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| − | Yep, that's right. Alternatively, you could consider subgraphs of the given graphs that consist of vertices of degree 4 and the edges connecting them. If the graphs are isomorphic, these subgraphs must be isomorphic too. However, these subgraphs are not isomorphic (draw them and check), hence, the graphs are not isomorphic | + | Yep, that's right. Alternatively, you could consider subgraphs of the given graphs that consist of vertices of degree 4 and the edges connecting them. If the graphs are isomorphic, these subgraphs must be isomorphic too. However, these subgraphs are not isomorphic (draw them and check), hence, the graphs are also not isomorphic.<br> |
--[[User:Asuleime|Asuleime]] 23:20, 10 November 2008 (UTC) | --[[User:Asuleime|Asuleime]] 23:20, 10 November 2008 (UTC) | ||
Latest revision as of 18:20, 10 November 2008
- I am not sure about this one. I think this is not isomorphic because the vertex with degree 4 is adjacent to another vertex with degree 4 while the other one is connected through a vertex with degree 3. Is this correct?
-Wooi-Chen Ng
Yep, that's right. Alternatively, you could consider subgraphs of the given graphs that consist of vertices of degree 4 and the edges connecting them. If the graphs are isomorphic, these subgraphs must be isomorphic too. However, these subgraphs are not isomorphic (draw them and check), hence, the graphs are also not isomorphic.
--Asuleime 23:20, 10 November 2008 (UTC)
