CONNECTED k-FORCING SETS OF GRAPHS AND SPLITTING GRAPHS

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. The notion of k-forcing number of a graph was introduced by Amos et al. For a given graph G and a given subset I of the vertices of the graph G, the vertices in I are known as initially colored black vertices and the vertices in V (G)− I are known as not initially colored black vertices or white vertices. The set I is a k-forcing set of a graph G if all vertices in G eventually colored black after applying the following color changing rule: If a black colored vertex is adjacent to at most k-white vertices, then the white vertices change to be colored black. The cardinality of a smallest k-forcing set is known as the k-forcing number Zk(G) of the graph G. If the sub graph induced by the vertices in I are connected, then I is called the connected k-forcing set. The minimum cardinality of such a set is called the connected k-forcing number of G and is denoted by Zck(G). This manuscript is intended to study the connected k-forcing number of graphs and the splitting graphs.


INTRODUCTION
Through out this manuscript, we consider graphs without loops and multiple edges. That is we consider only simple graph G = (V, E) with vertex set V (G) and edge set E(G). The to the definition of connected zero forcing set, Z c (G) (See [11]). In this article, we deal with connected k-forcing number of some graphs and their splitting graphs. We use the following definitions for the further development of this article.
• Corona Product: For any two graphs G and H, the Corona product G • H of the graphs G and H is the graph detemined by taking one copy of G and | V (G) | copies of H and by connecting each vertex of the j th copy of H to the j th vertex of G, 1 ≤ j ≤| V (G) |.
• Rooted Product: Let G be a connected graph with vertices v 1 , v 2 , . . . , v n and let H be a sequence of n-rooted graphs H 1 , H 2 , ..., H n . The rooted product of G and H is defined as the graph obtained by identifying the root of H i , 1 ≤ i ≤ n with the i th vertex of G for all i. This graph is denoted by G(H) and is known as the rooted product of G by H (See [8]).
• Square of a Graph: Let G be a simple graph with vertex set V (G) and edge set E(G).
Then the square of G, denoted by G 2 , is the graph having the vertex set same as that of G and such that two vertices in G 2 are adjacent if the distance between them is at most two in G.
• When the k-color changing rule is applied to an arbitrary vertex u to change the color of the vertex v, we say u, k-forces (if it is zero forcing, then we say u forces v) v and write u → v.
For more definitions on graphs, we refer to [9]. From the definitions above, we have the following proposition.
Proposition 3. Let P n , n ≥ 3 be a path on n vertices. Then Proof. Case 1 Assume that k = 1. It can be easily observed that if we color any two adjacent vertices as black, it is not possible to obtain a derived coloring. Therefore, Z c1 [S(P n )] ≥ 3. Now, let u 1 , u 2 , . . . , u n be the vertices of P n and u 1 , u 2 , . . . , u n be the corresponding vertices in S(P n ).
Color the vertices u 1 , u 2 and u 1 as black. Clearly, the vertex u 1 forces u 2 as black, the vertex u 2 forces the vertex u 3 as black, the vertex u 2 forces the vertex u 3 as black and so on. Therefore, Z = {u 1 , u 2 , u 1 } forms a connected zero forcing set for the path P n . So, Z c1 S(P n ) ≤ 3. Hence the result follows.
Case 2 Assume that k ≥ 2. In this case, the vertex u 1 forms a connected zero forcing set and hence the result follows.
It can be observed that any conncetd k-forcing set is a k-forcing set. Therefore, we have the following Proposition 4. For any simple graph G, and for any fixed k, Z k (G) ≤ Z ck (G), where Z k is the k-forcing number and Z ck (G) is the connected k-forcing number of G.
We consider the next proposition from [5] to prove the result concerning the splitting graph of the cycle C n . Proposition 5 ( [5]). If G is the cycle C n on n ≥ 4 vertices, then Z[S(G)] = 4.
In the succeeding proposition, we consider the spliting graph of the cycle C n , n ≥ 4. Proposition 6. Let S(C n ) be the splitting graph of the cycle C n . Then Proof. Case 1 Assume that k = 1. From Proposition-4 and Proposition-5, we have the following: To prove the reverse part, let us consider the vertices of the cycle C n as v 1 , v 2 , . . . , v n and v 1 , v 2 , . . . , v n be the corresponding vertices of v 1 , v 2 , . . . , v n in S(C n ). Consider the set of ver- 4 to black and so on. Therefore, we can obtain a derived coloring with the set of Hence the result follows.
