Is it possible to partition k8 into 2 planar subgraphs?
Any help would be appreciated.
Here is a theorem relating the number of edges to the number of vertices in a planar graph: $\displaystyle V \geqslant 3 \Rightarrow \quad E \leqslant 3V - 6$.
In $\displaystyle K_8$ there are 28 edges. If you partition that graph into two subgraph you get $\displaystyle \left\{ {E_1 ,V_1 } \right\}\,\& \,\left\{ {E_2 ,V_2 } \right\}$ where $\displaystyle \left\{ \begin{gathered} E_1 + E_2 = 28 \hfill \\ V_1 + V_2 = 8 \hfill \\ \end{gathered} \right.$.
Now try for a contradiction.
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
my email id 140580@gmail.com