The Generalization of Signed Domination Number of Two Classes of Graphs ()
1. Introduction
The graphs considered in this paper are finite, undirected, and simple (no loops or multiple edges). For notation and graph theory terminology, please refer to the reference [1]. Let G be a simple undirected graph. The vertex set and the edge set of G are denoted by
and
, respectively. For a vertex
, the neighbor set
is the set of vertices adjacent to v, the degree of v is the number of adjacent vertices of v and denoted by
.
and
is the minimum degree and maximum degree of G. When no confusion can occur, we shall write
, instead of
. For a subset
, the subgraph induced by U is denoted by
, which is the graph on U whose edges are precisely the edges of G with both ends in U.
For
and
, the distance between u and U, denoted by
, is the length of a shortest path from u to a vertex in U. When U consists of a single vertex v, we write
, instead of
. For a positive integer k, the k-th power
of a graph G is the graph
whose vertex set is
, two distinct vertices being adjacent in
if and only if their distance in G is at most k. If
,
. In particular, the graph
and the graph
are as the square of G and the cube of G. A cycle on
vertices is a graph whose vertices can be arranged in a cyclic sequence in such a way that two vertices are adjacent if they are consecutive in the sequence, and are nonadjacent otherwise, denoted by
. A path
is a simple graph whose vertices can be arranged in a linear sequence in such a way that two vertices are adjacent if they are consecutive in the sequence, and are nonadjacent otherwise.
In recent years, several kinds of signed domination problems in graphs have been investigated [2] [3] [4] [5]. Most of those belong to the vertex domination (or edge domination) of graphs, such as signed (edge) domination [6] [7], minus domination [8], cycle domination [9], signed roman (total) domination [10], weak roman domination [11], inverse signed total domination [12], etc. The signed domination number of cycles
, paths
, the square
of
and the square
of
were given in [13] [14] [15], respectively. In the present paper, we compute the exact values of signed domination number of
and
for
.
Let
be a graph and
be a real set. For a real function
and a nonempty subset
, we may assume that
. In the following, for the sake of simplicity, simply write
as
, denotes
. In addition,
and
denotes the smallest integer not less than x and the largest integer not greater than x.
2. Preliminaries
Definition 1. [4] Let
be a graph, a function
is said to be a Signed Dominating Function (SDF) if
holds for all
, the signed domination number:
.
By Definition 1, it is easy to see the following a conclusion.
Lemma 1. For any vertex
and an SED f of G, if
is even, then
. If
is odd, then
.
For some vertex
, it is called a optimal if
is reach it’s lower bound.
Lemma 2. [8] Let
be a path. For
, we have
.
Lemma 3. [13] Let
be a cycle. For
, we have
.
Lemma 4. [13] If
is a complete graph, then:
Lemma 5. [14] If
is a square of
, then:
Lemma 6. [15] For a square
of
, if
, then
. If
, then:
Lemma 7. [13] If G is a r-regular graph on n order, then
.
In this paper, we give an exact value of signed domination number of
and
for
as follows.
Theorem 1. Let
be a k-th power graph of cycle
. For
, we have: :
If
in Theorem 1, then it is equal to Lemma 3. If
in theorem 1, then it is equal to Lemma 5.
Theorem 2. Let
be a k-th power graph of path
. For
,
, we have the following results.
If
and k is odd, then:
If
and k is even, then:
If
, then:
If
,then
.
In fact, if
in Theorem 2, then it is equal to Lemma 2. If
in theorem 2, then it is equal to Lemma 6.
3. Proof of Theorems
3.1. Proof of Theorem 1
Proof. Let
be a k-th power graph of cycle
. By definition of k-th power graph, it is easy to see that
is a 2k-regular graph. By symmetry of cycle, it is
enough to show that
. It remains to show that it is true for
. By Lemma 1, we have
. We may assume that G has t vertices
with −1 label and s vertices with +1 label, then
,
. Let
,
.
If
, then
.
If
, then
.
If
, then
. If n is odd, then q is odd, and
is even. But
is odd and therefore:
. If n is even, then q is even, and
is odd. But
is even and therefore
.
