1 Introduction
Domination and its variants are one of the classical problems in graph theory. Let be a graph and (or ) be the open (respectively, closed) neighborhood of in . A set is called a dominating set of a graph if for all . Our goal is to find a dominating set of minimum cardinality which is known as domination number of and denoted by . For details the readers are refered to [HHS98b, HHS98a].
In this paper, we have studied one variant of domination problem, namely vertexedge domination problem, also known as vedomination problem. Given a graph , a vertex vedominates all edges incident to any vertex of . A set is a vertexedge dominating set (or simply a vedominating set) if for all edges , there exists a vertex such that vedominates . The minimum cardinality among all the vedominating sets of is called the vertexedge domination number (or simply vedomination number), and is denoted by . A set is called an independent vedominating set if is both an independent set and a vedominating set. The independent vedomination number of a graph is the minimum cardinality of an independent vedominating set and is denoted by .
The vertexedge domination problem was introduced by Peters[Pet86] in his PhD thesis in . However, it did not receive much attention until Lewis[Lew07] in introduced some new parameters related to it and established many new results in his PhD thesis. In his PhD thesis, Lewis has given some lower bound on for different graph class like connected graphs, regular graphs, cubic graphs etc. On the algorithmic side, Lewis has also proved that the vedomination problem is NPComplete for bipartite, chordal, planar and circle graphs and independent vedomination problem is NPComplete even when restricted to bipartite and chordal graph. Also approximation algorithm and approximation hardness results are proved in [Lew07]. In [LHHF10], the authors have characterized the trees with equal domination and vertexedge domination number. In [KVK14], both upper and lower bounds on vedomination number of a tree have been proved. Some upper bounds on and and some relationship between vedomination number and other domination parameters have been proved in [BCHH16]. In [Żyl19], Żyliński has shown that for any connected graph with , . Other variations of vedominations have also been studied in literature [BC18, KCV17].
In [Lew07], Lewis proposed a linear time algorithm for vedomination problem for trees. Basically, he proposed a linear time algorithm for finding minimum distance dominating set of a weighted tree. A set is called a distance dominating set of a graph if every vertex in is at most distance from some vertex in . In case of a weighted graph, the goal is to find distance3 dominating set with minimum weight. Lewis claimed that “Given any tree , define a new tree by subdividing each edge of . Then place a weight of one on each of and a weight of on each of . Now solve the weighted distance dominating set problem for , the result will give the .” But, we have found a counter example of this claim. In Figure 1, it is easy to see that . Now, in the new weighted tree , it is not possible to find any distance3 dominating set whose weight is .
This motivates us to study vedomination problem in trees and other graph classes. The rest of the paper is organized as follows. In Section 2, we have proposed a linear time algorithm for finding minimum vedominating set in block graphs, which is a superclass of trees. Section 3 deals with NPcompleteness of this problem in undirected path graphs. In Section 4, we have characterized the trees having equal vedomination number and independent vedomination number. Finally Section 5 concludes the paper.
2 Block graph
In this section, we propose a linear time algorithm for block graphs to solve vedomination problem. A vertex is called a cut vertex of if removal of increases the number of components in . A maximal connected induced subgraph without a cut vertex of is called a block of . A graph is called a block graph if each block of is a complete subgraph. The intersection of two distinct blocks can contain at most one vertex. Two blocks are called adjacent blocks if they contain a common cut vertex of . A block graph with one or more cut vertices contains at least two blocks, each of which contains exactly one cut vertex. Such blocks are called end blocks. The distance between two blocks and is defined as . The distance between a vertex and a block of a block graph is denoted as .
Our proposed algorithm is a labeling based algorithm. Let be a block graph where each vertex is associated with a label and each edge is associated with a label , where , . We call such graph a labelled block graph.
Definition 1.
Given a labelled block graph with labels and , we first define an optional vedominating set as a subset such that

