NDC Pebbling Number for Some Class of Graphs

Proof. The non-split domination number for a complete graph is 1 and hence by placing one pebble on any vertex, we are done. Conversely, if placing a pebble on any vertex produces a non-split domination cover solution, then it implies that the non-split domination number is 1 and hence the graph $G$ is complete. 

Proof. Let $V(W_n)=\{v_0,v_1,v_2,v_3,\cdots ,v_{n-1}\}$ where $v_0$ is called the hub of $W_n$. Then $E(W_n)=\{v_0{v}_i,v_j{v}_{j+1},v_{n-1}{v}_1\}$ where $1\le i\le n-1$ and $1\le \le n-2$. Consider the nonsplit dominating set $D=\{v_0\}$. Placing one pebble on each $n-3$ consecutive outer vertices leaves a vertex of $W_n$ undominated. If there are 2 pebbles on any vertex, shift it to the center. Thus, we cover $D$. Similarly, if there is a pebble on $v_0$, we can cover dominate $D$. Thus, consider all distribution containing pebbled vertices that each contain a pebble. If there are $n-2$ vertices having a pebble each, the 2 outer vertices of $W_n$ are dominated since there are only 3 vertices in all $W_n$ that do not have pebbles. Hence, ${\psi }_{ns}(W_n)=n-2$. 

Proof. Let $V(P_n)=\{v_1,v_2,v_3,\cdots ,v_n\}$. Consider the non-split dominating set $D=\{v_1,v_2,\cdots ,v_{n-3},v_n\}$ or $\{v_1,v_4,\cdots ,v_n\}$. In both the non-split dominating set $D$ there will be $(n-2)$ vertices. Then $V(<V-D>)=\{v_{n-2},v_{n-1}\}$ or $\{v_2,v_3\}$ and $<V-D>$ is connected. Though we have many ways to construct the non-split dominating set, we consider these two because they require minimum number of pebbles to cover the non-split dominating sets.

Now we prove the necessary condition. We place all the pebbles either on $v_1$ or $v_n$. Without loss of generality, place $2^{n-1}+2^{n-3}-2$ pebbles on $v_1$. To cover $v_n$ we need $2^{n-1}$ pebbles. Then with the remaining $2^{n-3}-2$ pebbles, it is not possible to cover $v_1$. Hence, ${\psi }_{ns}(P_n)\ge 2^{n-1}+2^{n-3}-1$.

Now we prove the sufficient condition. Consider a configuration $C$ with $2^{n-1}+2^{n-3}-1$ pebbles on the vertices of $P_n$.

1. Case 1: Let the source vertex be $v_n$.

   If we place all the pebbles on $v_n$, to cover $v_1$ we need $2^{n-1}$ pebbles. The set $\{v_2,v_3\}$ is connected but not a subset of $D$. Next we need to place a pebble each on $v_4,v_5,\cdots ,v_n$. Now using the remaining $2^{n-3}-1$ pebbles we cover the remaining vertices in the non-split dominating set $D$. We can shift pebbles as follows: $v_n\underrightarrow{2^{n-4}-1}{v}_{n-1}\underrightarrow{2^{n-)}-1}\cdots ,\underrightarrow{3}{v}_5\underrightarrow{1}{v}_4$ using $2^{n-)}+2^{n-)}+2^{n-6}+\cdots +2^2+2^1+1=2^{n-3}-1$ pebbles. Hence with the configuration $C$ we cover all the vertices in $D$.

2. Case 2: Let the source vertex be either $v_4$ or $v_{n-3}$.

   Let us place $2^{n-1}+2^{n-3}-1$ pebbles on $v_4$. By Theorem 3, we can cover a path of length $n-3$ from $v_4$ to $v_n$ using $2^{n-3}-1$ pebbles. Hence, we are left with $2^{n-1}$ pebbles. To cover $v_1$ we require only 8 pebbles. Thus, with a Configuration of at most $2^{(n-3)}+7$ pebbles we cover the vertices of $D$. By symmetry, the proof follows for the source vertex $v_{n-3}$.

