Pebbling on crisscross sequence of m complete graphs


Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3s (2025) 

 

648 https://internationalpubls.com 

Pebbling on Crisscross Sequence of 𝒎 Complete Graphs 
 

1J.Jenifer Steffi, 2M Gayathri Lakshmi, 3Dr. D. Vamsi  Priya, 4A.K.Bhuvaneswari , 
5M.Kannan,  6J. Juli Amala Rani, 7abDr.M.Elangovan  

1Malla Reddy Engineering College, Hyderabad,  Telangana, India. 1Email ID:drjenifersteffi@gmail.com 
2Assistant professor,Mathematics,Saveetha engineering college,Chennai. 2Email ID:gayathrilakshmi1804@gmail.com 

3Associate Professor,Basic Science and Humanities,Vignan’s Institute of Information and Technology,Duvvada, 

Visakhapatnam. 3Email ID: vamsipriyabagi@gmail.com 
4Associate Professor, Department of  Mathematics, Aarupadai Veedu Institute of Technology, VMRF(DU),Paiyanoor 

4Email ID:bhuvanabalaji13@gmail.com 
5Assistant professor, Department of mathematics,Sardar Vallabhbhai Patel International School of Textiles and 

Management, coimbatore. 5Email ID:kannan8383@gmail.com 
6Assistant Professor, Department of Mathematics,Panimalar Engineering College,Chennai. 

6Email ID: jesujuli@gmail.com 
7aDepartment of Biosciences, Saveetha School of Engineering. Saveetha Institute of Medical and Technical Sciences, 

Chennai - 602 105 
7bApplied Science Research Center. Applied Science Private University, Amman, Jordan.  

7abEmail ID:muniyandy.e@gmail.com 

 

Article History: 

Received: 26-09-2024 

Revised: 14-11-2024 

Accepted: 29-11-2024 

Abstract:  

This paper investigates the pebbling number, the two pebbling property, the t-pebbling 

number, and the 2t-pebbling property of a crisscross sequence comprised of m complete 

graphs. By analyzing these pebbling properties, we aim to gain insights into the efficiency 

and feasibility of pebbling operations within this specific graph structure. 

Keywords: Graph pebbling, Graph theory, crisscross sequence of m complete graphs, 

Pebbling properties, Graph Optimization 

 

Introduction 

Graph pebbling is a mathematical concept that involves manipulating configurations of pebbles on the 

vertices of graph in order to achieve a specific target vertex with the desired amount of pebbles. 

Initially, Lagarias and Saks introduced graph pebbling while attempting to answer a number theoretic 

question posed by Erdos and Lemke  concerning zero-sum sequences of finite group. The graph 

pebbling concept was formally introduced by Chung , who defined the pebbling number 𝑓(𝐺) for 

connected graph 𝐺. [11][12]Since then, the field of graph pebbling has become highly active, with 

numerous open problems and conjectures awaiting resolution. In the context of graph pebbling , 

pebbling move involves taking two pebbles from one vertex and placing one of them on a neighboring 

vertex. The second pebble is disregarded. If there are 𝑤 pebbles distributed among the vertices of graph 

𝐺, the distribution is considered solvable, if it is possible to manipulate the pebbles through a series of 

moves so that any given vertex 𝑣 ends up with at least one pebble. On the other hand, if the distribution 

cannot be solved, it is referred to as unsolvable. The pebbling number , denoted as 𝑓(𝐺),- represents 

the smallest value of 𝑚 for which all initial distributions of 𝑚 pebbles on the graph 𝐺 can be solved. 

The 𝑡 −pebbling number , 𝑓𝑡(𝐺), of a connected graph 𝐺, represents the minimum positive integer 

such that given distribution of 𝑓𝑡(𝐺) pebbles, it is possible to move 𝑡 pebbles to the chosen target 

vertex by performing series of pebbling moves. The connected graph 𝐺 said to possess the 2-pebbling 

property  if for any distribution of pebbles in 𝐺 where the number of pebbles exceeds 2𝑓(𝐺) − 𝑞, 

where 𝑞, the number of vertices with at least one pebble, it is feasible, through the execution of 

mailto:gayathrilakshmi1804@gmail.com
mailto:vamsipriyabagi@gmail.com
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mailto:Email%3Akannan8383@gmail.com
mailto:jesujuli@gmail.com
mailto:muniyandy.e@gmail.com


Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3s (2025) 

 

649 https://internationalpubls.com 

pebbling moves, to ensure the presence of two pebbles at the given vertex within 𝐺. [15]A graph 𝐺 

considered to possess the 2𝑡 −pebbling property  if for any distribution of pebbles in 𝐺 where the 

number of pebbles exceeds 2𝑓𝑡(𝐺) − 𝑞, through the execution of pebbling moves, it is possible to 

move 2𝑡 pebbles to any arbitrary vertex in 𝐺. In section 2, we calculate the pebbling number and 

𝑡 −pebbling number of crisscross sequence of 𝑚 complete graph and in section 3 and section 4, we 

demonstrate the graph 𝐶𝑚(𝐾𝑛) exhibits both 2-pebbling property and 2𝑡 −pebbling property[12]. 

Pebbling on crisscross sequence of 𝒎 complete graphs 

In this section, we aim to determine the pebbling number and the 𝑡 −pebbling number of crisscross 

chain graph consisting 𝑚 complete graphs. However, before getting into the calculations, it is crucial 

to comprehend the structure of crisscross chain graph of 𝑚 complete graph. To aid in this 

understanding, we introduce 𝐶𝑚(𝐾𝑛) with the help of zig-zag sequence of 𝑛 cycles which has been 

already studied in [8],[9],[10].  

Definition 1. [14][15] The zig-zag chain graph of even cycles denoted by 𝑍𝑍𝑛(𝐶2𝑘), is a graph which 

consists of zig-zag sequence of 𝑛 even cycles, 𝐶2𝑘 with 𝑘 ≥ 3. We have the following vertex set and 

edge set of 𝑍𝑍𝑛(𝐶2𝑘) for 𝑛 even as follows. 

𝑉(𝑍𝑍𝑛(𝐶2𝑘)) = {𝑎𝑖, 𝑏𝑖: 1 ≤ 𝑖 ≤ 𝑛(𝑘 − 1)} ∪ {𝑥, 𝑦} and 

𝐸(𝑍𝑍𝑛(𝐶2𝑘)) = {𝑎𝑖𝑎𝑖+1, 𝑏𝑖𝑏𝑖+1:1 ≤≤ 𝑛(𝑘 − 1) − 1} ∪ {𝑎(𝑘+1)𝑖−1𝑏(𝑘+1)𝑖−2, 𝑎(𝑘+1)𝑗𝑏(𝑘+1)𝑗+1:1 ≤ 𝑖 ≤
𝑛

2
, 

1 ≤ 𝑗 ≤ (
𝑛

2
− 1)} ∪ {𝑥𝑎1, 𝑥𝑏1, 𝑦𝑎𝑛(𝑘−1), 𝑦𝑏𝑛(𝑘−1)} 

For 𝑛 odd, we have the following vertex set and edge set. 

𝑉(𝑍𝑍𝑛(𝐶2𝑘)) = {𝑎𝑖, 𝑏𝑖: 1 ≤ 𝑖 ≤ 𝑛(𝑘 − 1)} ∪ {𝑥, 𝑦} and 

𝐸(𝑍𝑍𝑛(𝐶2𝑘)) = {𝑎𝑖𝑎𝑖+1, 𝑏𝑖𝑏𝑖+1: 1 ≤ 𝑖 ≤ 𝑛(𝑘 − 1) − 1} ∪ 

{𝑥𝑎1, 𝑥𝑏1, 𝑦𝑎𝑛(𝑘−1), 𝑦𝑏𝑛(𝑘−1)} ∪ {𝑎(𝑘+1)𝑖−1𝑏(𝑘+1)𝑖−2, 𝑎(𝑘+1)𝑗𝑏(𝑘+1)𝑗+1: 1 ≤ 𝑖, 𝑗 ≤
𝑛−1

2
}. 

The Figure 2.1(𝑎) depicts the graph 𝑍𝑍3 for 𝑘 = 3. 

 

Figure 2.1(a) 

Similarly, for the graph zig-zag chain graph of odd cycles, denoted by 𝑍𝑍𝑛(𝐶2𝑘+1) has defined in[4]. 

Definition 2.[5],[7],[9],[10].  The crisscross sequence of 𝑚 complete graphs is denoted by 𝐶𝑚(𝐾𝑛), is 

a graph which consists of crisscross sequence of 𝑚 copies of 𝐾𝑛, with 𝑛 ≥ 5. We define the graph 

𝐶𝑚(𝐾𝑛) as follows: The vertex set of 𝐶𝑚(𝐾𝑛) is 𝑉(𝐶𝑚(𝐾𝑛)) = 𝑉(𝑍𝑍𝑚(𝐶𝑛)) and the edge set of 

𝐶𝑚(𝐾𝑛) is the union of 𝐸(𝑍𝑍𝑚(𝐶𝑛)) and each vertex 𝑣 ∈ 𝐶𝑖 is adjacent to all the vertices of 𝐶𝑖. Here 

𝐶1, 𝐶2, . . . , 𝐶𝑚 is considered as 𝑚 number of cells of 𝐶𝑚(𝐾𝑛). The figure depicts the the crisscross 

sequence of 𝑚 copies of 𝐾6. 



Communications on Applied Nonlinear Analysis 

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Figure 2.2(b) 

The reader can easily view that the Figure 2.2(b) has 4 copies of 𝐾6. For 𝐶𝑚(𝐾𝑛) we have 𝑚 copies of 

𝐾6 and we label each 𝐾6 as 𝐶1, 𝐶2, . . . , 𝐶𝑚 in order from left to right. 

Pebbling number of crisscross sequence of 𝒎 complete graphs 

Theorem 1.  The pebbling number of the graph 𝐶2(𝐾𝑛) is, 𝑓(𝐶2(𝐾𝑛)) = 2𝑛 − 2. 

Proof. Put 3 pebbles on the vertex 𝑥 and one pebble each on vertices 𝑎𝑖 and 𝑏𝑖, excluding vertices 𝑎𝑘 

and 𝑏𝑘−1 where 𝑎𝑘, 𝑏𝑘 ∈ 𝑉(𝐶1) ∩ 𝑉(𝐶2). We are unable to transport a single pebble to the vertex 𝑦. So 

we have 𝑓(𝐶2(𝐾𝑛)) ≥ 2𝑛 − 2. 

