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

Yueqing Li; Yongsheng Ye

doi:10.1515/math-2020-0009

Article Open Access

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

Yueqing Li and Yongsheng Ye

Published/Copyright: March 10, 2020

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$Open Mathematics$

From the journal Open Mathematics Volume 18 Issue 1

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 .

Keywords: square of paths; pebbling number; pebbling move

MSC 2010: 05C70

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 2^d −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)∣, 2^D}, where ∣V(G)∣ is the number of vertices of G, and D is the diameter of G. Furthermore, f(K_n) = n and f(P_n) = 2ⁿ⁻¹ in [1], where K_n denotes a complete graph with n vertices and P_n 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

Let G be a connected graph. For u, v ∈ V(G), we denote by d_G(u, v) the distance between u and v in G. The k-th power of G, denoted by G^k, is the graph obtained from G by adding the edge uv to G whenever 2 ⩽ d_G(u, v) ⩽ k in G. That is, E(G^k) = {uv : 1 ⩽ d_G(u, v) ⩽ k}.

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

Definition 1.2

We call 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, where q is 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] Let P_n = 〈 v₁, v₂, ⋯, v_n 〉. If p(v₁) + 2p(v₂) + ⋯ + 2ⁱ⁻¹p(v_i) + ⋯ + 2ⁿ⁻²p(v_n−1) ⩾ 2ⁿ⁻¹, then one pebble can be moved to v_n from v₁.

Lemma 1.3

[3] Let P_2k+1 = 〈 v₁, v₂, …, v_2k+1〉. If P2k+12 has 2^k pebbles and v₁ has an even number of pebbles, then a pebble can be moved to v_2k+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 p͠(H) and p͠(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 P_n = 〈 v₁, v₂, ⋯, v_n〉. 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

P2k2 has the 2-pebbling property.

Proof

Suppose that 2^k+1 − q + 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.

p(v) = 1. Then P2k2 has 2^k+1 − q pebbles other than one pebble on v. Since 2^k+1 − q ≥ 2^k, by Lemma 1.1, we can move one additional pebble to v so that p͠(v) = 2.
p(v) = 0.
If q = 2k − 1, then p(P2k2) = 2^k+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 2^k+1 − 3k + 1 ≥ 2^k pebbles on P2k2 . By Lemma 1.1, we may move one additional pebble to v so that p͠(v) = 2.
New we assume q ≤ 2k − 2. Without loss of generality, we may assume that our target vertex v is not v_2k(otherwise, relabeling). We claim that p(v_2k) ≥ 2^k + 2^k−1 − q + 1 > 2^k. Otherwise,

p(P2k−12)≥2k+1−q+1−(2k+2k−1−q)=2k−1+1.

By Lemma 1.1, we can move one pebble to v. This leaves

2k+1−q+1−(2k−1+1)=2k+2k−1−q≥2k

pebbles on P2k2 . By Lemma 1.1, we can move one additional pebble to v so that p͠(v) = 2. Using 2^k pebbles on v_2k, we can move at least one pebble to v by Lemma 1.1. This leaves at least 2^k −q + 1 pebbles on P2k2 other than one pebble on v. Let p(P2k−12) = s and s ≤ 2^k−1. Thus p͠(v_2k) = 2^k − q − s + 1 ≥ 2. We can move 2k−1−⌈q+s−12⌉ pebbles to P2k−12 pebbles from v_2k so that P2k−12 has

s+2k−1−q+s−12=2k−1+s−q+12≥2k−1(s≥q−1)

pebbles other than one pebble on v and v_2k−1 has an even number of pebbles. By Lemma 1.3, we can move one additional pebble to v so that p͠(v) = 2.□

Theorem 2.2

P2k+12 has 2-pebbling property.

Proof

Suppose that 2^k+1 −q + 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 2^k+1 − q + 2 pebbles other than one pebble on v. Since 2^k+1 − q + 2 ≥ 2^k + 1, by Lemma 1.1, we can move one additional pebble to v so that p͠(v) = 2. Hence we assume p(v) = 0.

If q = 2k, then p(P2k+12) = 2^k+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 2^k+1 − 3k + 2 ≥ 2^k + 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 v_2k or v_2k+1 (otherwise, relabeling). We claim that p(v_2k) + p(v_2k+1) ≥ 2^k + 2^k−1 −q + 3 > 2^k + 1. Otherwise,

p(P2k−12)≥2k+1−q+3−(2k+2k−1−q+2)=2k−1+1.

