 If G is a connected graph with genus G, we found the minimum degree satisfied the inequality. Consequently, for most graphs, the minimum degree is less than or equal to 6. For the genus zero graphs, this guarantees the existence of a vertex of degree 5 or less in a planar graph, and by induction we were able to prove that all planar graphs are six-colorable. Can we extend this approach to graphs of greater genus? We hope so, since that's what the rest of the video is about. So first, let's consider genus one graphs. If G equals one, then our minimum degree will satisfy, and so the minimum degree is less than or equal to six. So we might need seven colors. Now clearly, if G has up to seven vertices, then a seven coloring exists, which, in our induction proof, establishes the base case. So I suppose all genus one graphs with K or fewer vertices are seven-colorable. Let G be a genus one graph with K plus one vertices. It has to have a vertex of degree 6 or less, so if we remove that vertex, we have K vertices, so a seven color exists, and since V is only adjacent to six other vertices, restoring it will require at worst the seven color. This proves the induction set, and so all graphs of genus one are seven-colorable. And with a little effort, we can actually produce a seven-colorable genus one graph. What about genus two graphs? If G equals two, we have... Now this does depend on the number of vertices, but if we have enough vertices, say, more than seven, then this quantity is less than or equal to seven, and so our minimum degree is less than or equal to seven for most genus two graphs. So any graph on a two-handled sphere has a vertex with degree at most seven, and all graphs with up to eight vertices are eight-colorable, which establishes our base case. So if all graphs with N equals K vertices are eight-colorable, then a graph with K plus one vertices has a vertex with degree at most seven. The graph with that vertex removed has K vertices, so it has an eight-coloring. And restoring the vertex gives us a point with at most seven neighbors, so we can use the eighth color for the point. And that completes our induction proof, and so we know the genus two graphs are eight-colorable. While we could repeat this for genus three, four, five, and so on graphs, let's try to find a general result. Using our proof, we note that we first found the minimum degree of the graph. We took C to be one more than the minimum degree. For graphs with C or fewer vertices, there's a C coloring, and that's our base case. Then we could use induction. So our statement is true for our base case. We'll assume the statement is true for all graphs with up to K vertices. A graph with K plus one vertices must have a vertex of degree C minus one, so its neighbors only use C minus one colors. If we remove it, we get a C colorable graph, and when we restore it, we can use the C color. So what's C? Our goal is to find an upper bound on C, so we'll assume our minimum degree is as large as possible, and see where that takes us. Since we'll eventually be making C equal to one more than the minimum degree, we'll start by assuming that C minus one is the minimum degree. So remember the greatest that the minimum degree could be is six plus 12G minus 12 divided by V. But since we'll have a C coloring if we have C or fewer vertices, we only care about the case where V is greater than C. So for V greater than C, the greatest the minimum degree can be is. And so the greatest C minus one could be is, and this gives us an equation in C. Solving this equation gives us, and while the quadratic equation generally gives us two solutions, we can disregard the negative solution in this case. Now, generally speaking, this will not be a whole number, so we need to round it. Since we took the greatest possible value for the minimum degree, this means the solution corresponds to the greatest possible real value for C. So the greatest possible whole number will be found by rounding down. This suggests that any connected graph with G minus G has a C coloring where C is given by, if we let G equals zero, we get, which proves the four-color theorem. Or does it? So certainly if G equals zero, C is equal to four. So we know that any graph with up to four vertices is four-colorable. So that gives us our base case. So now we'll suppose that every graph with up to n equals k vertices is four-colorable. Let G have k plus one vertices. Now, if G equals zero, the minimum degree of a vertex must be. So the minimum degree is less than or equal to five. And so we can uh-oh. The problem is that after we've removed the vertex, even if the remaining graph has a four-coloring, since the vertex has five neighbors, we might require six colors. And so the problem is our induction fails if the genus of the graph is zero. In fact, we might worry that our induction step fails for other values of G. And unfortunately, it doesn't. This is unfortunate because if it did, then someone could solve it for fame and fortune, or at least a publication in the graph theory journal. Since our bound is valid for G greater than zero, we get what's known as Hewitt's theorem. Any genus G graph with G greater than or equal to one is C-colorable, where C is this expression. While we could find Hewitt's theorem to find an upper bound for the number of colors we require, we can also use it in the other direction. So what is the minimum genus graph that could require a 10-coloring? So we can substitute into Hewitt's formula and find... Now since the genus has to be a whole number, the genus is either three or four. To decide, we can invoke a complicated inequality argument. But remember, concrete doesn't hurt. We could compute C if G equals three, and we find... Then our graph has a nine-coloring, so G equals four is the least genus for which a ten-coloring might be required.