 A coloring of a graph is an assignment of colors to the vertices so that adjacent vertices have different colors. For example, we could color k4 using the colors puce, taupe, mauve, and umber. However, since no one knows what those colors look like, we'd probably use red, blue, black, green. But since some of us have varying degrees of color blindness, including me, we could also use numbers. Suppose we can assign each vertex of a graph a color, a number from one to k, so that no two adjacent vertices have the same color. We say that the graph is k-colorable. Now we might be able to use fewer colors. The chromatic number of a graph, chi of g, is the minimum number of colors required. Also, so we don't have to keep saying color, which is to say number, we'll just say color from now on with the understanding that it's just a label. Before trying to find the chromatic number of different graphs, let's see if we can say anything about graphs with specific chromatic numbers. First, if the chromatic number of a graph is one, in other words, we could color every vertex the same color, about black, then the graph has to be totally disconnected. If any vertex had a neighbor, we'd need a second color. So we know that if the chromatic number of a graph is one, then the graph must be totally disconnected. And again, when learning mathematics, it's useful to consider the converse. In this case, if g is totally disconnected, will the chromatic number be one? The answer to that is a homework problem. At the other extreme, let's consider kn, the complete graph on n vertices where every vertex is joined to every other vertex. Since every vertex is adjacent to every other vertex, every vertex must be a different color. So our chromatic number is n. And so we have the chromatic number of the complete graph on n vertices is n. Now there's a couple of different converses of this theorem. For example, if the chromatic number of a graph is n, it must be kn, which probably isn't true. But if we wanted to find a converse worth considering, we might consider graphs with n vertices where the chromatic number is n, and then ask whether that graph must be the complete graph. And the answer to that is the second problem on the homework. Now it's important to distinguish between whether a graph is k-colorable and what its chromatic number is. To show a graph is k-colorable, we can produce a coloration using k-colors. So any graph with n vertices is n-colorable because we can give each vertex its own unique color. But to find the chromatic number, we need to show that fewer colors are not enough. Given an arbitrary graph, this is sometimes hard to do, but most of the time it's extremely hard to do. So let's try to find the chromatic number of C6, the cycle graph on six vertices. C6 is the graph with six vertices joined to form a single cycle. Consider any vertex. We'll assign it a color, how about remi? Or since no one knows what that looks like, we'll use the label r. The adjacent vertices have to be assigned a different color, how about burgundy? We'll use b. Since adjacent vertices have to have different colors, let's try to assign the vertices adjacent to those the same color as the first vertex, r, and this last vertex can be assigned to color b, and this gives us a coloration using two colors. So C6 is two-colorable. But remember, just because it's two-colorable doesn't mean the chromatic number is two. And so the question you've got to ask yourself is, can we use fewer colors? Since we know it's two-colorable, because we have a two-coloration, then we only have to consider the possibility it's one-colorable. But that would require our graph to be completely disconnected, which it isn't. So we know the chromatic number isn't one, so the chromatic number must be two. Let's try to find the chromatic number of C7, so the cycle with seven vertices. We can try the same strategy of alternating colors, but we'll need to introduce a third color. How about tau? So C7 is three-colorable, but again this isn't necessarily the chromatic number. Could it be two-colorable? We can try, but not succeed. But remember, just because you can't solve a problem doesn't mean it's unsolvable. So let's think about this a little bit more. Here's a graph is K-colorable. We can partition the vertices into sets v1, v2, and so on, where vi contains all vertices colored i. And because we started with the K-coloring, no edge joins two vertices in the same set. And this motivates, again, our definition of a K-partite graph, a graph is K-partite if its vertices can be partitioned into sets where no edge joins vertices in the same set. And for a K equals 2, we have a bipartite graph, which is also known as a bi-graph. And we do know something about these. A graph is bipartite if and only if all cycles are even. So let's go back to the chromatic number of C9. We can produce a three-coloring, so we know that the chromatic number is less than or equal to three. Now if the chromatic number is 2, then we can partition the vertices of C9 into two sets, v1 and v2, where no vertex joins another vertex in the same set. So C9 would be bipartite, but bipartite graphs can't have odd cycles, and C9 is an odd cycle, so it's impossible. Remember, that's what we actually wanted to prove. Since the chromatic number of C9 is less than or equal to three, and it's not two, then it must be equal to three. Unless it's equal to one. Nah, it can't be. The same proof can work for Cn for any odd value n, giving us the chromatic number of Cn is three for all odd n. We also know that trees are bipartite graphs. Consequently, if a graph is a tree, its chromatic number is two. And so now we know several theorems about the chromatic number for different types of graphs. Unfortunately, that exhausts the easy results. In general, finding the chromatic number for an arbitrary graph is challenging. So we'll tackle that next.