JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 10.1155/2014/304514 304514 Research Article The 2-Pebbling Property of the Middle Graph of Fan Graphs Ye Yongsheng Liu Fang Shi Caixia Tan Ying 1 School of Mathematical Sciences Huaibei Normal University, Huaibei, Anhui 235000 China hbcnc.edu.cn 2014 2172014 2014 29 04 2014 07 07 2014 22 7 2014 2014 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.

A pebbling move on a graph G consists of taking two pebbles off one vertex and placing one pebble on an adjacent vertex. The pebbling number of a connected graph G, denoted by f(G), is the least n such that any distribution of n pebbles on G 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 . 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 v in a graph G is the smallest number f(G,v) with the property that from every placement of f(G,v) pebbles on G, it is possible to move a pebble to v by a sequence of pebbling moves. The pebbling number of a graph G, denoted by f(G), is the maximum of f(G,v) over all the vertices of G.

In a graph G, if each vertex (except v) has at most one pebble, then no pebble can be moved to v. Also, if u is of distance d from v and at most 2d-1 pebbles are placed on u (and none elsewhere), then no pebble can be moved from u to v. So it is clear that f(G)max{|V(G)|,2D}, where |V(G)| is the number of vertices of G and D is the diameter of G.

Throughout this paper, let G be a simple connected graph with vertex set V(G) and edge set E(G). For a distribution of pebbles on G, denote by p(H) and p(v) the number of pebbles on a subgraph H of G and the number of pebbles on a vertex v of G, respectively. In addition, denote by p~(H) and p~(v) the number of pebbles on H and the number of pebbles on v after a specified sequence of pebbling moves, respectively. For uvE(G), umv refers to taking 2m pebbles off u and placing m pebbles on v. Denote by v1,v2,,vn the path with vertices v1,v2,,vn 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 Fn, is a path Pn-1 plus an extra vertex v0 connected to all vertices of the path Pn-1, where Pn-1=v1,v2,,vn-1.

Definition 2.

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

Now one creates the middle graph of Fn. Edges v1v2,v2v3,,v(n-2)(n-1) of Fn are the inserted new vertices u12,u23,,u(n-2)(n-1) in the sequence, and edges v0v1,v0v2,,v0vn-1 of Fn are the inserted new vertices u01,u02,,u0(n-1), respectively. By joining by edges those pairs of these inserted vertices which lie on adjacent edges of Fn, this obtains the middle graph of Fn (see Figure 1).

M ( F 4 ) .

Definition 3.

A transmitting subgraph is a path v0,v1,,vk such that there are at least two pebbles on v0, and after a sequence of pebbling moves, one can transmit a pebble from v0 to vk.

Lemma 4 (see [<xref ref-type="bibr" rid="B6">2</xref>]).

Let Pk+1=v0,v1,,vk. If (1)p(v0)+2p(v1)++2ip(vi)++2k-1p(vk-1)2k, then Pk+1 is a transmitting subgraph.

Definition 5.

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

Lemma 6 (see [<xref ref-type="bibr" rid="B4">3</xref>]).

If Kn is the complete graph with n  (n2) vertices, then ft(Kn)=2t+n-2.

Lemma 7 (see [<xref ref-type="bibr" rid="B3">4</xref>]).

Consider f(M(Pn))=2n+n-2.

Chung found the pebbling numbers of the n-cube Qn, the complete graph Kn, and the path Pn (see ). The pebbling number of Cn was determined in . In [6, 7], Ye et al. gave the number of squares of cycles. Feng and Kim proved that f(Fn)=n and f(Wn)=n (see ). Liu et al. determined the pebbling numbers of middle graphs of Pn, Kn, and K1,n-1 (see ). In , Ye et al. proved that f(M(C2n))=2n+1+2n-2(n2) and f(M(C2n+1))=2n+3/3+2n, where M(Cn) denotes the middle graph of Cn. Motivated by these works, we will determine the value of the pebbling number and the 2-property of middle graphs of Fn.

2. Pebbling Numbers of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M106"><mml:mi>M</mml:mi><mml:mo mathvariant="bold">(</mml:mo><mml:msub><mml:mrow><mml:mi>F</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:msub><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula>

In this section, we study the pebbling number of M(Fn). Let S={v0,u01,u02,,u0(n-1)}, and let A={v1,u12,v2,u23,,vn-1}. Obviously, the subgraph induced by S is a complete graph with n vertices. For n=3, M(F3)M(C3). Hence we have the following theorem.

