The optimal pebbling of spindle graphs

Ze-Tu Gao; Jian-Hua Yin

doi:10.1515/math-2019-0094

Article Open Access

The optimal pebbling of spindle graphs

Ze-Tu Gao and Jian-Hua Yin

Published/Copyright: November 13, 2019

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$Open Mathematics$

From the journal Open Mathematics Volume 17 Issue 1

Abstract

Given a distribution of pebbles on the vertices of a connected graph G, a pebbling move on G consists of taking two pebbles off one vertex and placing one on an adjacent vertex. The optimal pebbling number of G, denoted by π_opt(G), is the smallest number m such that for some distribution of m pebbles on G, one pebble can be moved to any vertex of G by a sequence of pebbling moves. Let P_k be the path on k vertices. Snevily defined the n–k spindle graph as follows: take n copies of P_k and two extra vertices x and y, and then join the left endpoint (respectively, the right endpoint) of each P_k to x (respectively, y), the resulting graph is denoted by S(n, k), and called the n–k spindle graph. In this paper, we determine the optimal pebbling number for spindle graphs.

Keywords: pebbling; optimal pebbling; spindle graph

MSC 2010: 05C35

1 Introduction

Graph pebbling was first introduced into the literature by Chung (see [1]). Pebbling has developed its own subfield (see [2]). Let G be a simple graph with vertex set V(G) and edge set E(G). Let D be a distribution of pebbles on the vertices of G, or a distribution on G. For any vertex v of G, D(v) denotes the number of pebbles on v in D. For S ⊆ V(G), we let D(S) = ∑_v∈S D(v) and |D| = ∑_v∈V(G) D(v). A pebbling move consists of removing two pebbles from one vertex and then placing one pebble at an adjacent vertex. For v ∈ V(G), v is reachable under distribution D if v has at least one pebble after some sequence of pebbling moves starting from D. A distribution D is solvable if all vertices of G are reachable under D. The pebbling number of G, denoted by π(G), is the smallest number m such that every distribution of m pebbles on G is solvable. The optimal pebbling number of G, denoted by π_opt(G), is the smallest number m such that some distribution of m pebbles on G is solvable. We say a distribution D is optimal if D has π_opt(G) pebbles and is solvable; that is, it is a solvable distribution of minimum size. We say a distribution D is smooth if it has at most two pebbles on each vertex of degree 2. A vertex v is unoccupied under a distribution D if D(v) = 0.

The optimal pebbling number of G was first introduced by Pachter, Snevily, and Voxman [3]. The optimal pebbling number has been determined for paths [3, 4], cycles [4], m-ary trees [5], caterpillars [6], and ladders [4]. Moews [7] used a probabilistic argument to show that the n-cube Qⁿ has π_opt(Qⁿ) = (4/3)^{n+O(log n)}. In [8], Xue and Yerger investigated the optimal pebbling number of grids and found the optimal pebbling number for the 3 by n grid. For graphs of diameter two (respectively, three), Muntz et al. [9] characterized the classes of graphs having π_opt(G) equal to a value between 2 and 4 (respectively, between 3 and 8). The lower and upper bounds on the optimal pebbling number were further studied in [4]. Milans and Clark [10] showed that computing optimal pebbling number is NP-hard on arbitrary graphs. Interestingly, exact values for optimal pebbling number are known only for paths, cycles, caterpillars, m-ary trees, ladders, and the 3 by n grid. A survey of results of optimal pebbling number can be found in [2].

Let P_k be the path on k vertices. Snevily defined the n–k spindle graph as follows: take n copies of P_k and two extra vertices x and y, and then join the left endpoint (respectively, the right endpoint) of each P_k to x (respectively, y), the resulting graph is denoted by S(n, k), and called the n–k spindle graph. In fact, the spindle graph S(n, k) is the graph that the vertices x and y are connected by n internally-disjoint paths of length k + 1 (Figure 1 is the graph S(3, k)). Snevily and Foster [11] proposed the following Problem 1.1, which appears to be quite difficult.

$Figure 1 S(3, k).$

Figure 1

S(3, k).

Problem 1.1

[11] Find π(S(3, k)).

By S(1, k) = P_k+2 and S(2, k) = C_2k+2, we have π(S(1, k)) = 2^k+1 and π(S(2, k)) = 2^k+1(see [4]). Recently, Gao and Yin [12] determined π(S(3, k)), which solves Problem 1.1.

