Abstract
A block of a graph is a nonseparable maximal subgraph of the graph. We denote by the number of block of a graph . We show that, for a connected graph of order with minimum degree , . The bound is asymptotically tight. In addition, for a connected cubic graph of order , . The bound is tight.
1. Introduction
We consider finite, undirected, simple graphs only. Let be a graph. The numbers of vertices and edges of are called the order and the size of and denoted by and , respectively. A vertex is called a cut vertex if , where denotes the number of components of . denotes the number of cut vertices of . Rao [1] proved that, for a connected graph of order and size ,characterized all extremal graphs. Rao and Rao [2] solved the corresponding problem for a strong digraph. Later, Achuthan and Rao [3] determined the maximum number of cut edges in a connected -regular graph of order .
Let is a connected -regular graph of order . Rao [4] determined for . Nirmala and Rao [5] showed that or for odd and have obtained an upper bound for for even .
Alberten and Berman [6] proved that, for a graph of order and minimum degree ,
This bound is asymptotically tight.
Hopkins and Staton [7] showed that every connected graph of order contains no more than cut vertices of degree . Some related results are referred to [8, 9].
A separation of a connected graph is a decomposition of the graph into two nonempty connected subgraphs which have just one vertex in common. The common vertex is called a separating vertex of the graph. Since the graph under consideration is simple, is a separating vertex if and only if it is a cut vertex. A block of a graph is a nonseparable maximal subgraph of the graph. We denote by the number of blocks of a graph .
It is clear that any two blocks of a graph have at most one vertex in common. Recall that the block tree of is the bipartite graph with bipartition , where is the set of blocks of and , the set of separating vertices of , and a block , and a separating vertex is joined by an edge in if and only if contains . It is easy to see that if is connected, is a tree. Each leaf of corresponds to an end block of .
Inspired from the bound for the cut vertices, in the present paper, we consider the upper bound for the number of blocks, a connected graph of order with given minimum degree. Let us begin with two easy cases when and .
Proposition 1. For a connected graph of order , , with equality if and only if is a tree.
Proof. Our proof is induction on . If , then ; thus, the result holds. Next, we assume that . If has no cut vertex, then . Now suppose has a cut vertex. Let be an end block of and be the cut vertex, which belongs to . Let . Clearly, is connected. By the induction hypothesis, . Since , , we have , with equality only if and . By the induction hypothesis, is a tree, implying that is a tree.
On the contrary, if is a tree, clearly, .
Proposition 2. For a connected graph of order with , , with equality if and only if is the graph obtained from identifying each end with a vertex of separate , as given in Figure 1.
Proof. If has no cut vertex, the result holds trivially. Next, we assume that has cut vertices, and thus, it has at least two end blocks and . Let be all end blocks of . Let be the cut vertex of , which belongs to for each . Clearly, for any . If for any two distinct , then and . Therefore, .
Otherwise, has at least two cut vertices. It follows that the order of is at least two. Hence, and . By Proposition 1, . Summing up the above, we haveFrom the above, if and only if and , , as we promised.
It is clear that, for a graph of order , decreases when increases. For a connected graph of order and minimum degree at least , we have the following result, which is asymptotically best possible.
Theorem 1. For a connected graph of order with , .
We show that the bound in the above theorem is asymptotically best possible. Let . Consider a tree of order with each vertex having degree or 1. By the handshaking lemma, the number of leaves of this tree is
Let be the graph obtained from identifying each leaf of with a vertex of a clique of order separately. Therefore,
So, we haveHowever, as gets larger, gets arbitrarily close to .
What happens for the -regular graphs? The situation becomes complicated. We are just able to get an exact bound for a cubic graph of order : (Theorem 2), whereas by Theorem 1, we have for a connected graph of order with .
Theorem 2. For a connected cubic graph of order ,
The bound is sharp.
To see the sharpness of the bound, we denote by the graph obtained from by replacing an edge with a path of length two, as drawn in Figure 2.
The graphs achieve the upper bound in Theorem 2, which are classified into three types in terms of (mod 6), (mod 6), and (mod 6), respectively.
For an integer (mod 6), is an even integer. Let be a tree in which every vertex has degree 1 or 3. It is clear that has exactly vertices of degree 1 (leaves) and vertices of degree 3. Let be a graph obtained from identifying each leaf of with the vertex of degree two of a separate , as shown in Figure 3.
(a)
(b)
(c)
For an integer (mod 6), let be a cubic graph obtained from a graph by replacing a vertex of degree three (not belongs to any ) with a triangle, as shown in Figure 3.