if , then ,

vedominates every edge with .
The optional vedomination number, denoted by , is the minimum cardinality among all the optional vedominating sets of .
Note if for all and for all , then the minimum optional vedominating set is nothing but a minimum vedominating set of . Given a labelled block graph with labels for each and for each , our proposed algorithm basically outputs a minimum optional vedominating set of .
Next we present the outline of the algorithm. Let be an end block of a block graph . Since, block graph has a tree like structures, we can view as a graph rooted at the end block . The height of is defined as . At each step, the algorithm processes one of the end blocks at maximum height. Moreover, out of all the blocks at the same maximum height, the blocks with more number of edges with , are processed first. Based on some properties of the edges having label in an end block , we decide whether to take some vertices from in the optional vedominating set or not and then delete that block (except the cut vertex) from . We also modify the labels of some of the vertices and edges of the new graph. In the next iteration, we process another end block and the process continues till we are left with the root block . For the root block, we directly calculate the optional ve dominating set. The outline of the algorithm is given in Algorithm 1. In Algorithm 1, for an end block , denotes the number of edges with in and denotes the set of noncut vertices of , i.e., , where is the set of vertices of and is the cut vertex of . Also let denote the unique cut vertex of in which has the minimum distance from . Note that if and only if is the cut vertex of .
Next we prove the correctness of Algorithm 1. Note that, the modification of the labels of the reduced graph is done in such a way that if for some , then for all edges incident to any vertex . Hence we have the following claim:
Claim 2.
After each iteration, in any block , the set of edges with label 1 forms a clique.
Proof.
We will prove this by showing that it is not possible to have an edge with and in any block. Suppose a block is having an edge with . Since , there must be a vertex or with .
Case 1.
(When ) : In this case all the edges incident to must be labeled since they are being vedominated by the vertex . But for the edge , which is a contradiction.
Case 2.
(When ) : In this case all the edges incident to must be labeled since they are being vedominated by the vertex . But for the edge , which is a contradiction.
Hence, if then either all edges incident to is labelled or all edges incident to is labelled . And hence whenever an edge is labelled all other edges incident to at least one of its end point is also labelled . It reduces the size of clique (formed by label1 edges) by at least . ∎
Lemma 3.
Let be a block graph with an end block as root and be another end block such that . Also assume that , where is the cut vertex of and denotes the number of edges with label 1 in . Then followings are true.
(a) If , and is new block graph results from by relabelling as , deleting all and relabelling all edges as . Then .
(b) If but the edge with label1 is not incident to and is new block graph results from by relabelling as , deleting all and relabelling all edges as . Then .
(c) Let the conditions in (a) and (b) are not satisfied but and is new block graph results from by relabelling as , deleting all and relabelling all edges as . Then .
(d) If and has many vertices with and is new block graph results from by deleting all . Then .
Proof.
(a) Let be set of . If then is also optional vedominating set of , where is considered as a vertex with label . So assume . Now pick any edge with in . There must be some such that vedominates . Also . Therefore is optional vedominating set of . Hence, .
Conversely, let be set of . Since . Pick any edge with from . If then obviously some vedominates . If then vedominates . So, is also optional vedominating set of . Hence .
(b) The proof is same as the proof in (a).
(c) Let be set of . If then is also optional vedominating set of . Since all edges private to are also vedominated by . So assume . Now pick any edge with from . Since there must be some such that vedominates . Also . Therefore, is optional vedominating set of . Hence, .
Conversely, let be set of . Since . So, all edges incident to is vedominated by . In block only one edge is labelled and is incident to . So, it is vedominated by and is optional vedominating set of . Hence, .
(d) Let be set of . . So . There are two cases and .
Case 1.
( ) Pick any edge with from . Since is set of . There must exist some such that vedominates and this . Therefore, and hence .
Conversely, let be set of . All the edges of , except the edges of the newly added block is vedominated by and none of the edges of block needs to be vedominated. Hence .
Case 2.
( ) Pick any edge with from . Since , is not incident to . There must be some vertex to vedominate . Hence, is optional vedominating set of . Hence .
Conversely, let be set of . All the edges of , except the edges of the newly added block is vedominated by and none of the edges of block needs to be vedominated. Since contains vertices with label , is optional vedominating set of . Hence, .
∎
Lemma 4.
Let be a complete graph, , . If , then . Otherwise, , where is the number of vertices of with .
Proof.
When then does not have any vetex with label. So, we need at least one vertex from block to vedominate all the edges with and only one vertex is sufficient to vedominate all the edges. Hence, .
When , none of the edges needs to be vedominated. So, all the vertices with forms an optional vedominating set and there are many such vertices. Hence, . ∎
Lemma 3 and Lemma 4 shows that the output of Algorithm 1 is minimum optional vedominating set. At each iteration, we are taking time. Hence the running time of Algorithm 1 is . Thus, we have the following theorem.
Theorem 5.
The vedomination problem can be solved in time for block graphs.
Remark 1.
We can also find the minimum independent vedominating set of a given block graph in a similar approach. The algorithm is similar to Algorithm 1, but with little modification, to ensure the output is an independent set. Thus, for block graphs, the independent vedomination problem can also be solved in linear time.
3 Undirected Path Graphs
In this section, we prove that the vedomination problem for undirected path graphs is NPcomplete by showing a polynomial time reduction from dimensional matching problem which is a wellknown NPcomplete problem [GJ90]. A graph is called an undirected path graph if is the intersection graphs of a family of paths of a tree. In [Gav75], Gavril proved that a graph is an undirected path graph if and only if there exists a tree whose vertices are the maximal cliques of and the set of all maximal cliques containing a particular vertex of forms a path in . This tree is called the clique tree of the undirected path graph . The dimensional matching problem is as follows: given a set , where and are disjoint set with , does contains a matching , i.e., a subset such that and no two elements of agree in any coordinate?
Theorem 6.
The vedomination problem is NPcomplete for undirected path graphs.
Proof.
It is easy to see that vedomination problem is in NP. Now, we describe polynomial reduction form dimensional matching problem to vedomination problem in undirected path graph. Let , and be an instance of dimensional matching problem. Now we construct a tree having vertices that becomes the clique tree of an undirected path graph . The vertices of the tree are maximal cliques of . The vertex set and the edge set are as follows:
For , each corresponds to cliques which are vertices of , namely , , , , , , , and . These vertices depend only on the triple itself but not on the elements within the triple. These eight vertices induces a subtree corresponding to as illustrated in Figure 2. Further, for each , , we take two cliques and which are vertices of forming a subtree as shown in Figure 2. Similarly, for each , and , , we add the cliques , and , , respectively to the tree as shown in Figure 2.Finally, is the last vertex of tree . The construction of is illustrated in Figure 2.
Hence, is the clique tree of the undirected path graph whose vertex set is
Claim 7.
The graph has a vedominating set of size if and only if dimensional matching has a solution.
Proof.
Let be a vedominating set of of size . For any , the only way to vedominate the edgeset of the subgraph induced by the vertex set corresponding to with two vertices is to choose and . Hence, to vedominate the edgeset of that induced subgraph by any larger vertex set, at least three vertices has to be taken. Note that, the set vedominates the edgeset of that induced subgraph. So, without loss of generality, we assume that consists of for many ’s and for many other ’s. Also to vedominate the edges of the form , and , contains at least many vertices (namely, or , or , or ). Hence, we have,
So, . i.e. must contain at least many . Picking the corresponding ’s form a matching of size .
Conversely, let be the solution of the dimensional matching problem of size . Then we can form the vedominating set as . Clearly, is a vedominating set of of size . ∎
Hence, the vedomination problem is NPcomplete for undirected path graph. ∎
4 Trees with Equal and
For every graph , the independent vedomination number is obviously at least as large as the vedomination number. In this section, we characterize the trees for which these two parameters and are equal. We start with some pertinent definitions.
Definition 8.
An atom is a tree with at least vertices with a vertex, say , designated as center of the atom such that distance of every vertex from is at most .
We denote an atom along with its center by . Note that the center vedominates all edges of the atom . Next, we define an operation for joining two atoms to construct a bigger tree.
Definition 9.
Let and be two atoms along with their centers and , respectively. For some , we define join, denoted by , as the addition of an edge between the vertices and such that and .
With slight abuse of notation, denotes the tree obtained by join between two atoms and , where is an atom in . Given a subset , a vertex has a private edge with respect to the set if vedominates the edge and no other vertex in vedominates the edge . An edge is called a distance private edge of with respect to the set if is a private edge of with respect to and . Next, we give the recursive definition of a family of trees, say , using the notion of atom and join.
Definition 10.
The recursive definition of the family of trees is as follows:

every atom and

Let and be an atom in and be the set of all atom centers in . Then if one of the following cases hold

is a join such that

has a neighbour such that all edges incident to , except , are pendent edges. Also, has no distance edge and at least two edges of is incident to .

has distance private edges with respect to and at least one distance edge of is not incident to .

has no distance private edge with respect to and at least one distance edge of is not incident to .


is a join such that

has distance edges and has a neighbour such that all edges incident to , except , are pendent and private edges of with respect to .

has distance edges and has a neighbour such that is a leaf vertex and .


is a join when at least one distance private edge of with respect to is not incident to and at least one distance edge of is not incident to .

is a join when all distance private edges of with respect to are not incident to , where , and at least one distance edge of is not incident to .

Now, we show that if , then vedomination number and independent vedomination number are same.
Lemma 11.
If , then the set of all atom centers of forms a minimum vedominating set.
Proof.
We prove this by induction on the number of atoms in . Clearly, when is an atom, the hypothesis is true. Let be a tree containing atoms and is obtained from by joining the atom with an atom of satisfying the joining rules. Let and be the atom centers of and , respectively. Clearly, . By induction hypothesis, is a set of . We show that is a set of for all the seven types of joining defined in Definition 10.

In this type of joining, is star and has a neighbour such that all edges incident to , except , are pendent edges. Since is a vedominating set of , is obviously a vedominating set of . If possible, let us assume that is not a set of . Let be a set of such that . Note that contains exactly one vertex from , say , to vedominate all edges of . Also, to vedominate the pendent edges which are incident to , must contain one vertex, say , from , where is any leaf adjacent to . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This contradicts the fact that is a vedominating set of of minimum cardinality. Hence, is a set of .