The focus of this paper is to investigate the optimal pebbling number of S(n, k). By π_opt(C_n) = π_opt(P_n) = ⌈2n3⌉ (see [4]), we have πopt(S(1,k))=⌈2k+43⌉ and πopt(S(2,k))=⌈4k+43⌉. For n ≥ 3, we further determine π_opt(S(n, k)) in this paper. That is the following Theorem 1.1.

Theorem 1.1

Let n ≥ 3 and p ≥ 0, and denote ℓ = max{t|2^t ≤ n}.

If k < 2ℓ, then π_opt(S(n, k)) = 2^⌊k/2⌋ + 2^⌈k/2⌉;
If k = 2ℓ + 3p, then π_opt(S(n, k)) = 2^ℓ+1 + 2np;
If k = 2ℓ + 3p + 1, then π_opt(S(n, k)) = 2^ℓ+1 + 2^ℓ + 2np;
If k = 2ℓ + 3p + 2 and 2n ≥ 2^ℓ+1 + 2^ℓ–1, then π_opt(S(n, k)) = 2^ℓ+2 + 2np;
If k = 2ℓ + 3p + 2 and 2n < 2^ℓ+1 + 2^ℓ–1, then π_opt(S(n, k)) = 2^ℓ–1 + 2^ℓ + 2n(p + 1).

2 Lemmas

In order to prove Theorem 1.1, we need the following lemmas.

Let D be a distribution on G, and let H ⊆ G. The restriction of D to H is a pebble distribution D_H which is defined as follows: D_H(u) = D(u) if u ∈ V(H) and D_H(u) = 0 if u ∉ V(H). For convenience, we write A₁, A₂, …, A_n for the n copies of P_k in S(n, k). For 1 ≤ i ≤ n and 1 ≤ s < t ≤ k, we write A_i = u_i,1u_i,2 … u_i,k and A_i[s, t] = u_i,s u_i,s+1 … u_i,t.

Lemma 2.1

[4] If G is a connected n-vertex graph, with n ≥ 3, then G has a smooth optimal distribution with all leaves unoccupied.

Lemma 2.2

[3, 4] Let P_k be the path on k vertices. Then π_opt(P_k) = ⌈ 2k/3⌉.

The following Lemma gives an upper bound on π_opt(S(n, k)).

Lemma 2.3

Let n ≥ 3 and p ≥ 0, and denote ℓ = max{t|2^t ≤ n}.

If k < 2ℓ, then π_opt(S(n, k)) ≤ 2^⌊k/2⌋ + 2^⌈k/2⌉;
If k = 2ℓ + 3p, then π_opt(S(n, k)) ≤ 2^ℓ+1 + 2np;
If k = 2ℓ + 3p + 1, then π_opt(S(n, k)) ≤ 2^ℓ+1 + 2^ℓ + 2np;
If k = 2ℓ + 3p + 2 and 2n ≥ 2^ℓ+1 + 2^ℓ–1, then π_opt(S(n, k)) ≤ 2^ℓ+2 + 2np;
If k = 2ℓ + 3p + 2 and 2n < 2^ℓ+1 + 2^ℓ–1, then π_opt(S(n, k)) ≤ 2^ℓ–1 + 2^ℓ + 2n(p + 1).

Proof

Clearly, 2^ℓ ≤ n < 2^ℓ+1 and ℓ ≥ 1.