3. Case 3: Let the source vertex be $v_k$ ,$5\le k\le n-4$.

   Consider $V(<V-D>)=\{v_2,v_3\}$. When $n$ is even either $v_{\lfloor \frac{n}{2}\rfloor }$ or $v_{\lfloor \frac{n}{2}\rfloor +1}$ can be taken as a source vertex. If $v_{\lfloor \frac{n}{2}\rfloor }$ is the source vertex, then to the right of $v_{\lfloor \frac{n}{2}\rfloor }$ we have to cover the vertices in a path of length $\lfloor \frac{n}{2}\rfloor +1$ and to the left of $v_{\lfloor \frac{n}{2}\rfloor }$ we have to cover the vertices in a path of length $\lfloor \frac{n}{2}\rfloor -3$ excluding the vertex $v_1$ and the vertices in the graph $<V-D>$. By Theorem 3 we require at most $2^{\lfloor \frac{n}{2}\rfloor +1}-1+2^{\lfloor \frac{n}{2}\rfloor -3}-2$ pebbles to cover the vertices in the above path of length $\lfloor \frac{n}{2}\rfloor +1$ and $\lfloor \frac{n}{2}\rfloor -3$. Now to cover $v_1$ we need $2^{\lfloor \frac{n}{2}\rfloor -1}$ pebbles. The number of pebbles used in this process is $2^{\lfloor \frac{n}{2}\rfloor +1}-1+2^{\lfloor \frac{n}{2}\rfloor -3}-2+2^{\lfloor \frac{n}{2}\rfloor -1}<2^{n-1}+2^{n-3}-1$ pebbles.

   If $v_{\lfloor \frac{n}{2}\rfloor +1}$ is the source vertex, then to the right of $v_{\lfloor \frac{n}{2}\rfloor +1}$ we have to cover the vertices in a path of length $\lfloor \frac{n}{2}\rfloor$ and to the left of $v_{\lfloor \frac{n}{2}\rfloor }$ we have to cover the vertices in a path of length $\lfloor \frac{n}{2}\rfloor -2$ excluding the vertex $v_1$ and the vertices in the graph $<V-D>$. By Theorem 3 we require at most $2^{\lfloor \frac{n}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor -2}-2$ pebbles to cover the vertices in the above path of length $\lfloor \frac{n}{2}\rfloor$ and $\lfloor \frac{n}{2}\rfloor -2$. Now to cover $v_1$ we need $2^{\lfloor \frac{n}{2}\rfloor }$ pebbles. The number of pebbles used in this process is $2^{\lfloor \frac{n}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor -2}-2+2^{\lfloor \frac{n}{2}\rfloor }<2^{n-1}+2^{(n-3)}-1$ pebbles.

   If $v_k(k<\lfloor \frac{n}{2}\rfloor ork>\lfloor \frac{n}{2}\rfloor +1)$ is the source vertex, then to the right of $v_k$ we have to cover the the vertices in a path of length $n-k$ and to the left of $v_k$ we have to cover the vertices in a path of length $k-3$ excluding the vertex $v_1$ and the vertices in the graph $<V-D>$. By Theorem 3 we require at most $2^{n-k}-1+2^{k-3}-2$ pebbles to cover the vertices in the above path of length $n-k$ and $k-3$. Now to cover $v_1$ we need $2^k$ pebbles. The number of pebbles used in this process is $2^{n-k}-1+2^{k-3}-2+2^k<2^{n-1}+2^{n-3}-1$ pebbles.

