DETOUR PEBBLING NUMBER OF ROOTED PRODUCT OF
SOME GRAPHS

A. LOURDUSAMY, I. DHIVVIYANANDAM*, AND SAGAYA SUGANYA A

ABSTRACT. A pebbling transformation is considered as the shifting of two pebbles from a vertex and the insertion of one of them onto an adjacent vertex, given a configuration of pebbles on the vertices of a connected graph G. The detour pebbling number, $f^*(G)$, of a simple connected graph G is the smallest positive integer such that a sequence of pebbling moves can transfer one pebbles to any target vertex for any distribution of $f^*(G)$ pebbles on G's vertices. The detour pebbling number, $f^*(G)$, of a graph G is obtained by the detour path. We discuss the detour pebbling number of rooted product of path, complete and fan graphs.

**Keywords:** pebbling moves, pebbling number, detour pebbling number, rooted product.

**2020 AMS Subject Classification:** 05C12, 05C25, 05C38, 05C76.

# 1. INTRODUCTION

Pebbling, introduced by Lagarias and Saks, has sparked a lot of interest. Chung[9] was the first to put it into the literature, and many others have followed suit, including Hulbert, who published an overview of graph pebbling[11]. Since Hulbert's survey first appeared in graph pebbling, a lot has happened. Graph pebbling has been an important instrument for the conveyance of consumable resources for the past 30 years. Through out this paper, we have considered a simple connected graph G.

Let us denote $G'$'s vertex and edge sets as $V(G)$and $E(G)$, respectively. Consider a graph with a fixed number of pebbles at each vertex. One pebble is thrown away and the other is placed on an adjacent vertex when two pebbles are removed from a vertex. This process is known as a pebble move. The pebbling number of a vertex $v$ in a graph $G$ is the smallest number $f(G, v)$ that allows us to shift a pebble to $v$ using continuous pebbling move, regardless of where these pebbles are located on $G$'s vertices. The pebble number, $f(G)$, is the maximum $f(G, v)$ over all the vertices of a graph.

The pebble concept is similar to network optimization for military forces or other resources that would progressively disappear after they are transmitted. However, if there are any blocks or disturbances while transmitting through a specific medium, it is natural to wonder how any node in a connected environment can be guaranteed to reach any other node with the available resources. As a result, it motivates the concept of the detour in graph pebbling to ensure that any vertex in the connected graph can be reached.
DETOUR PEBBLING NUMBER OF ROOTED PRODUCT OF
SOME GRAPHS

A. LOURDUSAMY, I. DHIVVIYANANDAM*, AND SAGAYA SUGANYA A

ABSTRACT. A pebbling transformation is considered as the shifting of two pebbles from a vertex and the insertion of one of them onto an adjacent vertex, given a configuration of pebbles on the vertices of a connected graph G. The detour pebbling number, $f^*(G)$, of a simple connected graph G is the smallest positive integer such that a sequence of pebbling moves can transfer one pebbles to any target vertex for any distribution of $f^*(G)$ pebbles on G's vertices. The detour pebbling number, $f^*(G)$, of a graph G is obtained by the detour path. We discuss the detour pebbling number of rooted product of path, complete and fan graphs.

**Keywords:** pebbling moves, pebbling number, detour pebbling number, rooted product.

**2020 AMS Subject Classification:** 05C12, 05C25, 05C38, 05C76.

# 1. INTRODUCTION

Pebbling, introduced by Lagarias and Saks, has sparked a lot of interest. Chung[9] was the first to put it into the literature, and many others have followed suit, including Hulbert, who published an overview of graph pebbling[11]. Since Hulbert's survey first appeared in graph pebbling, a lot has happened. Graph pebbling has been an important instrument for the conveyance of consumable resources for the past 30 years. Through out this paper, we have considered a simple connected graph G.

Let us denote $G'$'s vertex and edge sets as $V(G)$and $E(G)$, respectively. Consider a graph with a fixed number of pebbles at each vertex. One pebble is thrown away and the other is placed on an adjacent vertex when two pebbles are removed from a vertex. This process is known as a pebble move. The pebbling number of a vertex $v$ in a graph $G$ is the smallest number $f(G, v)$ that allows us to shift a pebble to $v$ using continuous pebbling move, regardless of where these pebbles are located on $G$'s vertices. The pebble number, $f(G)$, is the maximum $f(G, v)$ over all the vertices of a graph.

The pebble concept is similar to network optimization for military forces or other resources that would progressively disappear after they are transmitted. However, if there are any blocks or disturbances while transmitting through a specific medium, it is natural to wonder how any node in a connected environment can be guaranteed to reach any other node with the available resources. As a result, it motivates the concept of the detour in graph pebbling to ensure that any vertex in the connected graph can be reached.
There are many articles [1, 2, 4, 5, 6, 7] discussed about the pebbling number and t-pebbling number of some graphs.

## 2. PRELIMINARIES

For graph theoretic terminologies, the reader can refer to [10].

**Definition 2.1.** [3] A detour pebbling number of a vertex $v$ of a graph $G$ is the smallest number $f^*(G, v)$ that allows a pebble to be moved to $v$ via a detour path by a sequence of pebbling moves for any placement of $f^*(G, v)$ pebbles on the vertices of $G$. The detour pebbling number of a graph is denoted by $f^*(G)$, is the maximum $f^*(G, v)$ over all the vertices of $G$.

**Definition 2.2.** [8] A detour t-pebbling number of a graph $G$ vertex $v$ is the smallest number $f_v^*(G, v)$ for which $t$ pebbles can be moved to $v$ using a detour path by a sequence of pebbling moves for any placement of $f_v^*(G, v)$ pebbles on the vertices of $G$.

**Theorem 2.3.** [3] For any path $P_n$ with $n$ vertices, the detour pebbling number is $f^*(P_n) = 2^{n-1}$

**Theorem 2.4.** [3] Consider $f^*(K_n) = 2^{n-1}$ with $n$ vertices ($n > 1$).

**Theorem 2.5.** [3] Consider $f^*(F_n) = 2^{n-1}$. In the case of the fan graph $F_n$ with $n$ vertices.

**Definition 2.6.** Let $G$ be a connected graph with $n$ vertices. Let the vertex set be $V(G) = \{x_1, x_2, \cdots, x_n\}$ and let $H$ be also a connected graph with $m$ vertices whose vertex set be $V(H) = \{y_1, y_2, \cdots, y_m\}$. Fix any vertex from $H$. Let $y_i$ be the vertex, where $1 \le i \le m$. Let this be the root vertex $y$. The rooted product of two graphs $G$ and $H$ denoted by $G \odot_y H$ is a graph that is obtained from identifying the root vertex $y$ of $H$ with every $i^{th}$ vertex $y$ of $G$ is $i = 1, 2, \cdots, n$ the vertex set of $G \odot_y H$ is $V(G) \times V(H)$.

In $G \odot_y H$ the vertices of $G$ are labeled as $(x_i, y)$ for $1 \le i \le n$ and the vertices of $H$ are labeled as $(x_i, y_1), (x_i, y_2), \cdots, (x_i, y_{m-1})$ sequentially for every $i^{th}$ copy of $G$. Let $d_G^*(x_i, x_j)$ denote the length of the detour length of the path between $x_i$ and $x_j$ for $(x_i, x_j) \in G$. The detour distance between two vertices $(x_i, y_l)(x_j, y_k) \in V(G \odot_y H)$ is denoted by $D^*$ where $D^* = d_H^*(y_1, y) + d_G^*(x_i, x_j) + d_H^*(y, y_k)$ where $y$ is the root vertex of $H$. Let $p(x_i, y_j)$ denote the number of pebbles initially placed on the vertex $(x_i, y_j)$. For $\{(x_i, y_j)(x_j, y_k) \in E(G \odot_y H)\} (x_i, y_j) \xrightarrow{t} (x_l, y_k)$ refers taking off atleast 2t pebbles from $(x_i, y_j)$ and placing atleast $t$ pebbles on $(x_l, y_k)$. In this paper, we denote $S_i$ as the detour path and $S_i$ be the vertices or detour paths which are not on $S_i$, where $i$ is a non negative positive numbers.

**Remark 2.7.** Consider graph $G$, which has a pebble configuration on its vertices. From $G$, we select a destination vertex $w$. We can easily shift a pebble to $w$ if $p(w) = 1$ or $p(s) \ge 2$, where $w \in E(G)$. When $w$ is the destination vertex, we consider that $p(s) \le 1$ and $p(w) = 0$ for all $w \in E(G)$ because if the destination vertex has a pebble then there is nothing to discuss.
There are many articles [1, 2, 4, 5, 6, 7] discussed about the pebbling number and t-pebbling number of some graphs.

## 2. PRELIMINARIES

For graph theoretic terminologies, the reader can refer to [10].

**Definition 2.1.** [3] A detour pebbling number of a vertex $v$ of a graph $G$ is the smallest number $f^*(G, v)$ that allows a pebble to be moved to $v$ via a detour path by a sequence of pebbling moves for any placement of $f^*(G, v)$ pebbles on the vertices of $G$. The detour pebbling number of a graph is denoted by $f^*(G)$, is the maximum $f^*(G, v)$ over all the vertices of $G$.

**Definition 2.2.** [8] A detour t-pebbling number of a graph $G$ vertex $v$ is the smallest number $f_v^*(G, v)$ for which $t$ pebbles can be moved to $v$ using a detour path by a sequence of pebbling moves for any placement of $f_v^*(G, v)$ pebbles on the vertices of $G$.

**Theorem 2.3.** [3] For any path $P_n$ with $n$ vertices, the detour pebbling number is $f^*(P_n) = 2^{n-1}$

**Theorem 2.4.** [3] Consider $f^*(K_n) = 2^{n-1}$ with $n$ vertices ($n > 1$).

**Theorem 2.5.** [3] Consider $f^*(F_n) = 2^{n-1}$. In the case of the fan graph $F_n$ with $n$ vertices.

**Definition 2.6.** Let $G$ be a connected graph with $n$ vertices. Let the vertex set be $V(G) = \{x_1, x_2, \cdots, x_n\}$ and let $H$ be also a connected graph with $m$ vertices whose vertex set be $V(H) = \{y_1, y_2, \cdots, y_m\}$. Fix any vertex from $H$. Let $y_i$ be the vertex, where $1 \le i \le m$. Let this be the root vertex $y$. The rooted product of two graphs $G$ and $H$ denoted by $G \odot_y H$ is a graph that is obtained from identifying the root vertex $y$ of $H$ with every $i^{th}$ vertex $y$ of $G$ is $i = 1, 2, \cdots, n$ the vertex set of $G \odot_y H$ is $V(G) \times V(H)$.

In $G \odot_y H$ the vertices of $G$ are labeled as $(x_i, y)$ for $1 \le i \le n$ and the vertices of $H$ are labeled as $(x_i, y_1), (x_i, y_2), \cdots, (x_i, y_{m-1})$ sequentially for every $i^{th}$ copy of $G$. Let $d_G^*(x_i, x_j)$ denote the length of the detour length of the path between $x_i$ and $x_j$ for $(x_i, x_j) \in G$. The detour distance between two vertices $(x_i, y_l)(x_j, y_k) \in V(G \odot_y H)$ is denoted by $D^*$ where $D^* = d_H^*(y_1, y) + d_G^*(x_i, x_j) + d_H^*(y, y_k)$ where $y$ is the root vertex of $H$. Let $p(x_i, y_j)$ denote the number of pebbles initially placed on the vertex $(x_i, y_j)$. For $\{(x_i, y_j)(x_j, y_k) \in E(G \odot_y H)\} (x_i, y_j) \xrightarrow{t} (x_l, y_k)$ refers taking off atleast 2t pebbles from $(x_i, y_j)$ and placing atleast $t$ pebbles on $(x_l, y_k)$. In this paper, we denote $S_i$ as the detour path and $S_i$ be the vertices or detour paths which are not on $S_i$, where $i$ is a non negative positive numbers.

**Remark 2.7.** Consider graph $G$, which has a pebble configuration on its vertices. From $G$, we select a destination vertex $w$. We can easily shift a pebble to $w$ if $p(w) = 1$ or $p(s) \ge 2$, where $w \in E(G)$. When $w$ is the destination vertex, we consider that $p(s) \le 1$ and $p(w) = 0$ for all $w \in E(G)$ because if the destination vertex has a pebble then there is nothing to discuss.
### 3. MAIN RESULTS

**Theorem 3.1.** For the graph $P_n \odot_y P_m$, the detour pebbling number is $f^*(P_n \odot_y P_m) = 2^{2m+n-3} + (n-2)(2^{m-1}-1)$ where $n \ge 3, m \ge 3$ and $d(y) = 1$.