Theorem 8 (see [<xref ref-type="bibr" rid="B7">9</xref>]).

Consider f(M(F3))=7.

Lemma 9.

Let f(M(Fn-1))=p. If p+3 pebbles are placed on M(Fn), then one pebble can be moved to any specified vertex of S by a sequence of pebbling moves.

Proof.

Let v be our target vertex, and let p(v)=0, where vS. We may assume that vu01 (after relabeling if necessary). Let B={v1,u12,u01}. If p(B)5, then p~(u01)2 by Lemma 6, and we can move one pebble to v. If p(B)=4, then B1u02. We delete v1,u01,andu12 to obtain the subgraph M(Fn-1) with p pebbles, thus we can move one pebble to v. If p(B)3, then we delete v1,u01,andu12 to obtain the subgraph M(Fn-1) with at least p pebbles and we are done.

Theorem 10.

Consider f(M(F4))=11.

Proof.

We place 7 pebbles on v3 and one pebble on each vertex of the set {v0,u02,v2}, other vertices have no pebble, then no pebble can be moved to v1. So p(M(F4))11. We now place 11 pebbles on M(F4). We assume that v is our target vertex and p(v)=0. Recall S={v0,u01,u02,u03} and A={v1,u12,v2,u23,v3}.

Consider vS. By Theorem 8 and Lemma 9, we can move one pebble to v.

Consider v=v1 (or v=v3). Let A1=A-{v1}, let A2={u12,v2}, and let A3=A1-A2. If p(S)=t, then p(A1)=11-t. Thus we can move at least (8-t)/2 pebbles from A1 to S so that p~(S)=(8+t)/26 for t4. By Lemma 6, p~(u01)=2 and we can move one pebble to v1. If t2, then p(A)9. By Lemma 7, we can move one pebble to v1. If t=3, then at least one of u01 and u03 can obtain one pebble from every placement of 3 pebbles on S by a sequence of pebbling moves. If p(A3)7, then A33u03. So u03,u01,v1 is a transmitting subgraph. If 4p(A3)6, then 2p(A2)4. By Lemma 6, p~(u23)2 and p~(u12)1. So u23,u12,v1 is a transmitting subgraph. If p(A3)3, then p(A2)5. So v2,u12,v1 is a transmitting subgraph.

Consider v=v2. If p(S)4 or p(S)2, then we are done with (2). If p(S)=3, then p(v1)+p(u12)4 or p(u23)+p(v3)4. So v1,u12,v2 or v3,u23,v2 is a transmitting subgraph.

Consider v=u12 (or v=u23). If p(S)4 or p(S)2, then we are done with (2). If p(S)=3, then p(v1)+p(v2)+p(u23)+p(v3)=8. Obviously, we are done if p(v1)2 or p(v2)2. Next suppose that p(v1)1 and p(v2)1. Thus p(u23)+p(v3)6. So v3,u23,u12 is a transmitting subgraph.

Theorem 11.

Consider f(M(Fn))=3n-1(n4).

Proof.

We place 7 pebbles on vn-1 and one pebble on each vertex of M(Fn) except v1,u01,u12,u(n-2)(n-1),u0(n-1), and vn-1. In this configuration of pebbles, we cannot move one pebble to v1. So f(M(Fn))3n-1. Next, let us use induction on n to show that f(M(Fn))=3n-1. For n=4, our theorem is true by Theorem 10. Suppose that f(M(Fk))=3k-1 if k<n. Now 3n-1 pebbles are placed arbitrarily on the vertices of M(Fn). Suppose that v is our target vertex and p(v)=0.

(1) Consider vS. By induction and Theorem 8, we can move one pebble to v.

(2) Consider v=v1 (or v=vn-1). Obviously, p(u01)1. Otherwise, p(u01)>1. v1 can obtain one pebble. Let Bi={ui(i+1),u0(i+1),vi+1}  (1in-2).

If p(Bn-2)3, then we delete Bn-2 to obtain the subgraph M(Fn-1) with at least 3(n-1)-1 pebbles. By induction, we can move one pebble to v1. If p(Bn-2)=4, then Bn-21u0(n-2). Thus we delete Bn-2 to obtain the subgraph M(Fn-1) with 3(n-1)-1 pebbles. By induction, we are done.

