The 2-pebbling property of squares of paths and Graham’s conjecture

Yueqing Li
  • Fundamental Education College, Beijing Polytechnic College, Beijing, 100042, China
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and Yongsheng Ye
  • Corresponding author
  • School of Mathematical Sciences, Huaibei Normal University, Huaibei, Anhui, 235000, China
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Abstract

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 vertex by a sequence of pebbling moves. In this paper, we determine the 2-pebbling property of squares of paths and Graham’s conjecture on P2n2.

1 Introduction

Throughout this paper, G denotes a simple connected graph with vertex set V(G) and edge set E(G). Let p pebbles be distributed onto the vertices of a graph G. A pebbling move consists of removing two pebbles from one vertex and then placing one pebble at 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 vertex by a sequence of pebbling moves.

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. Furthermore, f(Kn) = n and f(Pn) = 2n−1 in [1], where Kn denotes a complete graph with n vertices and Pn denotes a path with n vertices.

Pebbling of graphs was first introduced by Chung [1]. In [2], Pachter et al. obtained the pebbling number of Pn2 and showed that the most graphs have the 2-pebbling property. Y. S. Ye et al. gave that the pebbling number of Cn2 [3, 4]. There are many known results about pebbling numbers [5, 6, 7, 8, 9]. Motivated by these results, we first show that squares of paths have the 2-pebbling property and then establish Graham’s conjecture holds for P2n2. We now introduce definitions and lemmas, which will be used in the subsequent proofs.

Definition 1.1

LetGbe a connected graph. Foru, vV(G), we denote bydG(u, v) the distance betweenuandvin G. Thek-th power ofG, denoted byGk, is the graph obtained fromGby adding the edge uv toGwhenever 2 ⩽ dG(u, v) ⩽ kinG. That is, E(Gk) = {uv : 1 ⩽ dG(u, v) ⩽ k}.

Obviously, Gk is the complete graph whenever k is at least the diameter of G. In particular, let Pn be a path with n vertices, written 〈 v1, v2, ⋯, vn〉, i.e., Pn = 〈 v1, v2, ⋯, vn〉. Pn2 is obtained from Pn by joining vivi+2 (i = 1, 2, ⋯, n − 2).

Definition 1.2

We call a graphGsatisfies 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, whereqis the number of vertices with at least one pebble.

Lemma 1.1

[2] f(P2k2)=2k,f(P2k+12)=2k+1.

Lemma 1.2

[5] LetPn = 〈 v1, v2, ⋯, vn 〉. Ifp(v1) + 2p(v2) + ⋯ + 2i−1p(vi) + ⋯ + 2n−2p(vn−1) ⩾ 2n−1, then one pebble can be moved tovnfromv1.

Lemma 1.3

[3] LetP2k+1 = 〈 v1, v2, …, v2k+1〉. IfP2k+12has 2kpebbles andv1has an even number of pebbles, then a pebble can be moved tov2k+1.

Define p(H) to be the number of pebbles on a subgraph H of G and p(v) to be the number of pebbles on a vertex v of G. Moreover, denote by (H) and (v) respectively, the number of pebbles on H and the number of pebbles on v after a specified sequence of pebbling moves, respectively.

2 The 2-pebbling property of Pn2

Let Pn = 〈 v1, v2, ⋯, vn〉. In this section, we first will show that P2k2 has the 2-pebbling property and then we shall prove that P2k+12 also has the 2-pebbling property.

Theorem 2.1

P2k2has the 2-pebbling property.

Proof

Suppose that 2k+1q + 1 pebbles are placed arbitrarily at the vertices of P2k2, where q is the number of vertices with at least one pebble. Suppose that our target vertex is v. We divide into two cases: p(v) = 1 p(v) = 0.