Assume that k < 2ℓ. Let D be a distribution such that D(x) = 2^⌊k/2⌋, D(y) = 2^⌈k/2⌉, and D(v) = 0 for each v ∈ V(S(n, k)) ∖ {x, y}. Then D is solvable and π_opt(S(n, k)) ≤ 2^⌊k/2⌋ + 2^⌈k/2⌉.
Assume that k = 2ℓ + 3p. Let D be a distribution such that D(x) = D(y) = 2^ℓ, D(u_i,j) = 2 for all i ∈ {1, 2, …, n}, j ∈ {ℓ + 2, …, ℓ + 3(p – 1) + 2}, and D(v) = 0 for each v ∈ V(S(n, k)) ∖ {x, y, u_1,ℓ+2, …, u_{1,(ℓ+3p–1)}, …, u_n,(ℓ+2), …, u_{n,(ℓ+3p–1)}}. It is clear to see that D is solvable. Then π_opt(S(n, k)) ≤ 2^ℓ+1 + 2np.
Assume that k = 2ℓ + 3p + 1. Let D be a distribution such that D(x) = 2^ℓ, D(y) = 2^ℓ+1, D(u_i,j) = 2 for all i ∈ {1, 2, …, n}, j ∈ {ℓ + 2, …, ℓ + 3(p – 1) + 2}, and D(v) = 0 for each v ∈ V(S(n, k)) ∖ {x, y, u_1,(ℓ+2), …, u_{1,(ℓ+3p–1)}, …, u_n,(ℓ+2), …, u_{n,(ℓ+3p–1)}}. Then D is solvable and π_opt(S(n, k)) ≤ 2^ℓ+1 + 2^ℓ + 2np.
Assume that k = 2ℓ + 3p + 2 and 2n ≥ 2^ℓ+1 + 2^ℓ–1. Let D be a distribution such that D(x) = D(y) = 2^ℓ+1, D(u_i,j) = 2 for all i ∈ {1, 2, …, n}, j ∈ {ℓ + 3, …, ℓ + 3p}, and D(v) = 0 for each v ∈ V(S(n, k)) ∖ {x, y, u_1,(ℓ+3), …, u_1,(ℓ+3p), …, u_n,(ℓ+3), …, u_n,(ℓ+3p)}. Then D is solvable and π_opt(S(n, k)) ≤ 2^ℓ+2 + 2np.
Assume that k = 2ℓ + 3p + 2 and 2n < 2^ℓ+1 + 2^ℓ–1. Let D be a distribution such that D(x) = 2^ℓ, D(y) = 2^ℓ–1, D(u_i,j) = 2 for all i ∈ {1, 2, …, n}, j ∈ {ℓ + 2, …, ℓ + 3p + 2}, and D(v) = 0 for each v ∈ V(S(n, k)) ∖ {x, y, u_1,(ℓ+2), …, u_1,(ℓ+3p+2), …, u_n,(ℓ+2), …, u_n,(ℓ+3p+2)}. Then D is solvable and π_opt(S(n, k)) ≤ 2^ℓ–1 + 2^ℓ + 2n(p + 1).

Let

g(n,k)=2⌊k/2⌋+2⌈k/2⌉ if k<2ℓ,2ℓ+1+2np if k=2ℓ+3p,2ℓ+1+2ℓ+2np if k=2ℓ+3p+1,min{2ℓ+2+2np,2ℓ−1+2ℓ+2n(p+1)} if k=2ℓ+3p+2.

Then π_opt(S(n, k)) ≤ g(n, k).

Lemma 2.4

Assume that n ≥ 3, k ≥ 2, and α₁ ≥ 0, α₂ ≥ 0, α₁ + α₂ ≤ k. Let D be a smooth solvable distribution on S(n, k). If there are at most 2^α₁ + β₁ pebbles can be put on x by a sequence of pebbling moves starting from D, and there are at most 2^α₂ + β₂ pebbles can be put on y by a sequence of pebbling moves, then |D| ≥ 2^α₁ + 2^α₂ + n⌈2(k – α₁ – α₂)/3⌉, where 0 ≤ β_i ≤ 2^α_i – 1 for i ∈ {1, 2}.

Proof

Clearly, for i ∈ {1, 2, …, n}, we can move at most one pebble to x from A_i as D is a smooth distribution. Similarly, we can move at most one pebble to y from A_i. Consider a smooth solvable distribution D₀ with D₀(x) = 2^α₁, D₀(y) = 2^α₂.

If |D₀| ≤ 2^α₁ + 2^α₂ + n⌈2(k – α₁ – α₂)/3⌉ – 1, then there are at most

|D0|−D0(x)−D0(y)≤n⌈2(k−α1−α2)/3⌉−1

pebbles on A₁[1 + α₁, k – α₂], …, A_n[1 + α₁, k – α₂]. For i ∈ {2, …, n}, by Lemma 2.2, we need at least ⌈2(k – α₁ – α₂)/3⌉ pebbles that ensure each vertex in A_i[1 + α₁, k – α₂] to be reachable. Thus, there is some vertex in A_i[1 + α₁, k – α₂] which is not reachable. Hence, |D₀| ≥ 2^α₁ + 2^α₂ + n⌈2(k – α₁ – α₂)/3⌉. Note that each pebbling move reduces the size of D. Therefore, |D| ≥ |D₀| ≥ 2^α₁ + 2^α₂ + n⌈2(k – α₁ – α₂)/3⌉.□

Lemma 2.5

Let a be a nonnegative integer, a = 3p + s, s ∈ {0, 1, 2}. Then 2^⌈a/2⌉ + 2^⌊a/2⌋ ≥ 4p + s + 2.

