tailieunhanh - Báo cáo toán học: "A short proof of a theorem of Kano and Yu on factors in regular graphs"

Tuyển tập các báo cáo nghiên cứu khoa học về toán học trên tạp chí toán học quốc tế đề tài: A short proof of a theorem of Kano and Yu on factors in regular graphs. | A short proof of a theorem of Kano and Yu on factors in regular graphs Lutz Volkmann Lehrstuhl II fiir Mathematik RWTH Aachen University 52056 Aachen Germany e-mail volkm@ Submitted Jul 13 2006 Accepted Jun 1 2007 Published Jun 14 2007 Mathematics Subject Classification 05C70 Abstract In this note we present a short proof of the following result which is a slight extension of a nice 2005 theorem by Kano and Yu. Let e be an edge of an r-regular graph G. If G has a 1-factor containing e and a 1-factor avoiding e then G has a k-factor containing e and a k-factor avoiding e for every k 2 1 2 . r 1g. Keywords Regular graph Regular factor 1-factor k-factor. We consider finite and undirected graphs with vertex set V G and edge set E G where multiple edges and loops are admissible. A graph is called r-regular if every vertex has degree r. A k-factor F of a graph G is a spanning subgraph of G such that every vertex has degree k in F. A classical theorem of Petersen 3 says Theorem 1 Petersen 3 1891 Every 2p-regular graph can be decomposed into p disjoint 2-factors. Theorem 2 Katerinis 2 1985 Let p q r be three odd integers such that p q r. If a graph has a p-factor and an r-factor then it has a q-factor. Using Theorems 1 and 2 Katerinis 2 could prove the next attractive result easily. Corollary 1 Katerinis 2 1985 Let G be an r-regular graph. If G has a 1-factor then G has a k-factor for every k 2 1 2 . rg. Proofs of Theorems 1 and 2 as well as of Corollary 1 can also be found in 4 . The next result is also a simple consequence of Theorems 1 and 2. Theorem 3 Let e be an edge of an r-regular graph G with r 2. If G has a 1-factor THE ELECTRONIC JOURNAL OF COMBINATORICS 14 2007 N10 1 containing e and a 1-factor avoiding e then G has a k-factor containing e and a k-factor avoiding e for every k 2 1 2 . r 1g. Proof. Let F and Fe be two 1-factors of G containing e and avoiding e respectively. Case 1 Assume that r 2m 1 is odd. According to Theorem 1 the .