  1. Case 1p(v) = 1. Then P2k2 has 2k+1q pebbles other than one pebble on v. Since 2k+1q ≥ 2k, by Lemma 1.1, we can move one additional pebble to v so that (v) = 2.
  2. Case 2p(v) = 0.
  3. Subcase 2.1If q = 2k − 1, then p(P2k2) = 2k+1 − 2k + 2. For k = 2, it is clearly that we can move 2 pebbles to v. For k ≥ 3, we can move one pebble to v using at most k + 1 pebbles. This leaves 2k+1 − 3k + 1 ≥ 2k pebbles on P2k2. By Lemma 1.1, we may move one additional pebble to v so that (v) = 2.
  4. Subcase 2.2New we assume q ≤ 2k − 2. Without loss of generality, we may assume that our target vertex v is not v2k(otherwise, relabeling). We claim that p(v2k) ≥ 2k + 2k−1q + 1 > 2k. Otherwise,
    p(P2k12)2k+1q+1(2k+2k1q)=2k1+1.
    By Lemma 1.1, we can move one pebble to v. This leaves
    2k+1q+1(2k1+1)=2k+2k1q2k
    pebbles on P2k2. By Lemma 1.1, we can move one additional pebble to v so that (v) = 2. Using 2k pebbles on v2k, we can move at least one pebble to v by Lemma 1.1. This leaves at least 2kq + 1 pebbles on P2k2 other than one pebble on v. Let p(P2k12) = s and s ≤ 2k−1. Thus (v2k) = 2kqs + 1 ≥ 2. We can move 2k1q+s12 pebbles to P2k12 pebbles from v2k so that P2k12 has
    s+2k1q+s12=2k1+sq+122k1(sq1)
    pebbles other than one pebble on v and v2k−1 has an even number of pebbles. By Lemma 1.3, we can move one additional pebble to v so that (v) = 2.□

Theorem 2.2

P2k+12has 2-pebbling property.

Proof

Suppose that 2k+1q + 3 pebbles are placed arbitrarily at the vertices of P2k+12. Let our target vertex be v. If p(v) = 1, then P2k+12 has 2k+1q + 2 pebbles other than one pebble on v. Since 2k+1q + 2 ≥ 2k + 1, by Lemma 1.1, we can move one additional pebble to v so that (v) = 2. Hence we assume p(v) = 0.

If q = 2k, then p(P2k+12) = 2k+1 − 2k +3. For k = 2, it is clearly that we can move 2 pebbles to v. For k ≥ 3, we can move one pebble to v using at most k + 1 pebbles. This leaves 2k+1 − 3k + 2 ≥ 2k + 1 pebbles on P2k+12. By Lemma 1.1, we may move one additional pebble. New we assume q ≤ 2k − 1. Without loss of generality, we may assume that our target vertex v is not v2k or v2k+1 (otherwise, relabeling). We claim that p(v2k) + p(v2k+1) ≥ 2k + 2k−1q + 3 > 2k + 1. Otherwise,

p(P2k12)2k+1q+3(2k+2k1q+2)=2k1+1.

By Lemma 1.1, we can move one pebble to v. This leaves 2k+1q + 3 − (2k−1 + 1) ≥ 2k + 1 pebbles on P2k+12. By Lemma 1.1, we can move one additional pebble to v so that (v) = 2. Using 2k + 1 pebbles on the set {v2k, v2k+1}, we can move at least one pebble to v by Lemma 1.1. This leaves at least 2kq + 2 pebbles on P2k+12 other than one pebble on v. Let p(P2k12) = s and s ≤ 2k−1. Thus (v2k) + (v2k+1) = 2kqs + 3 ≥ 3. We can move 2k1q+s22+1 pebbles to P2k12 from v2k and v2k+1 so that

p(P2k12)=s+2k1q+s22+1=2k1+sq+22+12k1+1(sq2)

pebbles other than one pebble on v. By Lemma 1.1, we can move one additional pebble to v so that (v) = 2.□

3 Graham’s conjecture on Pn2

Given two graphs G and H, the Cartesian product of them is denoted by G × H. Its vertex set

V(G×H)={(u,v)|uV(G),vV(H)},

and (u, v) is adjacent to (u′, v′) if and only if u = u′ and vv′ ∈ E(H) or v = v′ and uu′ ∈ E(G).