Case 2 Let us assume that k = 2 and Z c2 [S(C n )] = 2. Consider a connected 2-forcing set consisting of two vertices. Let u and v be the two adjacent vertices in the connected 2-forcing set of S(C n ). Then we have two sub cases: Since u and v are adjacent to three white neighbors, color changing rule is not applicable in this case, we get a contradiction to our assumption that Subcase 2.2 deg(u) = 2 and deg(v) = 4. In this case the vertex u can force one more adjacent vertex of degree 4 to black. Therefore, in this case it is not possible to obtain a derived coloring. Hence from subcases 2.1 and 2.2, we have Z c2 [S(C n )] ≥ 3.
It can be easily observed that the vertices {v 1 , v 1 , v 2 } forms a zero forcing set for S(C n ) and hence the result follows. For k = 3 the result is obvious.
The Friendship graph F p is the graph obtained by identifying p copies of the cycle graph C 3 with a common vertex.
Proof. Case 1 Assume that k is even and k < ∆ − 2. Let v be the vertex with maximum degree ∆. It can be noted that v should be a member of any connected zeroforcing set.
Otherwise, the zero forcing set will not be connected. Therefore, assume that the vertex v is there in any connected zero forcing set of F p . If we take one vertex from each of the p − k 2 − 1 triangles, then it is not possible to obtain a derived coloring since deg(v) = 2p and by using color changing rule we get 2(p − k 2 − 1) = 2p − k − 2 black vertices which are adjacent to the vertex v. Now we have 2p − (2p − k − 2) = k + 2 white vertices remains. It is not possible to force these k + 2 vertices by using the vertex v. Therefore, we must take one black vertex from each of the p − k 2 triangles since the vertex v is black, Let us take one vertex from each of the p − k 2 triangles as black. Since the vertex v is black, these p − k 2 vertices will force the remaining vertices in the p − k 2 triangles as black. Now we have 2(p − k 2 ) black vertices together with the black vertex v in the connected k-forcing set. It can be observed that at this stage we have 2p − (2p − k) = k white vertices adjacent to the vertex v. Now the vertex v can force these k-vertices as black. Therefore we get a derived Case 2 Assume that k is odd and k < ∆ − 2. Let v be the vertex with maximum degree ∆. It can be noted that v should be a member of any connected zeroforcing set. Otherwise, the zero forcing set will not be connected. Therefore, assume that the vertex v is there in any It can be easily obseved that if k ≥ ∆ − 2, then Z ck (F p ) = 1.
Theorem 8. Let G be a connected graph with | V (G) |= p 1 and let H be another connected graph with Z ck (H) = p 2 . Let G be the graph obtained by taking the corona product of G and Proof. With out loss of generality, assume that G is connected, | V (G) |= p 1 and Z ck (H) = p 2 .
Color all vertices of G black. To form the k-forcing set for the sub graph induced by v 1 ∪ H 1 , we need a maximum of 1 + p 2 black vertices. That is, the second copy of H corresponds to the vertex v 2 in G . Proceeding like this, we can observe Therefore, Z k (G ) ≤ (1 + p 2 ) + (1 + p 2 ) + . . . + (1 + p 2 )-p 1 times. This follows that Z k (G ) ≤ p 1 (1 + p 2 ). Since each vertex in G is connected to the vertices of all copies of H, the k-forcing set obtained here forms a connected k-forcing set. Therefore, Z ck (G ) ≤ p 1 (1 + p 2 ).
Proposition 9. Let G be the complete bipartite graph K m,n , and n ≥ 2, m ≥ 2. Then the con- Proof. Since G is a complete bipartite graph, therefore, the vertex set of G can be partitioned into two sets X and Y . Let u 1 , u 2 , . . . , u m be the vertices in X and v 1 , v 2 , . . . , v n be the vertices in Y . Note that the vertices in X are non-adjacent. The vertices in Y are also non-adjacent. To start the color changing rule , color any vertex, say u 1 , in X as black. Since each vertex in X is We use the following results from [2] and [11] to prove the next result Also we have from proposition-10 and proposition-11 that From (1) and (2) the result follows.
