Is it possible to partition k8 into 2 planar subgraphs?
Any help would be appreciated.
yes I understand that. But surely a subgraph G of H is obtained by deleting some vertices and/or edges.
so if it cut k8 in the middle , i delete all the crossing edges and i am left with the complete graph on four vertices which is planar.
Whats wrong with that argument?
thanks.
K_8 has 28 edges k_4 + k_4 has 6+6 12 edges
let the graph k_8 have vertices x1,x2,x3,x4,x5,x6,x7,x8
consider a
First graph k_4 with x1,x2,x3,x4 and the edges from x1,x2,x3 to the vertices x5,x6 and x4 to x7,x8
second graph k_4 with x5,x6,x7,x8 and the edges x4 to the vertices x5,x6 and the edges x1,x2,x3 to x7,x8
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