A Note on Edge-Domsaturation Number of a Graph

The edge-domsaturation number ds'(G) of a graph G = (V, E) is the least positive integer k such that every edge of G lies in an edge dominating set of cardinality k. In this paper, we characterize unicyclic graphs G with ds'(G) = q – Δ'(G) + 1 and investigate well-edge dominated graphs. We further define γ'-critical, γ'-critical, ds'-critical, ds'-critical edges and study some of their properties.


Introduction
Throughout this paper, G denotes a graph with order p and size q.By a graph we mean a finite undirected graph without loops or multiple edges.For graph theoretic terms we refer Harary [1] and in particular, for terminology related to domination theory we refer Haynes et al. [2].

Definition
Let G = (V, E) be a graph.A subset D of E is said to be an edge dominating set if every edge in E-D is adjacent to at least one edge in D. An edge dominating set D is said to be a minimal edge dominating set if no proper subset of D is an edge dominating set of G.The edge domination number γ'(G) of a graph G equals the minimum cardinality of an edge dominating set of G.An edge dominating set of G with cardinality γ'(G) is called a γ'(G)-set or γ'-set.Acharya [3] introduced the concept of domsaturation number ds(G) of a graph.For any graph G of order p, and for any integer r such that γ(G) ≤ r ≤ p, we call the set  A r the r-level domination core of G.We say that G is r-level domination-saturated (or in short, "r-domsaturated") if DC r (G) -V(G).The domsaturation number ds(G) is then defined by  ( ) min isr-domsaturated ds G r G   .Arumu-  gam and Kala [4] observed that for any graph G, ( )  and obtained several results on ds (G).We now extend the concept of domsaturation number of a graph to edges.

Definition
The least positive integer k such that every edge of G lies in an edge dominating set of cardinality k is called the edge-domsaturation number of G and is denoted by ds'(G).
If G is a graph with edge set E and D is a γ'-set of G, then for any edge e  E-D, is also an edge dominating set and hence Thus we have the following definition.

Definition
A graph G is said to be of class 1 or class 2 according as ( )

Definition
An edge ; 4) γ'-fixed if every γ'-set contains e; 5) γ'-free if there exists γ'-sets containing e and also γ'-sets not containing e; 6) γ'-totally free if there is no γ'-set containing e.We use the following theorem.

Theorem [5]
For any connected unicyclic graph with cycle C, if and only if one of the following holds. 1  for all vertices w not on C and for at most one vertex w not on C; all the vertices not on C adjacent to u 1 have degree at most 2 and all vertices whose distance from u 1 is 2 are pendent vertices; 4) and all vertices not on C are pendent vertices; 4 , either exactly one vertex of C has degree at least 3 and all vertices not on C are pendent vertices.

Lemma
An edge e of G is γ' --critical if and only if

Proof
For any edge e, we observe that ( )  .The converse is obvious.

Theorem
An edge e is γ' --critical if and only if for some γ'-set  by lemma 2.1.Let S be a γ'-set of G -e.If S contains an edge of N(e), then S will be an edge dominating set of G and hence ( )  , a contradiction.Thus S does not contain edge of N(e).Since (  , is a γ'-set of G and so Equation (1) holds.Conversely, suppose e is an edge such that (1) is true.Then G -e is an edge dominating set of G -e and hence   Thus e is γ' --critical.

Theorem
Let G be a graph without isolated edges.An edge e in G is γ' --critical if and only if 1) e is γ'-free, and 2) no γ'-set of G -e contains any edge of N(e).
by Lemma 2.1.As in theorem 2.2, if S is any γ'-set of G -e, then S will not contain any edge of N(e) and is a γ'set of G for every . This implies that e is γ'-free.Conversely, suppose (1) and ( 2 . Thus e is γ' --critical.

Theorem
Let G be a graph and ( ) e E G  .Then 1) e is γ'-fixed if and only if there exists no edge dominating set of G -e with γ'(G) edges which is also an edge dominating set of G.
2) e is γ'-totally free if and only if every Assume that e is γ'-fixed.Suppose there exists an edge dominating set S of G -e with ( )  which is also an edge dominating set of G. Then S is a γ'-set not containing e which is impossible as e is γ'-fixed.The converse is obvious.
2) Let e be γ'-totally free.Then e does not belong to any γ'-set of G and so every γ'-set , then by theorem 2.3, e is γ'-free and so The converse is obvious.

