Euler's Theorem

A graph divides the plane into several regions called faces. Let us denote the number of the faces by F, the number of the vertices by V, the number of the edges by E. On the Figure 3 below

  

Figure 3: Faces

the given graph divides the plane into 4 regions, so we have that F=4,V=6,E=8

Problem 5. Evaluate F, V and E for the complete graph of 5 vertices, i.e. the pentagon and its diagonals.

Answer. F=12,V=5,E=10

A graph that can be drawn in such a way that its edges do not intersect each other (except at their endpoints) is called planar. An example of planar graph is Figure 4 (a), since it can be drawn as Figure 4 (b).

  

Figure 4: Constructing a planar graph

Although these two graphs are different, they have equally many vertices and we can numerate the vertices of both graphs in such a way that any two vertices of the first graph are connected by an edge if and only if the corresponding vertices of the second graph (with the same numbers) are connected by an edge. Such graphs are called isomorphic. Actually we will not distinguish isomorphic graphs. So a graph may be planar even if some of its edges intersect in a given picture (see Figure 4). Try to make another couples of pictures describing planar graphs

Euler's Theorem. For a planar graph the equality V-E+F=2 always holds true.

Remark. Euler's Theorem is an example of topological obstruction. Problem 5 shows that the requirement for a graph to be a planar is important. An interesting problem is to show that a complete graph is planar if and only if V<5

Proof of Euler's Theorem. For the proof we need to introduce a class of special graphs. A tree G is a connected graph without cycles, i.e. any two vertices of G can be connected by a path (a sequences of edges) and there is no path whose starting and ending vertices coincide. On Figure 5 you can see an example of a tree (a), and a graph which is not a tree (b).

  

Figure 5: tree and non-tree

Obviously the trees are simplier then the other graphs. So our goal is to delete some edges until we get a tree, keeping the graph connected. Figure 6 shows how after deleting one edge the quantity V-E+F does nor change. Really, before the operation we have F=5,V=5,E=8 and after that F=4,V=5,E=7. So in both cases V-E+F=2. Since for trees we have that V-E=1 (prove it!) and F=1 the theorem is proved.

  

Figure 6: Deleting an edge

Problem 6. There are 7 lakes in Lakeland. They are connected by 10 canals so that one can swim through the canals from any lake to any other. Haw many islands are there in Lakeland?

Hint. Consider the planar graph whose vertices are the lakes, whose edges are the canals, and whose faces are the islands. So F=5. Answer: 4