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Five color theorem

Planar maps require at most five colors

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Through Map, Graph (discrete mathematics) and If and only if

The five color theorem is implied by the stronger four color theorem, but is considerably easier to prove. It was first proved by Percy John Heawood in 1890, based on a purported proof of the four color theorem given by Alfred Kempe in 1879. That proof had been considered correct for 11 years, but Heawood found an error in it; he did manage to give a correct proof of the weaker five color theorem, based on Kempe's work. The first correct proof of the four color theorem was found only in 1976.

Outline of the proof by contradiction

First of all, one associates a simple planar graph G to the given map, namely one puts a vertex in each region of the map, then connects two vertices with an edge if and only if the corresponding regions share a common border. The problem is then translated into a graph coloring problem: one has to paint the vertices of the graph so that no edge has endpoints of the same color.

Because G is a simple planar graph, i.e. it may be embedded in the plane without intersecting edges, and it does not have two vertices sharing more than one edge, and it does not have loops, then it can be shown (using the Euler characteristic of the plane) that it must have a vertex shared by at most five edges. (Note: This is the only place where the five-color condition is used in the proof. If this technique is used to prove the four-color theorem, it will fail on this step, since there are some graphs that do not have any vertex shared by at most four edges. A notable counterexample is icosahedral graph: as a 5-regular and planar graph it does not have any vertex shared by at most four edges). Find such a vertex, and call it v.

Now remove v from G. The graph G′ obtained this way has one fewer vertex than G, so we can assume by induction that it can be colored with only five colors. If the coloring did not use all five colors on the five neighboring vertices of v, it can be colored in G with a color not used by the neighbors. So now look at those five vertices v1, v2, v3, v4, v5 that were adjacent to v in cyclic order (which depends on how we write G). So we can assume that v1, v2, v3, v4, v5 are colored with colors 1, 2, 3, 4, 5 respectively.

Now consider the subgraph G1,3 of G′ consisting of the vertices that are colored with colors 1 and 3 only and the edges connecting them. To be clear, each edge connects a color 1 vertex to a color 3 vertex (this is called a Kempe chain). If v1 and v3 lie in different connected components of G1,3, we can swap the 1 and 3 colors on the component containing v1 without affecting the coloring of the rest of G′. This frees color 1 for v completing the task. If on the contrary v1 and v3 lie in the same connected component of G1,3, we can find a path in G1,3 joining them that consists of only color 1 and 3 vertices.

Now turn to the subgraph G2,4 of G′ consisting of the vertices that are colored with colors 2 and 4 only and the edges connecting them, and apply the same arguments as before. Then either we are able to reverse the 2-4 coloration on the subgraph of G2,4 containing v2 and paint v color 2, or we can connect v2 and v4 with a path that consists of only color 2 and 4 vertices. Such a path would intersect the 1-3 colored path we constructed before since v1 through v5 were in cyclic order. This clearly contradicts the planarity of the graph.

So G can in fact be five-colored, contrary to the initial presumption.

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