Introduction
The length of dominating cycles in graphs is usually discussed in control problems.1–9 In this article, we only consider finite and simple graphs, and for the notation and terminology not defined here, please refer to Bondy and Murty
10
The length of a longest cycle in a graph G is called the circumference of G. If a graph G contains no
induced subgraph, then G is claw-free. Let
and
. A dominating cycle of a graph G is a cycle C of G such that
is an independent set. Let
denote the vertex set
containing the minimum number of vertices for all nonadjacent vertices
, and
denote the vertex set
with the minimum number of vertices for all vertices
with
.
Since the theory of dominating cycles can be applied to many fields, a lot of researchers put focus on it and have obtained many good results as follows.
Theorem 1.1.
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If G is a 2-connected graph of order n with
, then the circumference of G is at least
, where
denotes the minimum sum degree of all three independent vertices in G.
Theorem 1.2.
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If G is a 2-connected graph of order n with
, then the circumference of G is at least
, where
denotes the minimum sum degree of all three independent vertices in G.
Theorem 1.3.
13
If G is a 2-connected graph of order n containing a dominating cycle, which is not the Petersen graph, then G contains a dominating cycle of length at least
.
Theorem 1.4.
13
If G contains a dominating cycle with order n and
, then G contains a dominating cycle of length at least
.
Shen and Tian
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made the following conjecture:
Conjecture 1.5
If G contains a dominating cycle with order n and
, then
G contains a dominating cycle of length at least
if n is even;
G contains a dominating cycle of length at least
if n is odd.
In this article, we confirm the above conjecture holds for claw-free graphs as follows.
Theorem 1.6
If G is a claw-free graph of order n containing a dominating cycle with
, then the length of a longest dominating cycle is at least
.
Proof of Theorem 1.6
For a cycle C with a positive direction and a vertex v on C, let
denote the successor of v along the positive direction with
, and
denote the predecessor of v with
. Similarly,
. uCv denotes the vertices on C from u to v along the positive direction, and
denotes the vertices on C from u to v along the negative direction. Denote
,
.
To the contrary, suppose C is a longest dominating cycle with length at most
for a non-Hamiltonian claw-free graph G with
. In the following proof, assume
and
. We can get the following results to prove Theorem 1.6.
Claim 1
For any vertex
,
.
Proof
If
, then there is a dominating cycle
longer than C, a contradiction. Similarly,
. Since
.□
Claim 2
For any two distinct vertices
,
.
Proof
Suppose
for some vertex
, then G contains a longer dominating cycle
, a contradiction. Thus,
. Similarly,
.□
By the definition of dominating cycle, we can get the following result.
Claim 3
and
.
By Claim 1, Claim 2, and the maximality of C, it is easy to get the following results.
Claim 4
and
for
.
Claim 5
For any
.
Proof
If
, then by the definition of dominating cycle,
. Assume
and without loss of generality, suppose
and
. Obviously,
. Suppose
. By Claim 3,
. Without loss of generality, assume
Clearly,
and
by Claim 1. Then, G contains a longer dominating cycle
, a contradiction. It follows that
. Then,
, a contradiction.□
Claim 6
For any vertex
,
.
Proof
Without loss of generality, suppose
. Choose
such that
contains minimum vertices, that is,
. Clearly,
. Define a mapping f from
to
by
For any vertex
, by the choice of
. If
, then
and
, otherwise G contains a longer dominating cycle
, a contradiction. If
, then
and
, otherwise G contains a longer dominating cycle
, a contradiction.
If
, then
and
, otherwise G contains a longer dominating cycle
, a contradiction. Similarly,
if
. If
, then
and
, otherwise G contains a longer dominating cycle
, a contradiction. Similarly,
if
.
Clearly, if
, then
, and by the proof of Claim 2,
. It follows that
. By Claim 5,
and
. By the definition of f and the maximality of C,
for some vertex
. Thus,
, a contradiction.□
Proof of Theorem 1.6
For
, let
, and
, for
. Clearly,
and
. By Claim 2 and Claim 4,
. It follows that
By Claim 6,
, then
, a contradiction. It follows that Theorem 1.6 holds.