Next, suppose that p(Bn-2)5. By Lemma 6, p~(u0(n-1))2. If p(u01)=1, then u0(n-1),u01,v1 is a transmitting subgraph. If p(v0)2, then v01u01, and we are done. If there exists some Bi with p(Bi)5  (in-2), then Bi1u01, and we are done. Thus we assume that p(u01)=0, p(v0)1, and p(Bi)4 for 1in-3.

Now, we consider Bi (1in-3). Clearly, if p(B1)=4, then we are done. Suppose that there exists some Bj with p(Bj)=4 (j1). It is clear that if one of the three cases ((i) p(u0j)1 (u0jBj-1), (ii) p(Bj-1)3, and (iii) p(vj)2 (vjBj-1)) happens, then we can move one pebble to v. Thus we assume that p(Bi)=4 (2in-3), p(Bi-1)2, p(u0i)=0, and p(vi)1. If there are r sets Bi1,Bi2,,Bir such that p(Bik)=4 for 1kr, then p(Bik-1)2 for 1kr. Let N1={i1,i2,,ir}, let N2={i1-1,i2-1,,ir-1}, and let N3={1,2,,n-3}-N1-N2. If p(Bj)=2 for all jN2 and p(Bk)=3 for all kN3, then p~(uj(j+1))=1 and p~(uk(k+1))=1. Recall that p(Bi)=4 for all iN1 and p(Bn-2)5. Then p~(ui(i+1))=1 and p~(u(n-2)(n-1))=2. Thus u(n-2)(n-1),u(n-3)(n-2),,u12,v1 is a transmitting subgraph. So there is at least some j in N2 such that p(Bj)1 or at least some k in N3 such that p(Bk)2. If there are two j and j′′ in N2 such that p(Bj)1 and p(Bj′′)1 or two k and k′′ in N3 such that p(Bk)2 and p(Bk′′)2 or some j in N2 such that p(Bj)1 and some k in N3 such that p(Bk)2, then p(Bn-2)9. By Lemma 6, p~(u0(n-1))=4. Hence u0(n-1),u01,v1 is a transmitting subgraph.

Finally, there are two remaining cases, (i) there is only some j in N2 such that p(Bj)1, and (ii) there is only some k in N3 such that p(Bk)2. So p(Bn-2)8. If p(u(n-2)(n-1))=0, then vn-1,u0(n-1),u01,v1 is a transmitting subgraph. If p(u(n-2)(n-1))0, then, in Bn-2, p~(u(n-2)(n-1))2 and p~(u0(n-1))2. For (i), we have p~(ui(i+1))1 for j+2in-3. By the transmitting subgraph u(n-2)(n-1),u(n-3)(n-2),,u(j+1)(j+2), p~(Bj+1)=5 and we are done. Suppose that (ii) holds. If p(Bk)=2, then we can move one pebble from u0(n-1) to u0(k+1) so that p(Bk)=3, and we are done. If p(Bk)1, then p(Bn-2)9 and we are done.

(3) Consider v=u12 (or v=u(n-2)(n-1)). Obviously, p(u01)1 and p(vi)1  (i=1,2). Otherwise, one pebble can be moved to u12. The proof is similar to (2).

(4) Consider v=vi(2in-2) (or v=uj(j+1)(2jn-3)) and p(vi)=0. Let B={v1,u12,u01}, and let B={vn-1,u(n-2)(n-1),u0(n-1)}. If p(B)3, then delete B to obtain the subgraph M(Fn-1) with at least 3(n-1)-1 pebbles. By induction, we can move one pebble to v. If p(B)=4, then we can move one pebble from B to u02, after deleting B to obtain the subgraph M(Fn-1) with 3(n-1)-1 pebbles. Hence we assume that p(B)5. According to symmetry, p(B)5. Therefore we are done.

3. The 2-Pebbling Property of <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M361"><mml:mi>M</mml:mi><mml:mo mathvariant="bold">(</mml:mo><mml:msub><mml:mrow><mml:mi>F</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:msub><mml:mo mathvariant="bold">)</mml:mo></mml:math></inline-formula>

For a distribution of pebbles on G, let q be the number of vertices with at least one pebble. We say a graph G satisfies the 2-pebbling property if two pebbles can be moved to any specified vertex when the total starting number of pebbles is 2f(G)-q+1. Next we will discuss the 2-pebbling property of M(Fn). For the convenience of statement, let S={x1,x2,,xn}, and let A={y1,y2,,y2n-3}, where x1=v0, x2=u01,,xn=u0(n-1), y1=v1, and y2=u12,,y2n-3=vn-1. In this section let q=qs+qa, where qs and qa are the number of vertices with at least one pebble in S and A, respectively.

