Research Article | Open Access

Blessings T. Fundikwa, Jaya P. Mazorodze, Simon Mukwembi, "An Upper Bound on the Radius of a 3-Vertex-Connected -Free Graph", *Journal of Mathematics*, vol. 2020, Article ID 8367408, 7 pages, 2020. https://doi.org/10.1155/2020/8367408

# An Upper Bound on the Radius of a 3-Vertex-Connected -Free Graph

**Academic Editor:**Elena Guardo

#### Abstract

We show that if is a 3-vertex-connected -free graph of order and radius , then the inequality holds. Moreover, graphs are constructed to show that the bounds are asymptotically sharp.

#### 1. Introduction

Let be a finite, connected, undirected graph with vertex set and edge set . The distance between two vertices of is the length of a shortest - path in . The eccentricity of a vertex is the maximum distance between and any other vertex in . The value of the minimum eccentricity of the vertices of is called the radius of denoted by . The degree of a vertex of is the number of edges incident with . The minimum degree is the minimum of the degrees of vertices in . The open neighbourhood of a vertex is the set of all vertices of adjacent to . The closed neighbourhood of is the set . A graph is triangle-free if it does not contain as a subgraph and if it does not contain as a subgraph. For notions not defined, here we refer the reader to [1].

In our research, we are concerned with upper bounds on radius in terms of order and each of the three classical connectivity measures, namely, minimum degree, edge-connectivity, and vertex-connectivity for general graphs and for graphs with forbidden subgraphs such as and . Several upper bounds on radius in terms of other graph parameters are known. For example, Erdős et al. [2] and independently Dankelmann et al. [3] proved that if is a connected graph of order and minimum degree , then

Erdős et al. [2] proved that if is a connected triangle-free graph of order and minimum degree , thenand that if is a connected -free graph of order and minimum degree , then

Bounds on the radius in terms of order and edge-connectivity were given by Dankelmann et al. [4]. They showed that if , thenand they constructed graphs that, apart from the additive constant, attain the bound. Mukwembi, in [5], showed that if is a triangle-free 3-edge-connected graph, thenand constructed graphs that, apart from the additive constant, attain the bound. Bounds on the radius in terms of order and vertex-connectivity were given by Harant and Walther [6]. If is even, then the well-known bound, , on the diameter is also sharp for the radius. For odd , Harant and Walther [6] showed that

In [7], Harant showed that for , the term can be replaced by a constant. It is worth noting that, while it has not been formally communicated, the bounds on the radius of triangle-free graphs with respect to order and vertex-connectivity are exactly the same as those for graphs without restrictions where subgraphs are concerned. To date, there are no known bounds reported on the radius of -free graphs with respect to order and vertex-connectivity. The main theorem of this paper seeks to prove the following:

Theorem 1. *Let be a 3-vertex-connected -free graph of order . Then, .*

To show that this bound is asymptotically sharp, consider the graph where is the graph in Figure 1 and is the graph formed by replacing each vertex of a -cycle by the graph and replacing an edge between consecutive vertices in the cycle by an edge connecting to and another edge connecting to , and this is between the two copies of that have replaced consecutive vertices in . As an example, consider in Figure 2. Note that is a 3-vertex-connected -free graph and that whenever , we have

#### 2. Results

Let be a 3-vertex-connected, -free graph, of order , and be a fixed centre vertex of , so that . For each , let and . Note that for each with and any , we have . We employ the notation and . Since , from now onwards fix a vertex . Form a spanning tree of that is distance preserving from . For a vertex , denote by , the path connecting and in . We say is related to if there exist vertices where and such that .

Lemma 1. *Let and as above. Then, there exists a vertex in which is not related to .*

*Proof. *Suppose to the contrary that every vertex is related to . Let be the vertex of which belongs to . Then, for any , we have . Since every vertex in is related to , for any , there exist , where and such that . Thus, , for any . Hence, , contradicting the fact that .

Throughout the rest of this paper, is a fixed vertex in which is not related to . Let (respectively, ) be a path from to (respectively, ) in . For each with () denote by (), , the set of all vertices in whose distance from () is at most in . Thus, for example, . For each with () denoted by (), , the set of all vertices in whose distance from () is exactly in . Thus, for example, . For each with () denoted by (), , the set of all vertices in whose distance from () is where in . Thus, for example, . For each denoted by , the set of all vertices in whose distance from is at most in . Thus, for example, . The following fact follows from the fact that is not related to :

*Fact 1. *.

Lemma 2. *Let . With the above notation, for each , (respectively, ).*

*Proof. *We show that the result holds for as it follows analogously on . Clearly, , and . Also, . Furthermore, , and each pair of vertices (in this case and ) shares at most one neighbour since is -free. There are two cases to consider: Case 1: suppose that and do not share a neighbour. Then, , and we are done. Case 2: assume that and share a neighbour, say . Note that since . From this, it becomes obvious that . Thus, since is adjacent to and , we realise that . Now, . Observe that since , has at least one neighbour . Again since and because is -free, we see that . Because and , we see that . Therefore, , which settles Lemma 2.For , let be an induced subgraph of such that (respectively, ) and be an induced subgraph of . Then, we have the following definitions.

*Definition 1. * is a scant subgraph of (or simply, is scant) if .

*Definition 2. *We say is an attached subgraph of (or simply, is attached) if there exists a path in connecting at least one vertex (respectively, ) to at least one vertex (respectively, ), otherwise is a detached subgraph of (or simply, is detached).

*Definition 3. * is said to be a detachable subgraph of (or simply, is detachable) if there exists a vertex such that is a detached subgraph of .