Think of a graph where the vertices are covered in at least 2𝑛 − 2 pebbles. Suppose 𝑣 is any target 

vertex. Let 𝑝(𝑣) = 0. We consider the following scenarios: 

Case (1) 𝑣 = 𝑎𝑘 or 𝑣 = 𝑏𝑘−1. 

Take the supposition that 𝑣 = 𝑎𝑘 without losing generality. Consider the case where 𝑝(𝐶1) ≥ 𝑛 or 

𝑝(𝐶2) ≥ 𝑛. After that we may move a pebble to 𝑎𝑘. Considering the way 𝐶1 ≅ 𝐶2 ≅ 𝐾𝑛. Also 𝑓(𝐾𝑛) =
𝑛. Assume that 𝑝(𝐶1) ≥ 𝑛 − 1 and 𝑝(𝐶2) ≥ 𝑛 − 1. If 𝑝(𝑉(𝐶1) − {𝑎𝑘}) ≥ 𝑛 − 1, at least two of the 

pebbles are present in one of the vertices of 𝑉(𝐶2) − {𝑎𝑘, 𝑏𝑘−1}. Assume that 𝑝(𝑦) ≥ 2 to maintain the 

generality. Due to the fact that 𝑦 is close to 𝑎𝑘, we can transfer a pebble there. 

Case (2) 𝑣 ∈ 𝑉(𝐶2) − {𝑎𝑘} or 𝑣 ∈ 𝑉(𝐶1) − {𝑏𝑘−1}. 

Take 𝑣 = 𝑦 and suppose that 𝑣 ∈ 𝑉(𝐶2) − {𝑎𝑘}. If 𝑝(𝐶2) ≥ 𝑛, then we can reach the vertex 𝑦 with one 

pebble. Given that 𝐶2 ≅ 𝐾𝑛. Thus we suppose that 𝑝(𝐶2) < 𝑛. Assume that 𝑝(𝑉(𝐶2) − {𝑎𝑘}) ≥ 𝑛 − 1. 

After that, we may move a pebble to the vertex 𝑦. Due to the fact that the graph induced by the vertices 

𝑉(𝐶2) − {𝑎𝑘} is isomorphic to 𝐾𝑛−1. After that, we presumptively know that 𝑝(𝑉(𝐶2) − {𝑎𝑘}) ≤ 𝑛 −
2. We consider the following subcases: 

Subcase (1a) 𝑝(𝑉(𝐶2) − {𝑎𝑘}) = 𝑛 − 2. 

Based on our assumptions, we obtain 𝑝(𝑉(𝐶1 − {𝑏𝑘−1}) = 𝑛. Then, at least any of the vertices 

belonging to 𝑉(𝐶1) − {𝑏𝑘−1} contain at least two pebbles. Assume that 𝑝(𝑥) ≥ 2 and we can move 

one pebble to 𝑏𝑘−1. This is because 𝑥 is adjacent to 𝑏𝑘−1. The subgraph induced by the vertices 𝑉(𝐶2) −
{𝑎𝑘} contains 𝑛 − 1 pebbles. Therefore, the pebble can be moved to 𝑦. The subgraph induced by 

vertices 𝑉(𝐶2) − {𝑎𝑘} is isomorphic to 𝑘𝑛−1. 

Subcase (1b) 𝑝(𝑉(𝐶2) − {𝑎𝑘}) < 𝑛 − 2. 



Communications on Applied Nonlinear Analysis 

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By assumption, 𝑝(𝑉(𝐶1) − {𝑏𝑘−1}) ≥ 𝑛 + 1 and any of the vertices in 𝑉(𝐶1) − {𝑏𝑘−1} has more than 

one pebble. Using exactly two pebbles, we can reach vertex 𝑎𝑘 with one pebble. Thus, the number of 

pebbles retained in 𝑉(𝐶1) − {𝑏𝑘−1} was at least 𝑛 − 1. We can then move an additional pebble to 𝑎𝑘. 

Therefore, one pebble can be moved to 𝑦 from 𝑎𝑘. 

Theorem 2.  The pebbling number of the graph 𝐶3(𝐾𝑛) is 𝑓(𝐶3(𝐾𝑛)) = 3𝑛 − 2. 

Proof. Put seven pebbles on vertex 𝑥 and put one pebble on each 𝑎𝑖′𝑠 and 𝑏𝑖′𝑠 except 

{𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}. Then, we cannot move the pebble to 𝑦. Thus, 𝑓(𝐶3(𝐾𝑛)) ≥ 3𝑛 − 2. Consider 

a graph with 3𝑛 − 2 pebbles distributed at its vertices. Let 𝑣 be a target vertex. Clearly 𝑣 ∈ 𝐶𝑖, for any 

𝑖 = 1,2,3. We consider the following cases. 

Case(1)  

If 𝑝(𝐶3) ≥ 𝑛 + 2, then the two pebbles can be moved to 𝑎2𝑘−2 and one pebble to 𝑣. 𝐶3 ≅ 𝐾𝑛 and the 

pebbling number of a complete graph 𝐾𝑛 is 𝑛. Therefore, we assume 𝑝(𝐶3) ≤ 𝑛 + 1. This implies that 

at least 2𝑛 − 3 pebbles were retained in 𝐶1 ∩ 𝐶2. If 𝑝(𝑉(𝐶3) − {𝑎2𝑘−2, 𝑏2𝑘−1}) ≥ 𝑛. Then, the two 

pebbles can be moved to 𝑦. We can reach vertex 𝑎𝑘+1 using one pebble. This results in the subgraph 

𝐶1 ∩ 𝐶2 obtaining 2𝑛 − 2 pebbles after moving one pebble to 𝑎𝑘+1. By Theorem 1 we move the pebble 

to 𝑣. 

Case (2) 𝑣 ∈ 𝐶1 or 𝑣 ∈ 𝐶3. 

Without loss of generality, assume that 𝑣 ∈ 𝐶3 and let us assume 𝑣 = 𝑦. If 𝑝(𝑉(𝐶3) −
{𝑎2𝑘−2, 𝑏2𝑘−1}) ≥ 𝑛 − 2, then we can move the pebble to 𝑦. The graph is induced by vertices 

⟨𝑉(𝐶3) − {𝑎2𝑘−2, 𝑏2𝑘−1}⟩ ≅ 𝐾𝑛−2. Therefore, we assume 𝑝(𝑉(𝐶3) − {𝑎2𝑘−2, 𝑏2𝑘−1}) ≤ 𝑛 − 3. Thus, 

the number of pebbles retained in 𝑉(𝐶1 ∩ 𝐶2) is at least 2𝑛 + 1. We may take 𝑝(𝑉(𝐶2) − {𝑎𝑘, 𝑏𝑘−1}) ≥
𝑛 − 2. We can then move the pebble to 𝑎𝑘. Additionally, the pebbles retained in 𝐶1 are at least 𝑛 + 3. 

In this case, we can move the two pebbles to 𝑎𝑘 and move an additional pebble to 𝑎2𝑘−2. Thus, the 

pebble can be moved to 𝑦. Therefore, we take 𝑝(𝐶2) ≤ 𝑛 − 4, 𝑝(𝑎2𝑘−2) = 0 and 𝑝(𝑏2𝑘−1) = 0. This 

implies that the number of pebbles retained in 𝑉(𝐶1) was at least 𝑛 + 5. Suppose 𝑝(𝑎𝑘) ≥ 2 or 

𝑝(𝑏𝑘−1) ≥ 2. Then, we can move one pebble to 𝑎2𝑘−2 and use exactly two pebbles from 𝑎𝑘 or 𝑏𝑘−1. 

In addition, 𝑛 + 3 pebbles settled in 𝐶1. Because 𝐶1 ≅ 𝐾𝑛 and by using the pebbling number of 

complete graph we can move two pebbles to 𝑎𝑘 and move an additional pebble to 𝑎2𝑘−2. We can then 

reach vertex 𝑦 with one pebble. We assume that 𝑝(𝑎𝑘) ≤ 1 and 𝑝(𝑏𝑘−1) ≤ 1. If 𝑝(𝑎𝑘) = 1, at least𝑛 −
3 pebbles are distributed on 𝑛 − 1 vertices. Thus, using the pigeonhole principle, we conclude that at 

least one of those vertices must contain two pebbles, and we can then move an additional pebble to 𝑎𝑘. 

After moving a pebble to 𝑎𝑘 we obtain that 𝑝(𝑉(𝐶1) − {𝑎𝑘}) is at least 𝑛 + 1. We assume that 𝑝(𝑦) =
1. Then, 𝑛 pebbles are settled in 𝑉(𝐶1) − {𝑎𝑘, 𝑏𝑘−1} with 𝑛 − 1 vertices. Again, using the pigeonhole 

principle, we can move the two pebbles to 𝑎2𝑘−2 from 𝑎𝑘 and 𝑏𝑘−1. Therefore, we assume that 𝑝(𝑎𝑘) =
0 and 𝑝(𝑏𝑘) = 0. Then, all 𝑛 + 4 pebbles are distributed only on 𝑉(𝐶1) − {𝑎𝑘, 𝑏𝑘−1}. Because the 

graph induced by vertex set ⟨𝑉(𝐶1) − {𝑎𝑘, 𝑏𝑘−1}⟩ is isomorphic to 𝐾𝑛−2, by referring the pebbling 

number of complete graph we can move four pebbles to 𝑎𝑘 and move two pebbles to 𝑎2𝑘−2. Then, we 

move the pebble to 𝑦 from 𝑎2𝑘−2. 

Lemma 1.  Let 𝐺 = 𝐶𝑚(𝐾𝑛) be a crisscross sequence of 𝑚 complete graphs. Let us define 𝑋1 =
𝐶𝑚1

(𝐾𝑛) = 𝐶1 ∪. . .∪ 𝐶𝑚1
 and 𝑋2 = 𝐶𝑚2

(𝐾𝑛) = 𝐶𝑚1+1 ∪. . .∪ 𝐶𝑚 be two subgraphs of 𝐺 where 𝑚1 +

𝑚2 = 𝑚. Suppose that the number of pebbles distributed on 𝑋2 is at least 2
𝑚1(2𝑚2 − 1) +

𝑚2(𝑛 − 4) + 2. Then, move two pebbles to 𝑎
𝑚1(

𝑛

2
−1)

, for 𝑚1 or 𝑎
𝑚1(

𝑛

2
−1)+1

, 𝑚1 is odd. 



