Skip to main content

Timeline for On planarity in two related graphs

Current License: CC BY-SA 3.0

11 events
when toggle format what by license comment
Jan 17, 2012 at 12:20 comment added Etsch Yes, sorry my fault. I reconstructed the graph now and indeed $\mathcal{G}_A$ is nonplanar, but $\mathcal{G}_B$ is. It seems that the paper i found is wrong.
Jan 17, 2012 at 7:26 comment added David Eppstein The matrix A in your SAGE code doesn't subdivide the edges of a hexagon in $K_{3,3}$ as I said to do, does it? If you did you should have a matrix of dimension 12, not 6.
Jan 16, 2012 at 21:18 comment added Etsch Its looks a bit strange, but the title of the paper is "Research on the Non-planarity about the Tensor Product of Graphs". If i test this using SAGE: A = matrix(ZZ,6,[0,0,0,1,1,1, 0,0,0,1,1,1, 0,0,0,1,1,1, 1,1,1,0,0,0, 1,1,1,0,0,0, 1,1,1,0,0,0]); G = Graph(A); BG = BipartiteGraph(A); print "planar(G): ",G.is_planar(); print "planar(BG): ",BG.is_planar(); i get two times "false" as the answer.
Jan 16, 2012 at 17:04 comment added David Eppstein I'm pretty sure the example in my last sentence is correct. Maybe the paper you found is wrong?
Jan 16, 2012 at 13:54 comment added Etsch But i am not sure about your last sentence. Meanwhile is found a paper which states that the bipartite double cover operation preserves non-planarity but does not preserve planarity. So the bipartite double cover of $K_{3,3}$ must be (and indeed it is) non-planar. Or do is missunderstand your sentence?
Jan 16, 2012 at 9:45 comment added Etsch Thanks David for your nice answer. It gives a good starting point for new ideas to work with.
Jan 15, 2012 at 7:17 history edited David Eppstein CC BY-SA 3.0
fix typo
Jan 15, 2012 at 7:04 history edited David Eppstein CC BY-SA 3.0
more examples
Jan 14, 2012 at 22:37 history edited David Eppstein CC BY-SA 3.0
added 337 characters in body
Jan 14, 2012 at 21:46 history edited David Eppstein CC BY-SA 3.0
added 49 characters in body
Jan 14, 2012 at 21:15 history answered David Eppstein CC BY-SA 3.0