Lemma 12.

Suppose that p(M(Fn))2(3n-1)-q and qa=2n-4. If p(S)=qs+t  (t=0,1,2) and p(yr)=0  (1r2n-3), then one can move 2 pebbles to yr.

Proof.

Let r=2k-1 (or r=2k). Since qa=2n-4 and p(S)=qs+t, so p(A)4n+2-2qs-t. Without loss of generality, there exists a positive integer j  (j>r) such that the corresponding vertex yj with p(yj)2 and p(yi)=1 for r+1ij-1. Thus yj1yj-111yr. Using the remaining 4n+2-t-2qs-(j-r+1) pebbles on vertices y1,y2,,yr-1,yj,yj+1,,y2n-3, we can move at least n+(5-t)/2-qs pebbles to S so that p~(S)n+(5+t)/2. By Lemma 6, p~(xk+1)=2. So we can move one additional pebble from xk+1 to yr so that p~(yr)=2.

Lemma 13.

Suppose that p(M(Fn))=2(3n-1)-q+1 and qa=2n-5. If p(S)=qs+t  (t=0,1) and p(yr)=0  (1r2n-3), then one can move 2 pebbles to yr.

Proof.

Let r=2k-1  (or r=2k). Since qa=2n-5, we see that there is only some vertex yi0  (i0r) with p(yi0)=0. Without loss of generality, there exists a positive integer j  (j>r) such that the corresponding vertex yj with p(yj)2 and p(yi)1 for r<i<j. If i0=2k0-1  (k0k) or i0{r+1,r+2,,j-1}, then we can move one pebble to yr by the transmitting subgraph yj,yj-2,,yr+1,yr or yj,yj-1,yj-3,,yr+1,yr. Now using the remaining at least 4n+4-t-2qs-(j-r+1) pebbles on the set A1={y1,y2,,yr-1,yj,yj,,y2n-3}, we can move n+(7-t)/2-qs pebbles from the A1 to S so that p~(S)=n+(7+t)/2. By Lemma 6, p~(xk+1)=2 and we can move one additional pebble from xk+1 to yr so that p~(yr)=2.

Suppose that i0=2k0  (k0k) and i0{r+1,r+2,,j-1}. If j=i0+1, then yj1yi0. Thus there must exist one vertex yj  (jj) with p(yj)2 and p(yi)1 for r<i<j. Using the transmitting subgraph yj,yj-2,,yr+1,yr or yj,yj-1,yj-3,,yr+1,yr, we can move one pebble to yr. Now, using the remaining 4n+4-t-2qs-(j-r+2) pebbles on the set {y1,y2,,yr-1,yj,yj+1,,y2n-3}, we can move n+(6-t)/2-qs pebbles from the set {y1,y2,,yr-1,yj,yj+1,,y2n-3} to S so that p~(S)n+(6+t)/2. By Lemma 6, p~(xk+1)=2 and we are done. Next, suppose that ji0+2.

(1) Consider p(y2k)=1. We divide into three subcases.

(1.1) Consider p(xk+2)=0. We delete vertices yr,yr+1,,y2k0,xk+2 to obtain the subgraph with two sets A2=A-{yr,yr+1,,y2k0} and S1=S-{xk+2}, and p(A2)=4n+4-2qs-t-(2k0-r-1) and p(S1)=qs+t. Thus we can move n+(10-t)/2-qs pebbles from A2 to S1 so that p~(S1)=n+(10+t)/2. By Lemma 6, p~(xk+1)=4 and we can move two pebbles from xk+1 to yr.