For an integer (mod 6), let be a cubic graph obtained from a graph by inserting a into an edge of (not belongs to any ), as shown in Figure 3.
It can be checked that and for any graph constructed as above.
2. The Proof of Theorem 1
Suppose the result is not true and let be a counterexample of minimum order , i.e., and , but for any connected graph of order with , .
If , . By Propositions 1 and 2, . Hence, . Since , we have , and thus, has at least two blocks.
Claim 1. Every end block of is a clique of order .
Proof. of Claim 1. If it is not, let be an end block of . Let be the graph obtained from by replacing with of order . Clearly, and . By the choice of ,Combining the above facts, we conclude that , contradicting the choice of .
Claim 2. No cut vertex of belongs to at least two end blocks of .
Proof. of Claim 2. Let and be two end blocks of containing the same cut vertex of . Let . By Claim 1, , and thus, and . By the minimality of ,Combining (9) with the fact that , we have a contradiction:
Claim 3. Let be a cut vertex lying on an end block . If is a block containing , then .
Proof. of Claim 3. It suffices to show that .
First suppose that . Let be the graph obtained from and joining to every vertex in . Clearly, is a connected graph with . Moreover, by Claim 1, . Again, by the minimality of ,Combining (11) with the fact that , we have a contradiction.Now assume that . Let , where . Since and , each is a cut vertex of . Let be the graph obtained from by identifying all vertices in . Clearly, , , and . By the choice of , . Thus,contradicting the choice of . This proves the claim.
Take a longest path of . Let be an end block of , which corresponds to a terminal vertex of , where be the unique cut vertex of which belongs to . By Claim 3, . Next, we consider three possible cases in terms of the order of .
2.1. Case 1:
Let be the graph obtained from and joining to each vertex of . It is clear that , and by Claim 1, . By the minimality of , . Since , we have
2.2. Case 2:
By the choice of , each block of containing is isomorphic to . In addition, the end block containing the other end of is a leaf of . Since , there are such end block , each of which are jointed with an edge. Let be the graph obtained from by identifying a vertex of a new clique of order with . It is clear that , , and . So,is a contradiction.
2.3. Case 3:
Since , each vertex of is a cut vertex of . We distinguish two subcases in terms of .
2.3.1. Case 3.1:
Since , every vertex has a neighbor not in , which belongs to distinct blocks of . Let be the graph obtained from by contracting to a vertex . It can be seen that , , and . So,
2.3.2. Case 3.2:
Let and , where and . Since , for any , there are at least blocks containing , each of which is isomorphic to , as illustrated in Figure 4.
(a)
(b)
Let be the graph obtained from joining each component of to or such that for each . In addition, add an edge if . Note that is a connected graph of order with and . By the minimality of , . Therefore, , contradicting the choice of .
The proof of Theorem 1 is completed.
3. Proof of Theorem 2
Suppose the result is not true and let be a counterexample of minimum order . The following fact is clear:(1) must contain cut vertex. Since no cubic graph of order has a cut vertex, .(2)Moreover, . If , it is not hard to check that .
Claim 4. Every end block of is a .
Proof. of Claim 4. If it is not, let be an end block of . Let be the graph obtained from by replacing with . Clearly, is a connected cubic graph of order and . By the minimality of ,Combining the above facts, we conclude that , contradicting the choice of .
Take a longest path of . Let be an end block of , which corresponds to a terminal vertex of , where be the unique cut vertex of which belongs to . Since is a cubic graph, . Next we consider three possible cases.
3.1. Case 1:
If is a cut vertex of , then belongs to another block which is isomorphic to . In addition, the end block containing the other end of the is a leaf of . We may assume has cut vertices except , which belong to , respectively, where .
Let be the graph obtained from by identifying with the vertex of degree two of a new . It is clear that is a cubic graph of order and . By the induction hypothesis, . Thus,
3.2. Case 2:
It follows that . Every vertex of is cut vertex. Let be the graph obtained by the same operation as in the proof of Case 1. We have and . Therefore,
3.3. Case 3:
By the choice of and , one can find another longest path of . Let be the graph obtained from identifying of with the vertex of degree two of a new . Note that and . By the induction hypothesis, . Therefore,
The proof is completed.
4. Conclusions and Future Work
By arguing the properties of a minimum counterexample to the assertion of the main theorems and by using several kinds of graph transformation, we arrive at a contradiction, and thereby, we show our results. However, the upper bound for remains open if is a -regular graph with . One of the referees pointed out the possibility of the obtained results to some real-life applications and other fields (see [10–12] for instance).
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.