Abstract
Simple finite connected graphs of vertices are considered in this paper. A connected detour set of is defined as a subset such that the induced subgraph is connected and every vertex of lies on a detour for some . The connected detour number of a graph is the minimum order of the connected detour sets of . In this paper, we determined for three special classes of graphs , namely, unicyclic graphs, bicyclic graphs, and cog-graphs for , , and .
1. Introduction
For basic definitions of the concepts of graphs we refer to [1–4], and for detour distance and related terminologies in graphs, we refer to [5–7]. Let be a connected simple graph of vertices and edges. We assume that is finite and . For the length of a maximum path is called detour distance . A path of length is called detour. For vertex the detour eccentricity is defined by
The detour radius and the detour diameter (or ) of are defined as
A vertex is said to lie on a detour if is a vertex of including and . A detour set (denoted ) is a subset of such that every vertex of lies on an detour of some . The detour number of is defined by
A detour basis of is a of order .
If is a detour set of and the induced subgraph is connected, then is called connected detour set (denoted ) of . The connected detour number of denoted as is defined as
A connected detour basis of is a of order (see [8, 9]).
A simple connected graph with is called unicyclic graph iff . The graph is called bicyclic iff .
The concept of connected detour number was introduced and studied by Santhakumaran and Athisayanathan in [9]. They determined for some special graphs such as , , , trees, and Hamilton graph. There are many research papers on connected detour number and edge detour graphs (see [10–14]). Moreover, the concept of connected detour number and other related concepts have interesting applications in the channel assignment problem in radio technologies. This motivated us to determine connected detour number for other classes of graphs. Therefore, in this paper we determine the connected detour numbers for unicyclic graphs and bicyclic graphs. Moreover, the class of graphs called cog-graphs will be explained and determined the if is a complete graph, tree, cycle graph, and complete bipartite graph.
2. The Connected Detour Number of Unicyclic Graphs
Let be a connected graph of order and the unique cycle in , and let be of length . It is clear that has no chords, and every vertex of , which is not on , is either a cut-vertex or an end-vertex. We shall determine the connected detour number of such graphs in terms of and . Let be the number of vertices of that are not cut-vertices. Denote and . Then, .
If , then so . If , then contains at least one cyclic cut-vertex. If , that is every vertex of is a cut-vertex, then by Theorem 1.4 [Ref. 2] . From now on, we assume .
Proposition 1. Let be a connected unicyclic graph of order , and with cycle, . Then, iff .
Proof. If , then contains exactly one vertex which is not a cut-vertex. It is clear that there is a detour joining the other two vertices of the triangle and lies on it. Thus, . If , let and be cycle vertices which are not cut-vertices. Clearly, there is no path in between two vertices of that contains or (see Figure 1). Thus, every c.d.b. B of must contain either or . Therefore, . If is the third vertex of the triangle and , then lies on the detour. Therefore, . Hence, .
To prove the converse, let , and is a c.d.b. of , then contains two vertices of the cycle, one of them is a cut-vertex . Thus, in view of Theorem 1.4 [9], the cycle has one or two vertices in , that is, or . Hence, the proof is completed.
Theorem 1. Let be a connected unicyclic graph of order and with cycle, . Then, iff the induced subgraph consists of exactly components.
Proof. Let be the number of components of . Let , then since for every c.d.b., and are connected, every connected component of contains at least one cycle cut-vertex, and , then contains at least vertices from . Therefore, .
Conversely, let , and is a c.d.b. of . Since is connected and consists of components and , then contains at least vertices from . Because is a connected detour set of minimum order, then contains exactly vertices from . Thus,From the hypothesisTherefore, .
Hence, the proof of the theorem is completed.
Example 1. For the unicyclic graph in Figure 2, we have , so .
Now, we have the following result for the connected detour number of the unicyclic graph with exactly one cycle cut-vertex.
Proposition 2. Let be a connected unicyclic graph of order and with exactly one cycle cut-vertex, say , then , where is the length of the unique cycle of .
Proof. Let be a vertex of adjacent to . Then, there is a detour consisting of all the vertices of .
Thus, is a c.d.s. of order .
It is clear that there are no detour containing vertices of for every pair . Therefore, is a connected detour basis of , and hence .
For connected unicyclic graphs having more than one cycle cut-vertex, we need the following definition.
