Dominator partitions of graphs

Dominator Partitions of Graphs Sandra M. Hedetniemi Stephen T. Hedetniemi Clemson University Clemson SC 29634 shedet@cs.clemson.edu hedet@cs.clemson.edu Renu Laskar Clemson University Clemson, SC29634 rclsk@clemson.edu Alice A. McRae Appalachian State University Boone, NC 28608 aam@cs.cs.appstate.edu Charles K. Wallis Western Carolina university Cullowhee,NC 28723 cwallis@email.wcu.edu October 15, 2008 Abstract A vertex v ∈ V in a graph G = (V, E) dominates a set S ⊆ V if it is adjacent to every vertex w ∈ S, in which case we say that v is a dominator of S. A partition π = {V1 , V2 , . . . , Vk } of V (G) is called a dominator partition if every vertex v ∈ V is a dominator of at least one block Vj of π. The dominator partition number of a graph G, denoted πd (G), is the minimum order of a dominator partition of G. In this paper we introduce the concept of dominator partitions and obtain tight bounds for πd (G) for any graph G. 1 Introduction Let G = (V, E) be a graph with vertex set V = {v1 , v2 , . . . , vn }. A set S ⊆ V is a dominating set of G if every vertex in V − S is adjacent to at least one vertex in S. The domination number γ(G) of G equals the minimum cardinality of a dominating set S in G; we say that such a set S is a γ-set. A vertex v ∈ V dominates a set S ⊆ V , written v ≻ S, if it is adjacent to every vertex w ∈ S, in which case we say that v is a dominator of S. A dominator partition of a graph G is a partition π = {V1 , V2 , . . . , Vk } of V (G) such that every vertex u ∈ V is a dominator of at least one block Vj of π, that is, for every u ∈ V there exists j, 1 ≤ j ≤ k, such that u ≻ Vj . Note that a vertex v ∈ Vi is a dominator of its own block if Vi ⊆ N [v], where N [v] = {u|uv ∈ E} ∪ {v} is the closed neighborhood of vertex v. Since every vertex dominates itself, that is, v ∈ N [v], it follows that the trivial partition π = {{v1 , }, {v2 }, . . . , {vn }} is a dominator partition. Thus, every graph has a dominator partition. 1 A dominator partition π = {V1 , V2 , . . . , Vk } is minimal if any partition π ′ obtained from π by forming the union of any two classes Vi ∪ Vj , i 6= j, as one class in π ′ , is no longer a dominator partition. The dominator partition number πd (G) and the upper dominator partition number Πd (G) of are the minimum and maximum k such that G has a minimal dominator partition of order k. If π = {V1 , V2 , . . . , Vk } is a dominator partition of order πd (G), then π is called a πd -partition. In this paper we initiate the study of the two dominator partition numbers πd (G) and Πd (G). 2 Bounds for πd (G) and Πd (G) The first few observations about dominator partitions are easy to verify, and provide some intuition about the two dominator partition numbers. Observation 2.1 For any graph G of order n, 1. 1 ≤ πd (G) ≤ Πd (G) ≤ n. 2. πd (G) = 1 = Πd (G) if and only if G = Kn , the complete graph of order n. 3. πd (G) = n = Πd (G) if and only if G = Kn , that is, G consists of n isolated vertices. 4. For the cycle C4 of length four, πd (C4 ) = Πd (C4 ) = 2. For the cycle C6 with vertices labelled v1 , v2 , v3 , v4 , v5 , v6 , it is easy to see that π = {{v1 , v2 }, {v3 , v4 }, {v5 , v6 }} is a dominator partition of minimum order, and hence, πd (C6 ) = 3. However, π ′ = {{v1 }, {v3 }, {v5 }, {v2 , v4 , v6 }} is a minimal dominator partition, and hence, Πd (C6 ) ≥ 4. Later on we will show that Πd (C6 ) = 4. The following result characterizes graphs for which πd (G) = 2. Theorem 2.2 For a graph G 6= Kn , πd (G) = 2 if and only if either (i) the complement G of G is disconnected, or (ii) G is bipartite. Proof. Assume that G is disconnected, and let τ = {V1 , V2 , . . . , Vk } be the partition of V (G) into its connected components. Then π = {V1 , V2′ }, where V2′ = V2 ∪ V3 ∪ . . . ∪ Vk , is a dominator partition of G, since every vertex in V1 is adjacent to every vertex in V2′ , and conversely. Since we have assumed that G 6= Kn , we know, therefore, that πd (G) = 2. Conversely, assume