By Lemma 1.1, we can move one pebble to v. This leaves 2^k+1 −q + 3 − (2^k−1 + 1) ≥ 2^k + 1 pebbles on P2k+12 . By Lemma 1.1, we can move one additional pebble to v so that p͠(v) = 2. Using 2^k + 1 pebbles on the set {v_2k, v_2k+1}, we can move at least one pebble to v by Lemma 1.1. This leaves at least 2^k − q + 2 pebbles on P2k+12 other than one pebble on v. Let p(P2k−12) = s and s ≤ 2^k−1. Thus p͠(v_2k) + p͠(v_2k+1) = 2^k − q − s + 3 ≥ 3. We can move 2k−1−⌈q+s−22⌉+1 pebbles to P2k−12 from v_2k and v_2k+1 so that

p(P2k−12)=s+2k−1−q+s−22+1=2k−1+s−q+22+1≥2k−1+1(s≥q−2)

pebbles other than one pebble on v. By Lemma 1.1, we can move one additional pebble to v so that p͠(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)|u∈V(G),v∈V(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 conjecture or any two graphs G and H, we have

f(G×H)≤f(G)f(H).

Lemma 3.1

[5] Let P_n be a path with n vertices and G be a graph with the 2-pebbling property. Then f(P_n × G) ≤ 2ⁿ⁻¹f(G).

Lemma 3.2

Let P_2k = 〈 v₁, v₂, ⋯, v_2k〉 (k ≥ 2) and p(v_i) = j_i (1 ≤ i ≤ 2k). If one of the following cases holds

j₁ + j₂ + 1 = 2^k and j₂ > 0;
j₁ + j₂ + 2 = 2^k, j₂ > 0 and p͠(v₁) = j₁ + 1;
j₁ + j₂ + 2 = 2^k, j₂ > 0 and p͠(v₂) = j₂ + 1;
j₁ + j₂ + 3 = 2^k, p͠(v₁) = j₁ + 1 and p͠(v₂) = j₂ + 1,

then we can move one pebble to the vertex v of P2k2 , where v ∈ {v₃, v₄, ⋯, v_2k}.

Proof

Let A = 〈 v₂, v₃, v₄, ⋯, v_2k〉. Then A² ≅ P2k−12 .

If j₁ is odd and j₂ is even, then we can move j1−12 pebbles to v₃ from v₁. Thus we have

p~(A2)=j1−12+j2=(j1+j2+1)+(j2−2)2≥2k−1.

By Lemma 1.3, we are done. If j₁ is even and j₂ is odd, then we can move j1−22 pebbles to v₃ and one pebble to v₂ from v₁ so that p͠(v₂) = j₂ + 1 and

p~(A2)=j1−22+j2+1=(j1+j2+1)+(j2−1)2≥2k−1.

By Lemma 1.3, we are done.
Let j₁ and j₂ be odd. If p͠(v₁) = j₁ + 1, then we can move one pebble to v₂ and j1−12 pebbles to v₃ from v₁ so that p͠(v₂) = j₂ + 1 and

p~(A2)=j2+1+j1−12=(j1+j2+2)+(j2−1)2≥2k−1.

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)+(j2−2)2≥2k−1.

By Lemma 1.3, we are done.
The proof is similar to (2).
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)+(j2−1)2≥2k−1.

And p͠(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 j1−12 pebbles to v₃ from v₁ so that p͠(v₂) = j₂ + 2 and

p~(A2)=j2+2+j1−12=(j1+j2+3)+j22≥2k−1.

By Lemma 1.3, we are done.□

Theorem 3.3

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

Proof

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

Suppose that 2^k+1 f(G) pebbles have been distributed arbitrarily on the vertices of P2k+22 × G. Let p_i = p(v_i(G)), where 1 ≤ i ≤ 2k + 2. Let q_i be the number of occupied vertices in v_i(G)(1 ≤ i ≤ 2k + 2). Suppose that v is our target vertex in P2k+22 × G. Suppose that v = (v_i, x) ∈ v_i(G) (3 ≤ i ≤ 2k + 2) for x ∈ V(G) (relabeling if necessary). For simplicity, let A = 〈 v₃, v₄, …, v_2k+1, v_2k+2〉. By the induction hypothesis, f(A² × G) ≤ 2^k f(G). We shall divide into the following two cases.

