The Matching Uniqueness of A Graphs

Abstract

In the paper, We discussed the matching uniqueness of graphs with degree sequence . The necessary and sufficient conditions for and its complement are matching unique are given.

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Shen, S. (2015) The Matching Uniqueness of A Graphs. Applied Mathematics, 6, 1189-1192. doi: 10.4236/am.2015.68109.

1. Introduction

All graphs considered in the paper are simple and undirected. The terminology not defined here can be found in  . 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  - . In the paper, we prove

and its complement are matching unique if and only if or

.

2. Basic Results

Lemma 1  The matching polynomial satisfies the following identities:

1).

2) if is an edge of G.

Lemma 2  Let G be a connected graph, and let H be a proper subgraph G.

Then.

Lemma 3  Let, if, then H are precisely the graphs of the following

types:

Lemma 4 1)  .

2)  .

3)  .

4)  ,

.

5)  .

6)  .

Lemma 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  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

tion. If, then, a contradiction. Let. If, then

, a contradiction. If s2 ≥ 2, then,

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,

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  , so the complement of are matching unique if and only if or.

Conflicts of Interest

The authors declare no conflicts of interest.

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