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Research Article | Open Access

Volume 2014 |Article ID 304514 | https://doi.org/10.1155/2014/304514

Yongsheng Ye, Fang Liu, Caixia Shi, "The 2-Pebbling Property of the Middle Graph of Fan Graphs", Journal of Applied Mathematics, vol. 2014, Article ID 304514, 5 pages, 2014. https://doi.org/10.1155/2014/304514

# The 2-Pebbling Property of the Middle Graph of Fan Graphs

Academic Editor: Ying Tan
Accepted07 Jul 2014
Published22 Jul 2014

#### Abstract

A pebbling move on a graph consists of taking two pebbles off one vertex and placing one pebble on an adjacent vertex. The pebbling number of a connected graph , denoted by , is the least such that any distribution of pebbles on allows one pebble to be moved to any specified but arbitrary vertex by a sequence of pebbling moves. This paper determines the pebbling numbers and the 2-pebbling property of the middle graph of fan graphs.

#### 1. Introduction

Pebbling on graphs was first introduced by Chung [1]. Consider a connected graph with a fixed number of pebbles distributed on its vertices. A pebbling move consists of the removal of two pebbles from a vertex and the placement of one of those pebbles on an adjacent vertex. The pebbling number of a vertex in a graph is the smallest number with the property that from every placement of pebbles on , it is possible to move a pebble to by a sequence of pebbling moves. The pebbling number of a graph , denoted by , is the maximum of over all the vertices of .

In a graph , if each vertex (except ) has at most one pebble, then no pebble can be moved to . Also, if is of distance from and at most pebbles are placed on (and none elsewhere), then no pebble can be moved from to . So it is clear that , where is the number of vertices of and is the diameter of .

Throughout this paper, let be a simple connected graph with vertex set and edge set . For a distribution of pebbles on , denote by and the number of pebbles on a subgraph of and the number of pebbles on a vertex of , respectively. In addition, denote by and the number of pebbles on and the number of pebbles on after a specified sequence of pebbling moves, respectively. For , refers to taking pebbles off and placing pebbles on . Denote by the path with vertices in order.

We now introduce some definitions and give some lemmas, which will be used in subsequent proofs.

Definition 1. A fan graph, denoted by , is a path plus an extra vertex connected to all vertices of the path , where .

Definition 2. The middle graph of a graph is the graph obtained from by inserting a new vertex into every edge of and by joining by edges those pairs of these new vertices which lie on adjacent edges of .

Now one creates the middle graph of . Edges of are the inserted new vertices in the sequence, and edges of are the inserted new vertices , respectively. By joining by edges those pairs of these inserted vertices which lie on adjacent edges of , this obtains the middle graph of (see Figure 1).

Definition 3. A transmitting subgraph is a path such that there are at least two pebbles on , and after a sequence of pebbling moves, one can transmit a pebble from to .

Lemma 4 (see [2]). Let . If then is a transmitting subgraph.

Definition 5. The -pebbling number, , of a connected graph, , is the smallest positive integer such that from every placement of pebbles, pebbles can be moved to a specified target vertex by a sequence of pebbling moves.

Lemma 6 (see [3]). If is the complete graph with āā vertices, then .

Lemma 7 (see [4]). Consider .

Chung found the pebbling numbers of the -cube , the complete graph , and the path (see [1]). The pebbling number of was determined in [5]. In [6, 7], Ye et al. gave the number of squares of cycles. Feng and Kim proved that and (see [8]). Liu et al. determined the pebbling numbers of middle graphs of , , and (see [4]). In [9], Ye et al. proved that and , where denotes the middle graph of . Motivated by these works, we will determine the value of the pebbling number and the 2-property of middle graphs of .

#### 2. Pebbling Numbers of

In this section, we study the pebbling number of . Let , and let . Obviously, the subgraph induced by is a complete graph with vertices. For , . Hence we have the following theorem.

Theorem 8 (see [9]). Consider .

Lemma 9. Let . If pebbles are placed on , then one pebble can be moved to any specified vertex of by a sequence of pebbling moves.