p₁ + p₂ ≤ (2^k+1 − 2)f(G). Then A² × G can obtain at least 12(2k+1−2)f(G)−2f(G) + 2f(G) = 2^kf(G) pebbles by a sequence of pebbling moves and we are done.
p₁ + p₂ ≥ (2^k+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 p_i = (j_i + α_i)f(G), where j_i ≥ 0 and 0 ≤ α_i < 1 (i = 1, 2). Now we claim that ∑i=12 q_i > (j₀ + α₀) f(G). Otherwise, we have ∑i=12 q_i ≤ (j₀ + α₀) f(G). So p͠(A² × G) ≥ 122k+1f(G)−(j0+α0)f(G)−(j0+α0)f(G) + (j₀ + α₀) f(G) = 2^kf(G) and we are done. Let ∑i=02 α_i = s ≤ 2. Then ∑i=02 j_i = 2^k+1 − s. Next, we discuss two following cases. Let B = 〈(v₂, x), (v₃, x), ⋯, (v_2k+2, x)〉. Then B² ≅ P2k+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 = (2^k+1 − j₀ − s + α₁)f(G). We move (2^k − 2)f(G) pebbles to v₃(G) so that p͠₃ ≥ (2^k − 2)f(G) using (2^k+1 − 4)f(G) pebbles on v₁(G), and (1 − α₂)f(G) to v₂(G) so that p͠₂ = f(G) using (2−2α₂)f(G) pebbles on v₁(G). So we can move one pebble to (v₂, x) and (2^k − 2) pebbles to (v₃, x) so that p͠(B²) = 2^k − 1. And the remaining (2 − j₀ − s + α₁ + 2α₂)f(G) pebbles on v₁(G), i.e., p͠₁ = (2 − j₀ − s + α₁ + 2α₂)f(G).

p͠₁ = (1 + α₁ + 2α₂)f(G). If α₀ ≤ α₂, then p͠₁ ≥ (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 p͠((v₂, x)) = 2. By Lemma 1.3, we are done. If α₀ > α₂, note that q₁ + q₂ > α₀f(G) and q₂ ≤ α₂f(G), then q₁ > (α₀ − α₂)f(G). Thus p͠₁ + 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 p͠((v₂, x)) = 2 and p͠(B²) = 2^k. By Lemma 1.3, we are done.
p͠₁ = (α₁ + 2α₂)f(G). If α₀ ≤ α₂, then p͠₁ ≥ (α₀ + α₁ + α₂)f(G) = sf(G) = 2f(G). If α₀ ≥ α₂, then q₁ > (α₀ − α₂)f(G). Thus p͠₁ + q₁ > (α₀ + α₁ + α₂)f(G) = 2f(G). By (1), we are done.
p͠₁ = (α₁ + 2α₂)f(G). Note that q₁ + q₂ > (1 + α₀)f(G) and q₂ ≤ α₂f(G), so q₁ > (1 + α₀ − α₂)f(G). Thus p͠₁ + 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.

α₀ = 0. Obviously, 0 ≤ s ≤ 1.

Suppose that j₀ = 0, p₁ + p₂ = 2^k+1f(G). By Lemma 3.1, we can move 2^k pebbles to the vertex (v₂, x) of v₂(G) so that p͠(B²) = 2^k. By Lemma 1.3, we are done. Suppose j₀ = 1, p₁ + p₂ = 2^k+1f(G) − f(G). If s = 0, then p₁ + p₂ = (j₁ + j₂)f(G). This implies that j₁ + j₂ + 1 = 2^k+1. For j₂ = 0, we can move at least (2^k − 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 j_i pebbles to the vertex (v_i, x) of v_i(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 = 2^k+1. For j₂ = 0, p₁ = (2^k+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 (α_if(G) + q_i) ≤ 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.
0 < α₀ < 1. Obviously, 1 ≤ s ≤ 2.

Suppose that j₀ = 0, p₁ + p₂ = 2^k+1f(G)-α₀ f(G). If s = 1, then α₁ + α₂ = 1 − α₀. Thus p₁ + p₂ = (j₁ + j₂ + 1)f(G) − α₀ f(G). This implies that j₁ + j₂ + 1 = 2^k+1. For j₂ = 0, p₁ = (2^k+1 − 1 + α₁)f(G). By Claim, we are done. For j₂ > 0, note that we can move at least j_i pebbles to the vertex (v_i, x) of v_i(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 = 2^k+1. For j₂ = 0, then p₁ = (2^k+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 (α_if(G) + q_i) ≤ 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 = 2^k+1f(G)−(1 + α₀ ) f(G) = (2^k+1 − 1)f(G) − α₀ f(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 = 2^k+1. For j₂ = 0, p₁ = (2^k+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 (α_if(G) + q_i) ≤ 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) − α₀ f(G). This implies that j₁ + j₂ + 3 = 2^k+1. Note that α_if(G) + q_i < 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 (α_if(G) + q_i) ≤ 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).

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Received: 2018-09-01

Accepted: 2020-01-18

Published Online: 2020-03-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2020-0009

Keywords for this article

square of paths; pebbling number; pebbling move

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