Proposition 13. Let G be the star graph k 1,n on n + 1 vertices and n > 2. Then Z c (G) = n. In Proof. Let u 1 , u 2 , . . . , u n be the vertices of the star graph k 1,n with degree 1. Assume that v is the vertex having degree n. We generate the connected zero forcing set as follows. Since deg(v) = n, to apply the color changing rule, we have to color n − 1 vertices in G adjacent to v as black.
Then v forces the only remaining white vertex to black. Therefore, Z c (G) = n−1+1 = n.
If k = 2,we can easily show that the connected zero forcing number of G is n − 2 + 1 = n − 1.
Proceeding like this, we obtain Z ck (G) = n − k + 1 for any positive integer n ≥ k ≥ 2.

CONNECTED k-FORCING NUMBER OF ROOTED PRODUCT OF GRAPHS
In this section, we deal with the connected k-forcing number of rooted product of cycle with paths, cycle with cycles.
Proposition 14. Let P 1 , P 2 , . . . , P n be n-paths (each path is of length n ≥ 3) rooted at the pendant vertex and C n be a cycle on n ≥ 3 vertices. Let G be the graph obtained by taking the rooted product of the cycle C n with the paths P 1 , P 2 , . . . , P n . Then Proof. Let u 1 , u 2 , . . . , u n be the vertices of the cycle C n , n ≥ 3 and P 1 , P 2 , . . . , P n be the paths rooted at the vertices u 1 , u 2 , . . . , u n respectively. Each path is of length n, n ≥ 3.
Represent the vertices of P 1 by p 1 1 , p 1 2 , . . . , p 1 n , the vertices of P 2 by p 2 1 , p 2 2 , . . . , p 2 n and the vertices of P n by p n 1 , p n 2 , . . . , p n n . Let u 1 be the vertex identified with the vertex p 1 1 in G, u 2 be the vertex identified with the vertex p 2 1 in G., . . . , u n be the vertex identified with the vertex p n 1 .
Case 1. Assume that k = 1. This case is similar to that of the connected zero forcing number of G. Color the vertices u 1 , u 2 , . . . , u n in G black. Now one can easily infer that It can be worth mentioning that if we start the color changing rule with vertices of P i , 1 ≤ i ≤ n other than the vertices identified with the vertices u 1 , u 2 , . . . , u n of C n , we cannot obtain a connected zero forcing set with at least n black vertices. Therefore, we need to consider the vertices in the cycle to force the remaining vertices in G.
Now assume that we have a connected zero forcing set consisting of n − 1 black vertices. From the above it can be noted that these vertices must be from the cycle C n . Without loss of generality, assume that the vertices are u 1 , u 2 , . . . , u n−1 . Clearly the black vertex u 2 can force the vertices of the path P 2 , u 3 can force the vertices of the path P 3 , . . ., the vertex u n−2 can force the vertices of the path P n−2 . Since the black vertex u 1 is adjacent to two white vertices u n and p 1 2 , u 1 cannot force the vertices u n and p 1 2 . Similarly the vertex u n−1 is adjacent to two white vertices u n and p n−1 2 . Therefore, the vertex u n−1 cannot force u n and p n−1 2 , this contradicts our assumption that Z c (G) = n − 1. Therefore, (3) and (4)  Proposition 15. Let D 1 , D 2 , . . . , D n be the cycles C n of order n ≥ 3 rooted at a vertex and C n be another cycle of order n > 3. Let G be the graph derived from the rooted product of C n with the cycles D 1 , D 2 , . . . , D n . Then Proof. Without loss of generality, assume that u 1 , u 2 . . . , u n be the vertices of the cycle C n in G  It can be easily infer that to form a minimum connected zero forcing set for G, we need to color the vertices u 1 , u 2 , . . . , u n as black and color at least one vertex from each of the cycles D i , 1 ≤ i ≤ n adjacent to each u i , 1 ≤ i ≤ n as black, otherwise we cannot form a connected zero forcing set with at least 2n black vertices . Clearly, (5) and (6) concludes the result.