Communications on Applied Nonlinear Analysis 

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Proof. Consider graph 𝐶𝑚(𝐾𝑛) with 2
𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2 pebbles distributed only on the 

vertices of 𝑋2. Two pebbles must be moved to 𝑎
𝑚1(

𝑛

2
−1)

, and we prove this result through induction on 

𝑚2. For 𝑚2 = 1, at least 2
𝑚1 + 𝑛 − 2 pebbles were distributed on 𝑋2. Since 𝑚 ≥ 4. This implies that 

at least 𝑛 + 6 pebbles were retained in 𝑋2. For 𝑚2 = 1, we have 𝑋2 ≅ 𝐾𝑛. In addition, the two pebbling 

numbers of 𝐾𝑛 are 𝑓2(𝐾𝑛) = 𝑛 + 2 according to the pebbling number of complete graph. Thus, the two 

pebbles can move 2 pebbles to 𝑎
𝑚1(

𝑛

2
−1)

. Now, we assume that the result is true for all 𝑚′2 < 𝑚2. Let 

2
𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2 pebbles be distributed on 𝑋2. Suppose the number of pebbles 

distributed on 𝐶𝑚1+1 is at least 𝑛 + 2. Then, the two pebbles can be moved to 𝑎
𝑚1(

𝑛

2
−1)

. Therefore, we 

assume that 𝑝(𝐶𝑚1+1) ≤ 𝑛 + 1. We have the following cases. 

Case (1) 𝑝(𝐶𝑚1+1) ≥ 𝑛. 

Based on our assumptions, we can move one pebble to 𝑎
𝑚1(

𝑛

2
−1)

. This implies that at least 

2
𝑚1(2𝑚2

− 1) + 𝑚2(𝑛 − 4) + 2 − 𝑛 pebbles are retained in ⟨𝑋2 − 𝐶𝑚1+1⟩. We need to make the 

following claim: 

Claim (1) 𝑝(⟨𝑋2 − 𝐶𝑚1+1⟩) ≥ 2
𝑚1(2𝑚2−1 − 1) + (𝑚2 − 1)(𝑛 − 4) + 2. 

We have, 𝑝(⟨𝑋2 − 𝐶𝑚1+1⟩) = 2
𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2 − 𝑛  

= 2
𝑚1(2.2𝑚2−1) + (𝑚2 − 1)(𝑛 − 4) + (𝑛 − 4) + 2 − 𝑛 

= 2
𝑚1(2𝑚2−1 − 1) + (𝑚2 − 1)(𝑛 − 4) − 2 

                        ≥ 2
𝑚1(2𝑚2−1 − 1) + (𝑚2 − 1)(𝑛 − 4) + 2, Since 𝑚1 + 𝑚2 = 𝑛 ≥ 4. 

Thus, claim (1) allows us to move two pebbles to 𝑎
(𝑚1+1)(

𝑛

2
−1)+1

. Through induction, we can move 

an additional pebble to 𝑎
𝑚1(

𝑛

2
−1)

. 

Case (2) 𝑝(𝐶𝑚1+1) < 𝑛.Suppose that any vertex belonging to 𝐶𝑚1+1 has at least two pebbles. Then, 

we can move a pebble to 𝑎
𝑚1(

𝑛

2
−1)

 and by Claim (1), we can move an additional pebble to 𝑎
𝑚1(

𝑛

2
−1)

. 

Therefore, we assume that no vertices in 𝐶𝑚1+1 contain two pebbles. Without loss of generality, we 

assume that most pebbles are distributed on the vertices of 𝐶1. For the least case scenario, we consider 

that all the 𝑎𝑖’s and 𝑏𝑖’s in 𝑋2 have one pebble on each and then place the remaining pebbles on 𝑦. 

This implies at least 

2
𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2 − 𝑚2(𝑛 − 4) = 2

𝑚1(2𝑚2 − 1) + 2 

= 2
𝑚1+𝑚2 − 2

𝑚1 + 2 > 2.2𝑚2 pebbles retained in the vertex 𝑦. Then, using 2
𝑚2 pebbles twice, the two 

pebbles can move to 𝑎
𝑚1(

𝑛

2
−1)

. 

Theorem 3.  For any 𝑛, 𝑚, the pebbling number of graph 𝐶𝑚(𝐾𝑛) is 𝑓(𝐶𝑚(𝐾𝑛)) = 2
𝑚 + 2(𝑛 − 3) +

(𝑚 − 2)(𝑛 − 4). 

Proof. Set 2
𝑚 − 1 pebbles on vertex 𝑥 and put one pebble on each 𝑎𝑖 and 𝑏𝑖 except for the vertices in 

{𝑎𝑖𝑎𝑖+1, 𝑏𝑖𝑏𝑖+1: 1 ≤ 𝑖 ≤ 𝑛(𝑘 − 1) − 1}. Thus, we cannot move one pebble to vertex 𝑦. So 

𝑓(𝐶𝑚(𝐾𝑛)) ≥ 2
𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4). Consider a graph with at least 2

𝑚 + 2(𝑛 − 3) +
(𝑚 − 2)(𝑛 − 4) pebbles, Let 𝑣 ∈ 𝐶𝑖, 𝑖 ∈ {1,2, . . . , 𝑚} be any target vertex. We prove this result by 

induction on 𝑚. For 𝑚 = 2 and 𝑚 = 3, the results follow from Theorems 1 and 2 respectively. We 

assume that the result is true for all 𝑚′ < 𝑚. We prove this result for all 𝑚. We consider the following 

cases: 



Communications on Applied Nonlinear Analysis 

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653 https://internationalpubls.com 

Case (1) 𝑣 ∈ 𝐶𝑚1
 for 2 < 𝑚1 < 𝑚.Let us assume that 𝑋 = 𝐶1 ∪. . .∪ 𝐶𝑚1

 and 𝑌 = 𝐶𝑚1+1 ∪. . .∪ 𝐶𝑚 are 

two disjoint subgraphs of 𝐶𝑚(𝐾𝑛). Here 𝑋 = 𝐶𝑚1
(𝐾𝑛) and 𝑌 = 𝐶𝑚2

(𝐾𝑛), where 𝑚 = 𝑚1 + 𝑚2. 

Clearly 𝑣 ∈ 𝑋. Suppose 𝑝(𝑋) ≥ 2
𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4). Then, we can move the pebble 

to 𝑣 by induction on 𝑚. Therefore, we assume 𝑝(𝑋) < 2
𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4) − 1. Also 

𝑝(𝐶𝑚1
) ≤ 𝑛 − 2. Suppose 𝑝(𝑋 − 𝐶𝑚1

) ≥ 2
𝑚1+1(21 − 1) + (𝑛 − 4) + 2. From Lemma 1 we can move 

a pebble to 𝑣. So assume that 𝑝(𝑋 − 𝐶𝑚1
) < 2

𝑚1−1(21 − 1) + (𝑛 − 4) + 2 and 𝑝(𝐶𝑚1
) ≤ 𝑛 − 2. 

Therefore 𝑝(𝑋) ≤ 2
𝑚1−1 + 2𝑛 − 4. This implies that the number of pebbles distributed on 𝑌 was at 

least 2
𝑚1(2𝑚2 − 1/2) + (𝑚1 + 𝑚2 − 2)(𝑛 − 4) − 1. We have 𝑝(𝑌) ≥ 2

𝑚1(2𝑚2 − 1/2) + (𝑚1 +
𝑚2 − 2)(𝑛 − 4) − 1 ≥ 2

𝑚1(2𝑚2 − 1) + +𝑚2(𝑛 − 4) + 2. Again, by Lemma 1, two pebbles can be 

moved to 𝑎
𝑚1(

𝑛

2
−1)

 or 𝑎
𝑚1(

𝑛

2
−1)+1

 when 𝑚1 is even and 𝑚1 is odd. Then, we can move one pebble to 

𝑣. 

Case (2) 𝑣 ∈ 𝐶1 or 𝑣 ∈ 𝐶𝑚. 

Without a loss of generality, we assume that 𝑣 ∈ 𝐶𝑚 and 𝑣 = 𝑦. Also 𝑝(𝐶𝑚) ≤ 𝑛 − 2. Now take 𝑋1 =
𝐶𝑚 ≅ 𝐶1(𝐾𝑛) and 𝑋2 = 𝐶1 ∪. . .∪ 𝐶𝑚−1 ≅ 𝐶𝑚−1(𝐾𝑛). 

Claim (1) 𝑝(𝑋2) ≥ 2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2.  

We have to prove  

𝑝(𝑋2) − 2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2 ≥ 0 

= 2
𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4) − 𝑛 + 2 − 2

𝑚 + 2 − (𝑚 − 1)(𝑛 − 4) − 2 

= 2(𝑛 − 3) − (𝑛 − 4) − 𝑛 + 2 = 0. 

From Claim (1), we obtain 𝑝(𝑋2) ≥ 2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2 pebbles. Again, by Lemma 

1, two pebbles can be moved to 𝑎
(𝑚−1)(

𝑛

2
+1)

 or 𝑎
(𝑚−1)(

𝑛

2
+1)+1

 when 𝑚 is odd or even, respectively. We 

can then move the pebble to 𝑣. 

The 𝒕- pebbling number of crisscross sequence of 𝒎 complete graphs 

Theorem 4.  The t-pebbling number of the graph 𝐶2(𝐾𝑛) is 𝑓𝑡(𝐶2(𝐾𝑛)) = 4𝑡 + 2(𝑛 − 3). 

Proof. Set 4𝑡 − 1 pebbles on vertex 𝑥 and one pebble on each 𝑎𝑖 and 𝑏𝑖 except for vertices 𝑎𝑘 and 𝑏𝑘−1 

where 𝑎𝑘, 𝑏𝑘 ∈ 𝑉(𝐶1) ∪ 𝑉(𝐶2). Therefore, we cannot move 𝑡 pebbles to the vertex 𝑦. 

