Some Results on ( 1 , 2 n – 1 )-Odd Factors

Let be a graph. If there exists a spanning subgraph G F such that     1,3, , 2 1 F d x n    , then F is called to be -odd factor of . Some sufficient and necessary conditions are given for to have    n   1,2 1 G G U – 1,2 1 n  -odd factor where is any subset of such that U   V G U k  .


Introduction
We consider finite undirected graph without loops and multiple edges .Let be a graph with vertex set and edge set Given , the set of vertexes adjacent to x is said to be the neighborhood of x , denoted by , and is called the degree of x , If there exists a spanning subgraph F such that 1, f -odd factor of , especially, if for every , then it is called -odd factor.especially,    2  1, 1 n  1  -odd factor is 1-factor when n = 1.For a subset , let denote the subgraph obtained from by deleting all the vertexes of together with the edges incident with the vertexes of

S o
denotes the number of odd components of .The sufficient and necessary condition for graph to have f -odd factor was given in paper [1] Ryjacek [2] introduced one kind of new closure operation: let be a graph,  , is perfect matching ,then we call the graph to be k-factor critical.Of course, 0-factor critical graph is perfect matching.Favaron popularized a series of the properties of perfect matching to k-factor critical, at the same time the sufficient and necessary conditions were given for the graph to be k-factor critical, more results in factor critical graphs were referred to [4,5].

G T  G
For   -extended, and is [8].We will popularize some results of k-factor critical to -odd factor, and gain several sufficient and necessary conditions for

Main Results
We start with some lemmas as following.Lemma 1 The sufficient and necessary condition for a graph to have n  -odd factor, by the sufficient and necessary condition for graph with -odd factor we have For any and the following that the set U with any vertexes, has -odd factor, i.e., for any , there and , we have Lemma 2 [9] Connected claw free graphs of even order have 1-factor.
Lemma 3 Connected claw free graphs of even order have -odd factor.

1,2n
Proof If , by lemma 2, the conclusion is proved.Assume that .By contradiction, we assume that has no -odd factor, i.e., such that then there exists x S  such that x connecting with three components of at least.If not, for G S  x S   , x connects with two components of at most, consequently , contradiction.Theorem 1 Let be graph with order, p , x y are a couple of nonadjacent vertexes and satisfy then the sufficient and necessary condition for removing any vertexes with -odd factor is that getting rid of any vertexes with -odd factor.
Proof The necessary condition is obvious, next we prove the sufficient condition.
By contradiction, let remove any vertexes with -odd factor, but there exist vertexes after getting rid of the vertexes of G without -odd factor.By lemma 1, there exists at the same time, by   mod It shows that , x y B are part of two odd components of On the other hand, by hypothesis Contradiction.
, x y are a couple of any nonadjacent vertexes of , and satisfy then the sufficient and necessary condition for removing any vertexes with -odd factor is is a spanning subgraph of , so the necessary condition is obvious.

G G xy 
Next we prove the sufficient condition.We suppose G xy  getting rid of any vertexes with k Be similar to the discussion of theorem 1 Copyright © 2012 SciRes.
thereby , x y are part of two odd components of respectively.
By hypothesis Combining ( 1) with ( 2) Proof is a spanning subgraph of , so the necessary condition is obvious.

G G
Next we prove the sufficient condition.Let G U   have -odd factor, have no , which is contradiction to the fact that x is a locally connected vertex, since k The proof is complete.
On the other hand, G is claw free, so G U