Research Article | Open Access

Huiming Duan, Yonghong Li, "On Connected *Journal of Discrete Mathematics*, vol. 2013, Article ID 983830, 7 pages, 2013. https://doi.org/10.1155/2013/983830

# On Connected *m-HPK*-Residual Graphs

**Academic Editor:**Pantelimon Stǎnicǎ

#### Abstract

We define *m*-*HPK*-residual graphs in which *HPK* is a hyperplane complete graph. We extend P. Erdös, F. Harary, and M. Klawe's definition of plane complete residual graph to hyperplane and obtain the hyperplane complete residual graph. Further, we obtain the minimum order of *HPK*-residual graphs and *m*-*HPK*-residual graphs. In addition, we obtain a unique minimal *HPK*-residual graphs and a unique minimal *m*-*HPK*-residual graphs.

#### 1. Introduction

A graph is said to be -residual graph [1], if for every vertex in , the graph obtained from by removing the closed neighborhood of is isomorphic [2–6] to . We inductively define a multiply -residual graph by saying that is -residual graph, if the removal of the closed neighborhood of any vertex of results in an -residual graph, where of course a -residual graph is simply said to be -residual graph.

It is natural to ask what is the minimum number of vertices that an -residual graph must contain? It is easy to prove that this number is and that the only -residual graph with this number of vertex is . In [1], Erdös et al. show that a connected -residual graph must have at least points, if . Furthermore, the Cartesian product is the only such graph with points for . They complete the result by determining all connected -residual graphs of minimal order for .

In [1], the following conjectures were stated.

Conjecture 1 (see [1]). *If , then every connected -residual graph has at least vertices.*

Conjecture 2 (see [1]). *For large, there is a unique smallest connected -residual graph. *

Theorem 3 (see [1]). *
(1) If is an -residual graph, then for any vertex in G, the degree . **
(2) Every -residual graph has at least vertices, and is the only -residual graph with vertices. **
(3) Every connected -residual graph has at least vertices, if . **
(4) If , then is a connected -residual graph of minimum order, and, except for and , it is the only such graph. *

We know that these supporting results are summarized in residual graph [7–16]. In this paper, we will study the important properties of hyperplane complete graph in relain to the minimum order of hyperplane complete graph and the minimum graph of hyperplane complete graph. We will investigate Erdös, Harary, and Klawe’s residual graph and obtain the minimum order of -residual graphs and --residual graphs. And we also show a unique minimal connected -residual graphs of order and a unique minimal connected --residual graphs of order .

In general, we follow the notation in [1]. In particular, is the number of vertices in a graph , is the closed neighborhood of the vertex , and are called as neighborhood and closed neighborhood of in .

#### 2. On Connected -Residual Graphs

*Definition 4. *A graph is a 3-dimensional hyperplane complete graph, if for every , , , , . Two vertices and are adjacent to each other, if and only if and for some . For convenience, it writes instead of 3-dimensional hyperplane complete graphs.

*Definition 5. *Let , a graph is -residual graph, if for every , it is . A multiply -residual graph is --residual graph, if the removal of the closed neighborhood of any vertex of results in an --residual graph.

By Definitions 4 and 5, let and , if for some , then is indeed a complete graph . In fact, , and is an empty graph with vertices.

*Definition 6. *Let and are simple graphs. In the composition [17], and are recorded as , , and is adjacent to in , if and only if , , is adjacent to in , but is adjacent to in .

By Definitions 4, 5, and Theorem 3, we have the following.

Lemma 7. *If is a connected noncomplete graph, then there is no disconnected -residual graph. *

*Proof. *Suppose that is an -residual graph, if is disconnected, let , where is a connected component of the . Then, for any , so
Since is connected, so and . Thus, is connected, and is not a complete graph; hence, there are two vertices and in , and is nonadjacent to , let
Since . Hence, is disconnected; however is connected, which is a contraction.

Theorem 8. *Assume that is an --residual graph, particularly, is a -residual graph.*

*Proof. *The proof follows using Definitions 4, 5, 6, and .

Lemma 9. *Let . is pairwise nonadjacent in , then .*

*Proof. *By Definitions 5 and 6 and let , , , . Let , , , , , , , is pairwise nonadjacent in , if and only if and for some , since is pairwise nonadjacent, then , , , . And , hence,
then .

Lemma 10. *Assume that G is an -residual graph, ,,, ,, is pairwise nonadjacent in G, then . *

*Proof. *The proof follows using Definitions 4, 5, and 6.

Lemma 11. *Assume that is an F-residual graph, , , , , , for any , then . *

*Proof. *For any , and is nonadjacent to in , let , then , by Definitions 6, then
hence, . Since is pairwise nonadjacent in , by Lemma 10, then , so
This completes the proof.

Theorem 12. *Let G is -residual graph, , then .*

*Proof. *For any , let , by Theorem 3 and Lemma 11, then

Theorem 13. *Let G is -residual graph, and , then .*

In order to proof Theorem 13, we have the following Lemma.

Lemma 14. *Assume that , . is pairwise nonadjacent in , let , , , , then **
(1) , and contains 54 set intersection, and
**(2)**(3)*

*Proof. *Part (1) follows from Definition 6 and Lemma 9.

Similarly,
where contains 54 set intersection, contains 54 set intersection, and contains 54 set intersection.

(2) Let , by Definition 6, and is not adjacent , hence, , and is adjacent to , ,, , , hence , , because is adjacent to , then , , hence, , , for any .

Part (3) follows from the proof of part (2).

*Proof of Theorem 13. *We are now ready to prove the following quantitative version of our theorem, we have four cases as follows.*Case 1.* For any , let , by Definition 6, , is adjacent to in , if and only if and for any , .*Case 2*. Let , by Lemma 14, using in the name of the vertex of in the naming of vertices. First, prove that , for any , , , , . For convenience, is the closed neighborhood of a vertex , , , , , , , , , , by Lemma 14, let
Lastly let , so far vertex already all named, the uniqueness of the name. By Lemma 14 and Definition 6, then , . Two vertices and are adjacent to each other, if and only if exist and for .*Case 3*. Let , to not named point naming, these point sets are composed of
for any , , . By Lemma 14 and (11), we can name
By Lemma 14 and (12), we can name
By Lemma 14 and (13), we can name
Similarly, we can name
For the point rename, by Lemma 14 and the uniqueness of the name get the name and in name is the same, then , and two vertices , are adjacent to each other in the , if and only if and, for any , for . *Case 4.* Named all unnamed point, this kind of composition is set . Let ,,,. Obviously, , , , , , . By Lemmas 9 and 14, then
As described in Case 3, in accordance with the Lemma 14, we name , the same as in . Similarly, by Lemma 14 and (11), we can name
Hence,
two vertices , are adjacent to each other in , if and only if, exist and for .

So far, we have the of each vertex named and . According to the naming rules, for any , , if and in the same , is adjacent to , if and only if and for any , for . Otherwise, for any , is not adjacent , . Let , then , the uniqueness of the name also get and is adjacent to in . So is adjacent to in , if and only if and for any , for .

With comprehensive discussed above, then . The proof is complete.

#### 3. Connected --Residual Graphs

Lemma 15. *Assume that graph is --residual graph, , , and is pairwise nonadjacent in , then . *

*Proof. *By Definitions 4, 5, 6, and , we can turn out vertex , , , let , , for convenience, instead of , let be pairwise nonadjacent. Let , hence
is pairwise nonadjacent in , by Lemma 9, then