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 \(E(W_n)=\{v_0v_i, v_jv_{j+1}, v_{n-1}v_1\}\) where \(1\leq i \leq n-1\) and \(1\leq \leq n-2\). 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, \(\psi_{ns}(P_n)\geq 2^{n-1}+2^{n-3}-1\).

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 \(v_k\) ,\(5\leq k \leq n-4\).

   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 \(v_{k} \ (k<\lfloor\frac{n}{2}\rfloor\ or \ k>\lfloor\frac{n}{2}\rfloor+1)\) 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 \(v_{k} \ (k<\lfloor\frac{n}{2}\rfloor+1\ or \ k>\lfloor\frac{n}{2}\rfloor+1)\) 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, \(\psi_{ns}(P_n)= 2^{n-1}+2^{n-3}-1\).

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

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

1. Case 1: \(n\) is even.

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

   Now we prove the sufficient condition. Consider a configuration of \(2^{\left\lfloor \frac{n-1}{2}\right\rfloor}-1 +2^{\left\lfloor \frac{n}{2}\right\rfloor} -2\) 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 \((\left\lfloor \frac{n}{2}\right\rfloor-2)\) and \((\left\lfloor \frac{n}{2}\right\rfloor-1)\), using \(2^{\left\lfloor \frac{n}{2}\right\rfloor-1}-1\) pebbles we can cover all the vertices of the path of length \(\left\lfloor \frac{n}{2}\right\rfloor-2\) and using \(2^{\left\lfloor \frac{n}{2}\right\rfloor}-2\) pebbles we cover all the vertices of the path of length \(\left\lfloor \frac{n}{2}\right\rfloor-1\). The total number of pebbles used to cover the vertices of \(D_1\) is \(2^{\left\lfloor \frac{n}{2}\right\rfloor}-1 +2^{\left\lfloor \frac{n}{2}\right\rfloor} -2\).

   If we distribute all the pebbles on the vertices of \(<V-D_1>\), we need at least \(2^{\left\lfloor \frac{n}{2}\right\rfloor}-2\) 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, \(\psi_{ns}(C_n)= 2^{\left\lfloor \frac{n-1}{2}\right\rfloor}-1 +2^{\left\lfloor \frac{n}{2}\right\rfloor} -2\).

2. Case 2: \(n\) is odd.

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

   Now we prove the sufficient condition. Consider a configuration \(C\) with \(2^{\left\lfloor \frac{n}{2}\right\rfloor+1}-3\) 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 \((\left\lfloor \frac{n}{2}\right\rfloor-1)\). Using \(2^{\left\lfloor \frac{n}{2}\right\rfloor}-1\) pebbles we can cover all the vertices of the path of length \(\left\lfloor \frac{n}{2}\right\rfloor-1\) and \(2^{\left\lfloor \frac{n}{2}\right\rfloor}-2\) pebbles we cover all the vertices of path of length \(\left\lfloor \frac{n}{2}\right\rfloor-1\). The total number of pebbles used to cover the \(D_2\) is \(2^{\left\lfloor \frac{n}{2}\right\rfloor+1}-3\).

   If we distribute all the pebbles on the vertices of \(<V-D_2>\), we need at least \(2^{\left\lfloor \frac{n}{2}\right\rfloor-1}-1 + 2^{\left\lfloor \frac{n}{2}\right\rfloor}-1\) 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, \(\psi_{ns}(C_n)= 2^{\left\lfloor \frac{n}{2}\right\rfloor+1}-3\).

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Proof. Let the hub vertex be denoted by \(v_0\) and the vertices which are adjacent to \(v_0\) be denoted by \(v_1, v_2,…,v_{2n}\)(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, \(\psi_{ns}(Fr_n)\geq 4(n-1)+1\).

Now we prove the sufficient condition by distributing \(4(n-1)+1\) pebbles on the vertices of \(Fr_n\), 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 \(p^{'}\) 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, \(\psi_{ns}(Fr_n)= 4(n-1)+1\). ◻

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

Consider the non-split dominating set \(D=\{u_1, u_2,\cdots, \ u_n\}\). Clearly, \(<V-D>\) is connected. Consider the configuration of \(\sum\limits_{i=0}^{n+1} 2^i-7\) 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, \(\psi_{ns}(P_n\odot K_1)\geq \sum\limits_{i=0}^{n+1} 2^i-6\).

