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ABSTRACT
A proper vertex colouring is called a 2-dynamic colouring, if for every vertex v with degree at least 2, the neighbours of v receive at least two colours. The smallest integer k such that G has a dynamic colouring with k colours denoted by . We denote the cartesian product of G and H by
. In this paper, we find the 2-dynamic chromatic number of cartesian product of complete graph with complete graph
, complete graph with complete bipartite graph
and wheel graph with complete graph
.
2000 Mathematics Subject Classification:
1. Introduction
Graphs in this note are simple and finite [Citation1,Citation2]. We denote the edge set and vertex set of G, by and
, respectively. The number of vertices of G is called order of G. The degree of a vertex
, in a graph G denoted by
or
, is the minimum number of edges incident in it, where
and
denotes minimum and maximum vertex degree in a graph G. A proper vertex colouring of G is a function
, with property: if
are adjacent, then
and
are different. A vertex k-colouring is a proper vertex colouring with
. The smallest integer k such that G has a vertex k-colouring is called the chromatic number of G and denoted by
. A proper vertex k-colouring is called dynamic [Citation3,Citation4] if for every vertex v with degree at least 2, the neighbours of v receive atleast two different colours. The smallest integer k such that G has a dynamic k-colouring is called 2-dynamic chromatic number of G and denoted by
. Moreover, 2-dynamic colouring is generally said to be dynamic colouring.
Generally, a proper vertex k-colouring is called r-dynamic [Citation5] if for every vertex v with degree at least r, the neighbors of v receive atleast r different colours. The smallest integer k such that G has a dynamic k-colouring is called r-dynamic chromatic number of G and denoted by .
For any denotes the neighbour set of v in G. Let c be a proper vertex colouring of G. For any
, we mean
the set of colours appearing in the neighbors of v in G.
Let G and H be two graphs. We call that the cartesian product of G and H, , is a graph with the vertex set
such that two vertices
and
are adjacent if and only if u = x and
or v = y and
. Clearly
.
Similarly, join of G and H, G + H, is a graph with the vertex set and edge set
.
A graph is said to be wheel on n-vertices, it is a join of cycle of length n−1 and . Moreover, the minimum number of vertices in any wheel is 4.
2. Preliminaries
We shall make use of the following lemmas, in order to prove our results,
Lemma 2.1
[Citation6]
For any positive integer n,
Lemma 2.2
[Citation7]
For i and
and
.
Lemma 2.3
[Citation7]
For any positive integer n, .
Lemma 2.4
[Citation8]
Let G and H be two graphs. If
Lemma 2.5
[Citation9]
Let G be a connected graph then,
Lemma 2.6
[Citation2]
For any positive integer n,
3. Sub results
For finding the dynamic chromatic number of wheel, we need the dynamic chromatic number of join of two graphs.
Theorem 3.1
For any two connected graphs G and H, then
Proof.
Let and
. By the definition of join of two graphs,
Consider the vertex colourings
and
Let
and define
. For every
, provide
and for every
,
Now, we claim that c is a dynamic colouring of G + H. Clearly, c is a proper colouring. Moreover, for vertices
having
,
even still
and
and it completes the proof.
Corollary 3.2
For any positive integer n,
Proof.
From the definition of wheel . Using Theorem 3.1,
and it completes the proof.
4. Main results
In the next theorem, we obtain the dynamic chromatic number of cartesian product between two complete graphs.
Theorem 4.1
For any two positive integers then
Proof.
Let and
. By the definition of cartesian product,
. For proving this theorem we need to consider two cases,
Case 1: When r = s = 2.
Clearly, . So,
.
Case 2: When both .
Consider the mapping
Define,
since
and
so, f produces a proper colouring for
. Next we need to show f is dynamic colouring. Since for every vertex
,
even r or s is 1 and the proof is complete.
Generally, consider the p-dynamic colouring,
whenever
, but it varies when
. This will not suit for
, since
where
in
.
Corollary 4.2
For positive integers and ∃ atleast one
then
where
.
Corollary 4.3
where we can take cartesian product in any number of times.
In the next two theorem, we obtain the dynamic chromatic number of cartesian product between complete graph and complete bipartite graphs.
Theorem 4.4
For any two positive integers and n, then
Proof.
Let
and
By the definition of cartesian product,
Consider a mapping
. For proving this theorem we need to consider three cases,
Case 1: When n = 1
Clearly, . So
where
in this case.
Case 2: When n = 2contains a subgraph of
. So,
To prove the reverse inequality, let us define a map f in such a way that
and
. Clearly f preserves dynamic colouring and using this
requires atmost 4 colours. It clearly says
.
Case 3: When
Using Lemma 2.4,
For, the reverse inequality,
contains
. so, it requires atleast n colours for dynamic colouring and this proves the theorem.
From the Theorem 4.4, when r = 1 then
Using Theorem 4.1,
.
Theorem 4.5
For any positive integers and n, then
Proof.
Let
and
By the definition of cartesian product,
Consider a mapping
We prove the theorem on the number of vertices in complete graph,
Case 1: When n = 1.
Clearly, . So,
, where k = 4 in this case.
Case 2: When n = 2
contains a subgraph of
. So,
To prove the reverse inequality, by Lemma 2.4,
It clearly says,
, where k = 4 in this case.
Case 3: When
Define the mapping f by and
For every
. so, f produces a proper colouring for
with
colours. Next we need to show f is dynamic colouring. Since for every vertex
where n is minimum 3 in this case. Thus f contributes dynamic colouring and the proof is complete.
In the next theorem, we obtain the dynamic chromatic number of cartesian product of wheel graph with complete graph.
Theorem 4.6
For any positive integer and n, then
Proof.
Let
and
where
is the centre vertex in the wheel
i.e.
is adjacent to remaining l−1 vertices. By the definition of cartesian product,
Consider a mapping
where
.
For proving the theorem, we need to consider two cases,
Case 1: When l−1 is even.
Define,
Case 2: When l−1 is odd.
Define,
For the above two cases, we claim that f is a dynamic colouring of
. Clearly, f is a proper colouring. Moreover, for every vertex
where c is dynamic colouring for
. Since for every vertex
,
and the proof is complete.
Disclosure statement
No potential conflict of interest was reported by the authors.
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