   When $n$ is odd $v_{\lfloor \frac{n}{2}\rfloor +1}$ can be taken as a source vertex. If $v_{\lfloor \frac{n}{2}\rfloor +1}$ is the source vertex, then to the right of $v_{\lfloor \frac{n}{2}\rfloor +1}$ we have to cover the the vertices in a path of length $\lfloor \frac{n}{2}\rfloor$ and to the left of $v_{\lfloor \frac{n}{2}\rfloor +1}$ we have to cover the vertices in a path of length $\lfloor \frac{n}{2}\rfloor -2$ excluding the vertex $v_1$ and the vertices in the graph $<V-D>$. By Theorem 3 we require at most $2^{\lfloor \frac{n}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor -2}-2$ pebbles to cover the vertices in the above path of length $\lfloor \frac{n}{2}\rfloor$ and $\lfloor \frac{n}{2}\rfloor -2$. Now to cover $v_1$ we need $2^{\lfloor \frac{n}{2}\rfloor }$ pebbles. The number of pebbles used in this process is $2^{\lfloor \frac{n}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor -2}-2+2^{\lfloor \frac{n}{2}\rfloor }<2^{n-1}+2^{n-3}-1$ pebbles.

   If $v_k(k<\lfloor \frac{n}{2}\rfloor +1ork>\lfloor \frac{n}{2}\rfloor +1)$ is the source vertex, then to the right of $v_k$ we have to cover the the vertices in a path of length $n-k$ and to the left of $v_k$ we have to cover the vertices in a path of length $k-3$ excluding the vertex $v_1$ and the vertices in the graph $<V-D>$. By Theorem 3 we require at most $2^{n-k}-1+2^{k-3}-2$ pebbles to cover the vertices in the above path of length $n-k$ and $k-3$. Now to cover $v_1$ we need $2^k$ pebbles. The number of pebbles used in this process is $2^{n-k}-1+2^{k-3}-2+2^k<2^{n-1}+2^{n-3}-1$. Hence, ${\psi }_{ns}(P_n)=2^{n-1}+2^{n-3}-1$.

Proof. Let $V(C_n)=\{v_1,v_2,v_3,\cdots ,v_n\}.$ When $n$ is even, the non-split dominating set $D_1=\{v_1,v_2,\cdots ,v_{\lfloor \frac{n}{2}\rfloor -1},v_{\lfloor \frac{n}{2}\rfloor +2},v_{\lfloor \frac{n}{2}\rfloor +3},\cdots ,v_{n-1},v_n\}$. When $n$ is odd, the non-split dominating set is $D_2=\{v_1,v_2,\cdots ,v_{\lfloor \frac{n}{2}\rfloor },v_{\lfloor \frac{n}{2}\rfloor +3},v_{\lfloor \frac{n}{2}\rfloor +4},\cdots ,v_{n-1},v_n\}$.

We observe that $V(<V-D_1>)=\{v_{\lfloor \frac{n}{2}\rfloor },v_{\lfloor \frac{n}{2}\rfloor +1}\}$ and $V(<V-D_2>)=\{v_{\lfloor \frac{n}{2}\rfloor +1},v_{\lfloor \frac{n}{2}\rfloor +2}\}$. Notice that there exist two paths from $v_1$ to $v_{\lfloor \frac{n}{2}\rfloor -1}$ and from $v_1$ to $v_{\lfloor \frac{n}{2}\rfloor +2}$. Let them be $P_{\lfloor \frac{n-1}{2}\rfloor }$ and $P_{\lfloor \frac{n}{2}\rfloor }$, when $n$ is even. When $n$ is odd, there exist two paths from $v_1$ to $v_{\lfloor \frac{n}{2}\rfloor }$ and from $v_1$ to $v_{\lfloor \frac{n}{2}\rfloor +3}$. Let them be $P_{\lfloor \frac{n}{2}\rfloor }$ and $P_{\lfloor \frac{n}{2}\rfloor }$ respectively.

1. Case 1: $n$ is even.

   Placing $2^{\lfloor \frac{n-1}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor }-3$ pebbles on the source vertex $v_1$ we cannot put one pebble each on all the vertices of $D_1$. Hence, ${\psi }_{ns}(C_n)\ge 2^{\lfloor \frac{n-1}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor }-2$.

