Edge-Vertex Dominating Sets and Edge-Vertex Domination Polynomials of Cycles ()
            
            
        
                     
      
1. Introduction
 
Let G = (V, E) be a simple graph of order |V| = n. A set S Í V(G) is a dominating set of G, if every vertex 
 is adjacent to at least one vertex in S. For any vertex v Î V, the open neighbourhood of n is the set 
 and the closed neighbourhood of n is the set
. For a set S Í V, the open neighbourhood of S is 
 and the closed neighbourhood of S is
. The domination number of a graph G is defined as the minimum size of a dominating set in G and it is denoted as
. A cycle is defined as a closed path, and is denoted by
.
 
Definition 1.1
 
For a graph G = (V, E), an edge
, ev-dominates a vertex 
 if
 
1) u = w or v = w (w is incident to e) or
 
2) uw or vw is an edge in G (w is adjacent to u or v).
 
Definition 1.2 [1]
 
A set 
 is an edge-vertex dominating set of G (or simply an ev-dominating set), if for all vertices
; there exists an edge 
 such that e dominates v. The ev-domination number of a graph G is defined as the minimum size of an ev-dominating set of edges in G and it is denoted as
.
 
Definition 1.3
 
Let 
 be the family of ev-dominating sets of a graph 
 with cardinality i and let 
. We call the polynomial 
 the ev-domination polynomial of the graph
.
 
In the next section, we construct the families of the ev-dominating sets of cycles by recursive method. As usual we use 
 for the largest integer less than or equal to x and 
 for the smallest integer greater than or equal to x. Also, we denote the set 
 by [en] and the set 
 by [n], throughout this paper.
 
2. Edge-Vertex Dominating Sets of Cycles
 
Let 
 be the family of ev-dominating sets of 
 with cardinality i. We investigate the ev-domina- ting sets of
. We need the following lemma to prove our main results in this section.
 
Lemma 2.1: [2] 
.
 
By Lemma 2.1 and the definition of ev-domination number, one has the following Lemma:
 
Lemma 2.2: 
if and only if 
 or
.
 
Lemma 2.3: If a graph G contains a simple path of length 4k − 1, then every ev-dominating set of G must contain at least k vertices of the path.
 
Proof: The path has 4k vertices. As every edge dominates at most 4 vertices, the 4k vertices are covered by at least k edges.
 
Lemma 2.4. If
, and there exists x Î [en] such that 
 then 
.
 
Proof: Suppose that
. Since
, Y contains at least one edge labelled en−5, en−6 or en−7.
 
If
, then
, a contradiction. Hence 
 or
, but then in this case 
 for any x Î [en], also a contradiction.
 
Lemma 2.5. [3]
 
1) If 
 then
.
 
2) If
, and 
 then 
 and
.
 
3) If
, 
,
 , and
, then
.
 
Proof: 1) Since
, by Lemma 2.2, 
or
. In either case, we have 
 and
.
 
2) Since 
 and
, by Lemma 2.2, we have 
 and
. Hence 
 and
. Therefore, 
 and
.
 
3) By hypothesis, 
or
. Therefore, 
or
. Therefore, 
or
. Therefore,
.
 
Lemma 2.6. [4] If
, then
 
1) 
and 
 if and only if n = 4k and i = k for some k Î N.
 
2) 
and 
 if and only if i = n.
 
3)
, 
, 
, 
, if and only if n = 4k + 2 and 
 for some k Î N.
 
4)
, 
, 
and 
 if and only if i = n ? 2.
 
5)
, 
, 
and 
 if and only if i = n ? 1.
 
6)
, 
, 
and 
 if and only if
.
 
Proof: 1) (Þ) Since
, by Lemma 2.2, 
or
. If
, then 
 and by Lemma 2.2, 
, a contradiction.
 
So
 
 
(2.1)
 
and since
, we have
 
 
(2.2)
 
From (2.1) and (2.2),
 
 
(2.3)
 
When n is a multiple of 4, 
and
. Therefore,
. Therefore, 
, we get
. Thus, when
, (2.3) holds good and
. When
, 
 and
. Therefore, 
, which is not possible.
 
Hence 
 and ![]()
 
(Ü) If n = 4k and i = k for some k Î N, then by Lemma 2.2,
. Therefore, 
, which implies
. Similarly, 
and
.
 
Now
. Therefore, 
, which implies
.
 
2) (Þ) Since
, by Lemma 2.2, 
or
. If 
 then by Lemma 2.2, 
, a contradiction.
 