Definition 1. Let be a connected unicyclic graph of order and with at least two cycle cut-vertices, and let be the unique cycle of length . Moreover, let be the number of components of the induced subgraph . These components divide the cycle vertices which are not cut-vertices into nonempty subsets , in successive order around , as illustrated in Figure 3.
Example 2. Consider the unicyclic graph shown in Figure 3. The set of cycle vertices which are not cut-vertices is . It is clear that , , and . The set is partitioned into , , , and .
The c.d.n. for unicyclic graphs having more than one cut-vertex is determined by the following theorem.
Theorem 2. Let be a connected unicyclic graph of order and with at least two cycle cut-vertices, and the induced subgraph consists of components. Then, if and only if
Proof. Let be as defined in (7) for the graph , and let be a c.d.b. for . By Theorem 1.4 of Ref [9], contains the set . Since the induced subgraph is connected, then must contain all the vertices of at least subsets from . Since is of minimum order, then does not contain the subset from that has maximum order, say . It is clear that there are two vertices which are adjacent on ; hence, there is an detour containing all the vertices of . Therefore, .
Thus, .
To prove the converse, let and let be a c.d.b. of . If is not equal to , then either is not of minimum order or the induced subgraph is disconnected, contradicting the definition of connected detour basis.
Thus, , and hence the proof of the theorem is completed.
Remark 1. Clearly , iff . Thus, Theorem 1 follows from Theorem 2.
3. The Connected Detour Numbers of Connected Bicyclic Graphs
A graph is bicyclic if and only if . Thus, if is a connected bicyclic graph, then contains either three cycles having some edges in common or contains exactly two cycles having no edges in common. The connected detour number for a block bicyclic graph is determined by the following result.
Proposition 3. Let be a 2-connected bicyclic graph of order as shown in Figure 4. Then, (i), iff and . (ii), if and they are different.
Proof. (i) If , then there are two detours, namely, and . It is clear that each vertex of lies on one of the two detours. Thus, is a c.d.b. of , so . Conversely, if , say and , then does not contain adjacent vertices such that is a detour set. Hence, the proof of Part (i) is completed.(ii)If are different, say , then it is clear that is a connected detour set of . So, . In view of Part (i), . Thus, . Hence, the proof of the proposition is completed.
Remark 2. If is a 2-connected bicyclic graph of order with a cycle and with exactly one chord, that is, an edge joining nonadjacent vertices of , then .
This section is divided into two subsections according to the types of the bicyclic graphs.
3.1. The Connected Detour Numbers of Bicyclic Graphs of Three Cycles
Now assume that is a connected bicyclic graph of order with one or more cut-vertices and with three cycles, that is, three paths which are internally vertex disjoints denoted by , and as shown in Figure 5. We assume without loss of generality that . Let be the set of all cycle vertices which are not cut-vertices in , and let .
We shall determine the connected detour number for three kinds of bicyclic graphs of three cycles.
Case 1. Assume that each , , contains at least one cut-vertex other than and . Moreover, let be the set of all cycle cut-vertices in . Then, we have the following proposition which determines the c.d.b. of such kind of bicyclic graph .
Proposition 4. Let be a connected bicyclic graph of three cycles and with one or more cut-vertices on each , , other than and as explained above and shown in Figure 5. Then,where is a subset of of minimum order such that the induced subgraph is connected.
Proof. Since and is connected, then the induced subgraph is connected. Because each , , contains a vertex of , then contains or and each contains two adjacent vertices from . Therefore, every vertex of the paths lies on an detour for some . Thus, is a c.d.s. of . Moreover, from the minimalist of we deduce that is a c.d.b. of . Therefore, . Hence, the proof is completed.
The following example illustrates Proposition 3.2.
Example 3. Consider the bicyclic graph shown in Figure 6.
It is clear that and :Thus,
Case 2. Assume that contains exactly one path that does not contain cut-vertices, other than and . So we have three possibilities for such bicyclic graph :(i)Let and each contains at least one internal cut-vertex, and does not contain an internal cut-vertex. Then, is a unicyclic graph. By Theorem 2, , in which is defined in Definition 1 for the graph . We can easily verify that if is a c.d.b. of , then it is a c.d.b. of because . Therefore,(ii)Let and each contains at least one internal cut-vertex and does not contain an internal cut-vertex. Then, is a unicyclic graph. By Theorem 2, , where is the number defined in Definition 1 for the unicyclic graph . Thus, as in (i),(iii)Let and each contains at least one internal cut-vertex, and does not contain an internal cut-vertices. Then, is a unicyclic graph. By Theorem 2, , where the number is explained in Definition 1. If , then every c.d.b. of is a c.d.b. of . Therefore,If , then any c.d.b. of is not c.d.s. of because each vertex of does not lie on a u- detour for every pair of vertices . But it is clear that either or . Thus, if then is a c.d.b. of ; and if then is a c.d.b. of . Therefore,The following example illustrates formulas (11)–(13b).
