On Domination Number and Distance in Graphs

On Domination Number and Distance in Graphs

Cong X. Kang
Texas A&M University at Galveston, Galveston, TX 77553, USA
kangc@tamug.edu
Abstract

A vertex set of a graph is a dominating set if each vertex of either belongs to or is adjacent to a vertex in . The domination number of is the minimum cardinality of as varies over all dominating sets of . It is known that , where denotes the diameter of . Define as the largest constant such that for any vertices of an arbitrary connected graph ; then in this view. The main result of this paper is that for . It immediately follows that , where and are respectively the average distance and the Wiener index of of order . As an application of our main result, we prove a conjecture of DeLaViña et al. that , where denotes the eccentricity of the boundary of an arbitrary connected graph .

Key Words: domination number, distance, diameter, spanning tree
2010 Mathematics Subject Classification: 05C69, 05C12

1 Introduction

We consider finite, simple, undirected, and connected graphs of order and size . For , we denote by the subgraph of induced by . For , the open neighborhood of is the set , and the closed neighborhood of is . Further, let and for . The degree of a vertex is . The distance between two vertices in the subgraph , denoted by , is the length of the shortest path between and in the subgraph . The diameter of a graph is .

A set is a dominating set (resp. total dominating set) of if (resp. ). The domination number (resp. total domination number) of , denoted by (resp. ), is the minimum cardinality of as S varies over all dominating sets (resp. total dominating sets) in ; a dominating set (resp. total dominating set) of of minimum cardinality is called a -set (resp. -set).

Both distance and (total) domination are very well-studied concepts in graph theory. For a survey of the myriad variations on the notion of domination in graphs, see [4].

It is well-known that (); a “proof ” to () can be found on p.56 of the authoritative reference [4]. However, the “proof ” contained therein is logically flawed. We provide a counter-example to a crucial assertion in the “proof ” and then present a correct proof to (). Upon some reflection, we see that () is the two parameter case of a family of inequalities existing between and the distances in , in the following way: . The inequality naturally brings up the question: What is the largest constant , such that , for all connected graphs and arbitrary vertices , where ? Taking this viewpoint, we have by .

The main result of this paper is that for . Since, for a graph of order , is the Wiener index of (see [6]) and is the average distance (per definition found in [1]), it follows that . As an application of our main result, we prove a conjecture in [3] by DeLaViña et al. that , where denotes the eccentricity of the boundary of an arbitrary connected graph (to be defined in Section 4).

This paper is motivated by the work of Henning and Yeo in [5], where they obtained similar inequalities for total domination number (rather than domination number ). Given the close relation between the two graph parameters, we expect the techniques used in [5] to be readily adaptable towards the results of this paper. However, in striking contrast to [5], we avoid the painstaking case-by-case, structural analysis employed there by making use of the easy and well-known Lemma 3.1; this results in a much simpler and shorter paper. Further, we are able to obtain (in domination) the exact value of for every , rather than only a bound (in total domination, c.f. [5]) for for all but the first few values of .

2 An Error in the proof of in FoDiG

For readers’ convenience, we first reproduce Theorem 2.24 and its incorrect proof as it appears on p.56 of [4], the authoritative reference in the field of domination titled Fundamentals of Domination in Graphs.

Theorem 2.1.

For any connected graph , .

“Proof ” (as found on p.56 of [4]). Let be a -set of a connected graph . Consider an arbitrary path of length . This diametral path includes at most two edges from the induced subgraph for each . Furthermore, since is a -set, the diametral path includes at most edges joining the neighborhoods of the vertices of . Hence, and the desired result follows.” 

Figure 1: a counter-example

Presumably, by a “diametral path”, the authors had in mind an induced path with length . Still, the assertion of the sentence beginning with “Furthermore” is incorrect, as seen by the example in Figure 1: notice that is a -set and the vertices form a diametral path containing edges joining with , whereas .

3 Domination number and distance in graphs

The following lemma can be proved by exactly the same argument given in the proof of Lemma 2 in [2]; it was also observed on p.23 of [1].

Lemma 3.1.

[1, 2] Let be a -set. Then there is a spanning tree of such that is a -set.

