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Actually, two sets of vertices should be {a,e,g} and {b,f,h}. It's K_3,3 after all :). | Actually, two sets of vertices should be {a,e,g} and {b,f,h}. It's K_3,3 after all :). | ||
--[[User:Asuleime|Asuleime]] 19:24, 1 December 2008 (UTC) | --[[User:Asuleime|Asuleime]] 19:24, 1 December 2008 (UTC) | ||
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| + | You're right, it's homeomorphic to K_3,3 so we just use Theorem 2 to prove it's not planar. | ||
| + | --[[User:Tsnowdon|Tsnowdon]] 01:26, 4 December 2008 (UTC) | ||
Latest revision as of 21:26, 3 December 2008
From what I know, this should not be a planar graph because it contains a subgraph homeomorphic ot K3,3. With (g,h) and (c,d) as the pairs. Correct me if I am wrong.
-ngw
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Actually, two sets of vertices should be {a,e,g} and {b,f,h}. It's K_3,3 after all :). --Asuleime 19:24, 1 December 2008 (UTC)
You're right, it's homeomorphic to K_3,3 so we just use Theorem 2 to prove it's not planar.
--Tsnowdon 01:26, 4 December 2008 (UTC)