*Proof.* The detour distance from $(x_1, y_{m-1})$ to any other vertex on $G \odot_y H$ is $d^*(G \odot_y H) \le 2m+n-3$. Let us consider $G = P_n$ and $H = P_m$ fix $y$ as the rooted vertex of $H$ and $d(y) = 1$. Placing $2^{2m+n-3}-1$ pebbles on $(x_n, y_{m-1})$ and distributing $(2^{m-1}-1)$ pebbles each on every $(n-2)$ copies of $H$ which are not on the detour path we cannot reach the target. Hence the detour pebbling number of $f^*(P_n \odot y P_m) \ge 2^{2m+n-3} + (n-2)(2^{m-1}-1)$
To prove the sufficient part.

**Case 1:** Let $(x_1, y_{n-1})$ be the target.
The detour distance from $(x_1, y_{m-1})$ to any other vertex is at most $(2m+n-3)$. Let the detour path $S_1$ be $(x_n, y_{m-1}) \rightarrow (x_n, y_{n-2}) \rightarrow \cdots (x_n, y) \rightarrow (x_{n-1}, y) \rightarrow \cdots \rightarrow (x_1, y) \rightarrow \cdots (x_1, y_{m-1})$. For this particular detour path $S_1$, there are $(n-2)$ copies of $P_{m-1}$ paths not on $S_1$. Since the length of $S_1 = d^*((x_1, y_{m-1}), (x_n, y_{m-1})) = (2m+n-3)$, then distributing $2^{2m+n-3}$ pebbles on $S_1$ and $(n-2)(2^{(m-1)}-1)$ pebbles on remaining $n-2$ copies of $P_{m-1}$ paths such a way that no pebble should reach on $S_1$ using the detour path, we can reach the target.

**Subcase 1.1:** Let $p(V(S_1) < 2^{2m+n-3}$ and $p(V(S_1) \ge (n-2)(2^{(m-1)}))$. If we transfer at least a pebble each from $n-2$ copies of $P_{m-1}$ paths of $S_1$ to $S_1$ and distributing at least $2^m$ pebbles on $(x_n, y_i)$ vertices, we can transfer $(x_n, y_i) \xrightarrow{2^m} (x_n, y) \xrightarrow{1} \cdots \xrightarrow{1} (x_1, y)$. Now we require at most $(2^{(m-1)}-1)$ pebbles on $(x_1, y_i)$ vertices to place a pebble on the target. Hence, We are done. If $p(x_n, y_i)$ has at least $2^{2m+n-3} + (n-2)(2^{m-1}-1) - (n-2)(2^m) = 2^{2m+n-3} - (n-2)(2^{m-1}+1)$ pebbles, we can transfer at least $(x_n, y_i) \le 2^{(m+n-2)} \xrightarrow{2^{(m+n-2)}} (x_n, y) \le 2^{(m+n-3)} \xrightarrow{2^{(m+n-3)}} \cdots \ge 2^{(m-1)} \xrightarrow{2^{(m-1)}} (x_1, y)$ pebbles. Hence, we are done. By symmetry for $(x_n, y_{m-1})$ we can reach the target.

**Case 2:** Let $(x_1, y)$ be the target.
The detour distance from $(x_1, y)$ to any other vertex is $(m+n-2)$. Let the detour path $S_2$ be $(x_n, y_{m-1}) \rightarrow \cdots \rightarrow (x_n, y) \rightarrow (x_{n-1}, y) \rightarrow \cdots \rightarrow (x_1, y)$. There are $(n-1)$ copies of $P_{m-1}$ paths which are not on this detour path. Let it be $S_2$. Hence distributing $2^{m+n-2}$ pebbles on $S_2$ and $(n-1)(2^{m-1}-1)$ pebbles on $S_2$, we can reach the target. When we distribute pebbles on $S_2$, we should see to it no pebble should reach $S_2$. Thus, each $P_{m-1}$ path receives $(2^{m-1}-1)$ pebbles. Hence the detour pebbling number of $f^*(P_n \odot y P_m, x_1, y) = 2^{(m+n-2)} + (n-1)(2^{(m-1)}-1)$ pebbles. By symmetry we can prove for the target $(x_n, y)$.

**Subcase 2.1:** Let $p(V(S_2) < 2^{m+n-2}$ and $p(V(S_2) \ge (n-1)(2^{(m-1)}))$. If we transfer at least a pebble each from $n-2$ copies of $P_{m-1}$ paths of $S_2$ to $S_1$ except from the vertices of $(x_1, y_i)$ and distributing at least $2^m$ pebbles on $(x_n, y_i)$ vertices, we can transfer $(x_n, y_i) \xrightarrow{2^m} (x_n, y) \xrightarrow{1} \cdots \xrightarrow{1} (x_1, y)$. Hence, we are done. Otherwise, if we distribute $2^{m-1}$ pebbles on $(x_1, y_i)$ vertices, we can put a pebble on the target without using pebbles from the remaining vertices. Let $p(V(S_2) = 0$. If we transfer pebbles from $S_2$ to $S_2$ and have at most $2^{(n-1)}$ pebbles, then we are done. By symmetry for $(x_n, y_1)$ we can reach the target.
### 3. MAIN RESULTS

**Theorem 3.1.** For the graph $P_n \odot_y P_m$, the detour pebbling number is $f^*(P_n \odot_y P_m) = 2^{2m+n-3} + (n-2)(2^{m-1}-1)$ where $n \ge 3, m \ge 3$ and $d(y) = 1$.

*Proof.* The detour distance from $(x_1, y_{m-1})$ to any other vertex on $G \odot_y H$ is $d^*(G \odot_y H) \le 2m + n - 3$. Let us consider $G = P_n$ and $H = P_m$ fix $y$ as the rooted vertex of $H$ and $d(y) = 1$. Placing $2^{2m+n-3}-1$ pebbles on $(x_n, y_{m-1})$ and distributing $(2^{(m-1)} - 1)$ pebbles each on every $(n-2)$ copies of $H$ which are not on the detour path we cannot reach the target. Hence the detour pebbling number of $f^*(P_n \odot y P_m) \ge 2^{2m+n-3} + (n-2)(2^{m-1}-1)$
To prove the sufficient part.

**Case 1:** Let $(x_1, y_{n-1})$ be the target.
The detour distance from $(x_1, y_{m-1})$ to any other vertex is at most $(2m+n-3)$. Let the detour path $S_1$ be $(x_n, y_{m-1}) \to (x_n, y_{n-2}) \to \cdots (x_n, y) \to (x_{n-1}, y) \to \cdots \to (x_1, y) \to \cdots (x_1, y_{m-1})$. For this particular detour path $S_1$, there are $(n-2)$ copies of $P_{m-1}$ paths not on $S_1$. Since the length of $S_1 = d^*((x_1, y_{m-1}), (x_n, y_{m-1})) = (2m+n-3)$, then distributing $2^{2m+n-3}$ pebbles on $S_1$ and $(n-2)(2^{(m-1)} - 1)$ pebbles on remaining $n-2$ copies of $P_{m-1}$ paths such a way that no pebble should reach on $S_1$ using the detour path, we can reach the target.

**Subcase 1.1:** Let $p(V(S_1) < 2^{2m+n-3}$ and $p(V(S_1) \ge (n-2)(2^{(m-1)}))$. If we transfer at least a pebble each from $n-2$ copies of $P_{m-1}$ paths of $S_1$ to $S_1$ and distributing at least $2^m$ pebbles on $(x_n, y_i)$ vertices, we can transfer $(x_n, y_i) \xrightarrow{2^m} (x_n, y) \xrightarrow{1} \cdots \xrightarrow{1} (x_1, y)$. Now we require at most $(2^{(m-1)} - 1)$ pebbles on $(x_1, y_i)$ vertices to place a pebble on the target. Hence, We are done. If $p(x_n, y_i)$ has at least $2^{2m+n-3} + (n-2)(2^{m-1}-1) - (n-2)(2^m) = 2^{2m+n-3} - (n-2)(2^{m-1}+1)$ pebbles, we can transfer at least $(x_n, y_i) \le 2^{(m+n-2)} \xrightarrow{2^{(m+n-2)}} (x_n, y) \le 2^{(m+n-3)} \xrightarrow{2^{(m+n-3)}} \cdots \ge 2^{(m-1)} \xrightarrow{2^{(m-1)}} (x_1, y)$ pebbles. Hence, we are done. By symmetry for $(x_n, y_{m-1})$ we can reach the target.

**Case 2:** Let $(x_1, y)$ be the target.
The detour distance from $(x_1, y)$ to any other vertex is $(m+n-2)$. Let the detour path $S_2$ be $(x_n, y_{m-1}) \to \cdots \to (x_n, y) \to (x_{n-1}, y) \to \cdots \to (x_1, y)$. There are $(n-1)$ copies of $P_{m-1}$ paths which are not on this detour path. Let it be $S_2$. Hence distributing $2^{m+n-2}$ pebbles on $S_2$ and $(n-1)(2^{m-1}-1)$ pebbles on $S_2$, we can reach the target. When we distribute pebbles on $S_2$, we should see to it no pebble should reach $S_2$. Thus, each $P_{m-1}$ path receives $(2^{m-1}-1)$ pebbles. Hence the detour pebbling number of $f^*(P_n \odot y P_m, x_1, y) = 2^{(m+n-2)} + (n-1)(2^{(m-1)} - 1)$ pebbles. By symmetry we can prove for the target $(x_n, y)$.

**Subcase 2.1:** Let $p(V(S_2) < 2^{m+n-2}$ and $p(V(S_2) \ge (n-1)(2^{(m-1)}))$. If we transfer at least a pebble each from $n-2$ copies of $P_{m-1}$ paths of $S_2$ to $S_1$ except from the vertices of $(x_1, y_i)$ and distributing at least $2^m$ pebbles on $(x_n, y_i)$ vertices, we can transfer $(x_n, y_i) \xrightarrow{2^m} (x_n, y) \xrightarrow{1} \cdots \xrightarrow{1} (x_1, y)$. Hence, we are done. Otherwise, if we distribute $2^{m-1}$ pebbles on $(x_1, y_i)$ vertices, we can put a pebble on the target without using pebbles from the remaining vertices. Let $p(V(S_2) = 0$. If we transfer pebbles from $S_2$ to $S_2$ and have at most $2^{(n-1)}$ pebbles, then we are done. By symmetry for $(x_n, y_1)$ we can reach the target.
**Case 3:** Let the target be $(x_k, y_l)$ where $k \neq 1$ or $n$ and $l = \{1, 2, \dots, m-1\}$. The detour distance between $(x_k, y_l)$ to any other vertex is $\le 2m+n-4$. There are $(n-3)$ copies of $P_{m-1}$ paths and a $P_m$ path which do not touch the detour path. Placing $2^{(2m+n-4)}$ pebbles on the detour path and distributing $(n-3)(2^{(m-1)}-1)$ pebbles on $(n-3)$ copies of $P_{m-1}$ paths and $2^{m-1}$ pebbles on one copy of $P_m$ such a way no pebble reaches the rooted vertex of $S_1$. The total number of pebbles used to reach the target is $2^{(2m+n-4)} + (n-1)(2^{(m-1)}) - (n-2)$ pebbles. Hence the detour pebbling number of $f^*(P_n \odot yP_m, x_k, y_l) = 2^{(2m+n-4)} + (n-1)(2^{(m-1)}) - (n-2)$.

**Subcase 3.1:** Let the target vertex be $(x_k, y_l)$ where $k \neq 1, n$ then $1 \le l \le m-1$ as $2 \le k \le n-1$. Then the detour distance from $x_k, y_l$ to any other vertex is $\le 2m+n-5$. There will be 2 copies of $P_m$ paths and $n-4$ copies of $P_{m-1}$ paths which are not on the detour path from $x_k, y_l$ to any other vertex. Hence the detour pebbling number of $f^*(P_n \odot yP_m, x_k, y_l) = 2^{2m+n-5} + 2^{m+1} + (n-4)(2^{m-1}-1)$.