We can depict G × H pictorially by drawing a copy of H at every vertex of G and joining each vertex in one copy of H to the corresponding vertex in an adjacent copy of H. We write u(H) (respectively, v(G)) for the subgraph of vertices whose projection onto V(G) is the vertex u, (respectively, whose projection onto V(H) is the vertex v). Obviously, u(H) ≅ H, v(G) ≅ G.

Graham’s conjectureor any two graphsGandH, we have

f(G×H)f(G)f(H).

Lemma 3.1

[5] LetPnbe a path withnvertices andGbe a graph with the 2-pebbling property. Thenf(Pn × G) ≤ 2n−1f(G).

Lemma 3.2

LetP2k = 〈 v1, v2, ⋯, v2k〉 (k ≥ 2) andp(vi) = ji (1 ≤ i ≤ 2k). If one of the following cases holds

  1. j1 + j2 + 1 = 2kandj2 > 0;
  2. j1 + j2 + 2 = 2k, j2 > 0 and(v1) = j1 + 1;
  3. j1 + j2 + 2 = 2k, j2 > 0 and(v2) = j2 + 1;
  4. j1 + j2 + 3 = 2k, (v1) = j1 + 1 and(v2) = j2 + 1,then we can move one pebble to the vertexvofP2k2, wherev ∈ {v3, v4, ⋯, v2k}.

Proof

Let A = 〈 v2, v3, v4, ⋯, v2k〉. Then A2P2k12.

  1. If j1 is odd and j2 is even, then we can move j112 pebbles to v3 from v1. Thus we have
    p~(A2)=j112+j2=(j1+j2+1)+(j22)22k1.
    By Lemma 1.3, we are done. If j1 is even and j2 is odd, then we can move j122 pebbles to v3 and one pebble to v2 from v1 so that (v2) = j2 + 1 and
    p~(A2)=j122+j2+1=(j1+j2+1)+(j21)22k1.
    By Lemma 1.3, we are done.
  2. Let j1 and j2 be odd. If (v1) = j1 + 1, then we can move one pebble to v2 and j112 pebbles to v3 from v1 so that (v2) = j2 + 1 and
    p~(A2)=j2+1+j112=(j1+j2+2)+(j21)22k1.
    By Lemma 1.3, we are done. For even numbers j1 and j2, we can move j12 pebbles to v3 from v1 so that
    p~(A2)=j2+j12=(j1+j2+2)+(j22)22k1.
    By Lemma 1.3, we are done.
  3. The proof is similar to (2).
  4. If j1 is even and j2 is odd, then we can move j12 pebbles to v3 from v1 so that
    p~(A2)=j2+1+j12=(j1+j2+3)+(j21)22k1.
    And (v2) = j2 + 1, we are done by Lemma 1.3. If j1 is odd and j2 is even, then we can move one pebble to v2 and j112 pebbles to v3 from v1 so that (v2) = j2 + 2 and
    p~(A2)=j2+2+j112=(j1+j2+3)+j222k1.
    By Lemma 1.3, we are done.□

Theorem 3.3

Let G be a graph with the 2-pebbling property. Then f(P2k2×G) ≤ 2kf(G).

Proof

For k = 1, P2k2 = P2. By Lemma 3.1, the conclusion is right. We assume that the inequality f(P2k2×G) ≤ 2kf(G) is right for k > 1. Next, we shall determine the above bound of the pebbling number of P2k+22 × G.

Suppose that 2k+1f(G) pebbles have been distributed arbitrarily on the vertices of P2k+22 × G. Let pi = p(vi(G)), where 1 ≤ i ≤ 2k + 2. Let qi be the number of occupied vertices in vi(G)(1 ≤ i ≤ 2k + 2). Suppose that v is our target vertex in P2k+22 × G. Suppose that v = (vi, x) ∈ vi(G) (3 ≤ i ≤ 2k + 2) for xV(G) (relabeling if necessary). For simplicity, let A = 〈 v3, v4, …, v2k+1, v2k+2〉. By the induction hypothesis, f(A2 × G) ≤ 2kf(G). We shall divide into the following two cases.

