1. Introduction
All graphs considered in the paper are simple and undirected. The terminology not defined here can be found in [1] . Let G be a graph with n vertices. An r-matching in a graph G is a set of r edges, no two of which have a vertex in common. The number of r-matching in G will be denoted by
. We set
and define the matching polynomial of G by

For any graph G, the roots of
are all real numbers. Assume that
, the
largest root
is referred to as the largest mathing root of G.
Throughout the paper, we denote by
and
the path and the cycle on n vertices, respectively.
denotes the tree with a vertex v of degree 3 such that
, and
denotes the tree obtained by appending a pendant vertex of the path
in
to a vertex with degree 2 of
.
is obtained by appending a cycle
to a pendant vertex of a path
. Two graphs are matching equivalency if they share the same matching polynomial. A graph G is said to be matching unique if for any graph H,
implies that H is isomorphic to G. The study in this ares has made great progress. For details, the reader is referred to the surveys [2] -[6] . In the paper, we prove
and its complement are matching unique if and only if
or ![]()
.
2. Basic Results
Lemma 1 [1] The matching polynomial
satisfies the following identities:
1)
.
2)
if
is an edge of G.
Lemma 2 [1] Let G be a connected graph, and let H be a proper subgraph G.
Then
.
Lemma 3 [2] Let
, if
, then H are precisely the graphs of the following
types:
![]()
Lemma 4 1) [1]
.
2) [2]
.
3) [2]
.
4) [3]
,
.
5) [4]
.
6) [5]
.
Lemma 5 [5] Let G be a tree and let
be obtained from G by subdividing the edge uv of G, then
1)
, if uv not lies on an internal path of G.
2)
, if uv lies on an internal path of G, and if G is not isomorphic to
.
Lemma 6 [6]
are matching unique.
Lemma 7
.
Proof. Direct computation (using Matlab 8.0), we immediately have the following:
,
.
![]()
By Lemma 2, 5, we get ![]()
.
3. Main Results
Theorem 1 Let
, then G are matching unique if and only if
or
.
Proof. The necessary condition follows immediately from Lemma 1. We have
![]()
![]()
![]()
![]()
Now suppose that
or
, H is a graph being matching equivalency with G. We proceed to prove that H must be isomorphic to G. By Lemma 3
![]()
Case 1. If
. By
, we know that
. Hence, the component of
in H may be only
. By Lemma 4,
and
. Let
, then
, a contradiction. Let
. If
, then
, a contradiction. If
, then
, a contradic-
tion. If
, then
, a contradiction. Let
. If
, then
, a contradiction. If s2 ≥ 2, then
,
a contradiction. Let
. If
, then
, a contradiction. If
, then
, a contradiction. Let
, then
, a contradiction.
Case 2 If
. By
, hence the component of
in H may be
only
. Let
. If
, then
, a contradiction. If
, then
, a contradiction. If
, then
, by Lemma 4, we get
, thus
. That is,
, then
, by Lemma 6,
has at least one equal to 6, a contradiction. If
,
by Lemma 4, 6, we have
, thus H be isomorphic to G. Let
. If
,
, a contradiction. If
, a contradiction. Let
, by
Lemma 4,
, a contradiction.
Case 3 If![]()
, by
, a contradiction. Combing cases 1 - 3, H is isomorphic to G.
The proof is complete. For a graph, its matching polynomial determine the matching polynomial of its Comple-
ment [6] , so the complement of
are matching unique if and only if
or
.