# Induction on Connected Graphs

• Apr 25th 2010, 11:53 AM
geometrywiz
Induction on Connected Graphs
If anyone could give any suggestions or help me solve this it would be greatly appreciated!

Problem Use induction to prove the following theorem. For the purposes of this question, a graph is a finite set of points in the plane, some of which are connected by straight line segments, none of which intersect. A
graph is connected if you can get from any point of the graph to any other by going on the line segments.

Theorem. Suppose G is a connected graph. Then
vG + rG = eG + 2 (1)
where
 vG is the number of points.
 rG is the number of regions (counting the in nite outside area as one region).
 eG is the number of line segments.

The proof may use any of the following facts:
Fact. If G is any connected graph, then at least one of the following is true:
1. G has only one point (and so has no lines and one region).
2. G has at least one edge with one free end, so that removing that edge and the free point will still leave
a connected graph (with the same regions as before).
3. G has an edge which can be deleted, without removing any points, leaving a connected graph and merging
two regions of the original graph into one region.

Thanks again.
• Apr 25th 2010, 12:37 PM
Plato
Quote:

Originally Posted by geometrywiz
If anyone could give any suggestions or help me solve this it would be greatly appreciated!

Problem Use induction to prove the following theorem. For the purposes of this question, a graph is a finite set of points in the plane, some of which are connected by straight line segments, none of which intersect. A
graph is connected if you can get from any point of the graph to any other by going on the line segments.

Theorem. Suppose G is a connected graph. Then
vG + rG = eG + 2 (1)
where
 vG is the number of points.
 rG is the number of regions (counting the in nite outside area as one region).
 eG is the number of line segments.

That is Euler's Formula for faces & edges of a planar graph.
That webpage has a complete disscussion.