Theorem
Let G be a connected graph.If a cut edge e of G is γ'fixed, then e is γ' + -critical Proof Let S be a γ'-set of G. Let e be a cut edge that is γ'fixed.Then e belongs to every γ'-set.Since e is a cut edge, G -e is a disconnected graph with at least two components G' and G .Let e' and e" be the neighbors of e in G' and G" respectively.Therefore  .Hence e is γ' + -critical.

Theorem
An edge e in a graph G is γ' + -critical if and only if 1) e is not isolated edge 2) e is γ'-fixed and 3) There is no edge dominating set for having γ'(G) edges can dominate N(e).By Theorem 2.4, e is γ'-fixed.The converse is obvious.
We now investigate relationships between, γ'-free edges, γ'-totally free edges and graphs which are class 1 and class 2.

Theorem
If G is a graph without isolated edges, then G is of class 2 if and only if G has γ'-totally free edges.
Proof Suppose G has a γ'-totally free edge e.By Theorem 2.4 (2), G is of class 2. Conversely, suppose G is of class 2. Then there exists an edge e which is not in any γ'-set.Hence every γ'-set of G is also a γ'-set of G -e so that e is γ'-totally free.

Proof
Let G be a connected graph.If G has a γ'-fixed edge, then it has a γ'-totally free edge.
Suppose G has a γ'-fixed edge e.Then e belongs to every γ'-set.
Claim: No neighbor of e belongs to any γ'-set of G. Suppose at least one of its neighbor say e' belongs to a γ'set D. Let and e' be incident with u.Then 1 , where e" is any edge incident with v is an edge dominating set of G -e with γ'-edges which is also an edge dominating set of G.But by Theorem 2.6, this is a contradiction, since e is a γ'-fixed edge.Therefore no neighbor of e belongs to any γ'-set of G. Thus neighbors of e are all γ'-totally free in G.
We now investigate the class of graphs which are ds' + , ds' --critical.

Let e  E(G).
If e is γ'-totally free and G -e is of class 1, then .
Since e is γ'-totally free, by Theorem 2.3, G is of class 2 and so Since G -e is of class 1, we have From Equations (1), ( 2) and (3), we have

Let
. If e is γ'-totally free and G -e is of class 2, then .

Proof
If e is γ'-totally free, then by Theorem 2.4, Since G and G -e are of class 2, we have 1), ( 2) and (3), we have

Lemma
Let e be an edge of G.If e is γ'-free and G -e is of class 1, then Proof Suppose e is a γ'-free edge.In any case G is either of class 1 or class 2.
Let S be a γ'-set of G -e.If S does not contain any neighbor of e, then every neighbor of e is γ'-totally free in G -e.This implies that G -e is of class 2. But this is a contradiction and so S must contain a neighbor of e. Then by theorem 2.4, Since G and G -e are of class 1, we have

Case (2). G is of class 2.
Since G -e is of class 1, then by a similar argument, S must contain a neighbor of e.Since G is of class 2, we have

Lemma
Let e be an edge of G.If e is γ'-free and G -e is of class 2, then Let S be a γ'-set of G -e.We have the following cases: Subcase (1).S contains a neighbor of e.Now

   
Since G is of class 1 and G -e is of class 2, we have .
By an argument similar to that in case (1), we have

Lemma
Let e be an edge of G.If e is γ'-fixed and G -e is of class 1, then

Proof
If e is γ'-fixed, then by Theorem 2.8, all of its neighbors are γ'-totally free.Then by Theorem 2.7, G is of e (1) class 2 and hence As e is γ'-fixed, by Theorem 2.4, , then e is γ'-critical.Then by Lemma 2.11, e is γ'-free and this is a contradiction.Therefore .Since G is of class 2 and G -e is of class 1, we have

Let
. If e is γ'-fixed and G -e is of class 2, then .
  By an argument analogous to that in Lemma 2.13, since G -e is of class 2, we have .

Theorem
Let G be a graph without isolated edges.An edge e in G is ds' --critical if and only if one of the following holds.
1) e is γ'-totally free and G -e is of class 1.
2) e is γ'-free, G is of class 2 and G -e is of class 1.
3) e is γ'-free and both G and G -e are of class 2. Proof Suppose e is ds' --critical.Then (1) Let S be a γ'-set of G. Then we have the following cases: Case (1).G and G -e are of class 1.By (1), .By theorem 2.3, e is γ'free and no γ'-set of G-e contains any edge of N(e).Now every neighbor of e is γ'-totally free in G -e. Therefore G -e is of class 2, which is a contradiction.
Case (2).G is of class 1 and G -e is of class 2. Then Equation (1) becomes .But this is not possible.
Case (3).G is of class 2 and G -e is of class 1.Then Equation (1) becomes .Then either e is γ'-free or γ'-totally free.

