 We've determined that k5 and k33 are not planar graphs. Consequently, any graph that includes these as subgraphs can't be planar. So what's the biggest graph we can create that has neither? This is the type of question asked in extremal graph theory. Before we answer this question, we have to decide what do we mean by bigger, more vertices, or more edges? Since edges live on vertices, it's probably easier to start by choosing how many vertices we have and maximizing the number of edges. And in general, a question in extremal graph theory has the form of all graphs with the vertices, which graph has the most? Extremal graph theory emerged from one of the darkest events in history. A populist denagogue blamed all his country's problems on immigrants, foreigners, liberals, and anyone who disagreed with him. Also, he claimed there was a vast conspiracy to discredit him. In 1923, he and his followers attempted a violent overthrow of the government. When that failed, his party ran on a platform of making Germany great again. Germans voted them into power in 1933. For many Germans, it would be the last election they ever voted in. In 1940, the Nazis sent Hungarian mathematician Pal Theron to a forced labour camp for the crime of being different. This was fortunate because Theron avoided becoming one of six million people murdered by the Nazis. While there, Theron kept his sanity by working on math problems. One question Theron worked on. Consider a graph with v vertices. What is the greatest number of edges it can have without having kn as a subgraph? So, remember the more important something is, the more ways we have to talk about it. A subgraph isomorphic to Kn is also called an n-clic, so we could also ask what is the largest graph with v vertices that contains no n-clics? Clearly, we could do this if the number of vertices is less than n, so let's assume we have at least n vertices. To prevent the formation of an n-clic, we need to have vertices not joined by edges. And a useful question. Where have we seen something like this before? And this actually happens in a bipartite graph. No vertices in the same part have an edge between them. And this suggests a generalization. G is a k-partite graph when its vertices can be partitioned into k-partite sets where no edge joins vertices in the same partite set. So now suppose G is a k-partite graph. Any set of k-plus-1 vertices must have two vertices in the same partite set. But no edge can be between such vertices. So if n is greater than or equal to k-plus-1, G cannot have kn as a subgraph. More elegantly, n greater than or equal to k-plus-1 means n strictly greater than k, so we can state our result as a k-partite graph has no subgraph kn for n greater than k. While we started with the question of finding a graph that did not include k5 as a subgraph, our analysis suggests a way to produce a graph without kn for any n. This raises a new question. Suppose a k-partite graph has v vertices. What is the largest number of edges it can have? Clearly, it depends on the distribution of the vertices into the partite sets. Let's investigate. Remember, it's the journey, not the destination. And so if we never answer our first question, we'll still have learned something interesting. So consider a 3-partite-tripartite graph where the partite sets have 5, 3 and 7 vertices. What's the maximum number of edges such a graph can have? Now in higher mathematics, it often helps not to do the arithmetic. So let's think about that. Since an edge could join vertices in different sets, we could have 5 times 3 edges from the first set to the second, 5 times 7 edges from the first set to the third, and 3 times 7 edges from the second set to the third. And so there are edges. And since we didn't do the arithmetic, we might notice that when we computed the number of edges when we had partite sets with 5, 3, and 7 vertices, we found a maximum of 5 times 3 plus 5 times 7 plus 3 times 7 edges. And if you look carefully, we're finding the sum of the products of the number of vertices in every set with the number of vertices in every set that follows it. And so this suggests our general theorem that if we have a k-partite set where the ith-partite set has vi vertices, then the number of edges is the sum of the products of the number of vertices in every set with the vertices in every later numbered set. And so we'll supply a proof. Well, actually, we'll let you do a proof. And remember, once you've proven something, prove it again in as many different ways as you can. There's a simple direct proof, but induction proofs are very important in graph theory, so it doesn't hurt to get a lot of practice constructing them. So if we know how the vertices are distributed among the partite sets, we can compute the number of edges. So suppose we don't know. Suppose we have vi vertices. How should we partition them to get a partite graph with the most edges? So remember, concrete doesn't hurt. Let's experiment. So let's consider all three partite graphs with seven vertices. What partition of the vertices will yield the most edges? To answer this, we'll need to find all the ways we can write seven as the sum of three whole numbers. Without loss of generality, we may assume the sum ends are in decreasing order. Or can we? Yeah, I think we can. Since each part has to have at least one vertex, we can start with partite sets having five, one, and one vertices, giving a maximum of five edges between the first and second set, five edges between the first and third, and one edge between the second and third set for a total of 11 edges. We can also use the partition four to one, giving us a maximum of eight edges between the first and second set, four edges between the first and third, and two edges between the second and third set for a maximum of 14 edges. Repeating this for the other partitions of seven gives us a maximum number of edges in the different possible three partite graphs. And so it seems that we can have as many as 16 edges in a three partite graph with seven vertices.