 Consider two roads. If they cross, then either everyone takes their chances of intersection, or traffic in one direction has to stop so traffic can pass in the other direction, or you can build a bridge or tunnel to allow continuous traffic flow. But bridges and tunnels are expensive, so if you don't have to build them, you might not want to. Since not all graphs are planar, we will need to build some bridges. But how many? Let's think about this. Suppose two edges hit a graph cross. While the edges must stay on the surface, we can add more surface. So we say that we add a handle. Remember that any graph on a plane can be embedded on a sphere. So if we add a handle to a plane, it's like adding a handle to a sphere. And consequently, we say that any graph can be embedded on a sphere with enough handles. So how many handles do we need? Suppose a graph has E edges. Since any two edges can cross, there are E choose two potential crossings. So if we want to eliminate those, then we could incorporate E choose two handles. And consequently, a graph with E edges can be embedded on a sphere with E choose two handles. And remember, the important question to ask is, can we do better? And in fact, we can on your homework. While we can find an upper bound, a particular graph might need far fewer handles. And so we define the genus of a graph is the least value k for which g can be embedded on a k-handled sphere. Now for simple graphs that can be embedded in the plane, which is to say on a sphere with zero handles, we must have Euler's formula V plus F equals E plus two. But what if a graph could be embedded on a k-handled sphere? Granted, this will need to find a suitable modification of Euler's formula. Our proof of Euler's formula relied on the result that V plus F minus E is constant for a planar graph, even if we remove an edge that isn't a bridge. But nothing in the proof required our graph E planar. It only required that the removed edge did not separate two parts of the same face. So if a graph is embedded on a k-handled sphere, V plus F minus E is still constant as long as we adhere to the requirement. Now on a plane, we were able to reduce the graph to a tree with a specific relationship between the edges, vertices and faces, and this allowed us to compute V plus F minus E directly. But if we're not on a plane, we can't reduce our graph to a tree since then we'd be producing a graph that could be embedded on the plane, and V plus F minus E would be equal to two in that case. Fortunately, the part of the proof that establishes the constancy of V plus F minus E does not change. So this means that for any genus k-graph, V plus F minus E is some constant, and to find it, we can take a genus k-graph and find V, F and E. In general, when we add handles to a sphere, we don't care about the size or shape of the handles, or for that matter, the size and shape of the sphere. So a sphere with one handle added is also a torus. So let's go back to the utility's problem. Remember a torus, a one-handled sphere, is a rectangle with opposite edges identified. A graph that doesn't cross opposite edges is not really on a torus, it's on a plane. And so we know that k-33 could not be embedded on a sphere, but could be embedded on a torus. So we can count the edges and faces of k-33 on a torus. There's six vertices and nine edges. And the number of faces is, well, let's see. What we'll want to do is we want to identify where we can wander without crossing an edge. So let's start over here, and we'll indicate our first face by where we can wander without crossing over an edge. Add crossover, and again, and again. So these four parts are really all part of the same face. Now for the middle section, because we could go from top to bottom, there's one face. And in the last section, we have one more face. And so there's three total faces. How do we find v plus f minus e is? And consequently, suppose a graph has genus 1, then v plus f equals e.