Proof

Firstly, assume that a = 2b. Then p = (2b – s)/3, and 2^⌈a/2⌉ + 2^⌊a/2⌋ = 2^b+1. Let f(t) = 2^t+1 – 2 – s – (8t – 4s)/3. If a = 0, then b = p = s = 0, and f(0) ≥ 0. If a = 2, then b = 1, p = 0, s = 2, and f(1) ≥ 0. We have that f′(t) = 2^t+1 ln 2 – 8/3 > 0 for t ≥ 1. Hence, f(t) > f(1) ≥ 0.

Now, we assume that a = 2b + 1. Then p = (2b + 1 – s)/3, and 2^⌈a/2⌉ + 2^⌊a/2⌋ = 3 × 2^b. If a = 1, then b = p = 0, s = 1, and 2^⌈a/2⌉ + 2^⌊a/2⌋ ≥ 4p + s + 2. Let f(t) = 3 × 2^t – 2 – s – (8t + 4 – 4s)/3. If a = 3, then b = p = 1, s = 0, and f(1) ≥ 0. We get f′(t) = 3 × 2^t ln 2 – 8/3 > 0 for t ≥ 1. Thus, f(t) > f(1) ≥ 0. Therefore, 2^⌈a/2⌉ + 2^⌊a/2⌋ ≥ 4p + s + 2.

3 Proof of Theorem 1.1

Proof of Theorem 1.1

Clearly, S(n, 1) is isomorphic to K_2,n, and π_opt(K_2,n) = 3. We assume k ≥ 2. By Lemma 2.3, we have that π_opt(S(n, k)) ≤ g(n, k). We now show that π_opt(S(n, k)) ≥ g(n, k). By Lemma 2.1, we can assume that D is a smooth optimal distribution. Now, we assume that there are at most 2^α₁ + β₁ pebbles can be put on x by a sequence of pebbling moves starting from D, and there are at most 2^α₂ + β₂ pebbles can be put on y by a sequence of pebbling moves, where α₁ ≥ 0, α₂ ≥ 0, 0 ≤ β_i ≤ 2^α_i – 1 for i ∈ {1, 2}. Thus, we assume D(x) = 2^α₁ + β₁ – m₁ and D(y) = 2^α₂ + β₂ – m₂, where 0 ≤ m_i ≤ min{n, 2^α_i + β_i} for i ∈ {1, 2}. Note that 2^ℓ ≤ n < 2^ℓ+1 and ℓ ≥ 1. Clearly, a ≥ min{a, b} for a and b are real numbers. We consider the following two cases.

α₁ + α₂ > k.

Then |D| = D(x) + D(y) + ∑i=1n D(A_i) ≥ 2^α₁ + β₁ – m₁ + 2^α₂ + β₂ – m₂ + m₁ + m₂ ≥ 2^α₁ + 2^α₂. Hence, |D| ≥ 2^α₁ + 2^α₂ ≥ 2^{⌈(α₁+α₂)/2⌉} + 2^{⌊(α₁+α₂)/2⌋} > 2^⌈k/2⌉ + 2^⌊k/2⌋ since 2^a + 2^b ≥ 2^a–1 + 2^b+1 for all integers satisfying a > b.

Assume that k = 2ℓ + 3p + s, s ∈ {0, 1, 2}. By Lemma 2.5, we have that

|D|>2⌈k/2⌉+2⌊k/2⌋=2⌈(2ℓ+3p+s)/2⌉+2⌊(2ℓ+3p+s)/2⌋=2ℓ(2⌈(3p+s)/2⌉+2⌊(3p+s)/2⌋)≥2ℓ(4p+s+2).