So
 
 
(2.4)
 
Since,
 
 
, (2.5)
 
From (2.4) and (2.5), we have
. Therefore,
. Therefore, ![]()
 
(Ü) If
, then by Lemma 2.2,
. Therefore, ![]()
 
Therefore, ![]()
 
3) (Þ) Since
, by Lemma 2.2, 
or
. If
, then
, by Lemma 2.2, 
a contradiction.
 
Therefore, 
, which implies,
 
 
(2.6)
 
Since, 
,
.
 
Hence,
 
 
(2.7)
 
Similarly,
 
 
(2.8)
 
and
 
 
(2.9)
 
From (2.6), (2.7), (2.8) and (2.9),
 
 
(2.10)
 
Therefore, (2.10) hold when 
 or 
 and
, for some
. Suppose
, then 
 and
. Therefore, from (2.10), we have, 
, which implies
. Suppose
, i.e.,
.
 
Case 1) When
.
 
From (2.10), we get 
 and
. Therefore, 
, which is not possible.
 
Case 2) When
. From (2.10), we get 
 and
. Therefore, 
, which is not possible.
 
Case 3) When
. From (2.10), we get 
 and ![]()
 
Therefore, 
, which is not possible. Therefore, ![]()
 
(Ü) If n = 4k + 2 and 
 for some k Î N, and
, then by Lemma 2.2, 
,
. Therefore,
. Therefore,
.
 
Also,
. Therefore, 
and 
 and
.
 
Hence
, 
,
.
 
4) (Þ) Since
, by Lemma 2.2,
 
 
 or
 (2.11) 
 
Since
, by Lemma 2.2,
 
 
(2.12)
 
Similarly, 
and
, by Lemma 2.2
 
 
(2.13)
 
and
 
 
(2.14)
 
From (2.11), we get 
 which is not possible.
 
Therefore,
 
 
(2.15)
 
From (2.12),
 
 
(2.16)
 
From (2.15) and (2.16), ![]()
 
(Ü) If
, 
then by Lemma 2.2, 
or
. Therefore,
. Also
, therefore,
;
, therefore,
; and
, therefore,
.
 
5) (Þ) Since 
 by Lemma 2.2,
 
 
 or
 (2.17) 
 
Since 
 by Lemma 2.2,
 
 
 or
 (2.18) 
 
Since 
 and
, by Lemma 2.2,
 
 
(2.19)
 
and
 
 
(2.20)
 
From (2.19) and (2.20), we have
. From (2.18), we have
. Therefore,
. But
. Therefore,
. Therefore,
.
 
(Ü) If
, 
then by Lemma 2.2, 
and 
 therefore, 
and
, therefore, 
and
, therefore,
.
 
6) (Þ) Since
, 
, 
, and
, by Lemma 2.2, 
, 
, 
, and
. So 
 and hence
.
 
(Ü) If
, then by Lemma 2.2 we have, 
, 
, 
,
.
 
Therefore, 
, 
, 
, and
.
 
Theorem 2.7 [5]
 
For every 
 and
,
 
1) If 
 and 
 then 
.
 
2) If 
 and 
 then 
.
 
3) If 
 and 
 then 
, where
 
 ![]()
 
 ![]()
 
 ![]()
 
4) If
, 
, and 
 then
 
 ![]()
 
5) If 
 then
 
 ![]()
 
Proof:
 
1) Since, 
and
, by Lemma 2.6 (i)
, and 
 for some
. The sets 
 have 
 elements and each one covers all vertices. Also, no other sets of cardinality 
 covers all vertices. Therefore, the collection of ev-dominating sets of cardinality 
 is
 
 ![]()
 
Hence,
.
 
2) We have 
 and
. By Lemma 2.6 (2), we have i = n. So,
.
 
3) We have 
 and
, by Lemma 2.6 (3), 
and
, for some
.
 
Let ![]()
 
 ![]()
 
 ![]()
 
We shall prove that
. It is clear that
, 
, and
. Therefore,
.
 
Conversely, Let
. Suppose, Y is of the form
, 
 then
. Now suppose, 
and Y is of the form 
 then
. Now suppose, 
, 
and
. We split 
 as four parts. If 
 with 
 then 
 and 
 and
. If 
 with 
 then 
 and 
 and
. If 
 with 
 then 
 and 
 and
. If 
 with 
 then 
 and 
 and
. In this case Y is of the form 
 then
. Therefore, 
. Thus, we have proved that
 
 ![]()
 
where ![]()
 
 ![]()
 
 
.
 