Therefore 𝑓𝑡(𝐶2(𝐾𝑛)) ≥ 4𝑡 + 2𝑛 − 6. Consider a graph with at least 4𝑡 + 2𝑛 − 6 pebbles distributed 

at the vertices of 𝐶2(𝐾𝑛). Let 𝑣 be a target vertex and 𝑝(𝑣) = 0. We prove this result by induction of 

𝑡. For 𝑡 = 1, the result follows from theorem 1. We assume that this result is true for all 𝑡′ < 𝑡. Without 

a loss of generality, we assume that 𝑣 ∈ 𝐶2 and 𝑣 = 𝑦. We consider the following cases: 

Case (1) 𝑛 ≤ 𝑝(𝐶2) < 2𝑡 + 𝑛 − 2. If 𝑝(𝐶2) ≥ 2𝑡 + 𝑛 − 2. By referring the pebbling number of 

complete graph, we can move 𝑡 pebbles to 𝑦. Therefore, we assume that 𝑝(𝐶2) ≥ 𝑛. We used 𝑛 pebbles 

for 𝑦. Also 𝑝(𝐶2) ≤ 2𝑡 + 𝑛 − 2. Suppose 𝑝(𝐶1) ≤ 4𝑡 + 𝑛 − 5. Then 𝑝(𝐶2) = 𝑛 + 1. This was because 

𝐶2 ≅ 𝐾𝑛. Therefore, any of the vertices in 𝐶2 contains two pebbles. Using exactly two pebbles, we can 

move a pebble to 𝑦. This leave at least 4𝑡 + 2𝑛 − 8 pebbles retained in 𝐶2(𝐾𝑛). Since 4𝑡 + 2𝑛 − 8 =
4(𝑡 − 1) + 2𝑛 − 6. So by induction we can move additional 𝑡 − 1 pebbles to 𝑦. 

Case2:𝑝(𝐶2) < 𝑛.It is assumed that vertices adjacent to 𝑦 have at most one pebble. Suppose 𝑝(𝑎𝑘) =
1 or 𝑝(𝑏𝑘) = 1. By using at least two pebbles from 𝐶1 and move an additional pebble to 𝑎𝑘. Then, we 

move the pebble to 𝑦. This left at least 4(𝑡 − 1) + 2𝑛 − 6 pebbles on 𝐶2(𝐾𝑛). Thus, by induction, we 

can move an additional 𝑡 − 1 pebbles to 𝑦. So 𝑝(𝐶2) ≤ 𝑛 − 3. This implies that the number of unused 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 32 No. 3s (2025) 

 

654 https://internationalpubls.com 

pebbles retained in 𝐶1 was at least 4𝑡 + 𝑛 − 2. Thus by 𝑓(𝐾𝑛) = 𝑛, we can move 2𝑡 pebbles to 𝑎𝑘 and 

then move 𝑡 pebbles to 𝑦. 

Theorem 5.  The t-pebbling number of the graph 𝐶3(𝐾𝑛) is 𝑓𝑡(𝐶3(𝐾𝑛)) = 8𝑡 + 3𝑛 − 10. 

Proof. Set 8𝑡 − 1 pebbles on vertex 𝑥 and place one pebble on each 𝑎𝑖 and 𝑏𝑖 except 𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2 

and 𝑏2𝑘−1. Thus, we cannot move 𝑡 pebbles to vertex 𝑦. So 𝑓𝑡(𝐶3(𝐾𝑛)) ≥ 8𝑡 + 3𝑛 − 10. 

Consider graph 𝐶3(𝐾𝑛) with at least 8𝑡 + 3𝑛 − 10 pebbles distributed on its vertices. Let 𝑣 ∈ 𝐶𝑖 for 

𝑖 = 1,2,3 be any target vertex. We consider the following cases: 

Case (1) 𝑣 ∈ 𝐶2. 

Suppose 𝑝(𝐶3) ≥ 4𝑡 + 𝑛 − 2. We can move 2𝑡 pebbles to 𝑎2𝑘−2 and move 𝑡 pebbles to 𝑣 by using 

pebbling number of complete graph. So assume that 𝑝(𝐶3) ≤ 4𝑡 + 𝑛 − 3. This results in at least 8𝑡 +
3𝑛 − 10 − (4𝑡 + 𝑛 − 3) = 4𝑡 + 2𝑛 − 7. We have 𝑝(𝐶1 ∪ 𝐶2) ≥ 4𝑡 + 2𝑛 − 7. If 𝑝(𝐶1 ∪ 𝐶2) ≥ 4𝑡 +
2𝑛 − 6, By previous theorem we can move 𝑡 pebbles to 𝑣. Therefore, we assume that 𝑝(𝐶1 ∪ 𝐶2) ≤
4𝑡 + 2𝑛 − 7. If 𝑝(𝐶1 ∪ 𝐶2) < 4𝑡 + 2𝑛 − 7, then we get 𝑝(𝐶3) ≥ 4𝑡 + 3𝑛 − 3. Otherwise, this 

contradicts the total number of pebbles distributed on 𝐶3(𝐾𝑛). So we take 𝑝(𝐶1 ∪ 𝐶2) = 4𝑡 − 2𝑛 − 7 

and 𝑝(𝐶3) ≥ 4𝑡 + 𝑛 − 3 ≥ 2(2(𝑡 − 1)) + 𝑛 − 3 = 4𝑡 + 𝑛 − 7. Again, by using 𝑓(𝐾𝑛) = 𝑛 we can 

move 2(𝑡 − 1) pebbles to 𝑎2𝑘−2 and 𝑡 − 1 pebble to 𝑣. Also 𝑝(𝐶1 ∪ 𝐶2) = 4𝑡 + 2𝑛 − 7 ≥ 2𝑛 − 2, 

since 𝑡 ≥ 2 and 𝐶1 ∪ 𝐶2 ≅ 𝐶2(𝐾𝑛). Thus, from Theorem 1 we can move an additional pebble to 𝑣. 

Case (2) 𝑣 ∈ 𝐶1 or 𝑣 ∈ 𝐶3. 

Fix 𝑣 ∈ 𝐶3 and 𝑣 = 𝑦. We assume that 𝑝(𝐶3) ≥ 𝑛. Subsequently, at least one of the vertices in 𝐶3 

contains at least two pebbles. Using exactly two pebbles, we can move one pebble to 𝑦. This results in 

at least 8𝑡 + 3𝑛 − 12 ≥ 8(𝑡 − 1) + 3𝑛 − 10 pebbles in 𝐶3(𝐾𝑛). By induction, we can move an 

additional 𝑡 − 1 pebbles to 𝑦. Thus, no vertices in 𝐶3 contained two pebbles. Thus 𝑝(𝑉(𝐶3) −
{𝑎2𝑘−2, 𝑏2𝑘−1}) ≤ 𝑛 − 3. This implies that at least 8𝑡 + 3𝑛 − 10 − 𝑛 + 3 = 8𝑡 + 2𝑛 − 7 pebbles are 

retained in 𝐶1 ∪ 𝐶2 because 8𝑡 + 3𝑛 − 7 ≥ 8𝑡 + 2𝑛 − 6 for 𝑛 ≥ 4. Thus, from Theorem 4, we can 

move 2𝑡 pebbles to 𝑎2𝑘−2 and then move 𝑡 pebbles to 𝑦. 

Theorem 6.  The t-pebbling number of the graph 𝐶𝑚(𝐾𝑛) is 𝑓𝑡(𝐶𝑚(𝐾𝑛)) = 𝑡2𝑚 + 2(𝑛 − 3) +
(𝑚 − 2)(𝑛 − 4). 

Proof. Set 𝑡2𝑚 − 1 pebbles on vertex 𝑥 and put one pebble on each 𝑎𝑖 and 𝑏𝑖 except for the vertices 

in {𝑎𝑖𝑎𝑖+1, 𝑏𝑖𝑏𝑖+1: 1 ≤ 𝑖 ≤ 𝑛(𝑘 − 1) − 1}. Thus, we cannot move 𝑡 pebbles to 𝑦. So 𝑓𝑡(𝐾𝑛) ≥ 𝑡. 2𝑚 +
2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4). 

Consider a graph with at least 𝑡2𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4) pebbles distributed at the vertices of 

𝐶𝑚(𝐾𝑛). Let 𝑣 ∈ 𝐶𝑖 for 𝑖 = 1,2, . . . , 𝑚 be any target vertex. We prove this result by induction of 𝑡 and 

𝑚. For 𝑡 = 1, 𝑚 = 2 and 𝑚 = 3, the results follow from Theorems 3, 4 and 2. We assume that this 

result is true for all 𝑡′ < 𝑡. We prove this result for all 𝑡. Let 𝑣 ∈ 𝐶𝑚1
. 

Let 𝑋1 = 𝐶1 ∪. . .∪ 𝐶𝑚1
 and 𝑋2 = 𝐶𝑚1+1 ∪. . .∪ 𝐶𝑚. Clearly 𝑋1 ∪ 𝑋2 ≅ 𝐶𝑚(𝐾𝑛). Suppose 𝑝(𝑋2) ≥

2
𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2. Then, the two pebbles can be moved to 𝑎𝑚1

(
𝑛

2
− 1) and move a 

pebble to 𝑣. This implies that the number of pebbles retained in 𝑋1 was at least 𝑡2𝑚 + 2(𝑛 − 3) +
(𝑚 − 2)(𝑛 − 4) − [2𝑚1(2𝑚2 − 1) + 𝑚1(𝑛 − 4) + 2] ≥ (𝑡 − 1)2𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4). 

Thus, by induction, we can move an additional 𝑡 − 1 pebbles to 𝑣. So assume that 𝑝(𝑋2) <
2

𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2. 

Claim (1) 𝑝(𝐶𝑚(𝐾𝑛)) − 𝑝(𝑋2) ≥ 𝑓𝑡 (𝐶𝑚1
(𝐾𝑛)). 



Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

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We have, 𝑝(𝐶𝑚(𝐾𝑛)) − 𝑝(𝑋2) 

= [𝑡. 2𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4)] − [2𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 1] 
= 𝑡. 2𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4) − 2

𝑚 + 2
𝑚1 − 𝑚2(𝑛 − 4) − 1 

= 𝑡. 2𝑚1(2𝑚2 − 1) + 𝑡. 2𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4) − 1 

= 𝑓𝑡 (𝐶𝑚1
(𝐾𝑛)) + 𝑡. 2𝑚 − 𝑡. 2𝑚1 − 2

𝑚 − 1 

= 𝑓𝑡 (𝐶𝑚1
(𝐾𝑛)) + 2

𝑚1(𝑡. 2𝑚2 − 𝑡 − 1) − 1 

≥ 𝑓𝑡 (𝐶𝑚1
(𝐾𝑛)), since 𝑡 ≥ 2 and 𝑚 ≥ 4. 