If s = 0, then |D| > (4p + 2)2^ℓ > 2^ℓ+1 + 2np. If s = 1, then |D| > 4p × 2^ℓ + 3 × 2^ℓ > 2^ℓ+1 + 2^ℓ + 2np. If s = 2, then |D| > 4p × 2^ℓ + 4 × 2^ℓ > 2^ℓ+2 + 2np ≥ min{2^ℓ+2 + 2np, 2^ℓ–1 + 2^ℓ + 2n(p + 1)}. Thus, |D| = π_opt(S(n, k)) > g(n, k). This is impossible as it would be greater than the known upper bound.
α₁ + α₂ ≤ k.

Denote ω = α₁ + α₂. By Lemma 2.4, we get

|D|≥2α1+2α2+n⌈2(k−ω)/3⌉≥2⌈ω/2⌉+2⌊ω/2⌋+n⌈2(k−ω)/3⌉ (3.1)

as 2^a + 2^b ≥ 2^a–1 + 2^b+1. Now we consider the following three subcases.
k < 2ℓ.

If ω = k, by (3.1), then |D| ≥ 2^⌈k/2⌉ + 2^⌊k/2⌋. If ω = k – 1, by (3.1), then

|D|≥2⌈ω/2⌉+2⌊ω/2⌋+n⌈2(k−ω)/3⌉≥2⌈(k−1)/2⌉+2⌊(k−1)/2⌋+2ℓ≥2⌈(k−1)/2⌉+2⌊(k−1)/2⌋+2⌈k/2⌉≥2⌈k/2⌉+2⌊k/2⌋.

If ω ≤ k – 2, then

|D|≥2⌈ω/2⌉+2⌊ω/2⌋+n⌈2(k−ω)/3⌉≥2⌈ω/2⌉+2⌊ω/2⌋+2×2ℓ≥2⌈k/2⌉+2⌊k/2⌋.
k = 2ℓ + 3p + s and 0 ≤ ω ≤ 2ℓ + s, where s ∈ {0, 1, 2}.

Let ω = 2ℓ + s – j, 0 ≤ j ≤ 2ℓ + s. By (3.1), we get

|D|≥2⌈ω/2⌉+2⌊ω/2⌋+n⌈(6p+2j)/3⌉≥2⌈(2ℓ+s−j)/2⌉+2⌊(2ℓ+s−j)/2⌋+2np+n⌈2j/3⌉. (3.2)

If s = 0 and 0 ≤ j ≤ 1, by (3.2), then |D| ≥ 2^ℓ+1 + 2np. If s = 0 and j ≥ 2, then

|D|≥2⌈(2ℓ−j)/2⌉+2⌊(2ℓ−j)/2⌋+2np+2ℓ⌈2j/3⌉>2ℓ+1+2np.

If s = 1 and j ≤ 3, then

|D|≥2⌈(2ℓ+1−j)/2⌉+2⌊(2ℓ+1−j)/2⌋+2np+2ℓ⌈2j/3⌉≥2ℓ+1+2ℓ+2np.

If s = 1 and j ≥ 4, then

|D|≥2⌈(2ℓ+1−j)/2⌉+2⌊(2ℓ+1−j)/2⌋+2np+2ℓ⌈2j/3⌉>2ℓ+1+2ℓ+2np.

Assume that s = 2. If j = 0, then |D| ≥ 2^ℓ+2 + 2np ≥ min {2^ℓ+2 + 2np, 2^ℓ–1 + 2^ℓ + 2n(p + 1)}. If j = 1, then |D| ≥ 2^ℓ+1 + 2^ℓ + 2np + n ≥ 2^ℓ+2 + 2np ≥ min {2^ℓ+2 + 2np, 2^ℓ–1 + 2^ℓ + 2n(p + 1)}. If 2 ≤ j ≤ 3, then |D| ≥ 2^ℓ + 2^ℓ–1 + 2np + 2n ≥ min{2^ℓ+2 + 2np, 2^ℓ–1 + 2^ℓ + 2n(p + 1)}. If j = 4, then |D| ≥ 2^ℓ–1 + 2^ℓ–1 + 2np + 3n ≥ 2^ℓ+2 + 2np ≥ min {2^ℓ+2 + 2np, 2^ℓ–1 + 2^ℓ + 2n(p + 1)}. If j ≥ 5, then |D| ≥ 2np + 4n ≥ 2^ℓ+2 + 2np ≥ min {2^ℓ+2 + 2np, 2^ℓ–1 + 2^ℓ + 2n(p + 1)}.
k = 2ℓ + 3p + s and 2ℓ + s < ω ≤ 2ℓ + 3p + s, where s ∈ {0, 1, 2}.

