EDGE OPEN PACKING SETS IN GRAPHS

. In a graph 𝐺 = ( 𝑉, 𝐸 ), two edges 𝑒 1 and 𝑒 2 are said to have a common edge if there exists an edge 𝑒 ∈ 𝐸 ( 𝐺 ) different from 𝑒 1 and 𝑒 2 such that 𝑒 joins a vertex of 𝑒 1 to a vertex of 𝑒 2 in 𝐺 . That is, ⟨ 𝑒 1 , 𝑒, 𝑒 2 ⟩ is either 𝑃 4 or 𝐾 3 in 𝐺 . A non-empty set 𝐷 ⊆ 𝐸 ( 𝐺 ) is an edge open packing set of a graph 𝐺 if no two edges of 𝐷 have a common edge in 𝐺 . The maximum cardinality of an edge open packing set is the edge open packing number of 𝐺 and is denoted by 𝜌 𝑜𝑒 ( 𝐺 ). In this paper, we initiate a study on this parameter.


Introduction
By a graph  = (, ), we mean a finite, undirected graph with neither loops nor multiple edges.For graph-theoretic terminology, we refer to [6].Throughout this paper, graphs are assumed to be connected and non-trivial.
The open neighborhood  () of an edge  ∈  is the set of all edges adjacent to  in , while the closed neighborhood of  in  is  [] =  () ∪ {}.The degree of an edge  =  of  is deg  = deg  + deg  − 2. A simple bipartite graph with bipartition (,  ) is (, )-biregular if every vertex in  has degree  and every vertex in  has degree  [2].
Coloring is one of the most important research areas in graph theory and umpteen number of coloring parameters have been introduced and well studied by several authors because of its numerous applications in various fields such as coding theory [13], biological networks [10], neural networks [3] and so on.Of these, the concept of injective edge coloring was introduced in [4] and obtained several results on it, we may refer to [5,11,12].In a graph , three edges  1 ,  2 and  3 (in this fixed order) are consecutive if  1 = ,  2 =  and  3 =  for some vertices , , ,  (where  =  is allowed).In other words, three edges are consecutive if they form a path or cycle of length three.A coloring,  : () → , where  is a set of colors, is an injective edge coloring (-edge coloring for short) if the edges  1 ,  2 and  3 are consecutive in , then  1 and  3 should receive different colors.The injective edge coloring number or injective edge chromatic index  ′  () of graph  is the minimum number of colors permitted in an -edge coloring.This concept is motivated by the problem of assigning channels between the stations in order to avoid secondary interference in the Packet Radio Network (PRN).
It is obvious that an -edge coloring of  partitioning the edge set () into edge subsets   having the property that no three edges in   to form a consecutiveness in .Of course, studying the nature of these subsets is more useful.For instance, suppose we have the position to find the maximum number of transmission lines of the given PRN in which the secondary interference does not occur.This situation can be interpreted by defining the parameter, we call as edge open packing sets in graphs as follows.

Standard graphs
In this section, we determine the value of the edge open packing number for some standard graphs such as paths, cycles, complete multipartite graphs, wheels and the Petersen graph.Proposition 2.1.Let   be a path of size  ≥ 2. Then

⌋︀
. Now, consider the sets Then any edge open packing set  ′ of   consisting the edges from exactly one of the sets  1 and ⌋︁ edges as any two consecutive edges of  1 have a common edge in   .Suppose ⌋︀ and this completes the proof.

Bounds and characterization results
In this section, we provide some bounds for the edge open packing number in terms of diameter, size, minimum degree, clique number and girth of a graph.Denote the set of edges incident at the vertex  by   ().A graph  is (ii)  is  1, -free, where  ≥ 3 and; (iii) for any two non-adjacent edges  1 =  and  2 =  such that  1 and  2 have no common edge in , every vertex in  () ∖ {, , , } is adjacent to at least two vertices in the set {, , , }.