By above Claim (1), 𝑝(𝑋1) ≥ 𝑓𝑡 (𝐶𝑚1
(𝐾𝑛)), and by induction on 𝑚 we can move 𝑡 pebbles to 𝑣. 

In the next sections, we show that the graph 𝐶𝑚(𝐾𝑛) has both the 2-pebbling and 2𝑡 −pebbling 

properties. 

The 2-pebbling Property of crisscross Sequence of 𝒎 − Complete Graphs 

Theorem 7.  The graph 𝐶2(𝐾𝑛) exhibits the two-pebbling property. 

Proof. Consider a graph with at least 2(2𝑛 − 2) − 𝑞 + 1 pebbles scattered at the vertices of 𝐶2(𝐾𝑛). A 

minimum of two pebbles must be pushed to the target vertex. We assume that 𝑣 ∈ 𝑉(𝐶2) without losing 

generality. We look at the following scenarios: 

Case (1) 𝑣 = 𝑎𝑘 or 𝑣 = 𝑏𝑘−1. 

We assume that 𝑣 = 𝑎𝑘 without losing generality. Because 𝑎𝑘 ∈ 𝑉(𝐶1) ∩ 𝑉(𝐶2). We have 

2𝑓(𝐶2(𝐾𝑛)) − +1 ≥ 4𝑛 − 4 − (2𝑛 − 2) + 1 = 2𝑛 − 1 ≥ |𝑉(𝐺)|. According to the pigeonhole 

principle, one of the vertices in 𝑉(𝐶2(𝐾𝑛)), say 𝑥, must have at least two pebbles. Thus, we can move 

one pebble to 𝑎𝑘 using precisely two pebbles. This is due to the fact that 𝑎𝑘 is adjacent to all of the 

graph’s vertices. This means that as 𝑞 ≤ 2𝑛 − 3, at least 2(2𝑛 − 2) − 𝑞 + 1 − 2 pebbles are 

maintained in 𝑉(𝐶2(𝐾𝑛)). As a result, 2(2𝑛 − 2) − 𝑞 − 1 ≥ 2𝑛 − 2 = 𝑓(𝐶2(𝐾𝑛)). We can transfer n 

more pebbles to 𝑎𝑘 using Theorem 1. 

Case (2) 𝑣 ∈ 𝑉(𝐶1) − {𝑎𝑘} or 𝑣 ∈ 𝑉(𝐶2) − {𝑏𝑘−1}. 

We assume that 𝑣 ∈ 𝑉(𝐶2) − {𝑏𝑘−1} and 𝑣 = 𝑦 and 𝑣 = 𝑦 without losing generality. By the 

pigeonhole principle, if 𝑝((𝐶2) − {𝑏𝑘−1}) ≥ 𝑛 − 1, at least one vertex, say 𝑎𝑘 + 1, contains two 

pebbles, and then pushes a pebble to 𝑦 from 𝑎𝑘+1 using precisely two pebbles. As a result, we suppose 

that 𝑝((𝑉2) − {𝑏𝑘−1}) ≤ 𝑛 − 2. The following subcases must be present: 

Subcase (1a) 𝑝(𝑉(𝐶2) − {𝑏𝑘−1}) = 𝑛 − 2. As by our assumption  

𝑝(𝑉(𝐶1) − {𝑏𝑘−1}) ≥ 2(2𝑛 − 2) − 𝑞 + 1 − 𝑛 + 2 = 2(2𝑛 − 2) − 𝑞 − 𝑛 + 3 

= 3𝑛 − 𝑞 − 1 ≥ 3𝑛 − (2𝑛 − 3) − 1 = 𝑛 + 2. 

The pigeonhole principle states that one of the vertices in 𝑉(𝐶1) − {𝑏𝑘−1} has at least two pebbles, i.e., 

𝑝(𝑎1) = 2. As a result, we can transfer one pebble to 𝑎𝑘 with only two pebbles. Hence, 

𝑝(𝑉(𝐶2) − {𝑏𝑘−1}) = 𝑛 − 1. This is due to the fact that |𝑉(𝐶2) − {𝑏𝑘−1}| = 𝑛 − 1 and 𝑝(𝑦) = 0. As a 

result, one of the vertices in 𝑉(𝐶2) − {𝑏𝑘−1}  has two pebbles. Then, with precisely two pebbles, we 

can move one pebble to 𝑦. Let 𝑞1 represent the number of occupied vertices in 𝑉(𝐶1) − {𝑏𝑘−1} and 𝑞2, 

and 𝑞2 represent the number of occupied vertices in 𝑉(𝐶2) − {𝑎𝑘}. 



Communications on Applied Nonlinear Analysis 

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Claim (1) 𝑝(𝐶2(𝐾𝑛)) − 𝑝(𝑉(𝐶2) − {𝑎𝑘}) − 𝑞2 − 2 ≥ 2𝑓(𝐾𝑛) − 𝑞2 + 1. We have  

𝑝(𝐶2(𝐾𝑛)) − [𝑝(𝑉(𝐶2) − {𝑎𝑘}) − 𝑞2] − 2 = 2(2𝑛 − 2) − 𝑞 + 1 − (𝑛 − 2) + 𝑞2 − 2 

= 4𝑛 − 4 − 𝑞1 − 𝑛 + 𝑞2 + 1 

= 3𝑛 − 𝑞3 − 3 

= 2𝑛 − 𝑞1 + 1 + 𝑛 − 4 

≥ 2𝑛 − 𝑞1 + 1, since 𝑛 ≥ 5 = 2𝑓(𝐾𝑛) − 𝑞1 + 1. 

Claim (1) states that at least 2𝑓(𝐾𝑛) − 𝑞1 + 1 pebbles are scattered in 𝐶1. This is due to the fact that 

𝐶1 ≅ 𝐾 − 𝑛 and 𝐾𝑛 satisfy the 2-pebbling property. Because 𝑎𝑘 is next to 𝑦, we can transfer two 

pebbles to 𝑎𝑘 and one more pebble to 𝑦 using the pebbling number of complete graph. 

Subcase (1b) 𝑝(𝑉(𝐶2) − {𝑏𝑘−1}) < 𝑛 − 2. 

Assume that at most 𝑛 − 3 pebbles are spread on 𝑉(𝐶2) − {𝑏𝑘−1} without losing generality. 

Claim (2) 𝑝(𝐶1) ≥ 2𝑓2(𝐾𝑛) − 𝑞1 + 1.Since  𝑝(𝐶2) − 𝑞2 = 0, then we have, 

2(2𝑛 − 2) − 𝑞1 − 𝑞2 + 1 = 4𝑛 − 4 − 𝑞1 + 1 − (𝑛 − 3) 

3𝑛 − 𝑞1 ≥ 2𝑛 − 𝑞1 + 5,Since 𝑛 ≥ 5,2(𝑛 + 2) − 𝑞1 + 1 

= 2𝑓2(𝐾𝑛) − 𝑞1 + 1. Because 𝐶1 ≅ 𝐾𝑛 and 𝑝(𝐶1) ≥ 2𝑓2(𝐾𝑛) − 𝑞1 + 1. Then we may shift four 

pebbles to 𝑎𝑘 and two pebbles to 𝑦. 

Theorem 8.  The graph 𝐶3(𝐾𝑛) exhibits the two-pebbling property[9][10].. 

Proof. Consider a graph containing at least 2(3𝑛 − 2) − 𝑞 + 1 pebbles at each vertex. The two pebbles 

must be moved to any vertex of the target. Let 𝑣 ∈ 𝐶𝑖 be the target vertex for 𝑖 = 1,2,3. We look at the 

following scenarios: 

Case (1) 𝑣 ∈ 𝐶2. 

We will suppose that 𝑝(𝐶2) ≥ 𝑛 + 2. We can transfer two pebbles to 𝑣 using the pebbling number of 

complete graph. As a result, we suppose that 𝑝(𝐶2) < 𝑛 + 2. If 𝑝(𝐶2) ≥ 𝑛, we can transfer a pebble to 

𝑣 for no more than two pebbles. Because 𝑝(𝑣) = 0 and 𝑞(𝐶2) ≤ 𝑛 − 1, one of the vertices of 𝐶2 holds 

at least two pebbles. This suggests that 𝑉(𝐶3(𝐾𝑛)) contained at least 2(3𝑛 − 2) − 𝑞 + 1 − 2 pebbles. 

We have 2(3𝑛 − 2) − (3𝑛 − 6) = 3𝑛 + 2 ≥ 3𝑛 − 2 = 𝑓(𝐶3(𝐾𝑛)) since 𝑞 ≤ 3𝑛 − 5. After that, the 

pebble may be transferred to 𝑣. Assume that 𝑝(𝑉(𝐶2) − {𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}) ≥ 𝑛 − 4. We can 

then transfer a pebble to 𝑣. As a result, we suppose that 𝑝(𝑉(𝐶2) − {𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}) ≤ 𝑛 − 5, 

which implies that at least 2(3𝑛 − 2) − 𝑞 − 1 − ((𝑛 − 5) − 𝑞2 − 1) pebbles are scattered on both 𝐶1 

and 𝐶3. 𝑞 = 𝑞1 + 𝑞2 + 𝑞3, where 𝑞1 = 𝑞(𝐶1), 𝑞2 = 𝑞(𝐶2 − {𝑎𝑘, 𝑏𝑘, 𝑎2𝑘−2, 𝑏2𝑘−1}), and 𝑞3 = 𝑞(𝐶3). As 

a result, in 𝑉(𝐶1) or 𝑉(𝐶3), which include at least 𝑛 + 2 pebbles, at least 6𝑛 − 4 − 𝑞1 − 𝑞2 − 𝑞3 + 1 −
𝑛 + 5 + 𝑞2 + 1 = 5𝑛 + 3 − 𝑞1 − 𝑞3 pebbles are kept. Assume that 𝑝(𝑉(𝐶1)) ≥ 𝑛 + 2. The two 

pebbles may then be moved to 𝑎𝑘, and the pebble can be moved to 𝑣𝑖𝑛𝐶2. This replaces at least 5𝑛 +
3 − 𝑞1 − 𝑞3 − (𝑛 + 2) = 4𝑛 + 1 − 𝑞1 − 𝑞3 pebbles left in 𝐶1 and 𝐶3. Assume 𝑝(𝐶1) ≥ 2𝑛 − 𝑞1 + 1. 