Example 4. Consider the graphs , , as shown in Figure 7.
It is easy to verify that:
(a)
(b)
(c)
(d)
Case 3. Assume that the connected bicyclic graph consists of two paths, and each path does not contain cut-vertices but only one path contains internal cut-vertices.
If contains at least two internal cut-vertices, and and have no cut-vertices, , then is a unicyclic graph, denoted . It is clear thatwhere is given for as defined in Definition 1. Thus,Similar results we have if has at least two internal cut-vertices and the other paths have no cut-vertices. Therefore,where is for the unicyclic graph and is for .
Remark 3. If the bicyclic graph depicted in Figure 5 has exactly one cycle cut-vertex which is a vertex of the path including and , and the other two paths have equal lengths, thenFrom now on, we assume that (see Figure 5). If contains internal cut-vertices and and contain no cut-vertices, then we may assume that the distance from to the first cut-vertex along the path is not more than the distance from to the last cut-vertex along . Let be the unicyclic graph constructed from , where is an end-edge incident to vertex of . It is clear contains vertices and in addition to vertices from , and soTherefore,If the distance from to the last cut-vertex along is less than the distance from to the first cut-vertex along , then we have the unicyclic graph . Hence,where is the number defined for (Definition 1).
We have results similar to (20a) and (20b) for the cases where has internal cut-vertices, the other two paths have no internal cut-vertices and . Namely, or for or and the unicyclic graphs or .
Remark 4. If the vertex or the vertex is the only cycle cut-vertex of the bicyclic graph shown in Figure 5 and , then
3.2. The Connected Detour Numbers of Bicyclic Graphs of Two Cycles
Let be a bicyclic graph containing exactly two cycles and , either having one vertex in common or there is a path joining a vertex of to a vertex of . Thus, is considered to consist of two unicyclic graphs and having exactly one vertex in common.
Let be a uncyclic graph obtained from by adding to it an end-edge . The connected detour number of is determined by the following theorem.
Theorem 3. Let be a connected bicyclic graph of order and consist of two edge-disjoint unicyclic graphs and having one vertex in common. Then, , in which is the number defined in Definition 1 for the unicyclic graph .
Proof. Let be a c.d.b. of , then contains . Moreover, let be the subset of consisting of the vertices of . It is clear that each vertex of lies on detour for some pair . Therefore, is a c.d.s. of , that is because the connectedness of the induced subgraph implies that is connected. Since is of minimum order, then is a c.d.b. of . Conversely, it is clear that if is a c.d.b. of , then is a c.d.b. of . By Theorem 2, , in which is the order of . Therefore, . Since , then .
4. The Connected Detour Numbers of Cog-Graphs
Let be a connected -graph, then is the graph constructed from the graph with additional vertices corresponding to the edges of and additional edges obtained from joining to the two vertices of for all . Such class of graphs are called cog-graphs of . For example, let be a star of order five, then is cog-star of order nine shown in Figure 8.
(a)
(b)
Clearly if is –graph then is -graph. The proofs of the following elementary results are obvious.
Proposition 5. (1) The cog-graph does not contain end-vertices.(2)If the graph has n end-vertices, then contains exactly vertices of degree 2.(3)For every vertex , .(4)Let , then is a cut-vertex in iff it is a cut-vertex in .Let and . If is an detour in , then is an detour in , in which for and is the vertex that corresponds to edge of . Therefore, , .
Moreover, if is an detour in , (as shown in Figure 9), thenAny way, if is a detour set of , then may not be a detour set of . Also for some graphs , . For example, if is an odd cycle graph with exactly one chord, then and . But there are special graphs such that , as given in the following proposition.
Proposition 6. Let be a connected graph. If is a tree or a cycle graph, then
Proof. It is obvious.
The following concepts were introduced by Santhakumaran and Athisayanathan in [12].