Now, we apply Lemma 3.1 to give a correct proof of Theorem 2.1.

Proof of Theorem 2.1. Given , take a spanning tree of such that . Suppose, for the sake of contradiction, . Since and , we have

(1)

Take a path of with length equal to . If (1) holds, there must exist a vertex of such that . Since is a path of (a tree), this is impossible. 

Theorem 3.2.

Given any three vertices of a connected graph , we have

(2)

Further, if equality is attained in (2), then for any pair .

Proof.

By Lemma 3.1, there exists a spanning tree of with . Since for any two vertices , it suffices to prove (2) on . If , and all lie on one geodesic, then the inequality (3) obviously holds by Theorem 2.1. Thus, let , , and , with , as shown in Figure 2. Then, the inequality (2) on becomes

(3)

Let be the vertex lying on the - path and adjacent to . Let and denote the - path and the - path, respectively. If there exists a -set Not containing , then must contain a neighbor of . Suppose, WLOG, . Then, inequality (3) follows immediately from applying Theorem 2.1 to and . If belongs to every -set , then , and (3) again follows.

Figure 2: case

Next, suppose equality is attained in (2). Again, let be a spanning tree with . Since and holds, we have . Thus, we deduce that and for each pair . With defined as above, the present assumption implies . Observe, in light of Theorem 2.1, that the equality is only possible if the following “optimal domination” of occurs: there is a -set which contains , a degree-three vertex in which dominates four or more vertices in ; every other vertex of dominates three or more vertices in ; no vertex of is dominated by more than one vertex of . (Note that Figure 2 only shows , which may be a strict subgraph of .) This “optimal domination” condition clearly implies that each member of must equal , which yields our second claim.   ∎

Next, we determine the largest for with the method deployed in [5]. However, rather than just getting a bound on in the case of total domination there, we obtain the exact value of for every .

Theorem 3.3.

For , .

Proof.

First, we prove . Let be a star with leaves labeled . Then and

So, .

Next, we show that . Notice that is given by Theorem 3.2. Thus, let be any arbitrary vertices of . Since holds for any triplet , we have

note that the last equality comes from the fact that there are triplets containing any given pair of vertices. Thus, as well.  ∎

4 Applying Theorem 3.2 to a Conjecture of DeLaViña et al.

We need a few more definitions. The eccentricity of a vertex in , denoted by , is . The boundary of is defined as the set ; we denote it simply as hereafter. The distance between a vertex and a set is defined as . Further, the eccentricity of is defined as .

In [3], DeLaViña et al. proved, for a tree , that . They further conjectured that the inequality holds for any connected graph . As an application of Theorem 3.2, we prove this conjecture. Our proof follows the arguments given by Henning and Yeo in [5] proving the analogous Graffiti.pc conjecture .

Theorem 4.1.

Let be a connected graph. Then .

Proof.

If , then and the desired inequality obviously holds. So, suppose ; this implies that and . Pick vertices and with ; then, . Let . Pick such that . We have and . Hence, we have . If equality holds in , then , and we can Not have both and be congruent to mod . In this case, by Theorem 3.2, we have that , which implies . On the other hand, if the inequality is strict, again by Theorem 3.2, we have that , which again implies .  ∎

References

  • [1] P. Dankelmann, Average distance and domination number. Discrete Appl. Math. 80 (1997) 21-35.
  • [2] E. DeLaViña, Q. Liu, R. Pepper, B. Waller, and D. West, Some conjectures of Graffiti.pc on total domination. Cong. Numer. 185 (2007), 81-95.
  • [3] E. DeLaViña, R. Pepper, B. Waller, Lower Bounds for the Domination Number. Discussiones Math. Graph Theory 30(3) (2010), 475-487
  • [4] T.W. Haynes, S.T. Hedetniemi, and P.J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, New York (1998).
  • [5] M.A. Henning, A. Yeo, A new lower bound for the total domination number in graphs proving a Graffiti.pc Conjecture, Discrete Appl. Math.(2014), http://dx.doi.org/10.1016/j.dam.2014.03.013
  • [6] H. Wiener, Structural determination of paraffin boiling points, J. Ameri. Chem. Soc. 69 (1947) 17-20.
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