that πd (G) = 2, and let π = {V1 , V2 } be a dominator partition of G. We can partition V1 = V11 ∪ V12 and V2 = V21 ∪ V22 , as follows: 2 1. V11 = {x|x ∈ V1 and x ≻ V1 }, 2. V12 = {x|x ∈ V1 − V11 and x ≻ V2 }, 3. V21 = {y|y ∈ V2 and y ≻ V1 }, 4. V22 = {y|y ∈ V2 − V21 and y ≻ V2 }. If V11 = ∅, that is, no vertex in V1 dominates V1 , then for every vertex v ∈ V1 , v ≻ V2 , that is, every vertex in V1 is adjacent to every vertex in V2 . This implies that in G there are no edges between V1 and V2 , i.e. G is disconnected. Similarly, if V22 = ∅, it follows that every vertex in V2 is adjacent to every vertex in V1 , i.e. G is disconnected. Assume, therefore, that V11 6= ∅ and V22 6= ∅. We have four subcases to consider. Case 1. V12 = ∅ and V21 = ∅. In this case both V1 and V2 induce complete subgraphs of G, which implies that G is bipartite. Case 2. V12 = ∅ and V21 6= ∅. In this case V1 and V22 induce complete subgraphs in G; every vertex in V22 is adjacent to every vertex in V21 , and every vertex in V21 is adjacent to every vertex in V1 . Therefore, consider the partition π = {V1 ∪ V22 , V21 }. In G there are no edges between V21 and V1 ∪ V22 , and hence G is disconnected. Case 3. V12 6= ∅ and V21 = ∅. This implies that V2 induces a complete subgraph in G, and every vertex in V12 is adjacent to every vertex in the set V11 ∪ V2 . Therefore, there are no edges in G between V12 and V11 ∪ V2 , that is, G is disconnected. Case 4. V12 6= ∅ and V21 6= ∅. In this case, consider the partition π = {V11 ∪ V22 , V12 ∪ V21 }. One can observe that every vertex in V12 is adjacent to every vertex in V11 ∪ V22 , as is every vertex in V21 . Therefore, in G there are no edges between V12 ∪ V21 and V11 ∪ V22 , which means that G is disconnected. If G is bipartite, and X and Y are partite sets of G, then X and Y are complete subgraphs of G and since G 6= Kn , Πd (G) = 2. ✷ The following results provides lower and upper bounds for πd (G) and Πd (G), respectively, in terms of the maximum degree ∆(G) and minimum degree δ(G) of a vertex in G. Lemma 2.3 Let π = {V1 , V2 , . . . , Vk } be a minimal dominator partition. Then there is at most one block Vi such that |Vi | > 1 + ∆(G). Proof. Let π = {V1 , V2 , . . . , Vk } be a minimal dominator partition. By definition, any partition π ′ obtained from π by forming the union of any two blocks Vi ∪ Vj , i 6= j, as one block in π ′ is no longer 3 a dominator partition. But this implies that for every pair i, j, i 6= j, there exists a vertex x ∈ V such that x does not dominate any block of π ′ , but dominates at least one of Vi or Vj in π, say Vi . But then |Vi | ≤ 1 + ∆(G). In other words, for a minimal dominator partition π = {V1 , V2 , . . . , Vk } of G, for every pair Vi , Vj , there exists at least one block, say Vi , such that |Vi | ≤ 1 + ∆(G). But this implies that at most one Vr can exist such that |Vr | > 1 + ∆(G). ✷ Theorem 2.4 For any graph G of order n, n ≤ πd (G) ≤ Πd (G) ≤ n − δ(G). 1 + ∆(G) Proof. Let π = {V1 , V2 , . . . , Vk } be a minimal dominator partition of G. Then the partition π ′ obtained from π by forming the union of any two classes Vi and Vj , i 6= j, as one class is no longer ′ } be so obtained from π. Therefore, there exists a a dominator partition. Let π ′ = {V1′ , V2′ , . . . , Vk−1 vertex u ∈ V such that u does not dominate any class of π ′ . But this implies that every class in π ′ has a vertex not adjacent to u. So, degG (u) ≥ k − 1, which implies that ∆(G) ≥ k − 1, which means that n − 1 − δ(G) ≥ k − 1, or k ≤ n − δ(G). Therefore, πd (G) ≤ Πd (G) ≤ n − δ(G). For the lower bound: Let π = {V1 , V2 , . . . , Vk } be a πd -partition of G, that is, k = πd (G). Case1. |Vi | ≤ 1 + ∆(G), for all i, 1 ≤ i ≤ k. This implies that n= k X |Vi | ≤ k(1 + ∆(G)), i=1 which in turn implies that k = πd (G) ≥ 4 n . 