Proof.Assume that    () = 2. Then by Proposition 3.1 that diam() ≤ 4. Also, for the graphs of diameter one, that is for complete graphs, the value of    is 1, this implies that here diam() ≥ 2. Thus (i) follows.Now, if  contains  ∼ =  1, , ( ≥ 3) as an induced subgraph, then the edges of  forms an edge open packing set of  and so    () ≥ 3, which is a contradiction to our assumption.Therefore, condition (ii) is satisfied.Finally, let  1 =  and  2 =  be two non-adjacent edges such that  1 and  2 have no common edge in  and let  = {, , , }.Consider an arbitrary vertex  in  () ∖  and assume that  has at most one neighbor in .If  is adjacent with exactly one vertex in , say , then the set { 1 ,  2 , } will become an edge open packing set of , we have    () ≥ 3. Suppose  has no neighbor in .Let  ′ ∈  () ∖  such that  ′ ∈ ().If  ′ has no neighbor in , then the set { 1 ,  2 ,  ′ } is an edge open packing set of , we have    () ≥ 3. On the other hand, the vertex  ′ is adjacent with either exactly one vertex or two adjacent vertices in  as  is  1, -free( ≥ 3).Certainly, the set { ′ ,  ′ , } will form an edge open packing of  and so    () ≥ 3.In all the cases, we arrive at a contradiction to    () = 2 and this proves condition (iii) is true.Conversely, a graph  satisfies the conditions from (i) to (iii) stated in the theorem.Let  be any maximal edge open packing set of .We claim that || ≤ 2. Now, by condition (iii), ⟨⟩ has at most two components.If ⟨⟩ has exactly one component  1 , then by condition (ii), either  1 ∼ =  1,1 or  1 ∼ =  1,2 .Otherwise, let  1 and  2 be two components of ⟨⟩.Then both  1 and  2 are isomorphic to  1,1 according to the condition (iii) and hence our claim.Thus    () = 2 follows from the condition (i).
In the following theorems, we characterize the graphs  for which    () =  − 1 and    () =  − 2. Theorem 3.3.Let  be a graph with size  ≥ 3. Then    () =  − 1 if and only if  is a graph obtained from a star  1,−1 by subdividing exactly one edge of  1,−1 by once.
Proof.Let  be a    -set of  such that || =  − 1 and let  =  ∈  − .If ⟨⟩ consists of two components, say  1 and  2 , then two vertices  and  do not lie in the same component.Without loss of generality, assume that  ∈  ( 1 ) and  ∈  ( 2 ).Now, let  1 =  and  2 =  be two edges in  1 and  2 respectively.Certainly,  is a common edge between the edges  1 and  2 , which is not possible and so ⟨⟩ has exactly one component.Let  ∼ =  1,−1 be the component of ⟨⟩.As  is connected, either  or  should be a pendant vertex in  and thus  is isomorphic to a graph obtained from a star  1,−1 by subdividing exactly one edge of  1,−1 by once.The converse is obvious.
Next, we characterize graphs  for which    () =  − 2. For this, we use the following definitions and describe new family of graphs as follows.(ii) A non-pendant edge  =  is a support edge of  if either  or  is a support vertex in .
Observation 3.5.If  is a graph of size at least 3 that is not a star and  is an edge open packing set of  such that ⟨⟩ contains  ≥ 2 components, then  −  contains at least  edges.
Let A 1 , A 2 and A 3 be the families of graphs obtained from  5 ,  6 and  7 respectively by attaching (≥ 0) number of pendant edges at each support vertex of  5 ,  6 and  7 .Let A 4 and A 5 be the families of graphs obtained from  3 and  4 respectively by attaching (≥ 0) pendant edges at exactly one vertex of  3 and exactly one vertex of  4 .Let A 6 be the family of graphs obtained from a star  1, , where  ≥ 3 by subdividing exactly two edges of  1, once at a time.Let A 7 be the family of graphs obtained from a star  1, , where  ≥ 3 by attaching two pendant edges at exactly one of the pendant vertices of  1, .
The seven families of graphs are shown in Figure 3. Proof.Suppose    () =  − 2 and let  be a    -set of .Then  −  has exactly two edges and so ⟨⟩ contains at most two components follows from Observation 3.5.Let  1 =  and  2 =  be the edges in  − .We prove this theorem in the following cases.In the following proposition, we present an upper bound for    in terms of the size and the minimum degree.
