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Chapter 11: Problem 2

Show that when any edge is removed from \(K_{5}\), the resulting subgraph is planar. Is this true for the graph \(K_{3,3}\) ?

Short Answer

Expert verified
When any edge is removed from \(K_{5}\), the resulting subgraph is indeed planar, it can be drawn in a plane without edges crossing. However, this is not the case with the \(K_{3,3}\) graph. Even after removing an edge from \(K_{3,3}\), the graph remains non-planar.

Step by step solution

01

Create a Subgraph of \(K_{5}\) by Removing an Edge

Choose any edge of the \(K_{5}\) graph and remove it. As a result get a \(K_{5}\) graph minus one edge, which comprises 5 vertices and 9 edges.
02

Draw the Subgraph of \(K_{5}\) in the Plane

Now, try drawing the resulting graph in the plane such that none of the edges intersect except at the vertices. It is possible to do so because removing an edge destroys the property of \(K_{5}\) that makes it non-planar, hence the graph becomes planar.
03

Create a Subgraph of \(K_{3,3}\) by Removing an Edge

Choose any edge of the \(K_{3,3}\) graph and remove it. Get a \(K_{3,3}\) graph minus one edge, which comprises 6 vertices and 8 edges.
04

Draw the Subgraph of \(K_{3,3}\) in the Plane

Try to draw the resulting graph in the plane such that none of the edges intersect except at the vertices. This, however, is impossible. Despite the fact that we removed an edge, the subgraph of \(K_{3,3}\) without an edge is still non-planar, as it still contains a homeomorphic copy of \(K_{3,3}\).

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