Nordhaus-Gaddum-type results for thegeneralized edge-connectivity of graphsSupported by NSFC No.11071130 and the “973” project.

Nordhaus-Gaddum-type results for the generalized edge-connectivity of graphs1

Abstract

Let be a graph, be a set of vertices of , and be the maximum number of pairwise edge-disjoint trees in such that for every . The generalized -edge-connectivity of is defined as . Thus . In this paper, we consider the Nordhaus-Gaddum-type results for the parameter . We determine sharp upper and lower bounds of and for a graph of order , as well as for a graph of order and size . Some graph classes attaining these bounds are also given.

Keywords

: edge-connectivity; Steiner tree; edge-disjoint trees; generalized edge-connectivity; complementary graph.
AMS subject classification 2010: 05C40, 05C05, 05C76.

1 Introduction

All graphs considered in this paper are undirected, finite and simple. We refer to the book [4] for graph theoretical notation and terminology not described here. For a graph and a set of at least two vertices, an -Steiner tree or an Steiner tree connecting (Shortly, a Steiner tree) is a subgraph of which is a tree such that . Two Steiner trees and connecting are edge-disjoint if . The Steiner Tree Packing Problem for a given graph and asks to find a set of maximum number of edge-disjoint -Steiner trees in . This problem has obtained wide attention and many results have been worked out, see [18, 19, 20]. The problem for is called the Spanning Tree Packing Problem. For any graph of order , the spanning tree packing number or number, is the maximum number of edge-disjoint spanning trees contained in . For the number, Palmer gave a good survey, see [17].

Recently, we introduced the concept of generalized edge-connectivity of a graph in [13]. For , the generalized local edge-connectivity is the maximum number of edge-disjoint trees in connecting . Then the generalized -edge-connectivity of is defined as . Thus . Set when is disconnected. We call it the generalized -edge-connectivity since Chartrand et al. in [5] introduced the concept of generalized (vertex) connectivity in 1984. There have been many results on the generalized connectivity, see [10, 11, 12, 13].

One can see that the Steiner Tree Packing Problem studies local properties of graphs, but the generalized edge-connectivity focuses on global properties of graphs. Actually, the number of a graph is just .

In addition to being natural combinatorial measures, the Steiner Tree Packing Problem and the generalized edge-connectivity can be motivated by their interesting interpretation in practice as well as theoretical consideration. For the practical backgrounds, we refer to [7, 8, 15].

From a theoretical perspective, both extremes of this problem are fundamental theorems in combinatorics. One extreme of the problem is when we have two terminals. In this case internally (edge-)disjoint trees are just internally (edge-)disjoint paths between the two terminals, and so the problem becomes the well-known Menger theorem. The other extreme is when all the vertices are terminals. In this case internally disjoint trees and edge-disjoint trees are just spanning trees of the graph, and so the problem becomes the classical Nash-Williams-Tutte theorem.

Theorem 1.

(Nash-Williams [14], Tutte [16]) A multigraph contains a system of edge-disjoint spanning trees if and only if

holds for every partition of , where denotes the number of crossing edges in , i.e., edges between distinct parts of .

Corollary 1.

Every -edge-connected graph contains a system of edge-disjoint spanning trees.

Let denote the class of simple graphs of order and the subclass of having edges. Give a graph theoretic parameter and a positive integer , the Nordhaus-Gaddum(N-G) Problem is to determine sharp bounds for: and , as ranges over the class , and characterize the extremal graphs. The Nordhaus-Gaddum type relations have received wide investigations. Recently, Aouchiche and Hansen published a survey paper on this subject, see [3].

In this paper, we study and for the parameter where and .

2 Nordhaus-Gaddum-type results in

The following observation is easily seen.

Observation 1.

If is a connected graph, then ;

If is a spanning subgraph of , then .

Let be a connected graph with minimum degree . If has two adjacent vertices of degree , then .