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 \(\sum\limits_{i=0}^{n} 2^i-3\) pebbles. Thus, the total number of pebbles used to cover the vertices of the non-split dominating set is \(\sum\limits_{i=0}^{n} 2^i-1\le \sum\limits_{i=0}^{n+1} 2^i-6\). 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 \(\sum\limits_{i=1}^{k} 2^i + \sum\limits_{i=1}^{n-(k-1)} 2^i\) 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 \(\sum\limits_{i=1}^{k+1} 2^i + \sum\limits_{i=1}^{n-(k-2)} 2^i +1\le\sum\limits_{i=0}^{n+1} 2^i-6\) pebbles. Hence, \(\psi_{ns}(P_n\odot K_1)= \sum\limits_{i=0}^{n+1} 2^i-6.\)

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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 \(E(B_{n,k})=\{v_0a_{k-1}^{j}, a_0^{j}a_i^{j}\}\), 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, \(\psi_{ns}(B_{n,k})\geq 64(n-1)(k-2)+32(n-1)+4(k-2)+3\).

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, \(\psi_{ns}(B_{n,k})= 64(n-1)(k-2)+32(n-1)+4(k-2)+3\).

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Proof. Let \(V(F_{n,k})=\{a_i, b_i, c_{ij} | i=1,2,\cdots,n,\ j=1,2,\cdots, k-2\} \ and \ E(F_{n,k})=\{a_ia_{i+1}|\ i=1,2,\cdots,n-1\}\cup \{a_ib_i,\ b_ic_{ij}\ | i=1,2,\cdots,n; \ j=1,2,\cdots, k-2\}\). 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 \(4(k-3)+2+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) pebbles on a pendant vertex of the first \(k\) star graph of the \((F_{n,k})\). Then, we require a minimum of \(16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) 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, \(\psi_{ns}(F_{n,k}\geq 4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\).

Now we prove the sufficient condition. Consider a configuration of \(4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) 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 \(4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) 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 \(8(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+4\left(\sum\limits_{i=1}^{n-1}2^i\right)\) pebbles. Thus, the total number of pebbles used to cover the vertices of the non-split dominating set is \(2(k-2)+1+8(k-2)+ \left(\sum\limits_{i=1}^{n-1}2^i\right)+4\left(\sum\limits_{i=1}^{n-1}2^i\right)\le 4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\). By symmetry, we can prove this for the source vertex \(b_n\).

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

   Let us place \(4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) 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 \(8(k-2) \left(\sum\limits_{i=1}^{n-s}2^i\right)+4\left(\sum\limits_{i=1}^{n-s}2^i\right)+ 8(k-2) \left(\sum\limits_{i=1}^{s-1}2^i\right)+4\left(\sum\limits_{i=1}^{s-1}2^i\right)\) pebbles. Thus, the total number of pebbles used to cover the vertices of non-split dominating set is \(2(k-2)+1+8(k-2)+ 8(k-2) \left(\sum\limits_{i=1}^{n-s}2^i\right)+4\left(\sum\limits_{i=1}^{n-s}2^i\right)+ 8(k-2) \left(\sum\limits_{i=1}^{s-1}2^i\right)+4\left(\sum\limits_{i=1}^{s-1}2^i\right)\le 4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\).

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 \(4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) 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 \(4(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+2\left(\sum\limits_{i=1}^{n-1}2^i\right)\) pebbles. Thus, the total number of pebbles used to cover the vertices of the non-split dominating set is \(4(k-2)+2+4(k-2)+ \left(\sum\limits_{i=1}^{n-1}2^i\right)+2\left(\sum\limits_{i=1}^{n-1}2^i\right)\le 4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\). By symmetry, we can prove the result for the source vertex \(a_n\).

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

   Let us place \(4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\) 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 \(4(k-2) \left(\sum\limits_{i=1}^{n-s}2^i\right)+2\left(\sum\limits_{i=1}^{n-s}2^i\right)+ 4(k-2) \left(\sum\limits_{i=1}^{s-1}2^i\right)+2\left(\sum\limits_{i=1}^{s-1}2^i\right)\) pebbles. Thus, the total number of pebbles used to cover the vertices of non-split dominating set is \(4(k-2)+2+4(k-2)+ 8(k-2) \left(\sum\limits_{i=1}^{n-s}2^i\right)+2\left(\sum\limits_{i=1}^{n-s}2^i\right)+ 4(k-2) \left(\sum\limits_{i=1}^{s-1}2^i\right)+2\left(\sum\limits_{i=1}^{s-1}2^i\right)\le 4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\). Hence, \(\psi_{ns}(F_{n,k}=4(k-3)+3+16(k-2) \left(\sum\limits_{i=1}^{n-1}2^i\right)+8\left(\sum\limits_{i=1}^{n-1}2^i\right)\).

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