   Now we prove the sufficient condition. Consider a configuration of $2^{\lfloor \frac{n-1}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor }-2$ pebbles. Let the source vertex be $v_1$. Using Theorem 3, we can cover the non-split dominating set $D_1$. Since $D_1$ has two paths of length $(\lfloor \frac{n}{2}\rfloor -2)$ and $(\lfloor \frac{n}{2}\rfloor -1)$, using $2^{\lfloor \frac{n}{2}\rfloor -1}-1$ pebbles we can cover all the vertices of the path of length $\lfloor \frac{n}{2}\rfloor -2$ and using $2^{\lfloor \frac{n}{2}\rfloor }-2$ pebbles we cover all the vertices of the path of length $\lfloor \frac{n}{2}\rfloor -1$. The total number of pebbles used to cover the vertices of $D_1$ is $2^{\lfloor \frac{n}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor }-2$.

   If we distribute all the pebbles on the vertices of $<V-D_1>$, we need at least $2^{\lfloor \frac{n}{2}\rfloor }-2$ pebbles whose pebbling process will result in another non-split dominating set $D$. The new $<V-D>$ is also connected. Similarly, we can prove the result for other source vertices. Hence, ${\psi }_{ns}(C_n)=2^{\lfloor \frac{n-1}{2}\rfloor }-1+2^{\lfloor \frac{n}{2}\rfloor }-2$.

2. Case 2: $n$ is odd.

   Placing $2^{\lfloor \frac{n}{2}\rfloor +1}-4$ pebbles on the source vertex $v_1$ we cannot put one pebble each on all the vertices of $D_2$. Hence, ${\psi }_{ns}(C_n)\ge 2^{\lfloor \frac{n}{2}\rfloor +1}-3$.

   Now we prove the sufficient condition. Consider a configuration $C$ with $2^{\lfloor \frac{n}{2}\rfloor +1}-3$ pebbles on the vertices of $C_n$. Let the source vertex be $v_1$. Using Theorem 3 of the cover pebbling number of path, we can cover the non-split dominating set $D_2$. Note that $D_2$ has two paths of equal length $(\lfloor \frac{n}{2}\rfloor -1)$. Using $2^{\lfloor \frac{n}{2}\rfloor }-1$ pebbles we can cover all the vertices of the path of length $\lfloor \frac{n}{2}\rfloor -1$ and $2^{\lfloor \frac{n}{2}\rfloor }-2$ pebbles we cover all the vertices of path of length $\lfloor \frac{n}{2}\rfloor -1$. The total number of pebbles used to cover the $D_2$ is $2^{\lfloor \frac{n}{2}\rfloor +1}-3$.

   If we distribute all the pebbles on the vertices of $<V-D_2>$, we need at least $2^{\lfloor \frac{n}{2}\rfloor -1}-1+2^{\lfloor \frac{n}{2}\rfloor }-1$ pebbles to form another non-split dominating set $D$. The new $<V-D>$ is also connected. Similarly, we can prove for the rest of the vertices assuming as source vertices. Hence, ${\psi }_{ns}(C_n)=2^{\lfloor \frac{n}{2}\rfloor +1}-3$.

Proof. Let the hub vertex be denoted by $v_0$ and the vertices which are adjacent to $v_0$ be denoted by $v_1,v_2,\dots ,v_{2n}$(clockwise manner).

Consider the non-split dominating set $D_1=\{v_1,v_3,\cdots ,v_{2n-1}\}$ or $D_2=\{v_2,v_4,\cdots ,v_{2n}\}$. Clearly the induced sub-graph $<V-D_1>$ and $<V-D_2>$ are connected. Without loss of generality, let us consider the non-split dominating set $D_1$. Consider the configuration of $4(n-1)$ pebbles on the vertex of $v_1$. Shifting $2(n-1)$ pebbles to the hub vertex we could cover maximum $n-1$ vertices of the non-split domination set. We are left with a vertex $v_1$ without a cover. Hence, ${\psi }_{ns}(F{r}_n)\ge 4(n-1)+1$.