**Case 4.** Let the target be $(x_k, y)$ where $k \neq 1$ or $n$
Then the detour distance from $(x_k, y)$ to any other vertex is $\le (m+n-3)$. There
will be at least one copy of $P_m$ paths and $(n-2)$ copies of $P_{m-1}$ paths which are not
on the detour path. Placing $2^{m+n-3}$ pebbles on the detour path and distributing
$(n-2)(2^{(m-1)}-1)$ pebble on $(n-2)$ copies of $(P_{m-1})$ paths such a way no pebbles
reach the target and placing $2^{m-1}$ pebbles on $P_m$ path, we get the detour pebbling
number as $2^{m+n-3} + (n-1)(2^{(m-1)}) - (n-2)$. Thus, We can reach the target.

**Subcase 4.1:** Let the target vertex be $(x_k, y)$ where $k \neq 1, n$.
The detour distance between $(x_k, y_l)$ to any other vertex is $\le (m+n-4)$. There
will be 2 copies of $P_m$ paths and $(n-4)$, $P_{m-1}$ paths such a way no pebbles reach
the detour path. Placing $2^{m+n-4}$ pebbles on the detour path and distributing
$(n-2)(2^{(m-1)}-1)$ pebble on $(n-2)$ copies of $(P_{m-1})$ paths such a way no pebbles
reach the target and placing $2^{m-1}$ pebbles on $P_m$ path, we get the detour pebbling
number as $2^{m+n-4} + (n-6)(2^{(m-1)}) - (n-4)$. Thus, we can reach the target.
Hence the detour pebbling number is $2^{m+n-4} + (n-6)(2^{(m-1)}) - (n-4)$ pebbles.
Henc the detour pebbling number for the $(P_n \odot yP_m)$, where $d(y) = 1$, is $2^{2m+n-3} +
(n-2)(2^{m-1}-1)$. $\square$

**Theorem 3.2.** For the graph $P_n \odot y[\lfloor \frac{m}{2} \rfloor]P_m$, the detour pebbling number is

$$
f^*(P_n \odot y[\lfloor \frac{m}{2} \rfloor]P_m) = \begin{cases} 2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1) & \text{if m is odd} \\ 2^{m+n-1} + (n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1) + n(2^{\lfloor \frac{m}{2} \rfloor} - 2) & \text{if m is even} \end{cases} \\ \text{where } d(y_{2^{\lfloor \frac{m}{2} \rfloor}}) = 2.
$$

Proof. 1: When m is odd.

The detour distance between $(x_1, y_m)$ to any vertex is $\le 2^{m+n-2}$. Let us consider the detour path $S_1$ be $(x_1, y_m) \rightarrow (x_1, y_{m-1}) \rightarrow \cdots \rightarrow (x_1, y_{\lceil \frac{m}{2} \rceil}) \cdots \rightarrow (x_n, y_{\lceil \frac{m}{2} \rceil}) \rightarrow \cdots \rightarrow (x_n, y_m)$ whose length is $m+n-1$. There are $(2n-2)$ copies of $P_{\lceil \frac{m}{2} \rceil-1}$ paths which have the root on the detour path. Let it be $S_1^-$.
Distributing $2^{m+n-2}-1$ pebbles on $(x_n, y_m)$ and $(2n-2)(2^{\lfloor \frac{m}{2} \rfloor}-1)$ pebbles on $S_1^-$ such a way no pebble reaches the root, we can not reach the target $(x_1, y_m)$.
**Case 3:** Let the target be $(x_k, y_l)$ where $k \neq 1$ or $n$ and $l = \{1, 2, \dots, m-1\}$. The detour distance between $(x_k, y_l)$ to any other vertex is $\le 2m+n-4$. There are $(n-3)$ copies of $P_{m-1}$ paths and a $P_m$ path which do not touch the detour path. Placing $2^{(2m+n-4)}$ pebbles on the detour path and distributing $(n-3)(2^{(m-1)}-1)$ pebbles on $(n-3)$ copies of $P_{m-1}$ paths and $2^{m-1}$ pebbles on one copy of $P_m$ such a way no pebble reaches the rooted vertex of $S_1$. The total number of pebbles used to reach the target is $2^{(2m+n-4)} + (n-1)(2^{(m-1)}) - (n-2)$ pebbles. Hence the detour pebbling number of $f^*(P_n \odot yP_m, x_k, y_l) = 2^{(2m+n-4)} + (n-1)(2^{(m-1)}) - (n-2)$.

**Subcase 3.1:** Let the target vertex be $(x_k, y_l)$ where $k \neq 1, n$ then $1 \le l \le m-1$ as $2 \le k \le n-1$. Then the detour distance from $x_k, y_l$ to any other vertex is $\le 2m+n-5$. There will be 2 copies of $P_m$ paths and $n-4$ copies of $P_{m-1}$ paths which are not on the detour path from $x_k, y_l$ to any other vertex. Hence the detour pebbling number of $f^*(P_n \odot yP_m, x_k, y_l) = 2^{2m+n-5} + 2^{m+1} + (n-4)(2^{m-1}-1)$.

**Case 4.** Let the target be $(x_k, y)$ where $k \neq 1$ or $n$
Then the detour distance from $(x_k, y)$ to any other vertex is $\le (m+n-3)$. There
will be at least one copy of $P_m$ paths and $(n-2)$ copies of $P_{m-1}$ paths which are not
on the detour path. Placing $2^{m+n-3}$ pebbles on the detour path and distributing
$(n-2)(2^{(m-1)}-1)$ pebble on $(n-2)$ copies of $(P_{m-1})$ paths such a way no pebbles
reach the target and placing $2^{m-1}$ pebbles on $P_m$ path, we get the detour pebbling
number as $2^{m+n-3} + (n-1)(2^{(m-1)}) - (n-2)$. Thus, We can reach the target.

**Subcase 4.1:** Let the target vertex be $(x_k, y)$ where $k \neq 1, n$.
The detour distance between $(x_k, y_l)$ to any other vertex is $\le (m+n-4)$. There
will be 2 copies of $P_m$ paths and $(n-4)$, $P_{m-1}$ paths such a way no pebbles reach
the detour path. Placing $2^{m+n-4}$ pebbles on the detour path and distributing
$(n-2)(2^{(m-1)}-1)$ pebble on $(n-2)$ copies of $(P_{m-1})$ paths such a way no pebbles
reach the target and placing $2^{m-1}$ pebbles on $P_m$ path, we get the detour pebbling
number as $2^{m+n-4} + (n-6)(2^{(m-1)}) - (n-4)$. Thus, we can reach the target.
Hence the detour pebbling number is $2^{m+n-4} + (n-6)(2^{(m-1)}) - (n-4)$ pebbles.
Henc the detour pebbling number for the $(P_n \odot yP_m)$, where $d(y) = 1$, is $2^{2m+n-3} +
(n-2)(2^{m-1}-1)$. $\square$

**Theorem 3.2.** For the graph $P_n \odot y[\lfloor \frac{m}{2} \rfloor]P_m$, the detour pebbling number is

$$
f^*(P_n \odot y[\lfloor \frac{m}{2} \rfloor]P_m) = \begin{cases} 2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1) & \text{if m is odd} \\ 2^{m+n-1} + (n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1) + n(2^{\lfloor \frac{m}{2} \rfloor} - 2) & \text{if m is even} \end{cases} \\
\text{where } d(y_{2^{\lfloor \frac{m}{2} \rfloor}}) = 2.
$$

Proof. 1: When m is odd.

The detour distance between $(x_1, y_m)$ to any vertex is $\le 2^{m+n-2}$. Let us consider the detour path $S_1$ be $(x_1, y_m) \rightarrow (x_1, y_{m-1}) \rightarrow \cdots \rightarrow (x_1, y_{\lceil \frac{m}{2} \rceil}) \cdots \rightarrow (x_n, y_{\lceil \frac{m}{2} \rceil}) \rightarrow \cdots \rightarrow (x_n, y_m)$ whose length is $m+n-1$. There are $(2n-2)$ copies of $P_{\lceil \frac{m}{2} \rceil-1}$ paths which have the root on the detour path. Let it be $S_1^-$.
Distributing $2^{m+n-2}-1$ pebbles on $(x_n, y_m)$ and $(2n-2)(2^{\lfloor \frac{m}{2} \rfloor}-1)$ pebbles on $S_1^-$ such a way no pebble reaches the root, we can not reach the target $(x_1, y_m)$.
Hence the detour pebbling number of $f^*(P_n \odot y_{\lfloor \frac{m}{2} \rfloor} P_m)$ where $d(y_{\lfloor \frac{m}{2} \rfloor}) = 2$ is
$\ge 2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$.

Consider $2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles on the vertices of $P_n \odot y_{\lfloor \frac{m}{2} \rfloor} P_m$.
Now we will prove the sufficient condition for the $P_n \odot y_{\lfloor \frac{m}{2} \rfloor} P_m$.

**Case 1.** Let the target vertex be $(x_n, y_m)$ or $(x_n, y_1)$. Without loss of generality, Let us consider $(x_n, y_m)$ be the target vertex. The detour path $S_2$ be $(x_1, y_m) \rightarrow (x_1, y_{m-1}) \rightarrow \cdots \rightarrow (x_1, y_{\lfloor \frac{m}{2} \rfloor}) \cdots \rightarrow (x_n, y_{\lfloor \frac{m}{2} \rfloor}) \rightarrow \cdots \rightarrow (x_n, y_m)$. The length of $S_2$ is $m+n-2$. There are $(2n-2)$ copies of $P_{\lfloor \frac{m}{2} \rfloor-1}$ paths which have the root on the detour path. Let it be $S_2$. Distributing $2^{m+n-2}$ pebbles on $(x_1, y_m)$ and $(2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles on $S_2$ such a way no pebble reaches the root, we can reach the target $(x_n, y_m)$.

**Subcase 1.1:** Let $p(V(S_2)) < 2^{m+n-2}$. If there exist at least $(2^{\lfloor \frac{m}{2} \rfloor})$ pebbles each on $(2n-2)$ copies of $P_{\lfloor \frac{m}{2} \rfloor-1}$, we can transfer 2 pebbles each to the $n-2$ roots.
Hence, we can shift a pebble to $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$. Now either placing $2^{\lfloor \frac{m}{2} \rfloor} - 1$ pebbles on $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$ or distributing at most $2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1) - (2n-2)(2^{\lfloor \frac{m}{2} \rfloor}) = 2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1) - (2n-2)$ pebbles on $(x_1, y_m)$ we can transfer $(x_1, y_m) < 2^{m+n-3} - (2n-2) \xrightarrow{\phantom{=}} (x_1, y_{m-1}) < 2^{m+n-4} - (2n-2) \cdots < 2^{\lfloor \frac{m}{2} \rfloor} + n-1 - (2n-2) \xrightarrow{\phantom{=}} (x_1, y_{\lfloor \frac{m}{2} \rfloor}) \cdots 2^{\lfloor \frac{m}{2} \rfloor} - 1 \xrightarrow{\phantom{=}} (x_n, y_{\lfloor \frac{m}{2} \rfloor})$. Using $2^{\lfloor \frac{m}{2} \rfloor} - 1$ and adding the pebbles shifted from $V(S_2)$ we get at least $2^{\lfloor \frac{m}{2} \rfloor}$. By which we can transfer $(x_n, y_{\lfloor \frac{m}{2} \rfloor}) \xrightarrow{\phantom{=} 2^{\lfloor \frac{m}{2} \rfloor}} \cdots \xrightarrow{\phantom{=} 1} (x_n, y_m)$. Hence, we are done. By symmetry, we can prove for $(x_1, y_m)$.