Proof. Let be our target vertex, and let , where . We may assume that (after relabeling if necessary). Let . If , then by Lemma 6, and we can move one pebble to . If , then . We delete to obtain the subgraph with pebbles, thus we can move one pebble to . If , then we delete to obtain the subgraph with at least pebbles and we are done.

Theorem 10. Consider .

Proof. We place 7 pebbles on and one pebble on each vertex of the set , other vertices have no pebble, then no pebble can be moved to . So . We now place 11 pebbles on . We assume that is our target vertex and . Recall and .(1)Consider . By Theorem 8 and Lemma 9, we can move one pebble to .(2)Consider (or ). Let , let , and let . If , then . Thus we can move at least pebbles from to so that for . By Lemma 6, and we can move one pebble to . If , then . By Lemma 7, we can move one pebble to . If , then at least one of and can obtain one pebble from every placement of 3 pebbles on by a sequence of pebbling moves. If , then . So is a transmitting subgraph. If , then . By Lemma 6, and . So is a transmitting subgraph. If , then . So is a transmitting subgraph.(3)Consider . If or , then we are done with (2). If , then or . So or is a transmitting subgraph.(4)Consider (or ). If or , then we are done with (2). If , then . Obviously, we are done if or . Next suppose that and . Thus . So is a transmitting subgraph.

Theorem 11. Consider .

Proof. We place 7 pebbles on and one pebble on each vertex of except , and . In this configuration of pebbles, we cannot move one pebble to . So . Next, let us use induction on to show that . For , our theorem is true by Theorem 10. Suppose that if . Now pebbles are placed arbitrarily on the vertices of . Suppose that is our target vertex and .
(1) Consider . By induction and Theorem 8, we can move one pebble to .
(2) Consider (or ). Obviously, . Otherwise, . can obtain one pebble. Let āā.
If , then we delete to obtain the subgraph with at least pebbles. By induction, we can move one pebble to . If , then . Thus we delete to obtain the subgraph with pebbles. By induction, we are done.
Next, suppose that . By Lemma 6, . If , then is a transmitting subgraph. If , then , and we are done. If there exists some with āā, then , and we are done. Thus we assume that , , and for .
Now, we consider (). Clearly, if , then we are done. Suppose that there exists some with (). It is clear that if one of the three cases ((i) (), (ii) , and (iii) ()) happens, then we can move one pebble to . Thus we assume that (), , , and . If there are sets such that for , then for . Let , let , and let . If for all and for all , then and . Recall that for all and . Then and . Thus is a transmitting subgraph. So there is at least some in such that or at least some in such that . If there are two and in such that and or two and in such that and or some in such that and some in such that , then . By Lemma 6, . Hence is a transmitting subgraph.
Finally, there are two remaining cases, (i) there is only some in such that , and (ii) there is only some in such that . So . If , then is a transmitting subgraph. If , then, in , and . For (i), we have for . By the transmitting subgraph , and we are done. Suppose that (ii) holds. If , then we can move one pebble from to so that , and we are done. If , then and we are done.
(3) Consider (or ). Obviously, and āā. Otherwise, one pebble can be moved to . The proof is similar to (2).
(4) Consider (or ) and . Let , and let . If , then delete to obtain the subgraph with at least pebbles. By induction, we can move one pebble to . If , then we can move one pebble from to , after deleting to obtain the subgraph with pebbles. Hence we assume that . According to symmetry, . Therefore we are done.

#### 3. The 2-Pebbling Property of

For a distribution of pebbles on , let be the number of vertices with at least one pebble. We say a graph satisfies the 2-pebbling property if two pebbles can be moved to any specified vertex when the total starting number of pebbles is . Next we will discuss the 2-pebbling property of . For the convenience of statement, let , and let , where , , , and . In this section let , where and are the number of vertices with at least one pebble in and , respectively.

Lemma 12. Suppose that and . If āā and āā, then one can move pebbles to .

Proof. Let (or ). Since and , so . Without loss of generality, there exists a positive integer āā such that the corresponding vertex with and for . Thus . Using the remaining pebbles on vertices , we can move at least pebbles to so that . By Lemma 6, . So we can move one additional pebble from to so that .

