eight=26.6 />, so the necessary condition is obvious.
Next we prove the sufficient condition. We suppose getting rid of any vertexes with - odd factor, but is not, i.e. there exist
Be similar to the discussion of theorem 1
thereby are part of two odd components of respectively.
Combining (1) with (2)
Theorem 3 Let be claw free graphs, be partial connection point. be graph obtained by locally fully on in point, then for, the sufficient and necessary condition for with -odd factor is with -odd factor.
Proof is a spanning subgraph of, so the necessary condition is obvious.
Next we prove the sufficient condition. Let have -odd factor, have no - odd factor. has -odd factor, , so.
On the other hand, is claw free, so is claw free.
By lemma 2, lemma 3, has two odd components at least.
If, let (is branch of). Now, has the same odd components as, therefore, has -odd factor. which is contradiction.
Next let, since has not odd components, for any odd components of,
Let be adjacent vertexes of in two odd components of respectively.
Then is nonadjacent in the induced subgraph of
, which is contradiction to the fact that
is a locally connected vertex, since
The proof is complete.