  1. Case 1p1 + p2 ≤ (2k+1 − 2)f(G). Then A2 × G can obtain at least 12(2k+12)f(G)2f(G) + 2f(G) = 2kf(G) pebbles by a sequence of pebbling moves and we are done.
  2. Case 2p1 + p2 ≥ (2k+1 − 2)f(G) + 1. Thus p(A2 × G) < 2f(G). Let p(A2 × G) = (j0 + α0)f(G), where 0 ≤ j0 ≤ 1 and 0 ≤ α0 < 1, and let pi = (ji + αi)f(G), where ji ≥ 0 and 0 ≤ αi < 1 (i = 1, 2). Now we claim that i=12qi > (j0 + α0) f(G). Otherwise, we have i=12qi ≤ (j0 + α0) f(G). So (A2 × G) ≥ 122k+1f(G)(j0+α0)f(G)(j0+α0)f(G) + (j0 + α0) f(G) = 2kf(G) and we are done. Let i=02αi = s ≤ 2. Then i=02ji = 2k+1s. Next, we discuss two following cases. Let B = 〈(v2, x), (v3, x), ⋯, (v2k+2, x)〉. Then B2P2k+12. We first prove the following claim.

Claim: Suppose j2 = 0. If one of the following cases holds:(1) j0 = 0, s = 1; (2) j0 = 0, s = 2; (3) j0 = 1, s = 1, then we can move one pebble to any vertex of B2.

In fact, since j2 = 0, p1 = (2k+1j0s + α1)f(G). We move (2k − 2)f(G) pebbles to v3(G) so that 3 ≥ (2k − 2)f(G) using (2k+1 − 4)f(G) pebbles on v1(G), and (1 − α2)f(G) to v2(G) so that 2 = f(G) using (2−2α2)f(G) pebbles on v1(G). So we can move one pebble to (v2, x) and (2k − 2) pebbles to (v3, x) so that (B2) = 2k − 1. And the remaining (2 − j0s + α1 + 2α2)f(G) pebbles on v1(G), i.e., 1 = (2 − j0s + α1 + 2α2)f(G).

  1. 1 = (1 + α1 + 2α2)f(G). If α0α2, then 1 ≥ (1 + α0 + α1 + α2)f(G) = (1+s)f(G) = 2f(G). we move two pebbles to (v1, x), and move one pebble to (v2, x) so that ((v2, x)) = 2. By Lemma 1.3, we are done. If α0 > α2, note that q1 + q2 > α0f(G) and q2α2f(G), then q1 > (α0α2)f(G). Thus 1 + q1 > (1 + α0 + α1 + α2)f(G) = 2f(G). By the 2-pebbling property, we move two pebbles to (v1, x), and move one pebble to (v2, x) so that ((v2, x)) = 2 and (B2) = 2k. By Lemma 1.3, we are done.
  2. 1 = (α1 + 2α2)f(G). If α0α2, then 1 ≥ (α0 + α1 + α2)f(G) = sf(G) = 2f(G). If α0α2, then q1 > (α0α2)f(G). Thus 1 + q1 > (α0 + α1 + α2)f(G) = 2f(G). By (1), we are done.
  3. 1 = (α1 + 2α2)f(G). Note that q1 + q2 > (1 + α0)f(G) and q2α2f(G), so q1 > (1 + α0α2)f(G). Thus 1 + q1 > (1 + α0 + α1 + α2)f(G) = (1 + s)f(G) = 2f(G). By (1), we are done.

Next, we continue to complete the proof of the theorem with the following two cases.

