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.