 The generalization of Euler's formula gives us a relationship between the number of edges, vertices, and genus of a connected graph. So consider the corollaries. If we take our relationship and solve for g, we find… And since g must be a whole number, we'll round this reaction up, and so this gives us the minimum genus of a graph with v vertices and e edges. So earlier we concluded that a graph with e edges can be embedded on a sphere with genus e choose 2. This could be very high, so can we find a lower bound? Well let's consider, suppose an edge crosses one or more other edges. We could actually avoid all intersections simultaneously by putting this edge on a handle. So this means we only need a handle for every edge, giving us the genus is less than or equal to the number of edges. While we might not know the number of edges, the number is limited by the number of vertices, and this gives us a sharper bound. A graph with n vertices can be embedded on a sphere with n choose 2 handles. And equivalently, the maximum genus of a graph with v vertices is… So let's find the bounds of the genus of k5. So k5 has 5 vertices and 10 edges. And so first we can find a lower bound, which will be… And so the genus of k5 is at least 1. Next, we can find an upper bound, which will be v choose 2, and so that will be… And so the genus of k5 is no more than 10. For bipartite graphs we have, and we can prove this in the same way that we did for genus 0 graphs. So let's find bounds of the genus of k3.3. So k3.3 has 6 vertices and 9 edges. Since k3.3 is bipartite, all cycles are even, so all faces have at least 3 sides. And so our formula gives us… And so the genus is greater than or equal to 1. Next, our upper bound v choose 2 will be… And so the genus is less than or equal to 15. At this point, remember Gauss' stake sum, solve a problem any way you can, then find a better solution. So let's think about this. The limit of v choose 2 was based on the possibility that any 2 vertices could be joined by an edge. But in k3.3, the complete bipartite graph, vertices in the bipartite sets can't be adjacent. So in fact, there are only 9 pairs of adjacent vertices, and putting each on its own bridge will give us a genus of 9. And in fact, we can leave one edge on the original surface, so we'd only need to add 8 handles. And so the bound of the genus is less than or equal to 8. Now, since every graph on n vertices is a subgraph of kn, then bounding the genus on kn is a useful result. So let's think about this. Kn will have n vertices, and n choose 2 edges. So we can substitute these into our formula. And let's clean this up a little bit. Let's get rid of that fraction by multiplying numerator and denominator by 2. And expanding, simplifying, factoring, and this gives us a useful result. The genus of kn satisfies. Now with some effort, we can make the inequality an equality. And this gives us an important theorem, first proved by Ringel, Young's, and Meyer in the 1960s, that the genus of kn is exactly. For essentially the same logic, we find the complete bipartite graph kmn must have. And again, with some effort, we can make this an equality. Ringel proved this a few years later that kmn has genus. So let's find the genus of k5 and k33, which are non-planar graphs, for n equals 5 vertices we have. And so the genus of k5 is 1. Since k33 is a bipartite graph we have, and so again, k33 also has genus 1, so both can be embedded on a torus.