Now we prove the sufficient condition by distributing $4(n-1)+1$ pebbles on the vertices of $F{r}_n$, that is, Hub vertex has minimum of $2(n-1)+2$ pebbles on it.

Using $2(n-1)$ pebbles we can cover dominate $n-1$ vertices of the non-split dominating set $D_1$. Further, using 2 pebbles we could cover the remaining vertex. Suppose, the hub vertex has less than $2(n-1)+2$. Let it be $p$ pebbles. Then using $p$ pebbles we could cover $\lfloor \frac{p}{2}\rfloor$ vertices. Assume that $p^{{}^{\prime }}$ is the number of vertices receiving pebbles in this way. Keep a maximum of two pebbles on each vertex and transfer the remaining to the hub vertex. Thus, Thus using at least $2n$ pebbles we can cover dominate $D_1$. Hence, ${\psi }_{ns}(F{r}_n)=4(n-1)+1$. 

Proof. Let the vertices of the path $P_n\odot K_1$ be denoted by $v_1,v_2,\dots ,v_n$ and the pendant vertices attached to each vertex $v_i,1\le i\le n$ on the path be $u_i,1\le i\le n$.

Consider the non-split dominating set $D=\{u_1,u_2,\cdots ,u_n\}$. Clearly, $<V-D>$ is connected. Consider the configuration of $\sum _{i=0}^{n+1}{2}^i-7$ pebbles placed on the vertex $u_1$. Then, a minimum of $1+2^3+2^4+\cdots +2^{n-1}+2^n+2^{n+1}$ pebbles are required in order to produce a non-split dominating set cover. But we lack one pebble to cover $u_1$. Therefore, ${\psi }_{ns}(P_n\odot K_1)\ge \sum _{i=0}^{n+1}{2}^i-6$.

1. Case 1: Let the source vertex be $v_1$  or $v_n$.

   Without loss of generality, let the source vertex be $v_1$. To cover the vertex $u_1$ we require 2 pebbles. Then, to cover the remaining vertices of $\{D-u_1\}$ we need $\sum _{i=0}^n{2}^i-3$ pebbles. Thus, the total number of pebbles used to cover the vertices of the non-split dominating set is $\sum _{i=0}^n{2}^i-1\le \sum _{i=0}^{n+1}{2}^i-6$. By symmetry, we can prove when $v_n$ is the source vertex.

2. Case 2: Let the source vertex be $v_k$, 1$\le k\le n$.

   Let $v_k$ be the source vertex. To cover the adjacent vertex $u_k$ we need 2 pebbles. Now to cover the remaining vertices $u_{k-1},u_{k-2},\cdots ,u_2,u_1,u_{k+1},\cdots ,u_{n-1},u_n$ of $D$ we need $\sum _{i=1}^k{2}^i+\sum _{i=1}^{n-(k-1)}{2}^i$ pebbles, which is less than the total number of available pebbles.

3. Case 3: Let the source vertex be $u_k$, 1$\le k\le n$.

   Let $u_k$ be the source vertex. To cover the vertex $u_k$ we need 1 pebble. Now to cover the remaining vertices $u_{k-1},u_{k-2},\cdots ,u_2,u_1,u_{k+1},\cdots ,u_{n-1},u_n$ of $D$ we need $\sum _{i=1}^{k+1}{2}^i+\sum _{i=1}^{n-(k-2)}{2}^i+1\le \sum _{i=0}^{n+1}{2}^i-6$ pebbles. Hence, ${\psi }_{ns}(P_n\odot K_1)=\sum _{i=0}^{n+1}{2}^i-6.$

Proof. Let $v_0$ be the vertex that joins all the $k$– star graphs. Let $V(B_{n,k})=\{v_0,a_1^j,a_2^j,\cdots ,a_{k-1}^j,a_0^j\}$, where $j=\{1,2,3,4,\cdots ,n\}$ and $E(B_{n,k})=\{v_0{a}_{k-1}^j,a_0^j{a}_i^j\}$, where $j=\{1,2,3,4,\cdots ,n\}$ and $1=\{1,2,3,4,\cdots ,k-1\}$. Let the non-split dominating set $D=\{v_0,a_1^j,a_2^j,\cdots ,a_{k-2}^j,a_0^j,a_{k-1}^1\}$, where $j=\{1,2,3,4,\cdots ,n\}$. Clearly, the induced sub-graph $<V-D>$ is connected.