Case 2. Let us suppose that k = 2. Color all vertices of C n in G black. Each vertex u i , 1 ≤ i ≤ n, is adjacent to exactly two white vertices of D i and k = 2. Therefore, these vertices forms a 2-forcing set for G. The sub graph induced by these black vertices are connected and hence it forms a connected 2-forcing forcing set for G. Therefore, It can be easily infer that to form a minimum connected 2-forcing set for G, we need to color the vertices u 1 , u 2 , . . . , u n as black, otherwise we cannot form a connected 2-forcing set with at least n black vertices. Clearly, Therefore from (7) and (8), the result follows.
Case 3. Let us suppose that k ≥ 3. In this case any arbitrary vertex from the cycle D i , 1 ≤ i ≤ n will k-forces the remaining vertices as black in G. Therefore, Z ck (G) = 1.
Proposition 16. Let G be the rooted product of P n P 2 (the Ladder graph) with P t ,t ≥ 3 rooted at the pendant vertex. Then Proof. Represent the vertices of the graph P n P 2 by u 1 ,u 2 ,. . . ,u n and v 1 ,v 2 ,. . . ,v n . Let P 1 ,P 2 ,. . . ,P n be the paths rooted at the vertices u 1 ,u 2 ,. . . ,u n respectively. Also let Q 1 ,Q 2 ,. . . ,Q n be the paths rooted at the vertices v 1 , v 2 , . . . , v n respectively. The vertices of the paths P 1 ,P 2 ,. . . ,P n and Q 1 ,Q 2 ,. . . ,Q n in G can be named as follows: Consider  It can be easily observed that if we take 2n vertices other than these three sets, then it will not form a minimum connected zero forcing set. Now Assume that there exists a connected zero forcing set consisting of 2n − 1 black vertices.     Proposition 17. Let G be the rooted product of P n P n (The Grid graph) with P t ,t ≥ 3 rooted at the pendant vertex. Then Proof. Case 1. Assume that k = 1. In this case color all the vertices of the Cartesian product P n P n in G as black. One can easily observe that these n 2 -black vertices forms a connected zero forcing set for G. Thus Z c (G) ≤ n 2 .
Case 2. Assume that k = 2. Let u 1 , u 2 , . . . , u n be the vertices of the path P n in P n P n of G. Color these vertices as black in G. Now one can easily verify that these vertices form a connected zero forcing set for G. Thus, Z c2 (G) ≤ n, if k = 2.
Case 3. Assume that 3 ≤ k ≤ 5. Let P t be the path identified at the vertex u 1 in G.
Now color the pendant vertex of the path P t in G as black. Let it be the vertex v. Clearly the vertex v forces the remaining vertices in G as black. Therefore we can form a derived coloring for G. Thus Z ck (G) = 1, as desired.
We strongly believe that the bounds in the above proposition is sharp.
Proposition 18. Let G be the rooted product of P n P 2 with the cycle C n . Then Proof. Case 1. Assume that k = 1. Let u 1 , u 2 , . . . , u n be the vertices of the path P n in G and let v 1 , v 2 , . . . , v n be the vertices corresponding to the copy of the path P n in G. Note that The remaining vertices of P n P 2 in G have degree 5. It can be noted that any connected zero forcing set of G must contain all the vertices of P n P 2 . Otherwise the zero forcing set will be disconnected. Without loss of generality, assume that we have a set consisting of 2n connected black vertices from P n P 2 in G. To force the white vertices in each cycle, we must select a vertex adjacent to the rooted vertex of each Therefore we need to choose 2n black vertices from the cycle C n . Now we have a set of 4n black vertices which forces the remaining vertices of G, which is connected.
Case 2. Assume that k = 2. It can be observed that the connected zero forcing set of G must contain all the vertices of P n P 2 , Otherwise, the zero forcing set will be disconnected.
If we take the 2n black vertices of P n P 2 in G, then these black vertices will 2-forces the remaining white vertices as black and hence Z ck (G) = 2n.
Case 3. Assume that 5 ≥ k ≥ 3. Consider the cycle identified with the vertex u 1 , say C 1 . Choose a vertex from C 1 of degree 2 as black. This vertex will 3-force the remaining vertices in G as black. Hence Z ck (G) = 1.
Definition 19 ([1]). A connected graph G is defined as a cycle-path graph (CP-graph) if it contains r vertex disjoint cycles that are connected by r − 1 edges of the path P r . Thus a CPgraph with n vertices contains m = n + r − 1 edges and edge between two cycles is a cut edge.