**Case 2.** Let the target vertex be $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$ or $(x_1, y_{\lfloor \frac{m}{2} \rfloor})$
Without loss of generality, let the target vertex be $(x_1, y_{\lfloor \frac{m}{2} \rfloor})$. Consider the detour path $S_3$ be $(x_n, y_{m-1}) \rightarrow (x_n, y_{m-2}) \rightarrow \cdots \rightarrow (x_n, y_{\lfloor \frac{m}{2} \rfloor}) \cdots \rightarrow (x_1, y_{\lfloor \frac{m}{2} \rfloor})$. The length of the detour path is $(\lfloor \frac{m}{2} \rfloor + n - 1)$. There are $(2n-1)$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths not on $S_3$. Let it be $S_3$. By distributing $2^{(\lfloor \frac{m}{2} \rfloor + n-1)}$ pebbles on $S_3$ and $(2n-1)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles on the vertices of $S_3$ such a way no pebble should reach the detour path. Thus, we can reach the target by using the pebbles on $S_3$ alone. The total number of pebbles used in this configuration is $2^{(\lfloor \frac{m}{2} \rfloor + n-1)} + (2n-1)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$

**Case 2.1:** Let $p(V(S_3)) < 2^{(\lfloor \frac{m}{2} \rfloor + n-1)}$.
If there exist at least $2^{(\lfloor \frac{m}{2} \rfloor)}$ pebbles on the $(x_n, y_i)$ vertices, where $(\lfloor \frac{m}{2} \rfloor + 1) \le i \le m$, we can shift a pebble from $(x_n, y_i)$ and another pebble from $(x_n, y_j)$, $(\lfloor \frac{m}{2} \rfloor - 1 \ge j \ge 1)$. Hence, we can transfer 2 pebbles to $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$ and a pebble each to the $(n-2)$ root vertices, which lies on the $V(S_3)$, by using $(2^{\lfloor \frac{m}{2} \rfloor})$ pebbles on either $(x_k, y_i)$ vertices, where $2 \le k \le n-1$, and $(\lfloor \frac{m}{2} \rfloor + 1) \le i \le m$, or $(x_k, y_j)$, where $2 \le k \le n-1$, and $(\lfloor \frac{m}{2} \rfloor - 1 \ge j \ge 1)$. Hence, we can shift a pebble to $(x_1, y_{\lfloor \frac{m}{2} \rfloor})$.
Hence the detour pebbling number of $f^*(P_n \odot y_{\lfloor \frac{m}{2} \rfloor} P_m)$ where $d(y_{\lfloor \frac{m}{2} \rfloor}) = 2$ is
$\ge 2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$.

Consider $2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles on the vertices of $P_n \odot y_{\lfloor \frac{m}{2} \rfloor} P_m$.
Now we will prove the sufficient condition for the $P_n \odot y_{\lfloor \frac{m}{2} \rfloor} P_m$.

**Case 1.** Let the target vertex be $(x_n, y_m)$ or $(x_n, y_1)$. Without loss of generality, Let us consider $(x_n, y_m)$ be the target vertex. The detour path $S_2$ be $(x_1, y_m) \rightarrow (x_1, y_{m-1}) \rightarrow \cdots \rightarrow (x_1, y_{\lfloor \frac{m}{2} \rfloor}) \cdots \rightarrow (x_n, y_{\lfloor \frac{m}{2} \rfloor}) \rightarrow \cdots \rightarrow (x_n, y_m)$. The length of $S_2$ is $m+n-2$. There are $(2n-2)$ copies of $P_{\lfloor \frac{m}{2} \rfloor-1}$ paths which have the root on the detour path. Let it be $S_2$. Distributing $2^{m+n-2}$ pebbles on $(x_1, y_m)$ and $(2n-2)(2^{\lfloor \frac{m}{2} \rfloor}-1)$ pebbles on $S_2$ such a way no pebble reaches the root, we can reach the target $(x_n, y_m)$.

**Subcase 1.1:** Let $p(V(S_2)) < 2^{m+n-2}$. If there exist at least $(2^{\lfloor \frac{m}{2} \rfloor})$ pebbles each on $(2n-2)$ copies of $P_{\lfloor \frac{m}{2} \rfloor-1}$, we can transfer 2 pebbles each to the $n-2$ roots.
Hence, we can shift a pebble to $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$. Now either placing $2^{\lfloor \frac{m}{2} \rfloor}-1$ pebbles on $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$ or distributing at most $2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor}-1) - (2n-2)(2^{\lfloor \frac{m}{2} \rfloor}) = 2^{m+n-2} + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor}-1) - (2n-2)$ pebbles on $(x_1, y_m)$ we can transfer $(x_1, y_m) < 2^{m+n-3} - (2n-2) < 2^{m+n-4} - (2n-2) \cdots < 2^{\lfloor \frac{m}{2} \rfloor} + n-1 - (2n-2) < (x_1, y_{\lfloor \frac{m}{2} \rfloor}) \cdots 2^{\lfloor \frac{m}{2} \rfloor} - 1 < (x_n, y_{\lfloor \frac{m}{2} \rfloor})$. Using $2^{\lfloor \frac{m}{2} \rfloor}-1$ and adding the pebbles shifted from $V(S_2)$ we get at least $2^{\lfloor \frac{m}{2} \rfloor}$. By which we can transfer $(x_n, y_{\lfloor \frac{m}{2} \rfloor}) (2^{\lfloor \frac{m}{2} \rfloor}) \cdots \mathbf{1} \rightarrow (x_n, y_m)$. Hence, we are done. By symmetry, we can prove for $(x_1, y_m)$.

**Case 2.** Let the target vertex be $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$ or $(x_1, y_{\lfloor \frac{m}{2} \rfloor})$
Without loss of generality, let the target vertex be $(x_1, y_{\lfloor \frac{m}{2} \rfloor})$. Consider the detour path $S_3$ be $(x_n, y_{m-1}) \rightarrow (x_n, y_{m-2}) \rightarrow \cdots \rightarrow (x_n, y_{\lfloor \frac{m}{2} \rfloor}) \cdots \rightarrow (x_1, y_{\lfloor \frac{m}{2} \rfloor})$. The length of the detour path is $(\lfloor \frac{m}{2} \rfloor + n - 1)$. There are $(2n-1)$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths not on $S_3$. Let it be $S_3$. By distributing $2^{(\lfloor \frac{m}{2} \rfloor + n-1)}$ pebbles on $S_3$ and $(2n-1)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles on the vertices of $S_3$ such a way no pebble should reach the detour path. Thus, we can reach the target by using the pebbles on $S_3$ alone. The total number of pebbles used in this configuration is $2^{(\lfloor \frac{m}{2} \rfloor + n-1)} + (2n-1)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$

**Case 2.1:** Let $p(V(S_3)) < 2^{(\lfloor \frac{m}{2} \rfloor + n-1)}$.
If there exist at least $2^{(\lfloor \frac{m}{2} \rfloor)}$ pebbles on the $(x_n, y_i)$ vertices, where $(\lfloor \frac{m}{2} \rfloor + 1) \le i \le m$, we can shift a pebble from $(x_n, y_i)$ and another pebble from $(x_n, y_j)$, $(\lfloor \frac{m}{2} \rfloor - 1 \ge j \ge 1)$. Hence, we can transfer 2 pebbles to $(x_n, y_{\lfloor \frac{m}{2} \rfloor})$ and a pebble each to the $(n-2)$ root vertices, which lies on the $V(S_3)$, by using $(2^{\lfloor \frac{m}{2} \rfloor})$ pebbles on either $(x_k, y_i)$ vertices, where $2 \le k \le n-1$, and $(\lfloor \frac{m}{2} \rfloor + 1) \le i \le m$, or $(x_k, y_j)$, where $2 \le k \le n-1$, and $(\lfloor \frac{m}{2} \rfloor - 1 \ge j \ge 1)$. Hence, we can shift a pebble to $(x_1, y_{\lfloor \frac{m}{2} \rfloor})$.
Now either placing $2^{\lfloor \frac{m}{2} \rfloor} - 1$ pebbles on $(x_1, y_i)$ or $(x_1, y_j)$, we are done. For this all these distribution we used at least $(2^{\lfloor \frac{m}{2} \rfloor} + 1) + (n-2)(2^{\lfloor \frac{m}{2} \rfloor}) + (n)(2^{\lfloor \frac{m}{2} \rfloor} - 1) = 2n(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles. Hence, we are done. By symmetry, we can prove for $(x_n, y_{\lceil \frac{m}{2} \rceil})$.

**Case 3.** Let the target vertex be $(x_k, y_l)$ where $k \neq 1$ or $n$, $1 \le l \le (\lfloor \frac{m}{2} \rfloor)$ or $(\lceil \frac{m}{2} \rceil) + 1 \ge l \ge m$.

The detour path $S_4$ be $(x_1, y_{\lceil \frac{m}{2} \rceil}) \to (x_2, y_{\lceil \frac{m}{2} \rceil}) \to \dots \to (x_{n-1}, y_{\lceil \frac{m}{2} \rceil}) \to (x_{n-1}, y_{\lfloor \frac{n}{2} \rfloor+1}) \dots (x_{n-1}, y_l)$, whose path length is $\le (n-2) + (\lfloor \frac{l}{2} \rfloor - 1)$. There are $2n-1$ $P_{\lfloor \frac{m}{2} \rfloor}$ paths which do not lie on the detour path having the length of $\lfloor \frac{m}{2} \rfloor$. Placing $2^{\lfloor \frac{m}{2} \rfloor-1}$ pebbles each on $P_{\lfloor \frac{m}{2} \rfloor}$ paths and distributing $2^{(n-3)+\lfloor \frac{m}{2} \rfloor}$ pebbles on the detour path we can reach the target. Hence the total number of pebbles used are $2^{(n-3)+\lfloor \frac{m}{2} \rfloor} + (2n-1)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$.

**Subcase 3.1:** Let the target vertex be $(x_k, y_l)$ where $k \neq 1, n$ then $1 \le l \le (\lfloor \frac{m}{2} \rfloor)$ or $(\lceil \frac{m}{2} \rceil) + 1 \ge l \ge m$ as $2 \le k \le n-1$.

The detour path $S_5$ be $(x_2, y_m \to x_2, y_{\lceil \frac{m}{2} \rceil}) \to \dots \to (x_{n-1}, y_{\lceil \frac{m}{2} \rceil}) \to (x_{n-1}, y_{\lfloor \frac{m}{2} \rfloor+1}) \dots (x_{n-1}, y_l)$, whose path length is $\le (n-3) + (\lfloor \frac{m}{2} \rfloor) + (l - \lceil \frac{m}{2} \rceil)$. There are $2n-6$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths and at least 2 copies of $P_m$ paths which do not lie on $S_5$. Let $2n-6$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ be $S_5^{\sim}$ and 2 copies of $P_m$ paths be $S_5^{\text{int}}$. Placing $2^{\lfloor \frac{m}{2} \rfloor-1}$ pebbles each on $2n-6$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths, and $S_5^{\text{int}}$, we can shift at least two pebbles each to every rooted vertices. Using this we could put two pebbles on $(x_{n-1}, y_{\lfloor \frac{m}{2} \rfloor})$. Distributing $2^{(n-3)+(l-\lfloor \frac{m}{2} \rfloor)+(\lfloor \frac{m}{2} \rfloor)}$ pebbles on $(x_2, y_m)$, we can shift at least $(l-\lceil \frac{m}{2} \rceil)+2$ to $(x_{n-1}, y_{(\lfloor \frac{m}{2} \rfloor+1)})$ which is sufficient to reach the target. Hence the total number of pebbles used are $2^{(n-3)+\lfloor \frac{m}{2} \rfloor} + (l-\lfloor \frac{m}{2} \rfloor) + (2n-2)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$.

**Case 4.** Let the target vertex be $(x_k, y_{\lceil \frac{m}{2} \rceil})$ where $k \neq 1$ or $n$
Let $(x_2, y_{\lceil \frac{m}{2} \rceil})$ be the target. Then, the detour path $S_6$ be $(x_n, y_m) \to \dots (x_n, y_{\lceil \frac{m}{2} \rceil}) \to \dots (x_2, y_{\lceil \frac{m}{2} \rceil})$. The length of $S_6$ is $(n-2) + \lfloor \frac{m}{2} \rfloor$. There are $2n-1$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths which are not on $S_6$. Let it be $S_6^{\sim}$. Hence, distributing $2^{(n-2)+\lfloor \frac{m}{2} \rfloor}$ pebbles on $S_6$ and $(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles each on $2n-1$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths such a way no pebbles should reach the root vertices $(x_k, y_{\lceil \frac{m}{2} \rceil})$, we can reach the target. The total number of pebbles used at most is $2^{(n-2)+\lfloor \frac{m}{2} \rfloor} + (2n-1)(2^{\lfloor \frac{m}{2} \rfloor} - 1)$.

