemma 1, we know vertices in D I are D -isolated. Now we are ready to provethe main theorem. Theorem 10.
For any graphs G and H , we haveΓ( G (cid:3) H ) ≥ Γ( G )Γ( H ) + min {| V ( G ) | − Γ( G ) , | V ( H ) | − Γ( H ) } . Proof. If G or H has no edge, then | V ( G ) | = Γ( G ) or | V ( H ) | = Γ( H ), andwe have Γ( G (cid:3) H ) = Γ( G )Γ( H ) = Γ( G )Γ( H )+ min {| V ( G ) | − Γ( G ) , | V ( H ) | − Γ( H ) } . Now we prove the case that every vertex of G and H has degreeat least 1, and then the case that G and H are nontrivial follows. Let D G and D H be the minimal dominating sets of G and H respectively. Define R G = V ( G ) \ D G , R H = V ( H ) \ D H , G ′ = G [ R G ], H ′ = H [ R H ]. If D PG and D PH are both empty, then D G and D H are maximal independent sets of G and H respectively. By Lemma 8, we have Γ( G (cid:3) H ) ≥ | D G || D H | + Γ( G ′ (cid:3) H ′ ) ≥| D G || D H | + min {| V ( G ) | − | D G | , | V ( H ) | − | D H |} .6uppose D PG is not empty. For each vertex ( u, v ) ∈ D PG × D H , we selecta corresponding vertex ( u, v ) ′ = ( u ′ , v ) such that u ′ is a private D G -neighborof u in G . Let D = { ( u, v ) ′ | ( u, v ) ∈ D PG × D H } . For each vertex ( u, v ) ∈ D IG × D PH , we select a corresponding vertex ( u, v ) ′′ = ( u, v ′′ ) such that v ′′ is a private D H -neighbor of v in H . Let D = { ( u, v ) ′′ | ( u, v ) ∈ D IG × D PH } .Since every vertex of G has degree at least 1, every vertex in D IG must havea neighbor in R G , and therefore R G is a dominating set of G . Similarly, R H is a dominating set of H . It is easy to check that D = D ∪ D ∪ ( D IG × D IH ) ∪ ( R G × R H ) is a dominating set of G (cid:3) H . Let D ′ ⊂ D be aminimal dominating set of G (cid:3) H . Notice that each vertex in D IG × D IH is D -isolated, and each vertex in D ∪ D has a private D -neighbor in G (cid:3) H .By Lemma 2, we have D ∪ D ∪ ( D IG × D IH ) ⊂ D ′ . There is no edge between D ∪ D ∪ ( D IG × D IH ) and D PG × R H . For each v ∈ R H , D ′ must have at leastone vertex in R G × { v } to dominate vertices in D PG × { v } . Therefore, we haveΓ( G (cid:3) H ) ≥ | D ′ | ≥ | D | + | D | + | D IG × D IH | + | R H | = | D G || D H | + | V ( H ) |−| D H | .It is easy to check that the above argument works no matter D IG , D IH , D PH are empty.In the above proof, if D PG and D PH are both empty, then we have Γ( G (cid:3) H ) ≥| D G || D H | + Γ( G ′ (cid:3) H ′ ). In case that G ′ and H ′ are nontrivial, such as G and H are not bipartite, then we have Γ( G (cid:3) H ) ≥ | D G || D H | + max {| V ( G ) | −| D G | , | V ( H ) | − | D H |} by Lemma 9. If D PG is not empty, we have Γ( G (cid:3) H ) ≥| D G || D H | + | V ( H ) | − | D H | . Therefore, if D PG and D PH are not empty, thenwe also have Γ( G (cid:3) H ) ≥ | D G || D H | + max {| V ( G ) | − | D G | , | V ( H ) | − | D H |} .The above discussion shows that the inequality in Theorem 10 can be im-proved to Γ( G (cid:3) H ) ≥ Γ( G )Γ( H )+ max {| V ( G ) | − Γ( G ) , | V ( H ) | − Γ( H ) } inmany cases. However, in general the minimum in the inequality can not bereplaced by maximum. Let G = K and H = K , . We know Γ( G ) = 1,Γ( H ) = 8, and Γ( G (cid:3) H ) = 12 by Proposition 6. It shows that the inequal-ity in Theorem 10 is sharp. We wonder if the lower bound of Γ( G (cid:3) H ) isclose to Γ( G )Γ( H )+ max {| V ( G ) | − Γ( G ) , | V ( H ) | − Γ( H ) } . For example,for G = K l and H = K m,n , l, m, n ≥
2, we have Γ( G (cid:3) H ) ≥ Γ( G )Γ( H )+max {| V ( G ) | − Γ( G ) , | V ( H ) | − Γ( H ) } − G (cid:3) H ) to end this paper. Thislower bound can be much better than Theorem 10 in some cases. For a graph G and S ⊂ V ( G ), we define N [ S ] = S ∪ { v ∈ V ( G ) | v has a neighbor in S } . Theorem 11. G and H are two arbitrary graphs, and D is a minimaldominating set of H . We have Γ( G (cid:3) H ) ≥ | V ( G ) || D P | + γ ( G ) | D I | .Moreover, if N [ D P ] ∪ D I = V ( H ), then we haveΓ( G (cid:3) H ) ≥ | V ( G ) || D P | + Γ( G ) | D I | . Proof.
Let D = V ( G ) × D P and D = V ( G ) × D I . Clearly, D ∪ D
7s a dominating set of G (cid:3) H , and suppose D ⊂ D ∪ D is a minimaldominating set of G (cid:3) H . Notice that vertices in D have private ( D ∪ D )-neighbors in G (cid:3) H , and therefore D ⊂ D . For v ∈ D I , D musthave at least γ ( G ) vertices in column v to dominate column v . Therefore,Γ( G (cid:3) H ) ≥ | D | ≥ | V ( G ) || D P | + γ ( G ) | D I | . In case N [ D P ] ∪ D I = V ( H ),let D ′ be a maximum minimal dominating set of V ( G ). It is easy to checkthat D ∪ ( D ′ × D I ) is a minimal dominating set of G (cid:3) H , and we haveΓ( G (cid:3) H ) ≥ | V ( G ) || D P | + Γ( G ) | D I | . Acknowledgements.
The author would like to thank the support ofNational Center for Theoretical Science, Taiwan.