No, you're wrong. I agree that my graph is homeomorphic to itself - but it doesn't contain K_5 as a subgraph! (Note that the graph doesn't contain the edge 4-5.)ReiVaX18 wrote:It's homeomorphic to itself, that contains the subgraph K_5, so the problem statement says it's not planar.misof wrote: ...
An example on where's the difference: Let N=9. Make a K_5 on vertices 1,2,3,4,5. Delete the edge 4-5. Add edges 4-6, 5-6. Make a K_4 on vertices 6,7,8,9. This graph cannot be reduced and thus it is NOT homeomorphic to any graph containing K_5 or K_{3,3}. But its subgraph (induced by vertices 1,2,3,4,5,6) IS homeomorphic to K_5 and THEREFORE this graph is NOT planar. (The problem statement says that it is.)
Does anybody know whether the test data was changed?
As the correct version of Kuratowski's theorem states, my graph isn't planar, because it contains K_5 as a minor -- in other words, it contains a subgraph homeomorphic to K_5.