In summary, we have:
By definition of singed domination number, we only need to give a singed domination function f of
. Let
be a cycle on n vertices
.
If
, then let:
It is easy to see that
has a
vertices with −1 label and
vertices with +1 label. It is follows that:
.
So, we have
.
If
, then let:
By the definition of f, for every
consecutive vertices on the
, there
are k vertices that can be signed as −1, and
of the remaining
vertices can be signed as −1. So, it is easy to see that
has a
vertices with −1 label and
vertices with +1 label. It is follows that:
.
So, we have
.
If
, then let:
Same as above, it is easy to see that
has a
vertices with −1 label and
vertices with +1 label. It is follows that:
.
So, we have
.
As mentioned above, we have:
3.2. Proof of Theorem 2
Proof. Let
be a k-th power graph of
, denoted by G. By definition of k-th power graph, we may assume that the vertex set of G are
, its edge set are
, where:
.
It is follows that:
If
, then
. So,
.
If
, then
(or
). So,
(or 1).
It remains to show that it is true for
.
Case 1.
.
We first prove that it is true for the lower bound of signed domination number of
. If
, then
. If n is odd, then:
. So,
. If n is even, then
. In such a case, since
is odd and
is even. By Lemma 1, we have
. So,
.
We only need to give a singed domination function f and prove that it is true for the upper bound of signed domination number of
.
Subcase 1.1.
.
Due to
,
and
is even. It follows that we are labeled −1 at
vertices in
and in
, and other vertices is label to +1. Namely, let:
It is easy to show that
for
and
. In fact, under this function f, we have:
. So,
.
In summary, we have
.
Subcase 1.2.
.
Because of
,
and
is odd. It follows that we are labeled −1 at
vertices in
and in
, and other vertices is label to +1. Namely, let:
It is easy to show that
for
and
. In fact, under this function f, we have:
. So
.
In summary, we have
.
Subcase 1.3.
.
Thanks to
,
and
is odd. It follows that we are labeled −1 at
vertices in
and labeled −1 at
vertices in
, and other vertices is label to +1. Namely, let:
It is easy to show that
for
and
. In fact, under this function f, we have:
. So,
.
In summary, we have
.
Subcase 1.4.
.
Since
,
and
is even. It follows that we are labeled −1 at
vertices in
and labeled −1 at
vertices in
, and other vertices is label to +1. Namely, let:
It is easy to show that
for
and
. In fact, under this function f, we have:
. So,
.
In summary, we have
.
Case 2.
.
Subcase 2.1.
.
We first prove that it is true for the lower bound of signed domination number of
.
If k is odd, then
is even. By Lemma 1, we have:
,
. It is follows that:
. So, we have:
.
If k is even, then
is odd. By Lemma 1, we have:
,
. It is follows that:
. So, we have:
.
On the other hand, we only need to give a signed domination function f.
If k is odd, then let:
where
. It is easy to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that
. So, we have
.
If k is even, then let:
where
. It is easy to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that
. So, we have:
.
Subcase 2.2.
.
We first prove that it is true for the lower bound of signed domination number of
.
If k is odd, then
is even. By Lemma 1, we have:
,
,
(otherwise, we have
, this is a contradiction to Definition 1). It is follows that
. So, we have
.
If k is even, then
is odd. By Lemma 1, we have
,
,
(otherwise, we have
, this is a contradiction to Definition 1). It is follows that
. So, we have
.
On the other hand, we only need to give a signed domination function f.
If k is odd, then let
where
. Specially, if
, then
and all such vertices are label to +1. Then G has a
vertices with −1 label and other vertex is label to +1. It is follows that:
. So, we have:
.
If k is even, then:
where
. Specially, if
, then
and all such vertices are label to +1. Then G has a
vertices with −1 and other vertex is label to +1. It is follows that
. So, we have
.
Subcase 2.3.
.
We first prove that it is true for the lower bound of signed domination number of
.
If k is odd, then
is even. By Lemma 1, we have:
,
,
. It is follows that
. So, we have
.
If k is even, then
is even. By Lemma 1, we have:
,
,
. It is follows that
. So, we have:
.