Then we may relocate the two pebbles to 𝑎𝑘 and add another pebble to 𝑣. This was due to 𝐶1 ≅ 𝐾𝑛. If 

not, at least 2𝑛 − 𝑞3 + 1 pebbles remained in 𝐶3. This was due to 𝐶3 ≅ 𝐾𝑛. Using the pebbling number 

of the entire graph, we can add another pebble to 𝑣 in 𝐶2. As a result, we suppose that 𝑝(𝐶1) ≤ 𝑛 + 1. 

This means that at least  𝑝 (𝑉(𝐶3(𝐾𝑛))) − 𝑝(𝐶1) = 6𝑛 − 4 − 𝑞 + 1 − (𝑛 + 1) − 𝑞15𝑛 − (𝑞2 + 𝑞3) −

4 is required that equals 𝑓(𝐶2(𝐾𝑛)) − 𝑞2 − 𝑞3 + 1. Because 𝐶2 ∪ 𝐶3 ≅ 𝐶2(𝐾𝑛). We can transfer two 

pebbless to 𝑣𝑖𝑛𝐶2 using Theorem 7. 

Case (2) 𝑣 ∈ 𝐶1 − {𝑎𝑘, 𝑏𝑘−1} or 𝑣 ∈ 𝐶3 − {𝑎2𝑘−2, 𝑏2𝑘−1}. 



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We assume 𝑣 ∈ 𝐶3 and take 𝑣 = 𝑦 without losing generality. Let 𝑞 = 𝑞1 + 𝑞2 + 𝑞3 − 4 and assume 

that all vertices of {𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}  contain at least one pebble. If 𝑝(𝐶1) ≥ 2(𝑛 + 2) − 𝑞1 + 1, 

the four pebbles can transfer to 𝑎𝑘. The number of pebbles maintained in 𝐶3(𝐾𝑛) must thus be at least 

2(3𝑛 − 2) − 𝑞 + 1 − (2(𝑛 + 2) − 𝑞1 + 1) ≥ 𝑓(𝐶2(𝐾𝑛)), and we may transfer an extra pebble to 𝑦 

using Theorem 1. As a result, we suppose that 𝑝(𝐶1) < 2(𝑛 + 2) − 𝑞1. If 𝑝(𝐶1) ≥ 𝑛 + 2, two pebbles 

can be transferred to 𝑎𝑘 and one pebble to 𝑎2𝑘−2. Then, using at least two pebbles from 𝑉(𝐶2) or 𝑉(𝐶3), 

we may transfer another pebble to 𝑎2𝑘−2 and one to 𝑦. This means that in 𝑉(𝐶2) and 𝑉(𝐶3), at least 

2𝑛 − 2 pebbles were kept. As a result, we can add another pebble to 𝑦. As a result, we suppose that 

𝑝(𝐶1) < 𝑛 + 2. 

Claim (1) 𝑝(𝐶3(𝐾𝑛)) − (𝑛 + 1) + 𝑞1 ≥ 2𝑓(𝐶2(𝐾𝑛)) − (𝑞2 + 𝑞3) + 1. 

𝑝(𝐶3(𝐾𝑛)) − (𝑛 + 1) + 𝑞1 = 2(3𝑛 − 2) − 𝑞 + 1 − 𝑛 − 1 − 𝑞1 

= (4𝑛 − 4) − (𝑞2 + 𝑞3) + 4 + 𝑛 − 1 

≥ (4𝑛 − 4) − (𝑞2 + 𝑞3) + 1 

= 2(2𝑛 − 2) − (𝑞2 + 𝑞3) + 1 

= 2𝑓(𝐶2(𝐾𝑛)) − (𝑞2 + 𝑞3) + 1. 

We can transfer two pebbles to 𝑦 using the following Theorem 7. Because 𝐶2 ∪ 𝐶3 ≅ 𝐶2(𝐾𝑛). Then 

we’re finished. 

Lemma 2.   

Let 𝐺 = 𝐶𝑚(𝐾𝑛) be a  crisscross sequence of 𝑚 complete graphs. Let 𝑋1 = 𝐶𝑚1
(𝐾𝑛) = 𝐶1 ∪. . .∪

𝐶𝑚−1 and 𝑋2 = 𝐶𝑚2
(𝐾𝑛) = 𝐶𝑚1+1 ∪ 𝐶𝑚1+2 ∪. . .∪ 𝐶𝑚 be two subgraphs of 𝐺, where 𝑚1 + 𝑚2 = 𝑚. 

Assume the number of pebbles scattered on 𝑋2 is more than 2(2𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2) − 𝑞2 +
1. Then we may relocate four pebbles to 𝑎

𝑚1(
𝑛

2
−1)

 for 𝑚1 is an even number, or 𝑎
𝑚1(

𝑛

2
−1)+1

, if 𝑚1 is 

an odd number.  Here, 𝑞 = 𝑞1 + 𝑞2, with 𝑞1 representing the number of occupied vertices of 𝑋1 and 

𝑞2 representing the number of occupied vertices of 𝑋2. 

Proof. Consider the graph 𝐶𝑚(𝐾𝑛) with the values 2
𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2 − 𝑞2 + 1 pebbles 

are exclusively found on the vertices of 𝑋2. We must relocate the four pebbles to 𝑎
𝑚1(

𝑛

2
−1)

. We 

demonstrate this lemma using induction on 𝑚2. At least 2(2𝑚1 + 𝑛 − 2) − 𝑞2 + 1 pebbles dispersed 

on 𝑋2 for 𝑚2 = 1. Because 𝑚 ≥ 4, at least 2(𝑛 + 6) − 𝑞2 + 1 pebbles must be kept in 𝑋2. If 𝑚2 = 1, 

we get 𝑋2 ≅ 𝐾𝑛. Also 2(𝑛 + 6) − 𝑞2 + 1 ≥ 2𝑓2(𝐾𝑛) − 𝑞2 + 1. Four pebbles can be moved to 𝑎
𝑚1(

𝑛

2
−1)

. 

This result is assumed to be true for every 𝑚2′ < 𝑚2. We demonstrate the findings for all 𝑚2. Allow 

2(2𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2) − 𝑞2 + 1 pebbles to be dispersed on 𝑉(𝑋2). 

Assume the number of pebbles dispersed on 𝐶𝑚1+1 is more than 2(𝑛 + 2) − 𝑞𝑚1+1 + 1. The four 

pebbles can then be transferred to 𝑎
𝑚1(

𝑛

2
−1)

. As a result, we suppose that 𝑝(𝐶𝑚1+1) is no more than 

2(𝑛 + 2) − 𝑞𝑚1+1. 

Case (1) 𝑝(𝐶𝑚1+1) ≥ 2𝑛 − 𝑞 + 1. 

We may relocate the two pebbles to 𝑎
𝑚1(

𝑛

2
−1)

 based on this assumption. This means that at least 

2
𝑚1(2𝑚2−1) + 𝑚2(𝑛 − 4) + 2 − 𝑞2 − 2(𝑛) + 𝑞𝑚1+1 − 1 pebbles are maintained in ⟨𝑋2 − 𝐶𝑚1+1⟩. 

 

 



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Claim(1) 

 𝑝(𝑋2 − 𝐶𝑚1+1) ≥ 2(2𝑚1(2𝑚2−1 − 1) + 𝑚2(𝑛 − 4) + 2) − 𝑞 + 1. 

We have  

𝑝(𝑋2 − 𝐶𝑚1+1) = 2(2𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4) + 2) − 𝑞 + 1 − 2𝑛 + 𝑞𝑚1+1 − 1 

= 2 (2𝑚1(2𝑚2−1 − 1) + 2
𝑚1(2𝑚2−1)+(𝑚2−1)(𝑛−4)+2) − 𝑞 + 1 − 2𝑛 − 𝑞𝑚1+1 − 1 

= 2(2𝑚1(2𝑚2 − 1) + (𝑚2 − 1)(𝑛 − 4) + 2) − 𝑞 + 1 + 𝑞𝑚1+1 + 2
𝑚1+𝑚2 − 𝑞 

> 2(2𝑚1(2𝑚2−1 − 1) + (𝑚2 − 1)(𝑛 − 4) + 2) − 𝑞 + 1, since 𝑚1 + 𝑚2 = 𝑛 ≥ 4 and we get the term 

𝑞𝑚1+1 + 2
𝑚1+𝑚2 − 𝑞. 

We can move four pebbles to 𝑎𝑚1+1 (
𝑛

2
+ 1) using Claim(1). Using induction on 𝑚2, two pebbles may 

be relocated to 𝑎𝑚1
(

𝑛

2
− 1). 

Case (2) 𝑝(𝐶𝑚1+1) < 2𝑛 − 𝑞𝑚1+1. 

If at least two pebbles are present at the two vertices in 𝐶𝑚1+1, they can be transferred to 𝑎
𝑚1(

𝑛

2
−1)

. We 

can move two additional pebbles to 𝑎𝑚1
(

𝑛

2
− 1) using Claim(1). Assume, for the sake of argument, 

that no two pebbles in 𝐶𝑚1+1 have two pebbles on each. We investigate the simplest example, in which 

the majority of the pebbles are placed on the vertices of 𝐶1. 𝑦, in particular, had a higher amount of 

pebbles. All other 𝑎𝑖 and 𝑏𝑖 have no more than one pebble. So the vertex 𝑦 has 

2(2𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 − 4)) − 𝑞 + 1 − 𝑚2(𝑛 − 4) ≥ 4(2𝑚2) pebbles, since 𝑚1 = 1. This means 

that at least 4(2𝑚2) pebbles were kept in 𝑦. As a result, the four pebbles can shift to 𝑎𝑚1
(

𝑛

2
− 1). 

Theorem 9.  The graph 𝐶𝑚(𝐾𝑛) satisfies two-pebbling property. 

Proof. Consider a graph that has at least 2
𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4) − 𝑞 + 1 pebbles scattered 

across its vertices 𝐶𝑚(𝐾𝑛). The remaining two pebbles must be moved to any desired vertex. For 1 ≤
𝑖 ≤ 𝑚, let 𝑣 ∈ 𝐶𝑖. We demonstrate this result using induction on 𝑚. Theorems 7 and 8 provide the 

following findings for 𝑚 = 2 and 𝑚 = 3, respectively. We suppose the result holds true for every 𝑚′ <
𝑚. We demonstrate the findings for all 𝑚. We have the following situations: 

Case (1) 𝑣 ∈ 𝐶𝑚1
 for 2 < 𝑚1 < 𝑚. 