1 + ∆(G) Case 2. From Lemma 2.3, there exists a unique index i such that |Vi | > 1 + ∆(G). Note that since |Vi | is greater than 1 + ∆(G), no vertex can dominate Vi . Construct a dominator partition π ∗ from π as follows: if any |Vr | ≥ 2, r 6= i, then move all the vertices of Vr except one to Vi . Note that the number of vertices moved from each such Vr is at most ∆(G). Thus, π ∗ = {V1∗ , V2∗ , . . . , Vk∗ }, where |Vj∗ | = 1, j 6= i, and |Vi∗ | > 1 + ∆(G). Let S1 be the set of all singleton sets of π ∗ , let S2 be Vi∗ , and let E(S1 , S2 ) denote the number of edges between S1 and S2 . Since each vertex of S1 is adjacent to at most ∆(G) vertices of S2 , and S1 has k − 1 vertices, we have that E(S1 , S2 ) ≤ (k − 1)∆(G). Again, since each vertex of S2 is adjacent to at least one vertex of S1 , we have that E(S1 , S2 ) ≥ |S2 |. Thus, |S2 | ≤ E(S1 , S2 ) ≤ (k − 1)∆(G), and n = |S1 | + |S2 | ≤ (k − 1) + (k − 1)∆(G) = (k − 1)(1 + ∆(G)). Thus, k − 1 ≥ n/(1 + ∆(G)), or, k ≥ 1 + n/(1 + ∆(G)). Combining the two cases we have the desired result. ✷ It can be seen that the lower bound for πd (G) is sharp for the graph G = 4K3 , consisting of four disjoint copies of the complete graph K3 . Here ∆(G) = 2, n = 12, and πd (G) = 4 = n/(1 + ∆(G)). The upper bound for Πd (G) is sharp for G = C5 , where n = 5, δ(G) = 2, and Πd (G) = 3 = n − δ(G). The next result provides a tight bound for πd (G) in terms of the domination number γ(G). Theorem 2.5 For any graph G, γ(G) ≤ πd (G) ≤ γ(G) + 1. Proof. Let π = {V1 , V2 , . . . , Vk } be a dominator partition of G of minimum order k = πd (G). Let xi ∈ Vi be an arbitrary vertex in Vi , and let S = {xi |xi ∈ Vi }, for 1 ≤ i ≤ k. It follows that S is a dominating set. Let w be any vertex of V (G). Since π is a dominator partition of V (G), by definition, w must dominate some class Vj , i.e. w must be adjacent to all vertices of Vj , and, in particular, w must be adjacent to xj . Hence, γ(G) ≤ |S| = πd (G). To prove the upper bound, let S = {u1 , u2 , . . . , uk } be a γ-set with k = γ(G). Construct a partition π = {V1 , V2 , . . . , Vk+1 }, where Vi = {ui }, for all i, 1 ≤ i ≤ k, and let Vk+1 = V − S. Clearly, π is a dominator partition and so πd (G) ≤ γ(G) + 1. ✷ Based on Theorem 2.5, let us say that a graph G is a dominator graph if πd (G) = γ(G), while a non-dominator graph is a graph for which πd (G) = γ(G) + 1. The following corollaries of Theorem 2.5 give us some insight into the nature of dominator graphs. 5 Corollary 2.6 If G is a dominator graph, then every vertex v ∈ V is in some γ-set of G. Graphs having the property that every vertex is in some γ-set are called γ-excellent in [4]. We note, however, that although every dominator graph is γ-excellent, not every γ-excellent graph is a dominator graph. For example, the 5-cycle C5 is γ-excellent, but πd (C5 ) = γ(C5 ) + 1 = 3. Thus, C5 is a non-dominator graph. Let π = {V1 , V2 , . . . , Vk } be any partition of V (G) into non-empty sets Vi ⊂ V . Let S ⊂ V be a set of cardinality k obtained by selecting one, arbitrary vertex Vi ∈ Vi , for each i, 1 ≤ i ≤ k. Such a set S is called a system of distinct representatives of π. Corollary 2.7 If G is a dominator graph, then every system of distinct representatives of every πd -partition of G is a γ-set of G. Corollary 2.8 If G is a dominator graph, then a path P3 = {u, v, w} with degG (u) = degG (w) = 1 is a forbidden subgraph of G. Corollary 2.9 If G is a dominator graph, and G has two vertices vi , vj ∈ V that are never in the same γ-set, then vi and vj must be in the same block of every πd -partition of G. Let π = {V1 , V2 , . . . , Vk } be a dominator partition. If a vertex u is a dominator of only one block, say Vi of π, then u will be called a private dominator of Vi , and we say that Vi has a private dominator. The following result is straightforward. Proposition 2.10 If G is a dominator graph, then every block in every πd -partition of G has a private dominator. Corollary 2.11 If a graph G has a πd -partition in which some block does not have a private dominator, then πd (G) = γ(G) + 1, and G is not a dominator graph. 