The zero forcing number of CPgraph was studied in some detail in [5]. Here we study the connected zeroforcing number of the CPgraph considered in [5].
Proof. Denote the cycles by C 1 ,C 2 , . . . ,C r . Let the vertex sets of the cycles in C 3 P r be Case 1. Assume that k = 1. We prove the result by mathematical induction on the number of cycles r on the CP-graph. Assume that r = 1. In this case G is the cycle C 3 therefore, Z c (C 3 ) = 2 and the result is true for r = 1.
Assume that the result is true for all C 3 P r graphs with r − 1 cycles C 3 , where r ≥ 2. Let C be the end cycle connected to the rest of the C 3 P r graph by an edge e = ab, where Assume that the result is true for the sub graph induced by V (C 3 P r ) − V (C) . That is Let W be the minimum zero forcing set of V (C 3 P r ) −V (C) with | W |= 2r − 2. Let u 1 and u 2 be two white neighbors of the vertex b in C. Since the vertex a is black it forces the vertex b to black. Since the vertex b has two white neighbors, further forcing is not possible. In order to make the zero forcing set connected, we have to include the black vertex b in the connected zero forcing set of G. Therefore, our new connected zero forcing set is W ∪ {b}. The set W ∪ {b} cannot force the remaining two white vertices ( u 1 and u 2 ) adjacent to b. Therefore, we need to include either u 1 or u 2 in the connected zero forcing set of G. Let it be u 1 . Hence by induction Case 2. Assume that either k = 2 or k = 3. In this case any vertex of degree 2 will form a connected k-forcing set.
The Cartesian product C n K 2 is known as the Prism graph or the circular ladder graph. The length of the shortest cycle in a graph G is called the girth of G. We recall the following observation from [6].
To establish the reverse inequality, we proceed as follows.
Without loss of generality, choose four vertices u 1 , u 2 , u 3 and v 1 . Allow these vertices to have black color. Then clearly the black vertex u 2 → v 2 to black. Now the black vertex Case 1 Assume that k = 2. In this case , clearly a set consisting of any two adjacent black vertices forms a connected zero forcing set for G. Hence, Z c2 (G) = 2.
Case 2. Assume that k = 3. It is obvious that any single black vertex gives a derived coloring for G. Therefore, the result follows. That is Z c3 (G) = 1.
Proposition 23. Let G be the rooted product of the circular ladder graph, C n K 2 with the path P t , a path of length t, t ≥ 4 rooted at the pendent vertex. Then, Z c (G) = 2n.
Proof. Represent the vertices of C n K 2 as u 1 , u 2 , . . . , u n , v 1 , v 2 , . . . , v n in G and the paths rooted at the pendent vertex by P 1 , P 2 , . . . , P n of length t. Let v 1 = p 1 1 , v 2 = p 2 1 , . . . , v n = p n 1 , where p 1 1 , p 2 1 , . . . , p n 1 are the pendent vertices of the paths identified at the vertices v 1 , v 2 , . . . , v n respectively, where  Note that the combination of the vertices u i and the vertices of P i is not considered, since that combination does not form a connected induced sub graph in G. It is easy to verify that none of the above combinations will never form a connected zero forcing set for G. Hence from the above cases, we can infer that To claim Z c (G) ≤ 2n, we proceed as follows. Select 2n black vertices u 1 , u 2 , . . . , u n , v 1 , v 2 , . . . , v n . Then the black vertex v 1 → p 1 2 to black, the black vertex p 1 2 → p 1 3 to black, . . . , p 1 t−1 → p 1 t to black. Similarly, all the vertices of the paths rooted at the black vertices v 2 , v 3 , . . . , v n are colored black. The same argument holds good for the vertices of the paths rooted at the black vertices u 1 , u 2 , . . . , u n . Therefore, Z = {u 1 , u 2 , . . . , u n , v 1 , v 2 , . . . , v n } generates a connected zero forcing set for G. Cardinality of Z is 2n. So, (12) Z c (G) ≤ 2n From (11) and (12), the result follows.
Proposition 24. Let G be the rooted product of the circular ladder graph with the cycle C k , k ≥ 4. Then Z c (G) ≤ 4n. Represent the vertices of cycles C 1 ,C 2 , . . . ,C n and D 1 , D 2 , . . . , D n as follows.