   
Case (4).G and G -e are of class 2.
In this case, Equation (1 Then by theorem 2.3

 
From Lemmas 2.9, 2.11 and 2.12, the converse is true.

Theorem
Let G be a graph without isolated edges.An edge e in G is ds' + -critical if and only if one of the following holds.
1) e is γ'-free, G is of class 1 and G -e is of class 2.
2) e is γ'-fixed and G -e is of class 2.

Proof
Suppose e in G is ds' Let S be a γ'-set of G. Then we have the following cases: Case (1).G and G-e are of class 1.From equation ( 1)   and so G is γ' + -critical.Hence by Theorem 2.6, e is γ'-fixed, which is a contradiction.
Case (2).G is of class 1 and G -e is of class 2. Now equation (1) becomes e .Then S must contain a neighbor of e.Since G is of class 1, e is γ'-free.
Case (3).G is of class 2 and G -e is of class 1.Then Equation (1) becomes , which is not possible.
Case (4).G is of class 2 and G -e is of class 2.
In this case, Equation (1 Then by Theorem 2.4  Conversely, suppose if (1) or ( 2) is true.Then by case (1) of Lemma 2.12 and Lemma 2.14, the result follows.

Edge-Domsaturation Number of a Graph Theorem
For any connected unicyclic graph with cycle C,  if and only if one of the following holds. 1) and there exists C C deg u  , exactly one vertex w not on C has and remaining vertices are pendent vertices.
 .Let S denote the set of all pendent edges of G and let S t  .
Claim 1:  is an edge dominating set for any edge e of C,   is an edge dominating set of G containing f.Here g is an edge adjacent to f and e is any edge of the cycle.Hence  , which is a contradiction.

Case (1). u or v lies on C.
3 Claim 3: is the union of P 1 and P 2 .Suppose not.Then, contains 1 2 . Suppose 1 lies on C K .Let 1 u be the maximal tree rooted at u 1 not containing any edge of C k .Clearly has at least ∆'(G) -2 pendent edges, say S. Then The converse is obvious.

Well-Edge Dominated Graph
A graph G is called well dominated if all minimal dominating sets have the same cardinality.This concept was introduced by Finbow, Hartnell and Nowakowski [6].
 , which is a contradiction.

Definition
A graph G is well-edge dominated if every minimal edge dominating set of G has the same cardinality.In this case, G has at least ∆' -2 pendent edges.Let W be the set of these pendent edges.Further  

Lemma
If G is a well-edge dominated graph and e is an edge of G, then there exists a minimum edge dominating set containing e and a minimum edge dominating set not containing e.

Subcase (1).
3 1 2 3 1 is the union of P 1 and P 2 .Also u or v lies on C.

Let
. Therefore 3 contains at least one P 2 .Since , no other vertex other than u and v has degree > 3.
If G -C 3 is the union of 2 P s  alone, then  1 2  , i j x x u u or i j is an edge dominating set and every edge lies in a γ'-set.Therefore .
 

P s  ds
To obtain an edge dominating set containing e, place e in the set D, delete   N e from G and continue in this greedy fashion until there are no edges left.Then D is minimal and since G is well-edge dominated, it is minimum.
To obtain a minimum edge dominating set not containing e, we use the same greedy method except that we use a neighbor of e as our initial edge in D.

If
3 is the union of and 2 , then from Theorem 1.5, . But pendent edges adjacent to u 1 does not lie in any γ'-set.Therefore .

 
,u , If G is well-edge dominated, then G is of class 1.As in subcase (1), G -C 4 also contains P 2 .Then is an edge dominating set of cardinality .Therefore .
From the above lemma, it is clear that every edge belongs to any one of the γ'-set.Therefore G is of class 1.
) are true.Let S be a γ'set of G -e.By (2) S does not contain any edge of N[e].Hence S cannot be an edge dominating set of G. But, for any edge [ f N e   f , is an edge dominating set of G. Since S is a minimum edge dominating set for G -e, is also a minimum edge dominating set for G and hence

Claim 2 :
e = uv is an edge with degree ∆'.Then either u or v lies on C k . of maximum degree ∆'.If e  C k , then for some edge k is a tree T of G with at least , 3