Alavi and Mitchem in [2] considered Nordhaus-Gaddum-type results for the connectivity and edge-connectivity parameters. In [13] we were concerned with analogous inequalities involving the generalized -connectivity and generalized -edge-connectivity. We showed that , but this is just a starting result and now we will further study the Nordhaus-Guddum type relations.

To start with, let us recall the Harary graph on vertices, which is constructed by arranging the vertices in circular order and spreading the edges around the boundary in a nice way, keeping the chords as short as possible. They have the maximum connectivity for their size and . Palmer [17] gave the number of some special graph classes.

Lemma 1.

[17] The number of a complete bipartite graph is .

The number of a Harary graph is .

Corresponding to of Observation 1, we can obtain a sharp lower bound for the generalized -edge-connectivity by Corollary 1. Actually, a connected graph contains spanning trees. Each of them is also a Steiner tree connecting . So the following proposition is immediate.

Proposition 1.

For a connected graph of order and , . Moreover, the lower bound is sharp.

In order to show the sharpness of this lower bound for , we consider the Harary graph . Clearly, . From of Lemma 1, contains spanning trees, that is, . So . For general , one can check that the cycle can attain the lower bound since .

The following proposition indicates that the monotone properties of , that is, , is true for .

Proposition 2.

For two integers and with , and a connected graph , .

Proof.

Assume . Set . For each with , we let , where . Since , there exist edge-disjoint trees connecting . These trees are also edge-disjoint trees connecting . So and . Combining this with of Observation 1, we get that for . ∎

Now we give the lower bounds of and .

Lemma 2.

Let . Then

;

.

Moreover, the two lower bounds are sharp.

Proof.

If , then , that is, and are all disconnected, which is impossible, and so .

By definition, and , and so . ∎

The following observation indicates the graphs attaining the lower bound of in Lemma 2.

Observation 2.

if and only if or is disconnected.

In [13] we obtained the exact value of the generalized -edge-connectivity of a complete graph .

Lemma 3.

[13] For two integers and with , .

For a connected graph of order , we know that . In [13] we characterized the graphs attaining the upper bound.

Lemma 4.

[13] For a connected graph of order with , if and only if for even; for odd, where is an edge set such that .

As we know, it is difficult to characterize the graphs with , even with . So we want to add some conditions to attack such a problem. Motivated by such an idea, we hope to characterize the graphs with . Actually, the Norhaus-Gaddum-type problems also need to characterize the extremal graphs attaining the bounds.

Before studying the lower bounds of and , we give some graph classes (Every element of each graph class has order ), which will be used later.

For , is a graph class as shown in Figure 1 such that and for , where ; is a graph class as shown in Figure 1 such that and for , where ; is a graph class as shown in Figure 1 such that and for , where ; is a graph class as shown in Figure 1 such that .

Figure 1. Graphs for Proposition 3 (The degree of a black vertex is ).

The following observation and lemma are some preparations for Proposition 3.

For , let and be two graphs obtained from the complete bipartite graph by adding one and two edges on the part having vertices, respectively.

Observation 3.

; , ; , .

Proof.

As shown in Figure 2 , .

As shown in Figure 2 , we have . Since , , which implies that . Since is connected, .

As shown in Figure 2 , it follows that . Let and be two parts of the complete bipartite graph . Choose . If there exists an -tree containing vertex , then this tree will use edges of , which implies that since . Suppose that there is no -tree containing vertex . Pick up a such tree, say . Then there exists a vertex of degree in , which implies that there is no other -tree in . So . Since is connected, . From Proposition 2, . ∎

Figure 2. Graphs for Observation 2.

Lemma 5.

Let be a connected graph. If and there exists a vertex such that , then for .

Proof.

Let be the connected components of . Since , it follows that . Let and . Then there exists an edge, without loss of generality, say such that is connected for . Thus contains a spanning tree, say . The trees and are two spanning trees of , that is, . Combining this with Proposition 2, for . ∎

Proposition 3.

if and only if (symmetrically, ) satisfies one of the following conditions:

or ;

and there exists a component of such that is a tree and ;

for and , or for , or for , or for , or for and , or for .