Proof.Let  be a    -set of  and let  ∈ .Since  is an edge open packing set, it follows that  − contains at least ( −1)-edges incident with  or  and which holds for every edge in .Thus |() − | ≥ ||(()−1) and hence the result follows.
Next, we characterize -regular graphs which attain the bound in the above proposition.For this purpose, we construct a family of -regular graphs  , as follows.
Let  () =  ∪  ∪  such that no two vertices of  or  or  are adjacent, where and || =  2 .Consider (,  − 1)-biregular graph with the vertex sets  and  in which the degree of each vertex in  is  and the degree of each vertex in  is  − 1.The existence of a biregular graph is possible by the condition that || = | |( − 1).Now, join each vertex in  to  vertices of  such that  () ∩  () =  for any two vertices ,  ∈  and let  be the resultant graph.It is obvious that  is an -regular graph with  vertices such that  ≡ 0(mod 2) (Fig. 4).Theorem 3.8.Let  be an -regular graph.Then    () =   if and only if  ∈  , .
Proof.Assume that  ∈  , .Obviously, the set of edges connecting the vertices of  to the vertices of  in the above construction forms an edge open packing set of  so that    () ≥  2  =   .Thus    () =   follows from Proposition 3.7.
Conversely, let  be an -regular graph of order  such that    () =   .Suppose  is a    -set of  with  ∼ =  1, ( < ) as one of its component, then | − | ≥    () −    () +  −  and which in turns that    () ≤ −(−)  <   , a contradiction.Therefore, every component of ⟨⟩ is isomorphic to  1, and there are  2 copies of such components in ⟨⟩.Let  and  be two sets of pendant vertices and support vertices of ⟨⟩ respectively.Then | | =  2 and || =  2 .Since  is a    -set, no two vertices of  are adjacent and no two vertices of  are adjacent.Let  be the set of vertices in  −   .Since  is -regular and every vertex in  has degree , it follows that each vertex in  is adjacent with  vertices in  alone and thus  ∈  , .Theorem 3.9.Let  be a graph with the size and the clique number .Then    () ≤  − (−1) 2 + 1 if and only if  is either   or a graph obtained from   by attaching any number of pendant edges at exactly one vertex of   .
Proof.Let  be a maximum clique and  be a maximal edge open packing set of .Then  contains at most one edge from  and hence we achieved the inequality.
As  ⊆ , the set  consists both the edges , , it follows that no edge of  belong to , which is not true.Thus, every vertex outside of  is pendant.Also, since  ⊆ ,  itself is an edge open packing set of .Thus ⟨ ⟩ is a star and hence either  ∼ =   or a graph obtained from   by attaching any number of pendant edges at exactly one vertex of   .Conversely, if  ∼ =   , then    () = 1 =  − (−1) 2 + 1. Suppose  is a graph obtained from   by attaching any number of pendant edges at exactly one vertex of   .Then the set of pendant edges of  together with exactly one edge of   , that is, adjacent with these pendant edges forms an edge open packing set of  and thus    () =  − (−

Conclusion and scope
In this paper, we have introduced an edge open packing number    () and determined the exact value of    for some standard graphs.Also, we have obtained several bounds on it and some characterizations were done.Even though, there is a wide scope for further research on this topic in the following directions.

Proposition 2 . 5 .
Let  be the Petersen graph.Then    () = 3.Proof.Consider the Petersen graph given inFigure 2. It is clear that the set { 1 ,  2 ,  1 } forms an edge open packing set of  and so    () ≥ 3. Also, since any maximal edge open packing set  has at most two edges from exactly one of the two cycles (outer and inner cycles) of  and one edge from { 1 ,  2 , . . .,  5 }.Thus    () ≤ || ≤ 3.

Definition 3 . 4 .
(i) An edge  =  has vertices  and  that are saturated by .Given a set of edges , denote the set of vertices saturated by edges of  by   .
then the set of all pendant edges is an edge open packing set of  so that    () =  − 2. Suppose  ∈ A 2 ∪ A 4 .Then the set of all pendant edges with exactly one of its support edges forms an edge open packing set of  and thus    () =  − 2. Suppose  ∈ A 3 ∪ A 5 .Then the set of all pendant and support edges is an edge open packing set of  and so    () =  − 2. If  ∈ A 6 ∪ A 7 , then the edges in  1, forms an edge open packing set of  and hence    () =  − 2.