**Subase 4.1:** Let the target vertex be $(x_k, y_{\lceil \frac{m}{2} \rceil})$ where $k \neq 1$ and $n$
Let $(x_2, y_{\lceil \frac{m}{2} \rceil})$ be the target. Then, the detour path $S_7$ be $(x_{n-1}, y_m) \to \dots (x_{n-1}, y_{\lceil \frac{m}{2} \rceil}) \to \dots (x_2, y_{\lceil \frac{m}{2} \rceil})$. The length of $S_7$ is $(n-3) + \lfloor \frac{m}{2} \rfloor$. There are $2n-1$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths which are not on $S_7$. Let it be $S_7^{\sim}$. Hence, distributing $2^{(n-3)+\lfloor \frac{m}{2} \rfloor}$ pebbles on $S_7$ and $(2^{\lfloor \frac{m}{2} \rfloor} - 1)$ pebbles each on $2n-1$ copies of $P_{\lfloor \frac{m}{2} \rfloor}$ paths such a way no pebbles should reach the root vertices $(x_k, y_{\lceil \frac{m}{2} \rceil})$, we can reach
Now either placing $2\lfloor\frac{m}{2}\rfloor - 1$ pebbles on $(x_1, y_i)$ or $(x_1, y_j)$, we are done. For this all these distribution we used at least $(2\lfloor\frac{m}{2}\rfloor + 1) + (n-2)(2\lfloor\frac{m}{2}\rfloor) + (n)(2\lfloor\frac{m}{2}\rfloor - 1) = 2n(2\lfloor\frac{m}{2}\rfloor - 1)$ pebbles. Hence, we are done. By symmetry, we can prove for $(x_n, y_{\lceil\frac{m}{2}\rceil})$.

**Case 3.** Let the target vertex be $(x_k, y_l)$ where $k \neq 1$ or $n$, $1 \le l \le (\lfloor\frac{m}{2}\rfloor)$ or $(\lceil\frac{m}{2}\rceil) + 1 \ge l \ge m$.

The detour path $S_4$ be $(x_1, y_{\lceil\frac{m}{2}\rceil}) \to (x_2, y_{\lceil\frac{m}{2}\rceil}) \to \dots \to (x_{n-1}, y_{\lceil\frac{m}{2}\rceil}) \to (x_{n-1}, y_{\lfloor\frac{n}{2}\rfloor+1}) \dots (x_{n-1}, y_l)$, whose path length is $\le (n-2) + (\lfloor\frac{l}{2}\rfloor - 1)$. There are $2n-1$ paths which do not lie on the detour path having the length of $\lfloor\frac{m}{2}\rfloor$. Placing $2\lfloor\frac{m}{2}\rfloor^{-1}$ pebbles each on $P_{\lfloor\frac{m}{2}\rfloor}$ paths and distributing $2^{(n-3)+\lfloor\frac{m}{2}\rfloor}$ pebbles on the detour path we can reach the target. Hence the total number of pebbles used are $2^{(n-3)+\lfloor\frac{m}{2}\rfloor} + (2n-1)(2\lfloor\frac{m}{2}\rfloor - 1)$.

**Subcase 3.1:** Let the target vertex be $(x_k, y_l)$ where $k \neq 1, n$ then $1 \le l \le (\lfloor\frac{m}{2}\rfloor)$ or $(\lceil\frac{m}{2}\rceil) + 1 \ge l \ge m$ as $2 \le k \le n-1$.

The detour path $S_5$ be $(x_2, y_m \to x_2, y_{\lceil\frac{m}{2}\rceil}) \to \dots \to (x_{n-1}, y_{\lceil\frac{m}{2}\rceil}) \to (x_{n-1}, y_{\lfloor\frac{m}{2}\rfloor+1}) \dots (x_{n-1}, y_l)$, whose path length is $\le (n-3) + (\lfloor\frac{m}{2}\rfloor) + (l - \lceil\frac{m}{2}\rceil)$. There are $2n-6$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ paths and at least 2 copies of $P_m$ paths which do not lie on $S_5$. Let $2n-6$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ be $S_5^{\sim}$ and 2 copies of $P_m$ paths be $S_5^{\to}$. Placing $2^{\lfloor\frac{m}{2}\rfloor-1}$ pebbles each on $2n-6$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ paths, and $S_5^{\to}$, we can shift at least two pebbles each to every rooted vertices. Using this we could put two pebbles on $(x_{n-1}, y_{\lfloor\frac{m}{2}\rfloor})$. Distributing $2^{(n-3)+(l-\lfloor\frac{m}{2}\rfloor)+(\lfloor\frac{m}{2}\rfloor)}$ pebbles on $(x_2, y_m)$, we can shift at least $(l-\lceil\frac{m}{2}\rceil)+2$ to $(x_{n-1}, y_{(\lfloor\frac{m}{2}\rfloor+1)})$ which is sufficient to reach the target. Hence the total number of pebbles used are $2^{(n-3)+\lfloor\frac{m}{2}\rfloor} + (l-\lfloor\frac{m}{2}\rfloor) + (2n-2)(2\lfloor\frac{m}{2}\rfloor - 1)$.

**Case 4.** Let the target vertex be $(x_k, y_{\lceil\frac{m}{2}\rceil})$ where $k \neq 1$ or $n$
Let $(x_2, y_{\lceil\frac{m}{2}\rceil})$ be the target. Then, the detour path $S_6$ be $(x_n, y_m) \to \dots (x_n, y_{\lceil\frac{m}{2}\rceil}) \to \dots (x_2, y_{\lceil\frac{m}{2}\rceil})$. The length of $S_6$ is $(n-2) + \lfloor\frac{m}{2}\rfloor$. There are $2n-1$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ paths which are not on $S_6$. Let it be $S_6^{\sim}$. Hence, distributing $2^{(n-2)+\lfloor\frac{m}{2}\rfloor}$ pebbles on $S_6$ and $(2\lfloor\frac{m}{2}\rfloor - 1)$ pebbles each on $2n-1$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ paths such a way no pebbles should reach the root vertices $(x_k, y_{\lceil\frac{m}{2}\rceil})$, we can reach the target. The total number of pebbles used at most is $2^{(n-2)+\lfloor\frac{m}{2}\rfloor} + (2n-1)(2\lfloor\frac{m}{2}\rfloor - 1)$.

**Subcase 4.1:** Let the target vertex be $(x_k, y_{\lceil\frac{m}{2}\rceil})$ where $k \neq 1$ and $n$
Let $(x_2, y_{\lceil\frac{m}{2}\rceil})$ be the target. Then, the detour path $S_7$ be $(x_{n-1}, y_m) \to \dots (x_{n-1}, y_{\lceil\frac{m}{2}\rceil}) \to \dots (x_2, y_{\lceil\frac{m}{2}\rceil})$. The length of $S_7$ is $(n-3) + \lfloor\frac{m}{2}\rfloor$. There are $2n-1$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ paths which are not on $S_7$. Let it be $S_7^{\sim}$. Hence, distributing $2^{(n-3)+\lfloor\frac{m}{2}\rfloor}$ pebbles on $S_7$ and $(2\lfloor\frac{m}{2}\rfloor - 1)$ pebbles each on $2n-1$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ paths such a way no pebbles should reach the root vertices $(x_k, y_{\lceil\frac{m}{2}\rceil})$, we can reach
the target. The total number of pebbles used at most is $2^{(n-2)-(k-1)}+\lfloor\frac{m}{2}\rfloor + (2n-1)(2^{\lfloor\frac{m}{2}\rfloor}-1)$ or $2^{(k-2)}+\lfloor\frac{m}{2}\rfloor + (2n-1)(2^{\lfloor\frac{m}{2}\rfloor}-1)$.

Hence, the detour pebbling number of $f^*(P_n \odot y_{\lfloor\frac{m}{2}\rfloor} P_m)$, where $d(y_{\lfloor\frac{m}{2}\rfloor}) = 2$ is
$$ \geq 2^{m+n-2} + (2n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1) $$

**Proof 2:** When $m$ is even:

The detour distance between $(x_1, y_m)$ to any vertex is $\le 2^{m+n-1}$. Let us consider the detour path $S_7$ be $(x_1, y_m) \rightarrow (x_1, y_{m-1}) \rightarrow \cdots \rightarrow (x_1, y_{\lfloor\frac{m}{2}\rfloor+1}) \cdots \rightarrow (x_n, y_{\lfloor\frac{m}{2}\rfloor+1}) \rightarrow \cdots \rightarrow (x_n, y_m)$ whose length is $m+n-1$. There are $(n-2)$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ and $n$ copies of $P_{\lfloor\frac{m}{2}\rfloor}-1$ paths which have the root vertices on $S_7$. Let it be $S_7'$. Distributing $2^{m+n-1}-1$ pebbles on $(x_n, y_m)$ and $(2n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1)$ pebbles on $S_1$, such a way no pebble reaches the root, we can not reach the target $(x_1, y_{m-1})$. Hence the detour pebbling number of $f^*(P_n \odot y_{\lfloor\frac{m}{2}\rfloor+1} P_m)$ where $d(y_{\lfloor\frac{m}{2}\rfloor+1}) = 2$ is
$$ \ge 2^{m+n-1} + (n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1) + n(2^{\lfloor\frac{m}{2}\rfloor}-2). $$

Consider $2^{m+n-1} + (n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1) + n(2^{\lfloor\frac{m}{2}\rfloor}-2)$. pebbles on the vertices of $P_n \odot y_{\lfloor\frac{m}{2}\rfloor+1} P_m$. We can prove the sufficient condition for the $P_n \odot y_{\lfloor\frac{m}{2}\rfloor+1} P_m$ using the previous method of **proof 1**.

□

**Theorem 3.3.** For the graph $K_n \odot y K_3$, the detour pebbling number is $f^*(K_n \odot y K_3) = 2^{(n+3)} + 3(n-2)$.

*Proof.* The detour distance from $(x_1, y_1)$ to any other vertex on $G \odot_y H$ is $d^*(G \odot_y H) \le (n+3)$. Let us consider $G = K_n$ and $H = K_3$ fix $y_3$ as the rooted vertex of $H$ and $d(y_3) = 2$. Let us consider the detour path $S_1$ be $(x_1, y_1) \rightarrow (x_1, y_2) \rightarrow (x_1, y) \rightarrow (x_2, y) \rightarrow \cdots \rightarrow (x_n, y) \rightarrow (x_n, y_2) \rightarrow (x_n, y_1)$ whose length is $n+3$. Placing $2^{n+3}-1$ pebbles on $(x_n, y_1)$ and distributing 3 pebbles each on any one of the vertices of $(n-2)$ copies of $H$ which are not on $S_1$, we cannot reach the target. Hence, the detour pebbling number of $f^*(K_n \odot y K_3) \ge 2^{(n+3)} + 3(n-2)$.

Let $C$ be the configuration of $2^{(n+3)} + 3(n-2)$ on the vertices of $K_n \odot y K_3$. Now we prove the sufficient condition.

**Case 1.** Let the target vertex be $(x_n, y_1)$ or $(x_n, y_2)$. Without loss of generality, let the target vertex be $(x_n, y_1)$. If there exists $2^{(n+3)}$ on $S_1$, we can reach the target without using the remaining pebbles.

**Subcase 1.1** Let $p(V(S_1)) < 2^{(n+3)}$. There are $(n-2)$ copies of $K_2$ which do not touch $S_1$. Placing one pebble and two pebbles each on every vertices of $n-2$ copies of $K_2$, we can shift one pebble each to the root vertices of $n-2$ copies of $K_3$. Now using the transmitting subgraph, we shift $(x_1, y_1) \le 2^{(n+2)} \rightarrow (x_1, y_2) \le 2^{(n+1)} \rightarrow (x_1, y) \overset{2^n}{\underset{2^2}{\rightarrow}} (x_2, y) \rightarrow \cdots \overset{2^2}{\underset{2^n}{\rightarrow}} (x_n, y) \overset{2^n}{\underset{2^n}{\rightarrow}} (x_n, y_2) \overset{2^n}{\underset{2^n}{\rightarrow}} (x_n, y_1)$. If there exist 4 pebbles each on any one of the vertices of $n-2$ copies of $K_2$, we can shift a pebble each to the root vertices. Hence, we can reach the target by using the previous process. If we shift three pebbles each to every $(x_i, y)$, where $2 \le i \le n-1$, by using at least 12 pebbles each on every (n-2) copies of the $K_2$, we can put a pebble to the target by using $2^{(n+3)} + 3(n-2) - 12(n-2) = 2^{(n+3)} - 9(n-2)$. If we use at most 19 pebbles on any one of the vertices of each $n-2$ copies of $K-2$, we can shift at most 4 pebbles to $(x_i, y)$. Hence, we can shift a pebble to the target by using at most $2^{(n+3)} + 3(n-2) - 19(n-2) = 2^{(n+3)} - 16(n-2)$. If $S_1$ has no pebbles on it then
the target. The total number of pebbles used at most is $2^{(n-2)-(k-1)}+\lfloor\frac{m}{2}\rfloor + (2n-1)(2^{\lfloor\frac{m}{2}\rfloor}-1)$ or $2^{(k-2)}+\lfloor\frac{m}{2}\rfloor + (2n-1)(2^{\lfloor\frac{m}{2}\rfloor}-1)$.