As the two subgraphs of 𝐶𝑚(𝐾𝑛), let us define 𝑊 = 𝐶1 ∪. . .∪ 𝐶𝑚1
 and 𝑍 = 𝐶𝑚1+1 ∪. . .∪ 𝐶𝑚. Assume 

𝑝(𝑊) ≥ 2(2𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4)) − 𝑞1 + 1. The two pebbles can then be inducted to 𝑣. 

Assume 𝑝(𝑊) < 2(2𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4) − 𝑞1). If 𝑝(𝑊 − 𝐶𝑚1
) ≥ 2 (2𝑚1−1(21 −

1) + (𝑛 − 4 + 2)) − 𝑞1 + 1, We can shift four pebbles to 𝑎
(𝑚1−1)(

𝑛

2
−1)

 or 𝑎
(𝑚1−1)(

𝑛

2
−1)+1

 and then two 

pebbles to 𝑣 using Lemma 2. So suppose 𝑝(𝑊 − 𝐶𝑚1
) < 2(2𝑚1−1(21 − 1) − (𝑛 − 4) + 2) − 𝑞1 + 1. 

This means that there were at least 2(2𝑚 + 2(𝑛 − 3) + (𝑚 − 2(𝑛 − 4)) + 2) − 𝑞2 + 1 pebbles 

dispersed on 𝑍. We can shift four pebbles using Lemma 2 to 𝑎
𝑚1(

𝑛

2
−1)

 or 𝑎
𝑚1(

𝑛

2
−1)+1

. The two pebbles 

can then be relocated to 𝑣. 

Case (2) 𝑣 ∈ 𝐶1 or 𝑣 ∈ 𝐶𝑚. 



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We assume that 𝑣 ∈ 𝐶𝑚 and 𝑣 = 𝑦 without losing generality. Also, 𝑝(𝐶𝑚) ≤ 𝑛 + 1. We assume 𝑊 =
𝐶𝑚 ≅ 𝐶1(𝐾𝑛) and 𝑍 = 𝐶1 ∪. . .∪ 𝐶𝑚−1 ≅ 𝐶𝑚−1(𝐾𝑛).  

Claim(1) 𝑝(𝑍) ≥ 2(2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2) − 𝑞2 + 1. We Have to prove  

𝑝(𝑍) − [2(2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2) − 𝑞2 + 1] ≥ 0 

= 2(2𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4)) − 𝑞 + 1

− [2(2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2) − 𝑞2 + 1] 

= 4(𝑛 − 3) + 2(𝑚 − 2)(𝑛 − 4) − 𝑞 + 1 − 2(𝑚 − 1)(𝑛 − 4) ≥ 4(𝑛 − 3 + 1) > 0. 

According to Claim(1), 𝑝(𝑍) ≥ 2(2(2𝑚−1 − 1) + (𝑚 − 1)(𝑛 − 4) + 2) − 𝑞 + 1. Again, using 

Lemma 2, we can move four pebbles to 𝑎
(𝑚−1)(

𝑛

2
+1)

 or 𝑎
(𝑚−1)(

𝑛

2
+1)+1

, and we may then transfer two 

pebbles to 𝑣. 

The 𝟐𝒕 −pebbling property of 𝑪𝒎(𝑲𝒏) 

Theorem 10.  The graph 𝐶2(𝐾𝑛) exhibits 2𝑡 − pebbling property. 

Proof. Consider a graph with at least 2(4𝑡 + 2𝑛 − 6) − 𝑞 + 1 pebbles scattered at the 𝐶2(𝐾𝑛) vertices. 

Assume 𝑣 ∈ 𝐶2 and 𝑣 = 𝑦. We demonstrate this result using induction on 𝑡. Theorem 7 yields the 

following result for 𝑡 = 1. Assume that the outcome holds true for every 𝑡′ < 𝑡. We take into account 

the following. 

Case (1) 𝑝(𝑣) = 0. 

To prove Case(1), we investigate the following subcases. 

Subcase (1) 𝑣 ∈ {𝑎𝑘, 𝑏𝑘−1}. Without loss of generality, assume that 𝑣 ∈ 𝑎𝑘. Since 

2(4𝑡 + 2𝑛 − 6) − 𝑞 + 1 − 4 ≥ 8𝑡 + 4𝑛 − 12 − (2𝑛 − 2) − 3 = 8𝑡 + 2𝑛 − 13 ≥ 5, since 𝑡 ≥ 2 and 

𝑛 ≥ 5. Thus, we may transfer two pebbles to 𝑎𝑘 by utilising precisely four pebbles from either 𝐶1 or 

𝐶2 since 𝑎𝑘 is adjacent to all the other vertices of 𝐶2(𝐾𝑛). This yields at least 2(4𝑡 + 2𝑛 − 6) − 𝑞 +
1 − 4, which is greater than 2(4(𝑡 − 1) + 2𝑛 − 6) − 𝑞 + 5. As a result, we can induct an extra 

2(𝑡 − 1) pebbles to 𝑎𝑘. As a result of induction, we can transfer one extra 2(𝑡 − 1) pebble to 𝑎𝑘. 

Subcase (2) 𝑣 ∈ 𝐶1 or 𝑣 ∈ 𝐶2. 

We assume that 𝑣 ∈ 𝐶2 and 𝑣 = 𝑦 without losing generality. Assume 𝑝(𝐶2) ≥ 𝑛 + 2. At least two 

pebbles must be present on each of its two vertices. As a result, we can move two pebbles to 𝑦 using 

precisely four pebbles. This means that in 𝑉(𝐶2(𝐾𝑛)), at least 2(4(𝑡 − 1) + 2𝑛 − 6) − 𝑞 + 1 pebbles 

were kept. So suppose 𝑝(𝐶2) = 𝑛 or 𝑝(𝐶2) = 𝑛 + 1. We can transfer one pebble to 𝑣 for no more than 

two pebbles from 𝑉(𝐶2). As a result, the number of pebbles kept in 𝑉(𝐶1) must be more than 

2(4𝑡 + 2𝑛 − 6) − 𝑞 + 1 − (𝑛 + 1) > 𝑛 + 5. This means that any two of the vertices must have at least 

two pebbles, or one of them must have four pebbles. Then, using precisely four pebbles, two pebbles 

may be relocated to 𝑎𝑘, and the number of pebbles kept in 𝑉(𝐶2(𝐾𝑛)) is at least 2(4𝑡 + 2𝑛 − 6) − 𝑞 +

1 − 6 ≥ 2(4(𝑡 − 1) + 2𝑛 − 6) − 𝑞 + 3 > 2(4(𝑡 − 1) + 2𝑛 − 6) − 𝑞 + 1. As a result of induction, we 

can transfer more 2(𝑡 − 1) pebbles to 𝑣. 

Case (2) 𝑝(𝑣) = 𝑥 for 𝑥 ≥ 1. 

We take into account the following subcases: 

Subcase (2a) 𝑥 is even. 



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Assume 𝑝(𝑣) = 𝑥 = 2𝑥′. We need to transport more 2(𝑡 − 𝑥′) pebbles to 𝑣. Because we have 

𝑝(𝐶2(𝐾𝑛)) − 2𝑥′ = 2(4𝑡 + 2𝑛 − 6) − 𝑞 + 1 − 2𝑥′ 

= 2(4(𝑡 − 𝑥′) + 4𝑥′ + 2𝑛 − 6) − 𝑞 + 1 − 2𝑥′ 

= 2(4(𝑡 − 𝑥′) + 2𝑛 − 6) − 𝑞 + 1 + 6𝑥′ 

> 2𝑓𝑡−𝑥′(𝐶2(𝐾𝑛)) − 𝑞 + 1. 

We can transfer more 2(𝑡 − 𝑥′) pebbles to 𝑣 via induction. 

subcase (2b) 𝑥 is odd. 

Assume 𝑝(𝑣) = 𝑥 = 2𝑥′ + 1. We need to transport more 2(𝑡 − 𝑥′) − 1 pebbles to 𝑣. Since 

𝑝(𝐶2(𝐾𝑛)) − 2𝑥′ − 1 = 2(4𝑡 + 2𝑛 − 6) − 𝑞 + 1 − 2𝑥′ − 1 

= 2(4(𝑡 − 𝑥′) + 4𝑥′ + 2𝑛 − 6) − 𝑞 − 2𝑥′ 

= 2(4(𝑡 − 𝑥′) + 2𝑛 − 6) − 𝑞 + 6𝑥 ′ > 2𝑓𝑡−𝑥′(𝐶2(𝐾𝑛)) − 𝑞 + 1Since𝑥′ ≥ 1 

So we can add more 2(𝑡 − 𝑥′) − 1 pebbles to 𝑣. 

Theorem 11.  The graph 𝐶3(𝐾𝑛) exhibits 2𝑡 −pebbling property[13][16]. 

Proof. Consider the graph 𝐶3(𝐾𝑛), which has at least 2(8𝑡 + 3𝑛 − 10) − 𝑞 + 1 pebbles on its vertices. 

We must relocate 2𝑡 pebbles to any target vertex. For 𝑖 = 1,2,3, let 𝑣 ∈ 𝐶𝑖. We look at the following 

scenarios: 

Case (1) 𝑝(𝑣) = 0. 

Theorem 8 yields the following result for 𝑡 = 1. Assume that the outcome holds true for every 𝑡′ < 𝑡. 

We take into account the following subcases: 

Subcase (1a)𝑣 ∈ 𝐶2. 