  1. Case 1α0 = 0. Obviously, 0 ≤ s ≤ 1.Suppose that j0 = 0, p1 + p2 = 2k+1f(G). By Lemma 3.1, we can move 2k pebbles to the vertex (v2, x) of v2(G) so that (B2) = 2k. By Lemma 1.3, we are done. Suppose j0 = 1, p1 + p2 = 2k+1f(G) − f(G). If s = 0, then p1 + p2 = (j1 + j2)f(G). This implies that j1 + j2 + 1 = 2k+1. For j2 = 0, we can move at least (2k − 1)f(G) pebbles to A2 × G from v1(G) so that
    p~(A2×G)=2kf(G).
    For j2 > 0, note that we can move at least ji pebbles to the vertex (vi, x) of vi(G) (i = 1, 2). By Lemma 3.2, we are done. If s = 1, then p1 + p2 = (j1 + j2 + 1)f(G). This implies that j1 + j2 + 2 = 2k+1. For j2 = 0, p1 = (2k+1 − 2 + α1)f(G) pebbles. By Claim, we are done. For j2 > 0, we claim that at least one of (α1f(G) + q1) and (α2f(G) + q2) is greater than f(G). For suppose not, we have i=12 (αif(G) + qi) ≤ 2f(G). But
    i=12(αif(G)+qi)=i=12αif(G)+i=12qi>i=12αif(G)+f(G)=(s+1)f(G)=2f(G).
    This is a contradiction. Suppose α1f(G) + q1 > f(G). By the 2-pebbling property we can move at least j1 + 1 pebbles to the vertex (v1, x) of v1(G). Note that we can move at least j2 pebbles to the vertex (v2, x) of v2(G). By Lemma 3.2, we are done. Suppose α2f(G) + q2 > f(G). Then we can move at least j2 + 1 pebbles to the vertex (v2, x) of v2(G). Note that we can move at least j1 pebbles to the vertex (v1, x) of v1(G). By Lemma 3.2, we are done.
  2. Case 20 < α0 < 1. Obviously, 1 ≤ s ≤ 2.Suppose that j0 = 0, p1 + p2 = 2k+1f(G)-α0f(G). If s = 1, then α1 + α2 = 1 − α0. Thus p1 + p2 = (j1 + j2 + 1)f(G) − α0f(G). This implies that j1 + j2 + 1 = 2k+1. For j2 = 0, p1 = (2k+1 − 1 + α1)f(G). By Claim, we are done. For j2 > 0, note that we can move at least ji pebbles to the vertex (vi, x) of vi(G) (i = 1, 2). By Lemma 3.2, we are done. If s = 2, then α1 + α2 = 2 − α0. Thus p1 + p2 = (j1 + j2 + 2)f(G) − α0f(G). This implies that j1 + j2 + 2 = 2k+1. For j2 = 0, then p1 = (2k+1 − 2 + α1)f(G). By Claim, we are done. For j2 > 0, now we claim that at least one of (α1f(G) + q1) and (α2f(G) + q2) is greater than f(G). For suppose not, we have i=12 (αif(G) + qi) ≤ 2f(G). But
    i=12(αif(G)+qi)=i=12αif(G)+i=12qi>i=12αif(G)+α0f(G)=(α1+α2+α0)f(G)=2f(G).
    This is a contradiction. By Case 1, we are done.Suppose that j0 = 1, p1 + p2 = 2k+1f(G)−(1 + α0 ) f(G) = (2k+1 − 1)f(G) − α0f(G). If s = 1, then α1 + α2 = 1 − α0. Thus
    p1+p2=(j1+j2+1)f(G)α0f(G).
    This implies that j1 + j2 + 2 = 2k+1. For j2 = 0, p1 = (2k+1 − 2 + α1)f(G). By Claim, we are done. For j2 > 0, we claim that at least one of (α1f(G) + q1) and (α2f(G) + q2) is greater than f(G). For suppose not, we have i=12 (αif(G) + qi) ≤ 2f(G). But
    i=12(αif(G)+qi)=i=12αif(G)+i=12qi>i=12αif(G)+(1+α0)f(G)=(1+α1+α2+α0)f(G)=2f(G).
    This is a contradiction. By Case 1, we are done.If s = 2, then α1 + α2 = 2 − α0. Thus p1 + p2 = (j1 + j2 + 2)f(G) − α0f(G). This implies that j1 + j2 + 3 = 2k+1. Note that αif(G) + qi < 2f(G) for 1 ≤ i ≤ 2. We claim that α2f(G) + q2 > f(G) and α1f(G) + q1 > f(G). For suppose not, we have i=12 (αif(G) + qi) ≤ 3f(G). But
    i=12(αif(G)+qi)=i=12αif(G)+i=12qi>i=12αif(G)+(1+α0)f(G)=(1+α1+α2+α0)f(G)=3f(G).
    This is a contradiction. By the 2-pebbling property we can move at least j1 + 1 pebbles to the vertex (v1, x) of v1(G) and at least j2 + 1 pebbles to the vertex (v2, x) of v2(G) By Lemma 3.2, we are done.□By Theorems 2.1, 2.2 and 3.3, we have the following