Proof.

Necessity. Let be a graph satisfying one of the conditions of , and . One can see that is connected and its complement is disconnected. Thus and . We only need to show that for each graph satisfying one of the conditions of , and . For , since we have by of Observation 1. For , it follows that by of Observation 1 since . Suppose and there exists a connected component of such that is a tree and . Set . We choose such that . Then . Since every spanning tree of uses edges of , there exists at most one spanning tree of , which implies that there is at most one tree connecting in . So . For , by of Observation 3. For , by of Observation 3, we have . For , . For , one can check that for or . From these together with , we have .

Sufficiency. Suppose . Then and , or and . By symmetry, without loss of generality, we let and . From these together with Proposition 1, and . So we have the following three cases to consider.

Case 1.  .

For , one can check that satisfies but . Now we assume . Since , there exists at least one cut edge in , say . Let and be two connected components of such that and . Set and , where . Suppose . For any , and are connected in since there exists a path in ; for any , and are connected in since there exists a path in ; for any and ( or ), . Clearly, the path connects and in . So is connected, a contradiction. Thus or . Without loss of generality, let . Then and . Clearly, is a graph obtained from by attaching the edge . Since , . If , then there exists one vertex such that , which results in , a contradiction. So and (See Figure 1 ).

Case 2.  .

For , the graph satisfies that and . Since , , , and , we have for ; for ; for . Now we assume . Since , there exists an edge cut such that . Let and be two connected components of , and , where . Clearly, or .

At first, we consider the case . Without loss of generality, let . Since , or . Without loss of generality, let . Clearly, any two vertices are connected in since there exists a path in . Furthermore, for any , or . So is connected and , a contradiction.

Next, we consider the case . Without loss of generality, let be the path of order . Since , there exist at least two vertices in . If and , then we can check that is connected, a contradiction. So we assume that or , that is, or .

For the former, . Since , . Clearly, , which implies that . Therefore, since is disconnected. Thus for each . So and (See Figure 1 ).

For the latter, let . First we consider the case . Since , we have . If and , then there exist two vertices and such that , which implies that is connected, a contradiction. So or . Without loss of generality, let . Thus (See Figure 1 ). Now we focus on the graph . Let be the connected components of and , where . If there exists some connected component such that , then (See Figure 1 ). So we assume . Then we prove the following claim and get a contradiction.

Claim 1. For each connected component of , if , or and , then for .

Proof of Claim . For an arbitrary with , we only prove for . The case for can be proved similarly. If there exists some connected component such that , then and has a spanning tree, say . It is also a Steiner tree connecting . Since is another Steiner tree connecting and are two edge-disjoint trees, we have . Let us assume now for . Let and . Clearly, and . Thus for each connected component such that , and for each connected component such that and . We will show that there are two edge-disjoint Steiner trees connecting in for each so that we can combine these trees to form two edge-disjoint Steiner trees connecting in . Suppose that is a connected component such that . Note that . Since , there exists a vertex, without loss of generality, say , such that . Clearly, contains a spanning tree, say . Thus is a Steiner tree connecting in . Since is another Steiner tree connecting . Clearly, and are edge-disjoint. Assume that is a connected component such that and . Note that . Then there exists an edge, without loss of generality, say such that contains a spanning tree of , say . Thus and are two edge-disjoint Steiner trees connecting . Now we combine these small trees connecting by the vertex to form two big trees connecting . Clearly, and are our desired trees, that is, . From the arbitrariness of , we have .

By Claim , we know that and there exists a connected component of such that and is a tree.

We next consider the case (See Figure 1 ). Thus . Since , , which results in since is disconnected. Thus for each . Let . If , then contains a subgraph , which implies that by of Observation 3. Combining this with Proposition 2, for , a contradiction. If , then and . From Observation 3 and Proposition 2, we have for and for , a contradiction. So for , or for , or