Hence, the detour pebbling number of $f^*(P_n \odot y_{\lfloor\frac{m}{2}\rfloor} P_m)$, where $d(y_{\lfloor\frac{m}{2}\rfloor}) = 2$ is
$$ \ge 2^{m+n-2} + (2n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1) $$

**Proof 2:** When $m$ is even:

The detour distance between $(x_1, y_m)$ to any vertex is $\le 2^{m+n-1}$. Let us consider the detour path $S_7$ be $(x_1, y_m) \to (x_1, y_{m-1}) \to \dots \to (x_1, y_{\lfloor\frac{m}{2}\rfloor+1}) \to \dots \to (x_n, y_m)$ whose length is $m+n-1$. There are $(n-2)$ copies of $P_{\lfloor\frac{m}{2}\rfloor}$ and $n$ copies of $P_{\lfloor\frac{m}{2}\rfloor}-1$ paths which have the root vertices on $S_7$. Let it be $S_7'$. Distributing $2^{m+n-1}-1$ pebbles on $(x_n, y_m)$ and $(2n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1)$ pebbles on $S_1$, such a way no pebble reaches the root, we can not reach the target $(x_1, y_{m-1})$. Hence the detour pebbling number of $f^*(P_n \odot y_{\lfloor\frac{m}{2}\rfloor+1} P_m)$ where $d(y_{\lfloor\frac{m}{2}\rfloor+1}) = 2$ is
$$ \ge 2^{m+n-1} + (n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1) + n(2^{\lfloor\frac{m}{2}\rfloor}-2). $$

Consider $2^{m+n-1} + (n-2)(2^{\lfloor\frac{m}{2}\rfloor}-1) + n(2^{\lfloor\frac{m}{2}\rfloor}-2)$. pebbles on the vertices of $P_n \odot y_{\lfloor\frac{m}{2}\rfloor+1} P_m$. We can prove the sufficient condition for the $P_n \odot y_{\lfloor\frac{m}{2}\rfloor+1} P_m$ using the previous method of **proof 1**.

□

**Theorem 3.3.** For the graph $K_n \odot y K_3$, the detour pebbling number is $f^*(K_n \odot y K_3) = 2^{(n+3)} + 3(n-2)$.

*Proof.* The detour distance from $(x_1, y_1)$ to any other vertex on $G \odot_y H$ is $d^*(G \odot_y H) \le (n+3)$. Let us consider $G = K_n$ and $H = K_3$ fix $y_3$ as the rooted vertex of $H$ and $d(y_3) = 2$. Let us consider the detour path $S_1$ be $(x_1, y_1) \to (x_1, y_2) \to (x_1, y) \to (x_2, y) \to \dots \to (x_n, y) \to (x_n, y_2) \to (x_n, y_1)$ whose length is $n+3$. Placing $2^{n+3}-1$ pebbles on $(x_n, y_1)$ and distributing 3 pebbles each on any one of the vertices of $(n-2)$ copies of $H$ which are not on $S_1$, we cannot reach the target. Hence, the detour pebbling number of $f^*(K_n \odot y K_3) \ge 2^{(n+3)} + 3(n-2)$.

Let $C$ be the configuration of $2^{(n+3)} + 3(n-2)$ on the vertices of $K_n \odot y K_3$. Now we prove the sufficient condition.

**Case 1.** Let the target vertex be $(x_n, y_1)$ or $(x_n, y_2)$. Without loss of generality, let the target vertex be $(x_n, y_1)$. If there exists $2^{(n+3)}$ on $S_1$, we can reach the target without using the remaining pebbles.

**Subcase 1.1** Let $p(V(S_1)) < 2^{(n+3)}$. There are $(n-2)$ copies of $K_2$ which do not touch $S_1$. Placing one pebble and two pebbles each on every vertices of $n-2$ copies of $K_2$, we can shift one pebble each to the root vertices of $n-2$ copies of $K_3$. Now using the transmitting subgraph, we shift $(x_1, y_1) \le 2^{(n+2)} \to (x_1, y_2) \le 2^{(n+1)} \to (x_1, y) \le 2^n \to (x_2, y) \to \dots \to 2^2 \to (x_n, y) \le 2 \to (x_n, y_2) \le 1 \to (x_n, y_1)$. If there exist 4 pebbles each on any one of the vertices of $n-2$ copies of $K_2$, we can shift a pebble each to the root vertices. Hence, we can reach the target by using the previous process. If we shift three pebbles each to every $(x_i, y)$, where $2 \le i \le n-1$, by using at least 12 pebbles each on every (n-2) copies of the $K_2$, we can put a pebble to the target by using $2^{(n+3)} + 3(n-2) - 12(n-2) = 2^{(n+3)} - 9(n-2)$. If we use at most 19 pebbles on any one of the vertices of each $n-2$ copies of $K-2$, we can shift at most 4 pebbles to $(x_i, y)$. Hence, we can shift a pebble to the target by using at most $2^{(n+3)} + 3(n-2) - 19(n-2) = 2^{(n+3)} - 16(n-2)$. If $S_1$ has no pebbles on it then
distributing at most $2^5-1$ pebbles each on $n-2$ copies of $K_2$, we can shift at least 5 pebbles each to $(x_i, y)$. Then $(x_j, y) \xrightarrow{3} (x_{n-1}, y) \xrightarrow{5+3} (x_n, y_0$, where $2 \le j \le n-2$. We used at most $(n-2)2^5-1$ pebbles. Similarly, we can prove for $(x_i, y_i)$, where $1 \le i \le n$ and $j = 1, 2$.

**Case 2.** Let the target vertex be $(x_n, y)$ or $(x_1, y)$. Without loss of generality, let the target vertex be $(x_n, y)$. Let us consider the detour path $S_2$ be $(x_1, y_1) \to (x_1, y_2) \to (x_1, y) \to (x_2, y) \to \dots \to (x_n, y)$ whose length is $n+1$. If there exists $2^{(n+1)}$ on $S_2$, we can reach the target without using the remaining pebbles.

**Subcase 2.1** Let $p(V(S_2)) < 2^{(n+1)}$ and $N(V(x_n, y) \le 1$.

There are $(n-1)$ copies of $K_2$ which do not touch $S_2$. Placing one pebble and two pebbles each on every vertices of $n-2$ copies of $k_2$ except on the adjacent vertices of the target, we can shift one pebble each to the root vertices $(x_i, y)$, where $2 \le i \le n-1$. Now using the transmitting subgraph, we shift $(x_1, y_1) \xrightarrow{4} (x_1, y_2) \xrightarrow{2} (x_1, y) \xrightarrow{1+1} (x_2, y) \to \dots \xrightarrow{1+1} (x_{n-1}, y) \xrightarrow{1} (x_n, y)$. The total number of pebbles used for this configuration is $3n+4$. If there exist 4 pebbles each on any one of the vertices of $n-2$ copies of $K_2$, we can shift a pebble each to the root vertices. Hence, we can reach the target by using the previous process. In this distribution we used at least $4n+2$ pebbles. If we shift three pebbles each to every $(x_i, y)$, where $2 \le i \le n-2$, by using at least 12 pebbles each on every (n-3) copies of the $K_2$, we can put a pebble to the target by using at most $12(n-3)+8 = 12n-28$. If we use at most 16 pebbles on any one of the vertices of each $n-3$ copies of $K-2$, we can shift at most 4 pebbles to $(x_i, y)$. Hence, we can shift a pebble to the target by using at most $16(n-3)$ pebbles. Similarly, we can prove for $(x_i, y)$, where $1 \le i \le n$. Hence, the detour pebbling number is $f^*(K_n \odot y K_3) = 2^{(n+3)} + 3(n-2)$ □

**Theorem 3.4.** For the graph $F_n \odot y K_{1,m}$, the detour pebbling number is

$$f^*(F_n \odot y K_{1,m}) = \begin{cases} 2^{n+3} + n(m+1) - 6 & \text{if } d(v) = 1 \\ 2^{n+1} + nm - 2 & \text{if } d(v) = m \end{cases}$$

*Proof.* Let the vertex set of $F_n$ be $\{x_0, x_1, x_2, \dots, x_{n-1}\}$ and $K_{1,m}$ be $\{y_0, y_1, y_2, \dots, y_m\}$. The star graph $K_{1,m}$ has a center vertex $y_0$ with degree $m$. Depending on the root vertex we get two different graphs for $F_n \odot y K_{1,m}$. Let the source vertex be some vertex $(x_1, y_1)$ and the target vertex $(x_0, y_1)$, $1 \le i \le m$.

**Proof 1:** When $d(v) = 1$

Let any vertex $y_i, i \in \{1, 2, \dots, m\}$ of $K_{1,m}$ be the root vertex in $F_n \odot y K_{1,m}$. Without loss of generality, let $(x_0, y_m)$ be the target and $y = y_1$ be the root vertex of $K_{1,m}$. Now, let us construct a detour path $S_1$ from $(x_1, y_m)$ to $(x_0, y_m)$. $S_1 = (x_1, y_m) \to (x_1, y_0) \to (x_1, y) \to (x_2, y) \to \dots \to (x_{n-1}, y) \to (x_0, y) \to (x_0, y_0) \to (x_0, y_m)$ whose length is $n+3$. Placing $2^{n+3}-1$ pebbles on $(x_1, y_m)$, 3 pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble each on the remaining vertices which are not on $S_1$, we cannot reach the target. Hence the detour pebbling number of $f^*(F_n \odot y K_{1,m}) \ge 2^{n+3} + n(m+1) - 6$, if $d(v) = 1$.

Let $C$ be the configuration of $2^{n+3}+n(m+1)-6$ on the vertices of $F_n \odot y K_{1,m}$. Now we prove the sufficient condition.

**Case 1.** Let the target vertex be $(x_i, y_k)$, $0 \le i \le n-1, 1 \le k \le m$.
distributing at most $2^5-1$ pebbles each on $n-2$ copies of $K_2$, we can shift at least 5 pebbles each to $(x_i, y)$. Then $(x_j, y) \xrightarrow{3} (x_{n-1}, y) \xrightarrow{5+3} (x_n, y_0$, where $2 \le j \le n-2$. We used at most $(n-2)2^5-1$ pebbles. Similarly, we can prove for $(x_i, y_i)$, where $1 \le i \le n$ and $j = 1, 2$.

**Case 2.** Let the target vertex be $(x_n, y)$ or $(x_1, y)$. Without loss of generality, let the target vertex be $(x_n, y)$. Let us consider the detour path $S_2$ be $(x_1, y_1) \to (x_1, y_2) \to (x_1, y) \to (x_2, y) \to \dots \to (x_n, y)$ whose length is $n+1$. If there exists $2^{(n+1)}$ on $S_2$, we can reach the target without using the remaining pebbles.

**Subcase 2.1** Let $p(V(S_2)) < 2^{(n+1)}$ and $N(V(x_n, y) \le 1$.

There are $(n-1)$ copies of $K_2$ which do not touch $S_2$. Placing one pebble and two pebbles each on every vertices of $n-2$ copies of $k_2$ except on the adjacent vertices of the target, we can shift one pebble each to the root vertices $(x_i, y)$, where $2 \le i \le n-1$. Now using the transmitting subgraph, we shift $(x_1, y_1) \xrightarrow{4} (x_1, y_2) \xrightarrow{2} (x_1, y) \xrightarrow{1+1} (x_2, y) \to \dots \xrightarrow{1+1} (x_{n-1}, y) \xrightarrow{1} (x_n, y)$. The total number of pebbles used for this configuration is $3n+4$. If there exist 4 pebbles each on any one of the vertices of $n-2$ copies of $K_2$, we can shift a pebble each to the root vertices. Hence, we can reach the target by using the previous process. In this distribution we used at least $4n+2$ pebbles. If we shift three pebbles each to every $(x_i, y)$, where $2 \le i \le n-2$, by using at least 12 pebbles each on every (n-3) copies of the $K_2$, we can put a pebble to the target by using at most $12(n-3)+8 = 12n-28$. If we use at most 16 pebbles on any one of the vertices of each $n-3$ copies of $K-2$, we can shift at most 4 pebbles to $(x_i, y)$. Hence, we can shift a pebble to the target by using at most $16(n-3)$ pebbles. Similarly, we can prove for $(x_i, y)$, where $1 \le i \le n$. Hence, the detour pebbling number is $f^*(K_n \odot y K_3) = 2^{(n+3)} + 3(n-2)$ □

**Theorem 3.4.** For the graph $F_n \odot y K_{1,m}$, the detour pebbling number is

$$f^*(F_n \odot y K_{1,m}) = \begin{cases} 2^{n+3} + n(m+1) - 6 & \text{if } d(v) = 1 \\ 2^{n+1} + nm - 2 & \text{if } d(v) = m \end{cases}$$

*Proof.* Let the vertex set of $F_n$ be $\{x_0, x_1, x_2, \dots, x_{n-1}\}$ and $K_{1,m}$ be $\{y_0, y_1, y_2, \dots, y_m\}$. The star graph $K_{1,m}$ has a center vertex $y_0$ with degree $m$. Depending on the root vertex we get two different graphs for $F_n \odot y K_{1,m}$. Let the source vertex be some vertex $(x_1, y_i)$ and the target vertex $(x_0, y_i)$, $1 \le i \le m$.

