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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. | 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. | ||
--[[User:Dakinsey|Dakinsey]] 09:12, 20 November 2008 (UTC) | --[[User:Dakinsey|Dakinsey]] 09:12, 20 November 2008 (UTC) | ||
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| + | b) I said that there is a Euler path whenever m=2 and n= odd or when m=odd and n=2. This way, there will always be two vertices with odd degree and the others will have degree 2 (which is even) as we want. --[[User:Kduhon|Kduhon]] 09:45, 21 November 2008 (UTC) | ||
Latest revision as of 04:45, 21 November 2008
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)
b) I said that there is a Euler path whenever m=2 and n= odd or when m=odd and n=2. This way, there will always be two vertices with odd degree and the others will have degree 2 (which is even) as we want. --Kduhon 09:45, 21 November 2008 (UTC)