4) If
, by Lemma 3.6 (iv)
.
 
Therefore
.
 
5) ![]()
 
Clearly, ![]()
 
Conversely, let
. Then
. If
, then we can write
, for some
. If 
 and 
 then we can write
, for some
. If 
 and 
 then we can write
, for some
. If
, 
, 
then we can write
, for some
.
 
Therefore we proved that
 
 ![]()
 
Hence, ![]()
 
3. Edge-Vertex Domination Polynomials of Cycles
 
Let 
 be the ev-domination polynomial of a cycle
. In this section, we derive the expression for
.
 
Theorem 3.1 [6]
 
1) If 
 is the family of ev-dominating sets with cardinality i of
, then
 
 ![]()
 
where
.
 
2) For every
,
 
 ![]()
 
with the initial values
 
 
,
 
 
,
 
 
,
 
 
.
 
Proof:
 
1) Using (1), (2), (3), (4) and (5) of Theorem 2.7, we prove (1) part.
 
Suppose, 
and 
 then, 
.
 
Therefore,
. In this case 
 and 
. Therefore, in this case the theorem holds.
 
Suppose, 
and
, then 
 Therefore,
. In this case
. Therefore, 
and
. Therefore, in this case the theorem holds.
 
Suppose,
. In this case,
 
 ![]()
 
where
 
 ![]()
 
 ![]()
 
 ![]()
 
Therefore,
. Also, 
and 
. Therefore, 
and 
. Therefore, in this case the theorem holds.
 
Suppose,
 
 ![]()
 
Then we have
.
 
Therefore,
. In this case, 
, 
and 
. Therefore,
 
 
.
 
Therefore, in this case the theorem holds.
 
Suppose,
. In this case, we have
 
 ![]()
 
Therefore,
 
 ![]()
 
Hence,
 
 ![]()
 
 ![]()
 
 ![]()
 
 ![]()
 
 ![]()
 
 
.
 
with the initial values
 
 ![]()
 
 ![]()
 
 ![]()
 
 ![]()
 
We obtain 
 for 1 ≤ n ≤ 16 as shown in Table 1.
 
In the following Theorem, we obtain some properties of
.
 
Theorem 3.2
 
The following properties hold for the coefficients of
;
 
1)
, for every n Î N.
 
2)
, for every n Î N.
 
3)
, for every n ³ 2.
 
4)
, for every n ³ 3.
 
5)
, for every n ³ 4.
 
6)
, for every n ³ 5.
 
7)
, for every n Î N.
 
Proof:
 
1) Since
, we have
.
 
2) Since
, we have the result 
 for every n Î N.
 
3) Since
, we have 
 for n ³ 2.
 
4) By induction on n. The result is true for
. 
(from Table 1)
. Therefore, the result is true for n = 3. Now suppose that the result is true for all numbers less than ‘n’ and we prove it for n. By Theorem 3.1
 
 
  ![]()
 
 Table 1.
, the number of ev-dominating set of 
 with cardinality i.
 
  
 ![]()
 
5) By induction on n, the result is true for
. 
(from Table 1). 
. Therefore, the result is true for
. Now suppose the result is true for all natural numbers less than n. By Theorem 3.1,
 
 ![]()
 
6) By induction on n, the result is true for n = 5. 
(from Table 1)
 
 ![]()
 
Therefore the result is true for
.
 
Now suppose that the result is true for all natural numbers less than n and we prove it for n. By Theorem 3.1,
 
 ![]()
 
7) From the table it is true.
 
Theorem 3.3
 
1) ![]()
 
2) For every
, ![]()
 
3) If
, then for every n ≥ 5, 
with initial values
, 
, 
, and
.
 
Proof:
 
1) We prove by induction on n.
 
First suppose that n = 2 then,
 
 ![]()
 
We have the result.
 
2) By Theorem 3.1, we have
 
 ![]()
 
Therefore, we have the result.
 
3) By Theorem 3.1, we have
 
 ![]()
 
4. Concluding Remarks
 
In [7] , the domination polynomial of cycle was studied and obtained the very important property, 
. It is interesting that we have derived an analogues relation for the edge-vertex domination of cycles of the form,
 
 
. One can characterise the roots of the polynomial 
 and identify whether they are real or complex. Another interesting character to be investigated is whether 
 is log-concave or not.