(1.2) Consider p(xk+2)=1. Suppose that j=2k or j=2k+1 (k>k). Let A3={y2k,y2k+1}. Obviously, p(A3)3. If p(A3)5, then (2)A32xk+21xk+21yr+11yr. We delete yr,yr+1,,y2k0,xk+2 to obtain the subgraph with two sets A2 and S1. So p(A2)=4n-2qs-t-(2k0-r-1) and p~(S1)=qs-1+t. We can move n+(6-t)/2-qs pebbles from A2 to S1 so that p~(S1)=n+(4+t)/2. By Lemma 6, p~(xk+1)=2 and we are done. If p(A3)=3,4 and p(xk+2)0, then (3)A31xk+21xk+21yr+11yr. We delete yr,yr+1,,y2k0,xk+2 to obtain the subgraph with two sets A2 and S1. So p(A2)=4n+2-2qs-t-(2k0-r-1) and p~(S1)=qs-2+t. We can move n+(8-t)/2-qs pebbles from A2 to S1 so that p~(S1)=n+(4+t)/2. By Lemma 6, p~(xk+1)=2 and we are done. If p(A3)=3,4 and p(xk+2)=0, then A31xk+1. We delete yr,yr+1,,y2k0,y2k,y2k+1,xk+2 to obtain the subgraph with two sets A4=A2-A3 and S2=S-{x2k+2}. So p(A4)4n-2qs-t-(2k0-r-1) and p~(S2)=qs+1+t. We can move n+(8-t)/2-qs pebbles from A4 to S2 so that p~(S2)=n+(10+t)/2. By Lemma 6, p~(xk+1)=4.

(1.3) Consider p(xk+2)=2 for t=1. Thus xk+21y2k1yr. We delete yr,yr+1,,y2k0,xk+2 to obtain the subgraph with two sets A2 and S1. So p(A2)=4n+3-2qs-(2k0-r-1) and p~(S1)=qs-1. n+4-qs pebbles can be moved from A2 to S1 so that p~(S1)=n+3. By Lemma 6, p~(xk+1)=3. So we can move one additional pebble from xk+1 to yr.

(2) Consider p(y2k)=0; that is, k=k0. We divide into three subcases.

(2.1) Consider p(x2k+2)=0. We delete yr,yr+1,yr+2,x2k+2 to obtain the subgraph with two sets A5=A-{yr,yr+1,yr+2} and S1. One has p(A5)=4n+3-2qs-t and p(S1)=qs+t. We can move n+(10-t)/2-qs pebbles from A5 to S1 so that p~(S1)=n+(10+t)/2. By Lemma 6, p~(xk+1)=4 and we can move two pebbles from xk+1 to yr.

(2.2) Consider p(xk+2)=1. We have (4)yj1yj-111yr+21xk+21xk+1. We delete vertices yr,yr+1,,yj-1,xk+2 to obtain the subgraph with two sets A1 and S1. So p(A1)=4n+4-2qs-t-(j-r) and p~(S1)=qs+t-1 (except one moved pebble on xk+1). We can move n+(8-t)/2-qs pebbles from A5 to S1 so that p~(S1)=n+(6+t)/2 (except one moved pebble on xk+1). By Lemma 6, we can move 3 additional pebbles to xk+1 so that p~(xk+1)=4.

(2.3) p(xk+2)=2 for t=1. Thus xk+21xk+1. Deleting yr,yr+1,yr+2,xk+2 to obtain the subgraph with two sets A5 and S1. One has p(A5)=4n+2-2qs and p~(S1)=qs-1 (except one moved pebble on xk+1). We can move n+4-qs pebbles from A4 to S1 so that p~(S1)=n+3 (except one moved pebble on xk+1). By Lemma 6, we can move 3 additional pebbles to xk+1 so that p~(xk+1)=4.

Theorem 14.

M ( F n ) has the 2-pebbling property.

Proof.

Suppose that v is our target vertex. If p(v)=1, then the number of pebbles on M(Fn) except one pebble on v is 2(3n-1)+1-q-1  (>3n-1). By Theorem 11, we can move one additional pebble to v so that p~(v)=2. Next we assume that p(v)=0.

(1) Consider v=xr (1rn). If there exists some vertex xi with p(xi)2  (ir), then xi1xr. Using the remaining 2(3n-1)+1-q-2>3n-1 pebbles, we can move one additional pebble to xr so that p~(xr)=2. If p(xi)1 for 1in, then p(A)=2(3n-1)-q+1-qs=6n-1-qa-2qs4n+2-2qs. Thus we can move at least n+2-qs pebbles from A to S so that p~(S)=n+2. By Lemma 6, we can move two pebbles to xr.

(2) Consider v=yr (1r2n-3). Let r=2k-1 (or r=2k). If p(xk+1)2, then we can put one pebble on yr. After that, the remaining 2(3n-1)-q+1-2  (>3n-1) pebbles on M(Fn) suffice to put one additional pebble on yr by Theorem 11. Next we assume p(xk+1)1.