Consider that 𝑝(𝑉(𝐶2) − {𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}) ≥ 𝑛 − 2. Then we may transfer two pebbles to 𝑣 for 

a total of four pebbles. We have 𝑝(𝑉(𝐶2) − {𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}) ≅ 𝐾𝑛−4. This indicates that 

𝑉(𝐶3(𝐾𝑛)) has at least 𝑝(𝐶3(𝐾𝑛)) − 4 ≥ 2(8(𝑡 − 1) + 3𝑛 − 10)𝑞 + 1 pebbles. We can transfer more 

2(𝑡 − 1) pebbles to 𝑣 via induction. We now suppose that 𝑝(𝑉2(𝐶2) − {𝑎𝑘, 𝑏𝑘−1, 𝑎2𝑘−2, 𝑏2𝑘−1}) ≤ 𝑛 −
3. As a result, the total number of pebbles maintained in 𝐶1 and 𝐶2 was 𝑝(𝐶3(𝐾𝑛) − (𝑛 − 3) + 𝑞2) ≥
4𝑛 + 12. Then, at least one 𝐶1 or 𝐶3 contains at least 2𝑛 + 6 pebbles. Make sure 𝑝(𝐶1) ≥ 2𝑛 + 6. We 

can transfer four pebbles to 𝑎𝑘 and two pebbles to 𝑣 ∈ 𝐶2 using at least eight pebbles from 𝐶1. Then 

we have 2(8𝑡 + 3𝑛 − 10) − 𝑞 + 1 − 8 > 2(8(𝑡 − 1) + 3𝑛 − 10) − 𝑞 + 1. We can transfer more 

2(𝑡 − 1) pebbles to 𝑣 via induction. 

Subcase (1b) 𝑣 ∈ 𝐶1 or 𝑣 ∈ 𝐶3. Allow 𝑣 ∈ 𝐶3 and set 𝑣 = 𝑦. Assume that 𝑝(𝐶3 − {𝑎2𝑘−2,𝑏2𝑘−1
}) ≥ 𝑛. 

Then, for the expense of four pebbles, we may shift two pebbles to 𝑣. The number of pebbles kept in 

𝑉(𝐶3(𝐾𝑛)) was obviously at least 2𝑓𝑡−1(𝐶3(𝐾𝑛)) − 𝑞 + 1. With induction, we can transfer 2(𝑡 − 1) 

additional pebbles to 𝑦. As a result, we can shift zero pebbles to 𝑦 by employing pebbles in 𝑉(𝐶3) −
{𝑎2𝑘−2, 𝑏2𝑘−1}. This implies that 𝑝(𝑉(𝐶3) − {𝑎2𝑘−2, 𝑏2𝑘−1}) ≤ 𝑛 − 4. 

Claim(1) 𝑝(𝐶3(𝐾𝑛)) − 𝑝(𝐶3) ≥ 2𝑓2𝑡(𝐶2(𝐾𝑛)) − 𝑞 + 1.We have  𝑝(𝐶3(𝐾 − 𝑛)) − 𝑝(𝐶3) = 2(8𝑡 +

3𝑛 − 10) − 𝑞 + 1 − 𝑛 + 4 

= 2(8𝑡 + 2𝑛 + 𝑛 − 10) − 𝑞 + 1 − 𝑛 + 4 

= 2(8𝑡 + 2𝑛 − 6) − 𝑞 + 1 + 2𝑛 − 8 − 𝑛 + 4 

= 2(8𝑡 + 2𝑛 − 6) − 𝑞 + 1 + 𝑛 − 4 

> 2(8𝑡 + 2𝑛 − 6) − 𝑞 + 1,Since𝑛 ≥ 5 = 2𝑓2𝑡(𝐶(𝐾𝑛)) − 𝑞 + 1 = 𝑝(𝐶1 ∪ 𝐶2). 



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According to Claim, we can transfer 4𝑡 pebbles to 𝑎2𝑘 using Theorem 4  because 𝐶3(𝐾𝑛) − 𝐶3 ≅ 𝐶1 ∪
𝐶2 ≅ 𝐶2(𝐾𝑛) and then move 2𝑡 pebble to 𝑦. 

Case (2) 𝑥 is even. 

We may express this as 𝑥 = 2𝑥′. We need to transport 2(𝑡 − 𝑥′) more pebbles to 𝑣. Since, 

𝑝(𝐶3) − 2𝑥′ = 2(8𝑡 + 3𝑛 − 10) − 𝑞 + 1 − 2𝑥′ = 2(8(𝑡 − 𝑥′) + 8𝑥′ + 3𝑛 − 10) − 𝑞 + 1 − 2𝑥′ >

2(8(𝑡 − 𝑥′) + 3𝑛 − 10) − 𝑞 + 1 + 16𝑥 ′ − 2𝑥 > 2𝑓𝑡−𝑥′(𝐶3(𝐾𝑛)) − 𝑞 + 1. 

Then we may add another 2(𝑡 − 𝑥′) pebbles to 𝑣. 

Subcase (2a) 𝑥 is odd. 

We may express this as 𝑥 = 2𝑥′ + 1. We need to transport another 2𝑡 − 2𝑥′ − 1 pebble to 𝑣. Because 

𝑝(𝐶3(𝐾𝑛) − 𝑝(𝐶3)) ≥ 2𝑓𝑡−𝑥′(𝐶2(𝐾𝑛)) − 𝑞 + 1. We can transfer more 2(𝑡 − 𝑥′) pebbles to 𝑣 via 

induction. 

Theorem 12.  The graph 𝐶𝑚(𝐾𝑛) exhibits the 2𝑡 − pebbling property. 

Proof. Consider a graph with at least 2(𝑡2𝑚 + 2(𝑛 − 3) + (𝑚 − 2)(𝑛 − 4)) − 𝑞 + 1 pebbles 

distributed at its vertices. The 2𝑡 pebbles must be moved to any target vertex. For 1 ≤ 𝑖 ≤ 𝑚, consider 

𝑣 ∈ 𝐶𝑖. We have identified the following cases. 

Case (1) 𝑝(𝑣) = 0. By demonstrating that this conclusion is true for all values of 𝑡 and 𝑚, we can 

verify it. Due to the fact that it follows from Theorems 10 and 11, we know the conclusion is true for 

𝑡 = 1, 𝑚 = 2, and 𝑚 = 3. For any values of 𝑡′ that are smaller than 𝑡, we assume that the conclusion 

is true. Then, we demonstrate that the conclusion must be true for 𝑡 if it is true for all values of 𝑡′ that 

are smaller than 𝑡. 

Take the value 𝑚 = 𝑚1 + 𝑚2. Undoubtedly, 𝑣 ∈ 𝐶𝑚1
. Establish the definitions of 𝑋1 = 𝐶1 ∪. . .∪

𝐶𝑚1
 and 𝑋2 = 𝐶𝑚1+1 ∪. . .∪ 𝐶𝑚. Consider the scenario where 𝑝(𝑋2) ≥ 2(2𝑚1(2𝑚2 − 1) + 𝑚2(𝑛 −

4) + 2) − 𝑞2 + 1. Based on Lemma 1, four pebbles can move to either 𝑎
𝑚1(

𝑛

2
−1)

 or 𝑎
𝑚1(

𝑛

2
)+1

, from 

which we can move two pebbles to 𝑣. As a result, we must transfer an extra 2(𝑡 − 1) pebbles to 𝑣. 

Claim (1) 𝑝(𝐶𝑚(𝐾𝑛)) − 𝑝(𝑋2) ≥ 2((𝑡 − 1)2𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4)) − 𝑞1 + 1. 

We  have 𝑝(𝐶𝑚(𝐾𝑛)) − 𝑝(𝑋2) 

= 2 ((𝑡 − 1)2𝑚1+𝑚2 + 2
𝑚1 + 2(𝑛 − 3) + (𝑚1 + 𝑚2 − 2)(𝑛 − 4) − 𝑚2(𝑛 − 4) − 2) − 𝑞1 

> 2((𝑡 − 1)2𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4)) − 𝑞1 + 1, since 𝑚1 ≥ 2 

> 2𝑓𝑡−1 (𝐶𝑚1
(𝐾𝑛)) − 𝑞1 + 1. 

We may transfer the 2(𝑡 − 1) pebbles to 𝑣 from claim (1). As a result, we presume that 𝑝(𝑋2) ≤

2(2𝑚1(2𝑚2−1) + 𝑚2(𝑛 − 4) + 2) − 𝑞2, which results in the claim that follows below. 

Claim(2)𝑝(𝐶𝑚(𝐾𝑛)) − 𝑝(𝑋2) ≥ 2𝑓𝑡 (𝐶𝑚1
(𝐾𝑛)).We have  

𝑝(𝐶𝑚(𝐾𝑛)) − 𝑝 = 2(𝑡2𝑚1+𝑚2 − 2
𝑚1+𝑚2 + 2

𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4) − 2) − 𝑞1 + 1 

> 2(𝑡2𝑚1 + 2(𝑛 − 3) + (𝑚1 − 2)(𝑛 − 4)) − 𝑞1 + 1 > 2𝑓𝑡(𝑋1) − 𝑞1 + 1. 

We can transfer 2𝑡 pebbles to 𝑣 through induction. We are then done. 

Case (2) 𝑝(𝑣) = 𝑥, for 𝑥 ≥ 1. 



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More 2𝑡 − 𝑥 pebbles need to be moved to 𝑣. The following subcases exist: 

Subcase (2a) 𝑥 is even. 

Now, 𝑥 can be written as 2𝑥′. Moving the 2(𝑡 − 𝑥′) pebbles to 𝑣 is necessary. It is evident that 

𝑝(𝐶𝑚(𝐾𝑛)) − 2𝑥′ ≥ 2𝑓𝑡−𝑥′(𝐶𝑚(𝐾𝑛)) − 𝑞 + 1. In order to use induction 𝑡, we can relocate the 

2(𝑡 − 𝑥′) pebbles to 𝑣. 

Subcase (2b) 𝑥 is odd. 

Let 𝑥 may be expressed as 2𝑥′ + 1. Moving 2𝑡 − 2𝑥′ − 1 pebbles to 𝑣 is necessary. We obtain 

𝑝(𝐶𝑚(𝐾𝑛)) − 2𝑥′ − 1 ≥ 2𝑓𝑡−𝑥′(𝐶𝑚(𝐾𝑛)) − 𝑞 + 1 without a doubt. Therefore, we may transfer 2𝑥′ −

1 more pebbles to 𝑣. 

Conclusion: 

In conclusion, this paper delivers a comprehensive exploration of the pebbling number, the two-

pebbling property, the t-pebbling number, and the 2t-pebbling property inside a crisscross sequence of 

m-complete graphs. By investigating these properties, we have expanded valuable perceptions into the 

productivity of pebbling operations in this structure. The study highlights the complications and 

variations in pebbling dynamics, offering a profounder understanding of how these properties impact 

the behavior of the graph. This research contributes to the broader field of graph theory, precisely in 

understanding resource distribution and optimization in networked systems. 

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