Proof. Let
forcing set for the graph G. For, consider the following cases.
Case 1. Select the pendant vertex of each path rooted at the vertices u 1 , u 2 , . . . , u n−1 .
Clearly they cannot form a connected zero forcing set for G.
Case 2: Form a set of n − 1 black vertices from the vertices of the paths rooted at the vertices u 1 , u 2 , . . . , u n . We can easily observe that this set will not form a connected zero forcing set for G.
Then we can see that the vertices of the paths rooted at the vertices u 1 , u 2 , . . . , u n−2 can be colored as black by applying color changing rule. Note that the forcing from the black vertex u n−1 is not possible, since u n−1 has two white neighbours. So the set Z cannot generate a zero forcing set for G. In view of the above cases, we have Z c (G) ≥ n.
To prove the reverse part, let Z 1 = {u 1 , u 2 , . . . , u n }. Assign black color to the vertices in the set Z 1 . Then it can be seen that the set Z 1 generates a connected zero forcing set for G.
Therefore, Z c (G) ≤ n. Hence the result follows.
Again, when k = 2, any black vertex of the graph G, other than the vertex having degree 3 , gives a derived coloring for G. Hence, Z c (G) = 1.
When, k = 3, any vertex of G forms a connected zero forcing set, as we wish.

CONNECTED k-FORCING NUMBER OF SQUARE OF GRAPHS
In this section, we deal with the connected k-forcing number of square of path graph P n , n ≥ 4, the cycle graph C n , n ≥ 5 .
Proposition 26. Let G denotes the square of the path P n ,n ≥ 3. Then the connected zero forcing number of G is 2.
Proof. Represent the vertices of G by u 1 , u 2 , . . . , u n and let u 1 and u n be the pendant vertices in G. The vertices in G and G 2 are the same. It is obvious that with one black vertex, we cannot get a derived coloring for G. Since δ (G) = 2 ≤ Z(G) ≤ Z c (G). So, Z c (G) ≥ 2.
On the other hand, without loss of generality, color the vertices u 1 and u 2 as black. Then the black vertex u 1 forces u 3 to black, u 2 forces u 4 to black, u 3 forces u 5 to black and so on till all the vertices of G are colored black. So, Z = {u 1 , u 2 } forms a connected zero forcing set for G.
Proposition 27. The connected zero forcing number of the square of a cycle C n , n ≥ 5, is 4.
Proof. Let G denotes the square of the cycle C n , n ≥ 5. It is clear that G is a 4-regular graph.
In order to establish the reverse inequality, choose any four connected vertices of G. Let they be u 1 , u 2 , u 3 and u n . Color them as black. In G, the white vertices adjacent to the vertex u 1 are u 2 , u 3 , u n and u n−1 . So the black vertex u 1 forces the vertex u n−1 to black. Now consider the black vertex u 2 . The adjacent vertices of u 2 are u 1 , u n , u 3 ,and u 4 . Of these vertices, u 1 , u n , u 3 are already black. So, the vertex u 2 forces u 4 to black. Again, consider the black vertex u 3 . At this stage, the vertex u 3 has only one white vertex u 5 . Hence u 3 forces u 5 to black and so on.
Finally, consider the black vertex u n−4 . The vertex u n−4 has 4 neighbours u n−5 , u n−6 , u n−3 , u n−2 of which the only one white vertex is u n−2 . Therefore, the vertex u n−4 forces u n−2 to black. The vertex u n−3 is already colored black by the vertex u n−5 . Therefore, the set Z = {u 1 , u 2 , u 3 , u n } yields a connected zero forcing set for the graph G. Hence, we have Z c (G) ≤ 4. This completes the proof.

CONCLUSION AND OPEN PROBLEMS
In this paper we addressed the problem of determining the connected k-forcing number of certain graphs. Also we found the exact value of connected zero forcing number of some classes of graphs. In Section 1, we found an upper bound of Z ck (G ) for the corona product of two graphs G and H. It is an open problem to charaterize the connected graphs for which Z ck (G ) = p 1 (1+ p 2 ). In Section 2, we found the exact values of the connected k-forcing number of rooted product of cycles with paths and cycle with cycles. Section 3, deals with the connected k-forcing number of square of graphs such as the paths and cycles.