Definition 1.1.An edge  of a graph  is a common edge of two distinct edges  1 ,  2 different from  if  joins a vertex of  1 to a vertex of  2 .That is, ⟨ 1 , ,  2 ⟩ forms either  4 or  3 .A non-empty set  ⊆ () is an edge open packing set if no two edges in  have a common edge in .The maximum cardinality among all edge open packings is the edge open packing number of  and is denoted by ().An edge open packing set of cardinality    () is called a    -set of .Example 1.2.Consider the graph  given in Figure 1.It is easy to observe that the set of pendant edges of  forms an edge open packing set with the maximum cardinality and so    () = 4. Remark 1.3.It is clear that the induced subgraph ⟨⟩ induced by an edge open packing set is the union of stars.
2 }︁ .Obviously, no two edges of  have a common edge and so it forms an edge open packing set of   and thus    (  ) ≥ || = +2 2 .On the other hand, let  be a maximal edge open packing set of   .Now, for 1 ≤  ≤ , define   = {  : 4 − 3 ≤  ≤ 4} and  * = (  ) ∖   .Then,   ∪  * = (  ) and | * | = 2. Obviously,  can have at most two edges from each   for 1 ≤  ≤  and all the edges from  * .Therefore,    (  ) ≤ || = 2 + 2 = 2 (︀ −2 let  1 be a maximal edge open packing set of   .Then  1 can contains at most two edges from each   (1 ≤  ≤ ) and contains at most ⌈︀  For the cycles   with size  ≥ 3, we have Let  =  1,2,...,  be a complete multipartite graph with partition ( 1 ,  2 , . . .,   ), where |  | =   , 1 ≤  ≤  and  1 ≤  2 ≤ . . .≤   .Then    () =   .Proof.Choose an arbitrary vertex  ∈   for some 1 ≤  ≤  − 1.Then the set of edges from the vertex  to the vertices belonging to   forms an edge open packing set of  and hence    () ≥ |  | =   .Also, it is clear that any maximal edge open packing set  of  can never contains non-adjacent edges of  and so || ≤   .Therefore,    () =   .Let { 1 ,  2 , . . .,  −1 } be the set of vertices on the rim and let  be the center vertex of   .Certainly, the set  = {  : 1 ≤  ≤  − 1 and  is odd} is a maximal edge open packing set of   , we get    (  ) ≥ || ≥ ⌊︀ 1, -free graph if it does not contain  1, as an induced subgraph.If  contains  1, as an induced subgraph, then () ≥ .It is obvious that for any graph , we have 1 ≤    () ≤ .Furthermore,    () = 1 if and only if  is complete and    () =  if and only if  is a star.Now, we characterize graphs  for which    () = 2. Before going to characterize, we prove the following proposition.Proposition 3.1.For any graph , we have    () ≥ ⌈︁ diam() 2 ⌉︁ .Proof.Suppose diam() = .In any diametral path  , we can choose at least ⌈︀  2 ⌉︀ edges with the property that no two edges have a common edge in  and so any maximal edge open packing set of  can have at least ⌈︀  2 ⌉︀ edges.Thus    () ≥ ⌈︀  2 ⌉︀ = ⌈︁ diam() 2 ⌉︁ .Theorem 3.2.   () = 2 if and only if the following conditions are true Case 1. ⟨⟩ has exactly one component.Let ⟨⟩ =  1, , where  =  − 2 and assume that  1 and  2 are adjacent in .Without loss of generality, let  =  and let ⟨ − ⟩ =  : (,  1 , ,  2 , ).As  is connected and | − | = 2, the set   should contains at least one vertex but at most two vertices of  .If   contains exactly one of the vertices ,  and , then none of the vertices ,  and  will become the center vertex of  1, .Otherwise, if any one of the vertices ,  and  is the center vertex of  1, , then the Families from A 1 to A 7 . ∈   and  = 2. Then  is isomorphic to the graph obtained from the star  1,3 by subdividing exactly one edge of  1,3 by once and so    () =  − 1 follows from