tailieunhanh - Báo cáo toán học: "ircular Chromatic Index of Generalized Blanuˇa s Snarks"

Tuyển tập các báo cáo nghiên cứu khoa học về toán học trên tạp chí toán học quốc tế đề tài: Circular Chromatic Index of Generalized Blanuˇa s Snarks. | Circular Chromatic Index of Generalized Blanusa Snarks Mohammad Ghebleh Department of Mathematics Simon Fraser University British Columbia Canada mghebleh@ Submitted Jul 27 2007 Accepted Mar 1 2008 Published Mar 12 2008 Mathematics Subject Classification 05C15 Abstract In his Master s thesis Jan Mazak proved that the circular chromatic index of the type 1 generalized Blanusa snark Bn equals 3 n This result provided the first infinite set of values of the circular chromatic index of snarks. In this paper we show the type 2 generalized Blanusa snark Bn has circular chromatic index 3 b I 3n 2C In particular this proves that all numbers 3 1 n with n 2 are realized as the circular chromatic index of a snark. For n 1 2 our proof is computer-assisted. 1 Introduction Let G be a graph and r 2. For all a 2 0 r let a r min a r a . For a b 2 0 r the r-circular interval a b r is defined by a b r ta b if a 6 b f a r u 0 b if a b. For a b 2 R a r and a b r are defined by first reducing a and b modulo r to a0 b0 2 0 r . An edge r-circular colouring or an edge r-colouring for short of G is a function c E G 0 r such that for any two adjacent edges e and e0 c e c e0 r 1. If G admits an edge r-colouring then G is edge r-colourable. The circular chromatic index of G is defined by xC G inf r 2 R G is edge r-colourable . 1 It is well-known see 9 for example that for every finite graph G the infimum in 1 is attained and that xC G is rational. It is also known that for every graph G x0 G dxC G l where x0 G is the chromatic index of G. Hence by Vizing s theorem A G 6 xC G 6 A G 1. Recall that a graph G is said to be class 2 if x G A G 1 or THE ELECTRONIC JOURNAL OF COMBINATORICS 15 2008 R44 1 G1 v x y G2 w x y y G1 G2 w Figure 1 The dot product construction equivalently if xC G A G . Afshani et al. 1 proved that if G is a bridgeless cubic graph then 3 6 x c G 6 11 3. The upper bound is attained by the Petersen graph. No bridgeless cubic graph other than the Petersen graph with .