**Proof 1:** When $d(v) = 1$

Let any vertex $y_i, i \in \{1, 2, \dots, m\}$ of $K_{1,m}$ be the root vertex in $F_n \odot y K_{1,m}$. Without loss of generality, let $(x_0, y_m)$ be the target and $y = y_1$ be the root vertex of $K_{1,m}$. Now, let us construct a detour path $S_1$ from $(x_1, y_m)$ to $(x_0, y_m)$. $S_1 = (x_1, y_m) \to (x_1, y_0) \to (x_1, y) \to (x_2, y) \to \dots \to (x_{n-1}, y) \to (x_0, y) \to (x_0, y_0) \to (x_0, y_m)$ whose length is $n+3$. Placing $2^{n+3}-1$ pebbles on $(x_1, y_m)$, 3 pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble each on the remaining vertices which are not on $S_1$, we cannot reach the target. Hence the detour pebbling number of $f^*(F_n \odot y K_{1,m}) \ge 2^{n+3} + n(m+1) - 6$, if $d(v) = 1$.

Let $C$ be the configuration of $2^{n+3}+n(m+1)-6$ on the vertices of $F_n \odot y K_{1,m}$. Now we prove the sufficient condition.

**Case 1.** Let the target vertex be $(x_i, y_k)$, $0 \le i \le n-1, 1 \le k \le m$.
Without loss of generality, let the target vertex be $(x_0, y_m)$. If there exists $2^{(n+3)}$ on $S_1$, we can reach the target without using the remaining pebbles.

**Subcase 1.1** Let $p(V(S_1)) < 2^{(n+3)}$.

There are $(n-2)$ copies of $K_{1,(m-1)}$ and $2(m-2)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y_0), (x_1, y)\} >$ and $< K_{1,m} - \{(x_0, y_m), (x_0, y_0), (x_0, y)\} >$ which do not touch $S_1$. Placing $2^{n+3}$ pebbles on $(x_1, y_m)$, 3 pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble each on the remaining vertices which are not on $S_1$, we can reach the target. If there exist four pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and two pebbles on any one of the vertices of the induced subgraph of $< K_{1,m} - \{(x_0, y_m), (x_0, y_0), (x_0, y)\} >$, we can reach a pebble each to the root vertices from $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble to $(x_0, y_0)$. Now using the transmitting subgraph, we shift $(x_1, y_m) \le 2^{(n+2)} \xrightarrow{\sim} (x_1, y_0) \le 2^{(n+1)} \xrightarrow{\sim} (x_1, y) \ge 2^n \xrightarrow{\sim} (x_1, y) \to \cdots \ge 2^2 \xrightarrow{\sim} (x_0, y) \ge 2 + 1 \xrightarrow{\sim} (x_0, y_0) \xrightarrow{\sim} (x_0, y_m)$. Let $S_1$ be the vertices which are not in $S_1$. If $p(V(S_1)) - (x_0, y), (x_0, y_1) > \xrightarrow{\sim} p(V(s_1)) \xrightarrow{\sim} y + p(x_0, y) = 4$, we are done. Similarly, we can prove for $(x_i, y_l)$, where $1 \le i \le n-1$ and $1 \le l \le m-1$.

**Case 2.** Let the target vertex be $(x_i, y_0)$, $0 \le i \le n-1$.

Without loss of generality, let the target vertex be $(x_0, y_0)$. Let us construct a detour path $S_2$ from $(x_1, y_m)$ to $(x_0, y_0)$. $S_2 = (x_1, y_m) \to (x_1, y_0) \to (x_1, y) \to (x_2, y) \to \cdots \to (x_{n-1}, y) \to (x_0, y) \to (x_0, y_0)$ whose length is $n+2$. If there exists $2^{(n+2)}$ on $S_2$, we can reach the target without using the remaining pebbles.

**Subcase 2.1** Let $p(V(S_2)) < 2^{(n+2)}$.

There are $(n-2)$ copies of $K_{1,(m-1)}$ and $(m-2)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y_0), (x_1, y)\} >$ and $(m-1)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_0, y_m), (x_0, y_0), (x_0, y)\} >$ which do not touch $S_2$. Placing $2^{n+2}$ pebbles on $(x_1, y_m)$, 3 pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble each on the remaining vertices which are not on $S_2$, we can reach the target. If there exist four pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$, we can reach a pebble each to the root vertices from $(n-2)$ copies of $K_{1,(m-1)}$. Now using the transmitting subgraph, we shift $(x_1, y_m) \le 2^{(n+1)} \xrightarrow{\sim} (x_1, y_0) \le 2^{(n)} \xrightarrow{\sim} (x_1, y) \ge 2^{n-1} \xrightarrow{\sim} (x_1, y) \to \cdots \xrightarrow{\sim} (x_n, y) \xrightarrow{\sim} (x_n, y_0)$. Let $S_2$ be the vertices which are not in $S_2$. If $p(V(S_2)) - (x_0, y) > \xrightarrow{\sim} p(V(s_2)) \xrightarrow{\sim} y + p(x_0, y) = 2$, we are done. Similarly, we can prove for $(x_i, y_j)$, where $1 \le i \le n$ and $j = 0, 1$. Hence, the detour pebbling number of $(F_n \odot y K_{1,m}) = 2^{n+3} + n(m+1) - 6$ if $d(v) = 1$.

**Proof 2:** When $d(v) = m$.

Let $y_0 \in K_{1,m}$ be the root vertex in $(F_n \odot y K_{1,m})$. Without loss of generality, let $(x_0, y_1)$ be the target and $y = y_0$ be the root vertex of $K_{1,m}$. Now, let us construct a detour path $S_3$ from $(x_1, y_m)$ to $(x_0, y_m)$. $S_3 = (x_1, y_m) \to (x_1, y) \to (x_2, y) \to (x_3, y) \to \cdots \to (x_{n-1}, y) \to (x_0, y) \to (x_0, y_m)$ whose length is $n+1$. Placing $2^{n+1}-1$ pebbles on $(x_1, y_m)$, and a pebble each on the remaining vertices which are not on $S_3$, we cannot reach the target. Hence the detour pebbling number of $f^*(F_n \odot y K_{1,m}) \ge 2^{n+1} + nm - 2$, if $d(v) = m$.

Let $C$ be the configuration of $2^{n+1}+nm-2$ on the vertices of $F_n \odot y K_{1,m}$. Now we prove the sufficient condition.

**Case 3.** Let the target vertex be $(x_i, y_k)$, $0 \le i \le n-1, 1 \le k \le m$.
Without loss of generality, let the target vertex be $(x_0, y_m)$. If there exists $2^{(n+3)}$ on $S_1$, we can reach the target without using the remaining pebbles.

**Subcase 1.1** Let $p(V(S_1)) < 2^{(n+3)}$.

There are $(n-2)$ copies of $K_{1,(m-1)}$ and $2(m-2)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y_0), (x_1, y)\} >$ and $< K_{1,m} - \{(x_0, y_m), (x_0, y_0), (x_0, y)\} >$ which do not touch $S_1$. Placing $2^{n+3}$ pebbles on $(x_1, y_m)$, 3 pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble each on the remaining vertices which are not on $S_1$, we can reach the target. If there exist four pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and two pebbles on any one of the vertices of the induced subgraph of $< K_{1,m} - \{(x_0, y_m), (x_0, y_0), (x_0, y)\} >$, we can reach a pebble each to the root vertices from $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble to $(x_0, y_0)$. Now using the transmitting subgraph, we shift $(x_1, y_m) \le 2^{(n+2)} \xrightarrow{\sim} (x_1, y_0) \le 2^{(n+1)} \xrightarrow{\sim} (x_1, y) \ge 2^n \xrightarrow{\sim} (x_1, y) \to \dots \xrightarrow{\sim} 2^2 \xrightarrow{\sim} (x_0, y) \ge 2 + \xrightarrow{\sim} (x_0, y_0) \xrightarrow{\sim} (x_0, y_m)$. Let $S_1$ be the vertices which are not in $S_1$. If $p(V(S_1)) - (x_0, y), (x_0, y_1) > \xrightarrow{\sim} p(V(s_1)) \xrightarrow{\sim} y + p(x_0, y) = 4$, we are done. Similarly, we can prove for $(x_i, y_l)$, where $1 \le i \le n-1$ and $1 \le l \le m-1$.

**Case 2.** Let the target vertex be $(x_i, y_0)$, $0 \le i \le n-1$.

Without loss of generality, let the target vertex be $(x_0, y_0)$. Let us construct a detour path $S_2$ from $(x_1, y_m)$ to $(x_0, y_0)$. $S_2 = (x_1, y_m) \to (x_1, y_0) \to (x_1, y) \to (x_2, y) \to \dots \to (x_{n-1}, y) \to (x_0, y) \to (x_0, y_0)$ whose length is $n+2$. If there exists $2^{(n+2)}$ on $S_2$, we can reach the target without using the remaining pebbles.

**Subcase 2.1** Let $p(V(S_2)) < 2^{(n+2)}$.

There are $(n-2)$ copies of $K_{1,(m-1)}$ and $(m-2)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y_0), (x_1, y)\} >$ and $(m-1)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_0, y_m), (x_0, y_0), (x_0, y)\} >$ which do not touch $S_2$. Placing $2^{n+2}$ pebbles on $(x_1, y_m)$, 3 pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$ and a pebble each on the remaining vertices which are not on $S_2$, we can reach the target. If there exist four pebbles each on any one of the vertices of $(n-2)$ copies of $K_{1,(m-1)}$, we can reach a pebble each to the root vertices from $(n-2)$ copies of $K_{1,(m-1)}$. Now using the transmitting subgraph, we shift $(x_1, y_m) \le 2^{(n+1)} \xrightarrow{\sim} (x_1, y_0) \le 2^{(n)} \xrightarrow{\sim} (x_1, y) \ge 2^{n-1} \xrightarrow{\sim} (x_1, y) \to \dots \xrightarrow{\sim} 2 \xrightarrow{\sim} (x_n, y) \xrightarrow{\sim} (x_n, y_0)$. Let $S_2$ be the vertices which are not in $S_2$. If $p(V(S_2)) - (x_0, y) > \xrightarrow{\sim} p(V(s_2)) \xrightarrow{\sim} y + p(x_0, y) = 2$, we are done. Similarly, we can prove for $(x_i, y_j)$, where $1 \le i \le n$ and $j = 0, 1$. Hence, the detour pebbling number of $(F_n \odot y K_{1,m}) = 2^{n+3} + n(m+1) - 6$ if $d(v) = 1$.

**Proof 2:** When $d(v) = m$.

Let $y_0 \in K_{1,m}$ be the root vertex in $(F_n \odot y K_{1,m})$. Without loss of generality, let $(x_0, y_1)$ be the target and $y = y_0$ be the root vertex of $K_{1,m}$. Now, let us construct a detour path $S_3$ from $(x_1, y_m)$ to $(x_0, y_m)$. $S_3 = (x_1, y_m) \to (x_1, y) \to (x_2, y) \to (x_3, y) \to \dots \to (x_{n-1}, y) \to (x_0, y) \to (x_0, y_m)$ whose length is $n+1$. Placing $2^{n+1}-1$ pebbles on $(x_1, y_m)$, and a pebble each on the remaining vertices which are not on $S_3$, we cannot reach the target. Hence the detour pebbling number of $f^*(F_n \odot y K_{1,m}) \ge 2^{n+1} + nm - 2$, if $d(v) = m$.