If possible, let us assume that is not a set of . Let be a set of such that . Clearly, contains exactly one vertex, say , from . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This is a contradiction. Hence, is a set of .

If possible, let us assume that is not a set of . Let be a set of such that . Clearly, contains exactly one vertex, say , from . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This is a contradiction. Hence, is a set of .

In this type of joining, has a neighbour such that all edges incident to , except , are pendent edges. If possible, let us assume that is not a set of . Let be a set of such that . Note that contains exactly one vertex from , say , to vedominate all edges of . Suppose is a leaf node adjacent to . So, to vedominate the pendent edge , must contain one vertex, say , from . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This is a contradiction. Hence, is a set of .

In this case, has a neighbour such that is leaf vertex. If possible, let us assume that is not a set of . Let be a set of such that . Note that contains exactly one vertex from , say , to vedominate all edges of . Since cannot vedominate the edge , must contain a vertex, say , to vedominate this edge. Note that because is leaf vertex and also vedominate the edge . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This is a contradiction. Hence, is a set of .
 join:

Since is a vedominating set of , is obviously a vedominating set of . If possible, let us assume that is not a set of . Let be a set of such that . Clearly, contains exactly one vertex, say , from . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This is a contradiction. Hence, is a set of .
 join:

Since is a vedominating set of , is obviously a vedominating set of . If possible, let us assume that is not a set of . Let be a set of such that . Clearly, contains exactly one vertex, say , from . Since the set of edges that are vedominated by can also be vedominated by , is also a set of . It is easy to see that is a set of . This is a contradiction. Hence, is a set of .
∎
Theorem 12.
For all , .
Proof.
By the definition of , the distance between any two atom centers in is at least . Hence, the set of these atom centers, say , forms an independent set. In Lemma 11, is also a set. Hence, for all . ∎
Next, we show that the converse of Theorem 12 is also true. For that, first we prove following lemmas that allow us to construct an independent vedominating set of a tree with some desirable properties.
Lemma 13.
For any tree , there exists an set which does not contain any leaf.
Proof.
Let be an set of . If does not contain any leaf, then we are done. Otherwise assume that contains a leaf, say . Let the neighbour of is and . Since is independent set, . Also, none of the are in , because if any of , then is also an independent vedominating set. Hence by replacing by , we get another set. Repeating this process, we can form an set of which does not contain any leaf. ∎
Lemma 14.
Let be an set of a rooted tree having depth , which does not have any leaf. If the vertex is at level and is the closest vertex to such that and , then is also an set without having any leaf, where is the parent of in .
Proof.
It is easy to see that all the edges that are vedominated by can also be vedominated by . So, is set. Also, since the minimum distance between and is at least , is independent set. Hence, is an set without having any leaf. ∎
Theorem 15.
If for a tree with , then .
Proof.
We prove this by induction on the size of set. As a base case when then is an atom, hence . Assume the induction hypothesis, if then and all the elements of set represent centers of the underlying atoms of .
Let a tree is rooted at a vertex and have levels as such that root is at level and most distant leaf from root is at level. Let us also assume that is an set of the tree such that . Clearly, has no vertices from level(Lemma 13). Consider a vertex that is at maximum distance from the root (if more than one vertices are at the same maximum level then consider any one who has descendent at level). So, is either at level or at level. Let is the nearest vertex to such that . So, or , otherwise, will not be an set. Now, we extract an atom with center to form a tree . Assume be the set and set of . By induction hypothesis and is set of all atom centers of . Clearly, is center of an atom, say , in . We show that the joining of atom with tree follows the joining rules defined in Definition 10. The whole proof is divided into three cases depending on the distance between and . Further, each case is divided into two subcases considering can be at level or at level. These different exhaustive cases are explored as follows:
Case 1.
:

is at level :
Since , can either be at level or at level. Both of these possibilities are considered as follows:

is at level : This case is not possible because we can find a new set such that is set and which violates the minimality of .

is at level : Only two possibilities are there, has some distance private edges with respect to or it has no distance private edge with respect to . Both of these cases are considered in following three points:

has a neighbour such that all other edges incident to are pendent edges : In this case, all edges incident to , except , are distance private edges of with respect to . Suppose is a subtree rooted at , we can think as an atom with center . Remove from to form tree . Assume is the center of an atom in . Now, it is very easy to see that the joining between and resembles the joining of type defined in Definition 10.

has no neighbour such that all other edges incident to are pendent edges, but it has distance private edges with respect to : In this case, edges incident to are distance private edges of with respect to . Clearly, none of the neighbours of , except , is in . So, is set as well as set with having no distance private edges. Suppose is the subtree rooted at vertex


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