 In general, finding the chromatic number of an arbitrary graph is difficult. However, we can try to find upper and lower bounds on the number. Clearly, if the graph has n vertices, the chromatic number must be less than or equal to n. This gives us an upper bound. Can we find a lower bound? Suppose we can find a k coloring of g. Then g is k part height with part vi consisting of all vertices with color i. Note that the subgraph induced by vi is totally disconnected. And this suggests the following definition. A set of points in g is independent if the induced subgraph is totally disconnected. The point independence number is the size of the largest independent set in our graph. So let's try to find an independent set. So again, what we're looking for is some set of vertices that are not connected. We'll talk a little bit later about how you might find independent sets in a graph. But for now, let's set some boundaries. Suppose a graph with n vertices has chromatic number chi g and independence number beta zero. The n vertices of the graph can be partitioned into chi g independent sets. And since each set can have at most beta zero vertices, then the number of vertices must be less than the product of the independence number and the chromatic number. And consequently, we have a lower bound for the chromatic number. So now suppose s is the independent set of beta zero points. Since a subgraph can't require more colors than the graph, the chromatic number of g minus s must be less than or equal to the chromatic number of g itself. And if we give the vertices of s a color different from any in the subgraph g minus s, we'll obtain a coloring of g. And so the chromatic number of g minus s plus one is greater than or equal to the chromatic number of the graph itself. And consequently, but since g minus s has n minus beta zero points, we know the chromatic number must be less than or equal to the number of points. And so, and this gives us an upper bound on a chromatic number. For example, let's find a bound on the chromatic number. So we've already found an independent set with six vertices. So we know the independence number has to be greater than or equal to six. Unfortunately, this place is an upper bound on our lower bound, which won't do us any good. Even if the table and the vase are both higher than the floor, you don't know if the table is higher than the vase or otherwise. On the other hand, our upper bound will be, which gives us an upper bound on the chromatic number. Intuitively, we'd expect the largest independent set to be only a small fraction of the total vertices. So in general, our upper bound will generally be very high. And so can we find a better upper bound? If you will. So let's think about that. Suppose we have a k coloring of a graph. Why would we need k colors? Since we want adjacent vertices to have different colors, one reason might be that there is a vertex with k minus one neighbors. If every neighbor was a different color, the vertex itself would have to be the k-th color. And this suggests that for any graph g, the chromatic number must be less than or equal to the greatest degree plus one. Well, let's prove that. First, we note that the chromatic number of the complete graph on n vertices is in fact the greatest degree plus one. So all we need to prove is that the chromatic number can't be even greater than this amount. So suppose we have a k coloring of g. Consider any vertex with color i. If j is not a color of an adjacent vertex, we can switch the color of v from i to j and still have a coloring. This is something that you should actually prove. So now suppose k is greater than the greatest degree plus one. Consider any vertex colored k. This vertex has at most delta g neighbors. So we can replace color k with some other color with i less than or equal to delta g plus one. Now, in case that's not obvious, remember, concrete never hurts. For example, suppose we have an eight coloring on g where the greatest degree is five and some vertex v uses color eight. Since the greatest degree is five, v has at most five neighbors using at most five different colors. So we can switch color eight to one of the first six colors. You should prove this. What this means is that we can always swap out that k-th color. So if we do this for all vertices colored k, we no longer need the k-th color and we obtain a k minus one coloring. Now, lather, rinse, repeat until k is not greater than delta g plus one. Consequently, every graph has a delta g plus one coloring. While that wasn't what we wanted to prove, remember it's the journey, not the destination, and we did find something useful. Now, to finish our journey, we note that if a graph has a k coloring, then the chromatic number has to be less than or equal to k. And so there's our theorem that any graph g has a chromatic number less than or equal to the greatest degree plus one.