Definition 2. [12, 15] “Any edge of is said to lie on an detour , if is an edge of . A set is called an edge detour set of if every edge of lies on a detour joining a pair of vertices of . The edge detour number of is the minimum order of its edge detour set. Any edge detour set of order is called an edge detour basis of . A graph which has an edge detour set is called a edge detour graph (denoted E.D. graph).”
There are graphs which are not E.D. graphs because they do not have edge detour sets [12]. For E.D. graphs we give the following definition.
Definition 3. Let be an edge detour set (will be denoted e.d.s.) of an E.D. graph . If is connected then is called a connected edge detour set (denoted c.e.d.s.). The connected detour number of is defined byAny c.e.d.s. of order is called connected edge detour basis (denoted c.e.d.b.) of .
It can easily be proved that if is an E.D. graph, then every c.e.d.s. of contains all the end-vertices and all the cut-vertices of . Thus, for every tree , .
Now, we shall determine c.e.d.n. for some special classes of connected graphs.
Proposition 7. For every cycle graph with ,
Proof. Let , then it is clear that every edge of other than lies on the detour. Moreover, the edge lies on the detour. Thus, is a c.e.d.b. of , and hence .
Proposition 8. Let be a complete graph of order , then for every pair of vertices, every edge other than lies on a detour of .
Proof. One can easily check that the statement is true for . Now assume that the statement is true for , and consider . Let be any pair of vertices of , and let and as shown in Figure 10.
By induction hypothesis for every pair of vertices of , every edge other than of lies on a detour in .
It is clear that the two edges , or with produce an detour in . This is true for all . Thus, every edge of other than lies on some detour in . Therefore, the proposition is true for . Hence, by induction on the proposition is true for , .
Corollary 1. For each complete graph with , .
Proof. Let be any three vertices in . By Proposition 8 every edge of other than (resp., ) lies on an detour (resp., detour). Thus, is a c.e.d.s. of . Clearly, , and hence .
Corollary 2. For every complete graph with , .
Proof. Let be a pair of vertices of , then by Proposition 8 every edge other than of lies on an detour in . Thus, every vertex of other than lies on an detour in , in which vertex corresponds to the edge in . Since vertex is adjacent to , then every vertex of lies on an detour. Therefore, is a c.d.b. of , and hence .
Proposition 9. Let , be a complete bipartite graph, then for any pair of adjacent vertices every edge other than lies on a detour in .
Proof. One can easily check that the proposition holds for , , and . Now assume that it holds for , and consider . Let be any edge of , and let as shown in Figure 11 in which its vertex set is , , and . By induction hypothesis, every edge of other than lies on detour in . Clearly, each detour in implies detour (namely, , detour, and ) in . Moreover, each edge of with edges and lie on . Since this holds for and , then every edge of (other than ) lies on an detour in . Thus, by induction the proposition holds for every .
Corollary 3. For every complete bipartite graph , then .
Proof. Consider the vertices of where and . Then, by Proposition 9 every edge of (other than ) lies on an detour, and lies on an detour in . Therefore, is a c.e.d.s. of , and hence .
Corollary 4. For every complete bipartite graph , , then .
Proof. Let be an edge of , then by Proposition 9 every edge other than lies on an detour in . From the definition of cog-graphs, every vertex other than lies on an detour in , where is the vertex that corresponds to the edge in . Adding the edge to every such detour in we obtain detours, and hence every lies on an detour in . Hence, .
Proposition 10. Let be an E.D. graph of order , then .
Proof. Let be a c.e.d.b. of . If and is an edge of , and is the vertex in that corresponds to the edge , then we interchange vertex to vertex in . We repeat such interchange for every edge to get the set of vertices in . By Definitions 2 and 3, is a c.d.s. of , and . Thus, .
5. Conclusions
The connected detour numbers for three classes of connected simple graphs are determined in this research paper. The three classes are unicyclic graphs, bicyclic graphs, and cog-graphs for , , and . We think that the methods used in proving the results in Section 3 can be used to determine the connected detour numbers for bridge graphs and chain graphs (defined in [16]) that are constructed from finite pairwise disjoint unicyclic graphs.
It is shown that is related to , and in view of Proposition 10 we suggest the following problem: characterize edge detour graphs such that .
Data Availability
The data used to support the findings of this study are available from the corresponding author upon request.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This paper was supported by the College of Computer Sciences and Mathematics, University of Mosul, Republic of Iraq.