I had a quite productive day yesterday when I flew from San Francisco to Tokyo, then travelled to Nagano because I read 50 pages of my Combinatorics book during my transit. I found one interesting theorem about planar graph so I’ll state here along with the proof: Euler’s formula: If is a connected planar graph with n (≥ 1) vertices, m edges, and r regions, then .
Proof by induction
Consider the case . In this case, n must be 1 since G is connected. With 1 vertex and no edges, there is exactly one outer region. Thus as desired.
Assume that the statement is true for any connected planar graph with at most k edges. In other words, for any connected planar graph with n vertices , edges, and r regions, . This is the inductive hypothesis.
Suppose is a connected planar graph with n vertices, vertices, and r regions. If G is a tree, then by the property of a tree. Notice that if G is a tree, then the only region is the outer region surround the tree so . Thus as desired.
If G is not a tree, then G has a cycle. Call the cycle C. Let e be any edge in C, and consider a subgraph of G. must be connected because removing an edge from a cycle breaks the cycle but does not break the connectivity of the graph. Furthermore, has one fewer region than G because removing an edge merges exactly two regions into one region. Because a subgraph of a planar graph is also planar and , is a connected planar graph with edges. Thus the inductive hypothesis applies to and yields . Therefore, .
Thus in either case, as desired. Q.E.D.
John M. Harris, J. L. H., & Mossinghoff, M. J. (2000). Combinatorics and Graph Theory. Springer.