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What is the minimum number of edges which must be removed from a complete bipartite graph of six nodes K(6) so that the remaining graph is a planar?

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It is $K_{3, 3}$ because other combinations result in planar graphs. To regain planarity, we need to remove only 1 edge.

Non-planarity is introduced when we try to connect two vertices which are otherwise unreachable(i.e., require intersection with other edges) with respect to each other.

Lets say we start with $K_{2, 3}$ which is planar. Lets name the partitions $A$(with 2 vertices) and $B$. One can observe that there is one(and only one) vertex(in B) say $h$ which is now hidden in the sense that it cannot be reached from any vertex in $A$. Now trying to add another vertex $n$ in A and connecting it with all vertices of B,  will require connecting $n$ with $h$, which is possible only through intersection with other edge(s) and hence the introduction of non-planarity.

My reasoning is informal and possibly sloppy, Would anyone please provide rigorous proof/explanation for this problem.

by Active (1.8k points)
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any graph which has 8 or less edge  is planar...so remove one edge
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As per question, input is a non planar complete bipartite graph with 6 vertices which I think can only be $K_{3,3}$.

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yes k33 has 9 edges.and we know that any graph less than 8 edge is planar..so remove 1 edge
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OK...you mean removing 1 edge is the answer....I thought you are asking me to remove one edge from my answer making answer to the original question as 2.
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yes