On Domination Number and Distance in Graphs
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
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 .
It is well-known that (); a “proof ” to () can be found on p.56 of the authoritative reference . 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 ) and is the average distance (per definition found in ), it follows that . As an application of our main result, we prove a conjecture in  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 , 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  to be readily adaptable towards the results of this paper. However, in striking contrast to , 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. ) 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 , the authoritative reference in the field of domination titled Fundamentals of Domination in Graphs.
For any connected graph , .
“Proof ” (as found on p.56 of ).
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.”
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
Proof of Theorem 2.1. Given , take a spanning tree of such that . Suppose, for the sake of contradiction, . Since and , we have
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.
Given any three vertices of a connected graph , we have
Further, if equality is attained in (2), then for any pair .
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
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.
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 . However, rather than just getting a bound on in the case of total domination there, we obtain the exact value of for every .
For , .
First, we prove . Let be a star with leaves labeled . Then and
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 , 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  proving the analogous Graffiti.pc conjecture .
Let be a connected graph. Then .
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 . ∎
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