(2.1) Suppose that p(xk+1)=1. If there is some vertex xi with p(xi)2  (ik+1), then xi1xk+11yr. The remaining 2(3n-1)-q+1-3  (>3n-1) pebbles on M(Fn) will suffice to put one additional pebble on yr so that p~(yr)=2. Next we assume that p(xi)1 for 1in. Obviously, p(S)=qs and p(A)=2(3n-1)-q+1-qs=6n-1-qa-2qs. If qa2n-5, then p(A)4n+4-2qs. Thus we can move at least n+5-qs pebbles from A to S so that p~(S)=n+5. By Lemma 6, we can move 3 additional pebbles to xk+1 so that p~(xk+1)=4 and we are done. If qa=2n-4, then, by Lemma 12, we are done.

(2.2) Suppose that p(xk+1)=0. If we can find some vertex xi with p(xi)4 or find two vertices xj with p(vj)2 and xj with p(xj)2, then using 4 pebbles on xi or two pebbles on xj and two pebbles on xj we can move one pebble to yr. Then the remaining 2(3n-1)-q+1-4  (>3n-1) pebbles on M(Fn) will suffice to put one additional pebble to yr so that p~(yr)=2.

Consider only some vertex xi with 2p(xi)3. If p(xi)=3, then xi1xk+1, p~(S)=qs, and p(A)=2(3n-1)-qs-qa+1-(qs+2)=6n-3-2qs-qa. When qa2n-5 and p(A)4n+2-2qs, we can move at least n+4-qs pebbles from A to S so that p~(S)n+4 except for one pebble on xk+1. By Lemma 6, we can put 3 additional pebbles on xk+1 so that p~(xk+1)=4. When qa=2n-4, we are done with Lemma 12. If p(xi)=2, then xi1xk+1, p~(S)=qs-1, and p(A)=2(3n-1)-qs-qa+1-(qs+1)=6n-2-2qs-qa. When qa2n-6 and p(A)4n+4-2qs, we can move at least n+5-qs pebbles from A to S so that p~(S)n+4 except for one pebble on xk+1. By Lemma 6, we can put 3 additional pebbles on xk+1 so that p~(xk+1)=4. When qa=2n-4 and qa=2n-5, we are done with Lemmas 12 and 13.

Consider p(xi)1 for 1in. Obviously, p(S)=qs and p(A)=6n-1-qa-2qs. When qa2n-6 and p(A)4n+5-2qs, we can move at least n+6-qs pebbles from A to S so that p~(S)n+6. By Lemma 6, p~(xk+1)=4 and we are done. When qa=2n-4 and qa=2n-5, 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).

Chung F. R. K. Pebbling in hypercubes SIAM Journal on Discrete Mathematics 1989 2 4 467 472 10.1137/0402041 MR1018531 Snevily H. S. Foster J. A. The 2-pebbling property and a conjecture of Graham’s Graphs and Combinatorics 2000 16 2 231 244 MR1771223 2-s2.0-0347480232 Lourdusamy A. Somasundaram S. The t-pebbling number of graphs Southeast Asian Bulletin of Mathematics 2006 30 5 907 914 MR2287011 Liu H. Y. Qin Q. Wang Z. P. Ma Y. G. Pebbling number, of middle graphs Journal of Dalian Maritime University 2006 32 4 125 128 Pachter L. Snevily H. S. Voxman B. On pebbling graphs Congressus Numerantium 1995 107 65 80 Ye Y. S. Zhai M. Q. Zhang Y. Pebbling number of squares of odd cycles Discrete Mathematics 2012 312 21 3174 3178 10.1016/j.disc.2012.07.013 MR2957937 2-s2.0-84865091892 Ye Y. Zhang P. Zhang Y. The pebbling number of squares of even cycles Discrete Mathematics 2012 312 21 3203 3211 10.1016/j.disc.2012.07.016 MR2957940 2-s2.0-84865064984 Feng R. Kim J. Y. Pebbling numbers of some graphs Science in China A 2002 45 4 470 478 10.1007/BF02872335 MR1912119 2-s2.0-0346426711 Ye Y. S. Liu F. Zhai M. Q. Pebbling numbers of middle graphs of cycles and Graham's conjecture Operations Research Transactions 2013 17 3 35 44 MR3186054