Corollary 3.4

f(P2k2×P2k2)22k,f(P2k2×P2k+12)22k+2k.

Acknowledgement

Supported by the Natural Science Foundation of Anhui Higher Education Institutions of China (KJ2016A633, KJ2016B001) and the Quality Engineering Foundation of Huaibei Normal University (2018jpkc04).

References

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    F. Chung, Pebbling in hypercubes, SIAM J. Discrete Math. 2 (1989), no. 4, 461–472.

  • [2]

    L. Pachter, H.S. Snevily, and B. Voxman, On pebbling graphs, Congr. Numer. 107 (1995), 65–80.

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    Y.S. Ye, M.Q. Zhai, and Y. Zhang, Pebbling number of squares of odd cycles, Discrete Math. 312 (2012), no. 21, 3174–3178.

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    Y.S. Ye, P.F. Zhang, and Y. Zhang, The pebbling number of squares of even cycles, Discrete Math. 312 (2012), no. 21, 3203–3211.

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    H.S. Snevily and J.A. Foster, The 2-pebbling property and conjecture of Graham’s, Graphs Combin. 16 (2000), 231–244.

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    D.S. Herscovici and A.W. Higgins, The pebbling number of C 5 × C 5, Discrete Math. 189 (1998), 123–135.

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    D. Moews, Pebbling graphs, J. Combin. Theory Ser. B 55 (1992), 244–252.

  • [8]

    R.Q. Feng and J.Y. Kim, Graham’s pebbling conjecture on product of complete bipartite graphs, Sci. China Ser. A 44 (2001), no. 7, 817–822.

  • [9]

    R.Q. Feng and J.Y. Kim, Pebbling number of some graphs, Sci. China Ser. A 45 (2002), no. 4, 470–478.

If the inline PDF is not rendering correctly, you can download the PDF file here.

  • [1]

    F. Chung, Pebbling in hypercubes, SIAM J. Discrete Math. 2 (1989), no. 4, 461–472.

  • [2]

    L. Pachter, H.S. Snevily, and B. Voxman, On pebbling graphs, Congr. Numer. 107 (1995), 65–80.

  • [3]

    Y.S. Ye, M.Q. Zhai, and Y. Zhang, Pebbling number of squares of odd cycles, Discrete Math. 312 (2012), no. 21, 3174–3178.

  • [4]

    Y.S. Ye, P.F. Zhang, and Y. Zhang, The pebbling number of squares of even cycles, Discrete Math. 312 (2012), no. 21, 3203–3211.

  • [5]

    H.S. Snevily and J.A. Foster, The 2-pebbling property and conjecture of Graham’s, Graphs Combin. 16 (2000), 231–244.

  • [6]

    D.S. Herscovici and A.W. Higgins, The pebbling number of C 5 × C 5, Discrete Math. 189 (1998), 123–135.

  • [7]

    D. Moews, Pebbling graphs, J. Combin. Theory Ser. B 55 (1992), 244–252.

  • [8]

    R.Q. Feng and J.Y. Kim, Graham’s pebbling conjecture on product of complete bipartite graphs, Sci. China Ser. A 44 (2001), no. 7, 817–822.

  • [9]

    R.Q. Feng and J.Y. Kim, Pebbling number of some graphs, Sci. China Ser. A 45 (2002), no. 4, 470–478.

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