Theorem 3.3.This is a contradiction and hence  ≥ 3. Therefore,  ∈ A 7 when  ≥ 3 and  ∈   .Suppose   contains two vertices of  .Now, we first claim that  / ∈   .If not, then either  ∪ {} or  ∪ {} forms an edge open packing set according as  ∈   or  ∈   .This is a contradiction and hence ,  ∈   .If both  and  are the pendant vertices of  1, , then  ∈ A 4 and  ∈ A 5 according as  = 1 and  ≥ 2. If any one of  and  is the center vertex of  1, , then  ∈ A 4 .Now, let us consider that the edges  1 =  and  2 =  are independent in ⟨ − ⟩.Since  is connected and  is edge open packing, the component  1, of ⟨⟩ should consist of exactly one vertex from  1 and one vertex from  2 , and both which cannot be the center vertex of  1, .Certainly, these two vertices are the pendant vertices of  1, and in which case  ∈ A 6 .Case 2. ⟨⟩ has two components.  as  is an edge open packing set of  and so exactly one of the vertices of  and  belongs to  1 and the other vertex in  2 .Suppose  ∈  ( 1 ) and  ∈  ( 2 ).Then  ∈ A 1 , when both  and  are the center vertices of  1 and  2 respectively.If both  and  are the pendant vertices of  1 and  2 , then (i)  ∈ A 1 when  =  = 1.(ii)  ∈ A 2 when  = 1 and  > 1 or  > 1 and  = 1.(iii)  ∈ A 3 when  > 1 and  > 1.Finally, if  is the center vertex of  1 and  is a pendant vertex of  2 , then we have (i)  ∈ A 1 when  ≥ 1 and  = 1.(ii)  ∈ A 2 when  ≥ 1 and  > 1. Conversely, assume that  ∈ ∪ 7 =1 A  .Since    () =  only when  is a star and    () =  − 1 only when  is a graph stated in Theorem 3.3, which implies that edge(s) on  incident at that center vertex together with  forms an edge open packing set of , which produces a contradiction to the maximality of .Therefore, if   contains exactly one vertex of  , then it is a pendant vertex of  1, .Further, if  = 1, then either  is  4 or a star, which implies that    () ≥  − 1, a contradiction and this yields that  ≥ 2. Now, if   contains exactly one of  and , then  ∈ A 1 .Suppose Figure 3. Let  1 =  1, and  2 =  1, be the components of ⟨⟩, where  +  =  − 2. Since  is connected and no two edges of  have a common edge in  − , it follows that  1 and  2 should be adjacent in ⟨ − ⟩.Without loss of generality, let  = .Certainly,  / ∈ Moreover, the equality holds if and only if  is either complete or a cycle of size  with  ̸ ≡ 0(mod 4).Proof.Consider a cycle  in  such that the length of  is equal to ().Then by Proposition 2.2, we have⌉︁ − 1.If () = 3, then    () =1 and thus  is complete.Assume that () ≥ 4. Suppose  is not a cycle.Let  be an edge incident at the vertex on , say  and label  1 ,  2 , . . .,  () to the edges of  such that  1 and  2 are incident at .Then the set  = ⌉︁ , a contradiction and hence  is a cycle.Now, by Proposition 2.2, we have  is a cycle with () ̸ ≡ 0(mod 4).The converse is just a verification.An alternate triangular snake (  ) is obtained from a path  1 ,  2 , . . .,   by joining   and  +1 (alternately) to a new vertex   .That is every alternate edge of a path is replaced by  3 .(ii) An alternate quadrilateral snake (  ) is obtained from a path  1 ,  2 , . . .,   by joining   ,  +1 (alternately) to a new vertex   ,   respectively and then joining   and   .That is every alternate edge of a path is replaced by  4 .Theorem 3.14.Given any two positive integers ,  ≥ 1 and an integer  ≥ 3 with Proof.If  =  = 1, then  is a complete graph on  + 1 vertices.If ,  > 1 and  = , then  is obtained from  1, by subdividing exactly  edges in  1, .Suppose ,  > 1 and  ̸ = .Then consider the following cases.