Let $C$ be the configuration of $2^{n+1}+nm-2$ on the vertices of $F_n \odot y K_{1,m}$. Now we prove the sufficient condition.

**Case 3.** Let the target vertex be $(x_i, y_k)$, $0 \le i \le n-1, 1 \le k \le m$.
Without loss of generality, let the target vertex be $(x_0, y_m)$. If there exists $2^{(n+1)}$ on $S_3$, we can reach the target without using the remaining pebbles.

**Subcase 3.1** Let $p(V(S_3)) < 2^{(n+1)}$.

There are $(n-2)$ copies of $< K_{1,m} - v_0 >$ and $2(m-1)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y)\} >$ and $< K_{1,m} - \{(x_0, y_m), (x_0, y)\} >$ which do not touch $S_3$. Let it be $S_3^\sim$. Placing $2^{n+1}$ pebbles on $(x_1, y_m)$, one pebble each on the remaining vertices which are not on $S_3$, we can reach the target. If there exist two pebbles each on any one of the vertices of $S_3^\sim$ except on the $N(x_0, y))$, we can reach a pebble each to the root vertices from $S_3^\sim$. Now using the transmitting subgraph, we shift $(x_1, y_m) \xrightarrow{< 2^{(n)}>} (x_1, y) \xrightarrow{< 2^{(n-1)}>} (x_2, y) \xrightarrow{2^{n-2}} (x_3, y) \rightarrow \cdots \xrightarrow{2^2} (x_{n-1}, y) \xrightarrow{1+1} (x_0, y) \xrightarrow{1} (x_0, y_m)$. If $p(V(S_3)) < p(x_0, y) = 2$, we are done. Similarly, we can prove for $(x_i, y_i)$, where $1 \le i \le n-1$ and $1 \le l \le m-1$.

**Case 4.** Let the target vertex be $(x_i, y_i)$, $0 \le i \le n-1$.

Without loss of generality, let the target vertex be $(x_0, y)$. Let us construct a detour path $S_4$ from $(x_1, y_m)$ to $(x_0, y)$. $S_4 = (x_1, y_m) \rightarrow (x_1, y_0) \rightarrow (x_2, y) \rightarrow (x_3, y) \rightarrow \cdots \rightarrow (x_{n-1}, y) \rightarrow (x_0, y)$ whose length is $n$. If there exists $2^n$ on $S_4$, we can reach the target without using the remaining pebbles.

**Subcase 4.1** Let $p(V(S_4)) < 2^n$.

There are $(n-1)$ copies of $K_{1,(m-1)}$ and $(m-1)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y_0)\} >$ which do not touch $S_4$. Let it be $S_4^\sim$. Placing $2^n$ pebbles on $(x_1, y_m)$, one pebble each on $S_4^\sim$, we can reach the target. If there exist two pebbles each on any one of the vertices of $S_4^\sim$ except on the $N(x_0, y))$, we can reach a pebble each to the root vertices from $S_4^\sim$. Now using the transmitting subgraph, we shift $(x_1, y_m) \xrightarrow{2} (x_1, y) \xrightarrow{1+1} (x_2, y) \xrightarrow{1+1} (x_3, y) \rightarrow \cdots \xrightarrow{2^2} (x_{n-1}, y) \xrightarrow{1+1} (x_0, y)$. If $p(V(S_4)) - (x_0, y_m) \xrightarrow{x} p(V(S_4^\sim)) \xrightarrow{y} p(x_{n-1}, y) = 2$, we are done. Similarly, we can prove for $(x_i, y_i)$, where $1 \le i \le n-1$. Hence, the detour pebbling number of $F_n \odot y K_{1,m} = 2^{n+1} + nm - 2$ if $d(v) = m$. $\square$

REFERENCES

[1] A. Lourdusamy, *t-pebbling the product of graphs*, Acta Ciencia Indica, XXXII (M.No.1) (2006), pp. 171 - 176.

[2] A. Lourdusamy, *t-pebbling the graphs of diameter two*, Acta Ciencia Indica, XXLX (M. No.3) (2003), pp. 465-470.

[3] A. Lourdusamy, S. Saratha Nellainayaki, *Detour pebbling on Graphs*, Advances in Mathematics: Scientific Journal.

[4] A. Lourdusamy and A. P. Tharani, *On t-pebbling graphs*, Utilitas Math., 87(2012), pp. 331 - 342.

[5] A. Lourdusamy, S. S. Jeyaseelan and A. P. Tharani, *t-pebbling the product of fan graphs and the product of wheel graphs*, International Mathematical Forum, 32(2009), pp. 1573-1585.

[6] A. Lourdusamy and T. Mathivanan, *The t-pebbling number of squares of cycles*, Journal of Prime Research in Mathematics, Vol.11, No. 2 (2015) pp. 61-76.

[7] A. Lourdusamy, and T. Mathivanan, *The t-pebbling number of the Jahangir graph J3,m*, *Proyecciones Journal of Mathematics*, Vol 34 (2), June 2015, pp. 161-174.

[8] A. Lourdusamy, S. Saratha Nellainayaki, *Detour t-pebbling Graphs*, Communicated.

[9] F.R.K. Chung, *Pebbling in hypercubes*, SIAMJ. Disc. Math., 2(4) (1989), pp. 467-472.

[10] G. Hurlbert, *A survey of graph pebbling*, Congressus Numerantium 139 (1999), pp. 41-64.

[11] Harary.F, *Graph Theory*, Narosa Publishing House, New Delhi. 6.
Without loss of generality, let the target vertex be $(x_0, y_m)$. If there exists $2^{(n+1)}$ on $S_3$, we can reach the target without using the remaining pebbles.

**Subcase 3.1** Let $p(V(S_3)) < 2^{(n+1)}$.

There are $(n-2)$ copies of $< K_{1,m} - v_0 >$ and $2(m-1)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y)\} >$ and $< K_{1,m} - \{(x_0, y_m), (x_0, y)\} >$ which do not touch $S_3$. Let it be $S_3^\sim$. Placing $2^{n+1}$ pebbles on $(x_1, y_m)$, one pebble each on the remaining vertices which are not on $S_3$, we can reach the target. If there exist two pebbles each on any one of the vertices of $S_3^\sim$ except on the $N(x_0, y))$, we can reach a pebble each to the root vertices from $S_3^\sim$. Now using the transmitting subgraph, we shift $(x_1, y_m) \xrightarrow{< 2^{(n)}>} (x_1, y) \xrightarrow{< 2^{(n-1)}>} (x_2, y) \xrightarrow{2^{n-2}} (x_3, y) \rightarrow \dots \xrightarrow{2^2} (x_{n-1}, y) \xrightarrow{1+1} (x_0, y) \xrightarrow{1} (x_0, y_m)$. If $p(V(S_3)) < p(x_0, y) = 2$, we are done. Similarly, we can prove for $(x_i, y_i)$, where $1 \le i \le n-1$ and $1 \le l \le m-1$.

**Case 4.** Let the target vertex be $(x_i, y_i)$, $0 \le i \le n-1$.

Without loss of generality, let the target vertex be $(x_0, y)$. Let us construct a detour path $S_4$ from $(x_1, y_m)$ to $(x_0, y)$. $S_4 = (x_1, y_m) \rightarrow (x_1, y_0) \rightarrow (x_2, y) \rightarrow (x_3, y) \rightarrow \dots \rightarrow (x_{n-1}, y) \rightarrow (x_0, y)$ whose length is $n$. If there exists $2^n$ on $S_4$, we can reach the target without using the remaining pebbles.

**Subcase 4.1** Let $p(V(S_4)) < 2^n$.

There are $(n-1)$ copies of $K_{1,(m-1)}$ and $(m-1)$ vertices from the induced subgraph of $< K_{1,m} - \{(x_1, y_m), (x_1, y_0)\} >$ which do not touch $S_4$. Let it be $S_4^\sim$. Placing $2^n$ pebbles on $(x_1, y_m)$, one pebble each on $S_4^\sim$, we can reach the target. If there exist two pebbles each on any one of the vertices of $S_4^\sim$ except on the $N(x_0, y))$, we can reach a pebble each to the root vertices from $S_4^\sim$. Now using the transmitting subgraph, we shift $(x_1, y_m) \xrightarrow{2} (x_1, y) \xrightarrow{1+1} (x_2, y) \xrightarrow{1+1} (x_3, y) \rightarrow \dots \xrightarrow{2^2} (x_{n-1}, y) \xrightarrow{1+1} (x_0, y)$. If $p(V(S_4)) - (x_0, y_m) \xrightarrow{x} p(V(S_4^\sim)) \xrightarrow{y} p(x_{n-1}, y) = 2$, we are done. Similarly, we can prove for $(x_i, y_i)$, where $1 \le i \le n-1$. Hence, the detour pebbling number of $F_n \odot y K_{1,m} = 2^{n+1} + nm - 2$ if $d(v) = m$. $\square$

## REFERENCES

[1] A. Lourdusamy, *t-pebbling the product of graphs*, Acta Ciencia Indica, XXXII (M.No.1) (2006), pp. 171 - 176.

[2] A. Lourdusamy, *t-pebbling the graphs of diameter two*, Acta Ciencia Indica, XXLX (M. No.3) (2003), pp. 465-470.

[3] A. Lourdusamy, S. Saratha Nellainayaki, *Detour pebbling on Graphs*, Advances in Mathematics: Scientific Journal.

[4] A. Lourdusamy and A. P. Tharani, *On t-pebbling graphs*, Utilitas Math., 87(2012), pp. 331 - 342.

[5] A. Lourdusamy, S. S. Jeyaseelan and A. P. Tharani, *t-pebbling the product of fan graphs and the product of wheel graphs*, International Mathematical Forum, 32(2009), pp. 1573-1585.

[6] A. Lourdusamy and T. Mathivanan, *The t-pebbling number of squares of cycles*, Journal of Prime Research in Mathematics, Vol.11, No. 2 (2015) pp. 61-76.

[7] A. Lourdusamy, and T. Mathivanan, *The t-pebbling number of the Jahangir graph J$_{3}$$_{,m}$*, *Proyecciones Journal of Mathematics*, Vol 34 (2), June 2015, pp. 161-174.

[8] A. Lourdusamy, S. Saratha Nellainayaki, *Detour t-pebbling Graphs*, Communicated.

[9] F.R.K. Chung, *Pebbling in hypercubes*, SIAMJ. Disc. Math., 2(4) (1989), pp. 467-472.

[10] G. Hurlbert, *A survey of graph pebbling*, Congressus Numerantium 139 (1999), pp. 41-64.

[11] Harary.F, *Graph Theory*, Narosa Publishing House, New Delhi. 6.
DEPARTMENT OF MATHEMATICS,
ST. XAVIER'S COLLEGE (AUTONOMOUS), PALAYAMKOTTAI-627002,
TAMIL NADU, INDIA.

HTTPS://ORCID.ORG/0000-0001-5961-358X.

E-mail address: lourdusamy15@gmail.com

DEPARTMENT OF MATHEMATICS
NORTH BENGAL ST. XAVIER'S COLLEGE, RAJGANJ-735134,
JALPAIGURI, WEST BENGAL, INDIA

HTTPS://ORCID.ORG/0000-0002-3805-6638.

E-mail address: * divyanasj@gmail.com

DEPARTMENT OF MATHEMATICS AND ACTUARIAL SCIENCE,
B. S. ABDUR RAHMAN CRESCENT INSTITUTE OF SCIENCE & TECHNOLOGY,
CHENNAI, INDIA.

E-mail address: sugi.exodus@gmail.com
DEPARTMENT OF MATHEMATICS,
ST. XAVIER'S COLLEGE (AUTONOMOUS), PALAYAMKOTTAI-627002,
TAMIL NADU, INDIA.

HTTPS://ORCID.ORG/0000-0001-5961-358X.

E-mail address: lourdusamy15@gmail.com

DEPARTMENT OF MATHEMATICS
NORTH BENGAL ST. XAVIER'S COLLEGE, RAJGANJ-735134,
JALPAIGURI, WEST BENGAL, INDIA

HTTPS://ORCID.ORG/0000-0002-3805-6638.

E-mail address: * divyanasj@gmail.com

DEPARTMENT OF MATHEMATICS AND ACTUARIAL SCIENCE,
B. S. ABDUR RAHMAN CRESCENT INSTITUTE OF SCIENCE & TECHNOLOGY,
CHENNAI, INDIA.

E-mail address: sugi.exodus@gmail.com