 Earlier we proved that any planar graph can be vertex colored using at most six colors. But it seems we can't find a planar graph that actually requires six colors. So perhaps we're not clever enough to find such a graph, or perhaps we only need five colors. Well, we're definitely clever enough because we're on the internet, so it must be that we only require five colors. A statement we suspect to be true, but it doesn't yet have a proof is called a conjecture. And graph theory has a number of conjectures. In this case we may conjecture that five colors are enough to color any planar graph. Let's try to prove it. As with the six-color theorem, let's try an induction proof. Suppose every planar graph with K vertices has a five coloring. Let G have K plus one vertices. Now we know it must have at least one vertex of degree five or less. We note that if we remove that vertex, the graph G minus V has K vertices, so by our induction assumption it has a five coloring. So what happens when we put that vertex back? We'll dispense with the easy cases first. If the degree of V is four or less, then we'll only need five colors. So that's because when we remove the vertex we only need four colors to color the adjacent vertices, and so when we restore the vertex we can assign it the fifth color. So we'll assume V has degree five with adjacent vertices V1 through V5. Now we know the graph minus that vertex has a five coloring. Remember that's our induction hypothesis. And if any two adjacent vertices have the same color, for example if V2 and V4 end up both being colored red, then there is a fifth color available for V. So we'll assume that the adjacent vertices all have different colors. In other words, after we've removed that degree five vertex and colored the remaining graph, we have to assign different colors to the vertices V1 through V5. So an important observation, any path from V1 to V3 must cross every path from V2 to V4. So there are two possibilities. Either there are no paths from V2 to V4. In other words, the removed vertex was actually a cut point, or there are paths from V2 to V4 that cut through paths from V1 to V3. Let's consider these cases separately. So if there are no paths from V2 to V4, then in G minus V, V2 and V4 are in separate components. We'll call them G2 and G4. But that means the coloring in G2 is independent of the coloring in G4. So we can swap the colors around so that V2 and V4 have the same color. Consequently, the vertices adjacent to V only need four colors. So when we restore V, it can be given a fifth color. On the other hand, suppose there are paths from V2 to V4. Necessarily any such path has to cut through the path between V1 and V3. So let's consider the subgraph induced by the vertices having the same color as V1 or V3. We'll just cover up the vertices with different colors. And there are two possibilities. Now V1 and V3 are in separate components. So remember the covered up vertices have different colors than these two. So as long as these vertices are either color, we still have a coloring of the graph. So V1 and V3 are in separate components. We can create the colors so that V1 and V3 now have the same color. So the vertices adjacent to V only require four colors. And again, when we restore V, it can be assigned a fifth color. But if there's a path from V1 to V3 in the induced subgraph, it must alternate between the two colors and have a common vertex with any path from V2 to V4. Consequently any path from V2 to V4 must have a vertex whose color is the same as that of either V1 or V3. So now consider the subgraph of G minus V induced by vertices having the same color as V2 or V4. Again we'll just cover up the vertices that have different colors. Since every path between V2 and V4 includes at least one vertex with the same color as V1 or V3, then V2 and V4 are in separate components in this subgraph. So again we can permute the colors so V2 and V4 have the same colors. You know the rest. Consequently, regardless of the initial coloring of the five adjacent vertices, we can always find a coloring where two vertices have the same color. So when we restore V, we can assign it a fifth color and get a five-color graph proving any later graph is five-colorable. Now once you're be careful with this, remember that if a graph is n-colorable, we can color its vertices with n colors so that adjacent vertices have different colors. But if we have an actual map, we don't need to worry about the vertex colors. Instead, we want to assign colors to the faces so that adjacent regions have different colors. This leads to the four-color conjecture. The faces of every planar graph can be colored using four colors. And the thing to realize is that the four-color conjecture and the five-color theorem are actually about different colorations. One is a coloration of faces and the other is a coloration of the vertices. The four-color conjecture was proven in 1976 by Kenneth Appel and Wolfgang Hocken. We'll present a proof. Just kidding, the full proof is a few hundred pages and relies on a computer analysis of hundreds of unavoidable configurations. It was, in fact, the first proof done largely by a computer. But while we can't reasonably give the complete proof, we'll outline how the proof was found. Well, eventually, watch this space.