On the other hand, we only need to give a signed domination function f.
If k is odd, then let:
where
. It is easy to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that
. So, we have
.
If k is even, then let:
where
. Specially, if
, then
and all such vertices are label to +1. It is easy to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that:
. So, we have:
.
Subcase 2.4.
.
We first prove that it is true for the lower bound of signed domination number of
.
Claim. Let
be a path and f be an SED of
. If every vertex
is optimal, then
, where
.
Proof: If k is odd, then
is even. Since every vertex
is
optimal, by Lemma 1, we have
. It is follows that
has
vertices with −1 label,
has
vertices with −1 label,
,
has
vertices with −1 label. Because of
,
and
is optimal and therefore
has k vertices with −1 label. So, we have
.
If k is even, then
is even. Since every vertex
is optimal, by Lemma 1, we have
. It is follows that
has
vertices with −1
label,
has
vertices with −1 label,
,
has
vertices with −1 label. Due to
, and
is optimal and
therefore
has k vertices with −1 label. So, we have
.
We consider that every vertex
is optimal and the following facts.
…
where
It is follows that:
…
where
So, we have:
.
Subsubcase 2.4.1. If
, then
.
If k is odd, then
is even and
. So,
. It is follows that:
. Hence,
.
If k is even, then
is even and
. So,
. It is follows that:
. Hence,
.
Subsubcase 2.4.2. If
, then
.
It is follows that
. Owing to
(Otherwise, since r is odd and therefore
. In order to obtain the
, we have to consider that all vertices are as optimal as possible. By Claim, we have
, namely,
. So,
, this is a contradiction to Definition 1).
Hence,
.
On the other hand, we only need to give a signed domination function f.
If k is odd, then we give a signed domination function f by the following rules.
If
and
, then let:
If
, then let:
where
. It is trivial to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that:
. So, we have:
.
If k is even, then we give a signed domination function f by the following rules.
If
and
, then let:
If
, then let:
where
. It is trivial to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that:
. So, we have:
.
Subcase 2.5.
.
We first prove that it is true for the lower bound of signed domination number of
.
Subsubcase 2.5.1 If
, then
.
If k is odd, then
is even and
. Hence,
. It is follows that:
. So, we have
.
If k is even, then
is even and
,
. Hence,
. It is follows that:
. So, we have:
.
Subsubcase 2.5.2. If
, then
.
It is follows that
. Because of
(Otherwise, we have
. In order to obtain the
, we have to consider that all vertices are as optimal as possible. By Claim, we have
, namely,
. It follows that:
, this is a contradiction to Definition 1).
Hence,
.
On the other hand, we only need to give a signed domination function f.
If k is odd, then we give a signed domination function f by the following rules.
If
and
, then let:
If
, then let:
where
. It is trivial to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that:
. So, we have
.
If k is even, then we give a a signed domination function f by the following rules.
If
and
, then let:
If
, then let:
where
. It is trivial to see that G has a
vertices with −1 label and other vertex is label to +1. It is follows that:
. So, we have
.
As mentioned above, we have:
If
and k is odd, then:
If
and k is even, then:
If
, then:
If
, then
.
4. Concluding Remarks
As the dominance theory of graphs becomes more and more diversified, the dominance problem of graphs is NP-complete. Therefore, studying the upper and lower bounds of domination numbers of graphs and even accurate estimation and mutual relations are the main aspects of current research. In this paper, we determine the exact value of the Signed Domination Number of k-th power graphs
and
for
, these conclusions are the basis for the study of the bounds of the signed domination number of the k-th power graphs of generally connected graphs, which have important meanings in the structural theory of graph theory. From the signed domination number of the k-th power graph, other types of domination numbers can be further studied, which will be the key research object in the future.
Supported
This research is supported by NSFC (11701257); (2020xjgj016); 2020ZKYB04, Basic Research Foundation of Henan Educational Committee (20ZX004), the Young Backbone Teachers in Henan Province (Grant 2020GGJS194, 2019GGJS202) (2020-JSJYYB-053); the Young Backbone Teachers in Luoyang Normal College (2019XJGGJS-10).