3 Dominator partitions of trees In this section we present a characterization of dominator trees. Theorem 3.1 A tree T of order n ≥ 3 is a dominator tree if and only if the leaves of T form a γ-set of T , or equivalently, if and only if every interior vertex of T is adjacent to exactly one leaf. In order to prove this theorem, we need the following lemmas. 6 Lemma 3.2 If T is a dominator tree, then no block of a πd - partition of T can be a singleton set. Proof. Suppose, to the contrary, that a πd -partition π of a dominator tree T contains a singleton block, V0 = {u}. Case 1. Suppose that u has two neighbors, say v1 , and v2 , that are in the same block Vj of π. Then Vj has no private dominator; u is not a private dominator of Vj since it dominates Vi , and no other vertex can be a private dominator of Vj since no other vertex can be adjacent to both v1 and v2 , since T is a tree. This contradicts Proposition 2.10. Case 2. Suppose that each of the neighbors v1 , v2 , . . . , vk of u are in different blocks, say V1 , V2 , . . . , Vk , of π, respectively. Then we can form a dominating set of T by selecting a system of distinct representatives S, by selecting one vertex from each block of π. In particular, we select vi ∈ Vi , for 1 ≤ i ≤ k, and one arbitrary vertex from each of the remaining blocks. The resulting set S is a dominating set, but is not a minimal dominating set, because the removal of u from S results in a smaller dominating set of T , which contradicts the assumption that πd (T ) = γ(T ). ✷ Lemma 3.3 If T is a dominator tree, then the blocks of every πd -partition of T form a perfect matching in T . Proof. Let π = {V1 , V2 , . . . , Vk } be a πd -partition of a dominator tree T . Let T be rooted, and let u be a vertex at a maximum distance from the root of T , which is in a block Vi not consisting of two adjacent vertices. Since vertex u is a dominator of at least one block, let vertices x and y be elements of a block Vj dominated by the vertex u. By Lemma 3.2, it follows that |Vi | > 1 and Vi 6= Vj . Since xy ∈ / E(T ), otherwise u, x, and y will form a triangle in T , Vj is a block not consisting of precisely one edge. However, at least one of x and y must be a child of u, since both are adjacent to u, contradicting the choice of vertex u. ✷ Lemma 3.4 If T is a dominator tree, then each block of every πd -partition of T contains a leaf. Proof. By Lemma 3.3, each block of every πd -partition π consists of two adjacent vertices. Suppose Vi = {u, v} is a block in which neither u nor v is a leaf. Then u and v are adjacent to distinct vertices x, and y, respectively. Since vertices x and y cannot be adjacent, they are in different blocks of π. Selecting a system of distinct representatives from each block of π determines a dominating set S. If vertices x and y are both selected from their respective blocks, then no vertex needs to be selected from the block Vi , since vertices u and v are dominated by the representative vertices from their own blocks, yielding a dominating set of size less than γ(T ), a contradiction. ✷ 7 Lemma 3.5 If V (T ) can be partitioned into a perfect matching in such a way that each block contains exactly one leaf, then γ(T ) = πd (T ) and T is a dominator tree. Proof. Assume that a tree T has a partition π of V (T ) into a perfect matching, in such a way that each block contains a leaf. Then any γ-set for T must contain a member of each block of the partition π, since each block contains a leaf. Thus, the size of the partition is a lower bound for γ(T ). However, such a matching forms a dominator partition, and thus, the size of the partition is an upper bound for πd (T ). It follows, therefore, that γ(T ) = πd (T ) and hence, T is a dominator tree. ✷ The proof of Theorem 3.1 is obtained by combining Lemmas 3.1, 3.2, 3.3, 3.4 and 3.5. The corona of a graph G, denoted GoK1 , is the graph obtained from G by attaching a leaf to every vertex of G. Thus, every dominator tree is a corona. 