Lemma 3. *If is scant, then it is detachable.*

*Proof. *We show that the result holds for as it follows analogously on . Note that if , , then is detached. Thus, if , we immediately conclude that is detachable. So, suppose that whenever . Now, because of our supposition, if for any , then contradicting the condition of this lemma. Thus, whenever , and consequently, . Accordingly, and . For , let be the other vertex in which is not . Note that a path such that does not exist since would be disconnected, thereby contradicting . If or , then is detachable since or is detached. So, suppose and . If then since and can share at most one neighbour and , immediately contradicting . Thus, . From a similar argument, we obtain . Note that is necessarily adjacent to at least one vertex in . Observe that , otherwise would form a -subgraph, which is illegal. Thus, can only be adjacent to . Consider . Note that , otherwise would form a -subgraph, which is forbidden. From the above, we see that is detached, and thus, is detachable and Lemma 3 is settled.

Lemma 4. *If is scant, then (respectively, ).*

*Proof. *We show that the result holds for as it follows analogously on . Suppose to the contrary that there exists . There are two cases to consider:(i)Case 1: suppose . This means or . If (), then () yielding (). Thus, since (), we obtain () contradicting our assertion that is scant. Case 2: suppose If this is so, then is necessarily adjacent to some . More specifically, . Observe that . Thus, . Taking into account , we obtain contradicting the condition of the lemma. Thus, Lemma 4 is settled.

Lemma 5. *If (respectively, ) and (respectively, ) are both scant, then .*

*Proof. *We show that the result holds for , as it follows analogously on , . By Lemma 4, .

*Claim 1. *.

*Proof of Claim 1. *Suppose to the contrary that there exists . Remember that is the set of all vertices in whose distance from is at most in . Since and is -free, we have . Note that . Hence, . There are two cases to consider: Case 1: suppose that . Note that . Thus, upon making use of Lemma 2, we arrive at immediately contradicting the condition that is scant. Case 2: now, suppose that . Then, , and since , we obtain, as a consequence of Lemma 2, which contradicts the condition that is scant. Thus, and our claim is settled.By Lemma 3, and are both detachable. In other words, there exist such that and are both detached.

*Claim 2. * and

*Proof of Claim 2. *Suppose to the contrary that or Observe that since () is a detached subgraph of (), every path connecting some vertex in () to some vertex in () contains (). This means that every path connecting some vertex in to some vertex in contains and . Thus, if (), since and are detached subgraphs of and , respectively, and also noting that , we see that there is no path connecting () to in contradicting . Hence, and , settling our claim.

We now consider . If for , then is detached, contradicting Claim 2. Thus, for . If , we obtain contradicting the condition of the lemma. Thus, . If , we are done since , so suppose . This means that , and thus, it remains to prove the following claim.

*Claim 3. *

*Proof of Claim 3. *Suppose to the contrary that . This means that and . This yields and . Let , be the vertex in which is not . If , then is detached contradicting Claim 2. Thus, . Now if , then since and can share at most one neighbour, we have , contradicting . A similar argument leads us to conclude that . Observe that is necessarily adjacent to at least a vertex in . Note that if , then forms a -subgraph which is illegal. Thus, can only be adjacent to leading us to conclude that . Now, consider . If , then would form -subgraph which is illegal. Thus, . Hence, is detached, contradicting Claim 2. Hence, Claim 3 is settled.

Thus, since and , we have . Therefore, since is scant, we conclude . A similar set of arguments leads us to conclude that , thereby settling Lemma 5.

Lemma 6. (a)*If and are both scant, then is not scant.*(b)*If and are both scant, then is not scant.*

*Proof. *We show that the result holds for (a) as it follows analogously for (b). Suppose to the contrary that , , and are all scant. By Lemma 4, . By Claim 1, ; therefore, . Since , , and are all scant, we know, by Lemma 3, that , , and are all detachable. Observe that, by Claim 2, and are both detached. Thus, since , we have it that there is no path connecting to in . Thus, is disconnected, contradicting . This settles Lemma 6.

Throughout the remainder of this paper, we assume that , , and we define the set .

Lemma 7. *Let such that .*(a)*If and are both scant and is not scant whenever , then there exists (where ) such that or or .*(b)*If and are both scant and is not scant whenever , then there exists (where ) such that or or .*

*Proof. *We show that the result holds for (a) as it follows analogously for (b). From Lemma 4, we have . Suppose there exists a vertex . Remember that . Consider . If , then and there would be no path from to contradicting the condition that is connected. Thus, . Since , we have and consequently . Note that ; furthermore, is scant. Thus, making use of Lemma 2, we obtain . This gives us , and using Lemma 2 again, we obtain and we are done. So, for the rest of this lemma, we will assume .

*Claim 4. *There exists a vertex .

*Proof of Claim 4. *Suppose to the contrary that , then . Since and are both scant, we have it, by Lemma 2, that and are both detachable. Thus, there exist and such that and are both detached. Thus, there is no path in connecting to . Thus, is disconnected, contradicting and therefore Claim 4 is settled.

Now we have it, by Claim 4, that there exists such that . With this in mind, there are four cases to consider. Case 1: suppose . Since and , we have, by Lemma 2 and , that Case 2: . It is obvious that is nonempty; hence, there exists a vertex . Observe that . Thus, since , we have and thus . Furthermore, and . Hence, by Lemma 2, Case 3: . Note that is necessarily adjacent to at least one vertex, say , in . More specifically, . Observe that and . Thus, if and do not share any neighbours, we are done, by Lemma 2, since