a) When m and n are even and positive Km,n will be a Euler circuit. --Jahlborn 09:08, 20 November 2008 (UTC)
I started out by drawing simple examples of bipartite graphs like K(1,1), K(1,2), K(2,2), etc. I got that there is a circuit whenever both m and n are even. Additionally, there is a path whenever |m-n| = 1, or when m=n=1. I don't know if this is right or not, but neither K(3,3) nor K(1,3) have a path, but K(2,4) has a circuit. That's the best i got. --Dakinsey 09:12, 20 November 2008 (UTC)
