(New page: For homeomorphic graphs one can disregard vertices of degree 2 (forget the vertex was ever there, but keep the edges). If you do that here, you get a graph with 4 vertices only: a,...) |
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We clearly see that it cannot be homeomorphic to K(3,3) as we need atleast 3 edges from each verted connecting to the other set | We clearly see that it cannot be homeomorphic to K(3,3) as we need atleast 3 edges from each verted connecting to the other set | ||
of vertices. | of vertices. | ||
Latest revision as of 06:13, 7 December 2008
We clearly see that it cannot be homeomorphic to K(3,3) as we need atleast 3 edges from each verted connecting to the other set
of vertices.