Consider the distribution of $64(n-1)(k-2)+32(n-1)+4(k-2)+2$ pebbles on any one of the pendant vertices. Let it be $a_1^1$. Then we could cover all the vertices of the dominating set $D$ except one vertex. Hence, ${\psi }_{ns}(B_{n,k})\ge 64(n-1)(k-2)+32(n-1)+4(k-2)+3$.

Now consider distributing $64(n-1)(k-2)+32(n-1)+4(k-2)+3$ pebbles on the vertices of $(B_{n,k})$.

1. Case 1: Let $v_0$ be the source vertex.

   There are $n$ copies of $(k-2)$ star graph pendant vertices at a distance of 3 from $v_0$ and $n$ copies of the hub vertex of the $k$-star at a distance of 2. Thus, using $8n(k-2)+4n$ pebbles, we could cover all the vertices of $D$ except $a_{k-1}^1$ which requires further 2 pebbles. Hence, using $8n(k-2)+4n+2$ pebbles, we can dominate the set $D$.

2. Case 2: Let any one of the hub vertex of the $k$– star graph be the source vertex.

   Without loss of generality, let $a_0^1$ be the source vertex. To cover the dominating set of vertices that are adjacent to $a_0^1$ we require $2(k-1)$ pebbles. The rest of the vertices are at distances of 5 and 4. There are $(n-1)$ copies of $(k-2)$ star graph of pendant vertices is at a distance of 5 and $n-1$ copies of the hub vertex of the star graph are at a distance of 4. Thus, using $32(n-1)(k-2)+16(n-1)+2(k-1)$ pebbles, we can cover the non-split dominating set $D$. Hence, ${\psi }_{ns}(B_{n,k})=64(n-1)(k-2)+32(n-1)+4(k-2)+3$.

Proof. Let $V(F_{n,k})=\{a_i,b_i,c_{ij}|i=1,2,\cdots ,n,j=1,2,\cdots ,k-2\}andE(F_{n,k})=\{a_i{a}_{i+1}|i=1,2,\cdots ,n-1\}\cup \{a_i{b}_i,b_i{c}_{ij}|i=1,2,\cdots ,n;j=1,2,\cdots ,k-2\}$. Consider the non-split dominating set $D=\{b_i,c_{ij}|i=1,2,\cdots ,n;j=1,2,\cdots ,k-2\}$. Clearly, $<V-D>$ is connected.

Consider the distribution of placing $4(k-3)+2+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles on a pendant vertex of the first $k$ star graph of the $(F_{n,k})$. Then, we require a minimum of $16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles to cover vertices of the non-split dominating set in $(n-1)$ k-star graphs. And the vertices in the first $(k-1)$ star graph are not covered. There are $k-3$ vertices at a distance 2 and one vertex at a distance one in the first $k-1$ star graph. Hence, we require further $4(k-3)+3$ pebbles. But the number of pebbles remaining is $4(k-3)+2$. Thus, ${\psi }_{ns}(F_{n,k}\ge 4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$.

Now we prove the sufficient condition. Consider a configuration of $4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles.