4 Complexity results The algorithmic complexity of the dominator partition numbers is predictable. We provide a sample theorem. DOMINATOR PARTITION INSTANCE: graph G = (V, E), positive integer k ≤ |V | QUESTION: does G have a dominator partition of order at most k? Theorem 4.1 DOMINATOR PARTITION is NP-complete, even when restricted to chordal, planar or bipartite graphs. Proof. Clearly the DOMINATOR PARTITION problem is in the class N P , since it is easy, in polynomial time to produce a guessed partition π of order at most k, and in polynomial time it can be verified if π is a dominator partition. We construct a transformation from the following, well known NP-complete problem. DOMINATING SET INSTANCE: graph G = (V, E), positive integer k ≤ |V | QUESTION: does G have a dominating set of size at most k? Let G = (V, E) be any graph of order n. Form a new graph G∗ from G by performing the following operation at each of the n vertices u ∈ V . Add an edge between vertex u and one of the leaves v of a K1,3 , whose leaves are labelled v, y, z and whose central vertex is labelled x; that is, K1,3 has edges vx, yx and zx. Notice that in the graph G∗, one can find at least n instances of a vertex x joined to two leaves, y and z. But from Corollary 2.8 we know that this is a forbidden subgraph of a dominator graph. Therefore, we know that the graph G∗ is not a dominator graph, and therefore that πd (G∗) = γ(G∗) + 1. 8 On the other hand, it is easy to see that γ(G∗) = γ(G) + n. Therefore, πd (G∗) = γ(G) + n + 1, and G has a dominating set of size at most k if and only if G∗ has a dominator partition of order at most k + n + 1. It is well known that DOMINATING SET remains NP-complete, even when restricted to chordal, planar or bipartite graphs. But the construction of graph G∗ from graph G has the property that if G is chordal, planar, or bipartite, then G∗ is also chordal, planar, or bipartite, respectively. Therefore, DOMINATOR PARTITION remains NP-complete for chordal, planar, and bipartite graphs. ✷ Finally, as a result of Theorem 3.1, we have the following: Proposition 4.2 There is a polynomial algorithm for deciding whether an arbitrary tree T is a dominator tree. 5 Open problems 1. Can you characterize other classes of dominator graphs than dominator trees? For example, we can provide, without proof, the following dominator graphs: 1. The only paths that are dominator trees are the paths P1 , P2 , and P4 . 2. The only cycles that are dominator graphs are the cycles C3 = K3 and C4 . 3. The only grid graphs Gm,n = Pm ✷Pn that are dominator graphs are the grid graphs G1,1 = P1 , G1,2 = P2 , G1,4 = P4 , and G2,2 = C4 . 4. For any graph G, the corona GoK1 is a dominator graph. 2. When is the complement G of a dominator graph G a dominator graph? 3. What can you say about the upper dominator partition number Πd (G)? This number appears to be much more difficult to determine. For example, consider the following. For the graph C6 = {v1 , v2 , v3 , v4 , v5 , v6 }, the partition Π = {{v1 }, {v3 }, {v5 }, {v2 , v4 , v6 }} is a minimal dominator partition and hence Πd (C6 ) ≥ 4, whereas Πd (C6 ) ≤ n − δ(G) = 4, from the upper bound of Theorem 2.4. So Πd (G) = 4. Similarly, consider the following partition of the path P7 , with vertices labeled in order, v1 , v2 , v3 , v4 , v5 , v6 , v7 . Let Π = {{v1 }, {v2 , v3 }, {v4 }, {v5 , v6 }, {v7 }}. It can be seen, with some thought, that this is a minimal dominator partition of order five, which achieves the upper bound of n−δ(P7 ) = n−1 = 6. Therefore, we can establish that Πd (P7 ) = 5. 9 References [1] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, Inc. New York, 1998. [2] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater, editors, Domination in Graphs: Advanced Topices, Marcel Dekker, New York, 1998. [3] E. J. Cockayne and S. T. Hedetniemi, Towards a theory of domination in graphs. Networks, 7:247-261, 1977. [4] G. H. Fricke, T. W. Haynes, S. M. Hedetniemi, S. T. Hedetniemi and R. C. Laskar, Excellent trees, Bull. ICA, 34:27-38, 2002. [5] O. Ore. Theory of Graphs. Amer. Math. Soc. Colloq. Publ., 38 (Amer. Math. Soc., Providence, RI), 1962. 10