Lemma 13. Suppose that and . If āā and āā, then one can move pebbles to .

Proof. Let āā(or ). Since , we see that there is only some vertex āā with . Without loss of generality, there exists a positive integer āā such that the corresponding vertex with and for . If āā or , then we can move one pebble to by the transmitting subgraph or . Now using the remaining at least pebbles on the set , we can move pebbles from the to so that . By Lemma 6, and we can move one additional pebble from to so that .
Suppose that āā and . If , then . Thus there must exist one vertex āā with and for . Using the transmitting subgraph or , we can move one pebble to . Now, using the remaining pebbles on the set , we can move pebbles from the set to so that . By Lemma 6, and we are done. Next, suppose that .
(1) Consider . We divide into three subcases.
(1.1) Consider . We delete vertices to obtain the subgraph with two sets and , and and . Thus we can move pebbles from to so that . By Lemma 6, and we can move two pebbles from to .
(1.2) Consider . Suppose that or (). Let . Obviously, . If , then We delete to obtain the subgraph with two sets and . So and . We can move pebbles from to so that . By Lemma 6, and we are done. If and , then We delete to obtain the subgraph with two sets and . So and . We can move pebbles from to so that . By Lemma 6, and we are done. If and , then . We delete to obtain the subgraph with two sets and . So and . We can move pebbles from to so that . By Lemma 6, .
(1.3) Consider for . Thus . We delete to obtain the subgraph with two sets and . So and . pebbles can be moved from to so that . By Lemma 6, . So we can move one additional pebble from to .
(2) Consider ; that is, . We divide into three subcases.
(2.1) Consider . We delete to obtain the subgraph with two sets and . One has and . We can move pebbles from to so that . By Lemma 6, and we can move two pebbles from to .
(2.2) Consider . We have We delete vertices to obtain the subgraph with two sets and . So and (except one moved pebble on ). We can move pebbles from to so that (except one moved pebble on ). By Lemma 6, we can move 3 additional pebbles to so that .
(2.3) for . Thus . Deleting to obtain the subgraph with two sets and . One has and (except one moved pebble on ). We can move pebbles from to so that (except one moved pebble on ). By Lemma 6, we can move 3 additional pebbles to so that .

Theorem 14. has the 2-pebbling property.

Proof. Suppose that is our target vertex. If , then the number of pebbles on except one pebble on is āā. By Theorem 11, we can move one additional pebble to so that . Next we assume that .
(1) Consider (). If there exists some vertex with āā, then . Using the remaining pebbles, we can move one additional pebble to so that . If for , then . Thus we can move at least pebbles from to so that . By Lemma 6, we can move two pebbles to .
(2) Consider (). Let (or ). If , then we can put one pebble on . After that, the remaining āā pebbles on suffice to put one additional pebble on by Theorem 11. Next we assume .
(2.1) Suppose that . If there is some vertex with āā, then . The remaining āā pebbles on will suffice to put one additional pebble on so that . Next we assume that for . Obviously, and . If , then . Thus we can move at least pebbles from to so that . By Lemma 6, we can move 3 additional pebbles to so that and we are done. If , then, by Lemma 12, we are done.
(2.2) Suppose that . If we can find some vertex with or find two vertices with and with , then using 4 pebbles on or two pebbles on and two pebbles on we can move one pebble to . Then the remaining āā pebbles on will suffice to put one additional pebble to so that .
Consider only some vertex with . If , then , , and . When and , we can move at least pebbles from to so that except for one pebble on . By Lemma 6, we can put 3 additional pebbles on so that . When , we are done with Lemma 12. If , then , , and . When and , we can move at least pebbles from to so that except for one pebble on . By Lemma 6, we can put 3 additional pebbles on so that . When and , we are done with Lemmas 12 and 13.
Consider for . Obviously, and . When and , we can move at least pebbles from to so that . By Lemma 6, and we are done. When and , we are done with Lemmas 12 and 13.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work is supported by National Natural Science Foundation of China (no. 10971248) and Anhui Provincial Natural Science Foundation (nos. 1408085MA08 and KJ2013Z279).

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Copyright © 2014 Yongsheng Ye et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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