1. Case 1: Let $b_1$ or $b_n$ be a source vertex.

   Without loss of generality, consider $b_1$ be the source vertex. Let us place $4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles on $b_1$. Since there are $k-2$ vertices of the non-split dominating set that are adjacent to $b_1$, we need $2(k-2)$ pebbles to cover them and one more pebble needed to cover $b_1$. Now we are left with vertices in the $(n-1)$ parts of  $(k-1)$ star graphs that are to be covered. To cover all those vertices we require $8(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+4\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles. Thus, the total number of pebbles used to cover the vertices of the non-split dominating set is $2(k-2)+1+8(k-2)+\left(\sum _{i=1}^{n-1}{2}^i\right)+4\left(\sum _{i=1}^{n-1}{2}^i\right)\le 4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$. By symmetry, we can prove this for the source vertex $b_n$.

2. Case 2: Let $b_s(2\le s\le n-1)$ be a source vertex.

   Let us place $4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles on $b_s$. Since there are $k-2$ vertices of the non-split dominating set that are adjacent to $b_s$, we need $2(k-2)$ pebbles to cover them and one more pebble to cover $b_s$. Now we are left with the vertices that are not covered in the $n-1$ parts of $(k-1)$ star graphs. To cover all those vertices we require $8(k-2)\left(\sum _{i=1}^{n-s}{2}^i\right)+4\left(\sum _{i=1}^{n-s}{2}^i\right)+8(k-2)\left(\sum _{i=1}^{s-1}{2}^i\right)+4\left(\sum _{i=1}^{s-1}{2}^i\right)$ pebbles. Thus, the total number of pebbles used to cover the vertices of non-split dominating set is $2(k-2)+1+8(k-2)+8(k-2)\left(\sum _{i=1}^{n-s}{2}^i\right)+4\left(\sum _{i=1}^{n-s}{2}^i\right)+8(k-2)\left(\sum _{i=1}^{s-1}{2}^i\right)+4\left(\sum _{i=1}^{s-1}{2}^i\right)\le 4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$.

3. Case 3: Let $a_1$ or $a_n$ be a source vertex.

   Without loss of generality, consider $a_1$ the source vertex. Let us place $4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles on $a_1$. Since there are $k-2$ vertices of the non-split dominating set that are at distance 2 and one vertex that is adjacent to $a_1$, we need $4(k-2)+2$ pebbles to cover them. Now we are left with the vertices that are not covered in the $n-1$ parts of $(k-1)$ star graphs. To cover all those vertices, we require $4(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+2\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles. Thus, the total number of pebbles used to cover the vertices of the non-split dominating set is $4(k-2)+2+4(k-2)+\left(\sum _{i=1}^{n-1}{2}^i\right)+2\left(\sum _{i=1}^{n-1}{2}^i\right)\le 4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$. By symmetry, we can prove the result for the source vertex $a_n$.

4. Case 4: Let $a_s(2\le s\le n-1)$ be a source vertex.

   Let us place $4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$ pebbles on $a_s$. Since there are $k-2$ vertices of the non-split dominating set that are at distance 2 and one vertex that is adjacent to $a_s$, we need $4(k-2)+2$ pebbles to cover them. Now we are left with vertices that are not covered in the $n-1$ parts of the $(k-1)$ star graphs. To cover all those vertices, we require $4(k-2)\left(\sum _{i=1}^{n-s}{2}^i\right)+2\left(\sum _{i=1}^{n-s}{2}^i\right)+4(k-2)\left(\sum _{i=1}^{s-1}{2}^i\right)+2\left(\sum _{i=1}^{s-1}{2}^i\right)$ pebbles. Thus, the total number of pebbles used to cover the vertices of non-split dominating set is $4(k-2)+2+4(k-2)+8(k-2)\left(\sum _{i=1}^{n-s}{2}^i\right)+2\left(\sum _{i=1}^{n-s}{2}^i\right)+4(k-2)\left(\sum _{i=1}^{s-1}{2}^i\right)+2\left(\sum _{i=1}^{s-1}{2}^i\right)\le 4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$. Hence, ${\psi }_{ns}(F_{n,k}=4(k-3)+3+16(k-2)\left(\sum _{i=1}^{n-1}{2}^i\right)+8\left(\sum _{i=1}^{n-1}{2}^i\right)$.