Note that p ≥ 1. Let ω = 2ℓ + 3p + s – j, 0 ≤ j ≤ 3p – 1. By (3.1), we have

|D|≥2ℓ(2⌈(3p+s−j)/2⌉+2⌊(3p+s−j)/2⌋)+n⌈2j/3⌉. (3.3)

Assume that 3p + s – j = 1. Then |D| ≥ 3 × 2^ℓ + n⌈2j/3⌉ = 2^ℓ+1 + 2^ℓ + 2np + n⌈(2s – 2)/3⌉. If s = 0, then |D| > 2^ℓ+1 + 2np. If s = 1, then |D| ≥ 2^ℓ+1 + 2^ℓ + 2np. If s = 2, then |D| ≥ 2^ℓ+1 + 2^ℓ + 2np + n > 2^ℓ+2 + 2np.

Assume that 3p + s – j = 2. Then |D| ≥ 2 × 2^ℓ+1 + n⌈2j/3⌉ = 2^ℓ+2 + 2np + n⌈(2s – 4)/3⌉. If s = 0, then |D| > 2^ℓ+1 + 2np. If s = 1, then |D| > 2^ℓ+1 + 2^ℓ + 2np. If s = 2, then |D| ≥ 2^ℓ+2 + 2np.

Assume that 3p + s – j = 3. Then |D| ≥ 6 × 2^ℓ + n⌈2j/3⌉ = 2^ℓ+2 + 2^ℓ+1 + 2np + n⌈(2s – 6)/3⌉. If s = 0, then |D| > 2^ℓ+1 + 2np. If s = 1, then |D| ≥ 2^ℓ+1 + 2^ℓ + 2np. If s = 2, then |D| > 2^ℓ+2 + 2np.

Assume that 3p + s – j = 4. Then |D| ≥ 2^ℓ+3 + 2np + n⌈(2s – 8)/3⌉. If s = 0, then |D| > 2^ℓ+1 + 2np. If s = 1, then |D| > 2^ℓ+1 + 2^ℓ + 2np. If s = 2, then |D| > 2^ℓ+2 + 2np.

Assume that 3p + s – j = 2c and c ≥ 3. By (3.3), and

p+s+⌈2j/3⌉−j≥(j+1−s)/3+s+⌈2j/3⌉−j=⌈2j/3⌉−2j/3+(1+2s)/3>0,

we have that

|D|≥2ℓ(2⌈(3p+s−j)/2⌉+2⌊(3p+s−j)/2⌋)+n⌈2j/3⌉=2ℓ+12c+n⌈2j/3⌉≥2ℓ+1(2+2c)+n⌈2j/3⌉=2ℓ+2+(3p+s−j)2ℓ+1+n⌈2j/3⌉>2ℓ+2+(3p+s−j)n+n⌈2j/3⌉≥2ℓ+2+2np+n(p+s+⌈2j/3⌉−j)>2ℓ+2+2np.

Now, we assume that 3p + s – j = 2c + 1 and c ≥ 2. By (3.3), and

p+3s+1−3j4+⌈2j/3⌉≥j+1−s12+3s+1−3j4+⌈2j/3⌉=⌈2j/3⌉−2j/3+1+2s3>0,

we have that

|D|≥2ℓ(2⌈(3p+s−j)/2⌉+2⌊(3p+s−j)/2⌋)+n⌈2j/3⌉=3×2ℓ×2c+n⌈2j/3⌉≥3×2ℓ(2+c)+n⌈2j/3⌉=2ℓ+2+2ℓ+1+3c×2ℓ+n⌈2j/3⌉>2ℓ+2+n(1+3c/2+⌈2j/3⌉)>2ℓ+2+n(1+9p+3s−3j−34+⌈2j/3⌉)>2ℓ+2+2np.

Therefore, π_opt(S(n, k)) ≥ g(n, k). The proof of Theorem 1.1 is completed.□

Acknowledgement

The authors are grateful to the referees for their helpful suggestions and comments. This research was supported by National Natural Science Foundation of China (Nos. 11461017, 11961019, 11561017).

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Received: 2018-10-26

Accepted: 2019-08-29

Published Online: 2019-11-13

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

Articles in the same Issue

https://doi.org/10.1515/math-2019-0094

Keywords for this article

pebbling; optimal pebbling; spindle graph

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