 Euler's formula relates to the number of vertices v, edges e, and faces f in a planar graph by... This tells us something if we have a planar graph, but if we don't know if a graph is planar, it isn't useful. But remember, consider the contrapositive. The contrapositive tells us that if vertices plus faces is not equal to edges plus 2, then the graph can't be planar. There's just one problem. We can easily determine v and e, but what about f? For example, we might try to determine if k5 is planar. Remember, this is the complete graph on five vertices. Now, we might try to see if we can move the vertices to get a planar graph, but even if we fail to produce a planar graph, that might only mean we're not clever enough to make one. Well, let's think about this. k5 will have five vertices and ten edges, so if it is a planar graph, Euler's formula has to hold. So if we substitute in the number of edges and vertices, and that tells us that we have to have seven faces. But how can we determine how many faces the graph has without drawing it as a planar graph? We might consider a different approach. Since we always know the number of vertices and edges, let's see if we can rewrite Euler's formula without the faces. Now in a planar graph, each edge separates two faces, and it takes at least three edges to define a face. So how can we use this information? Suppose we number the faces of the graph and let the ith face have mi edges. Now if we have at least two faces, then each face will require at least three edges, and since each mi is greater than or equal to three, then the sum has to be greater than or equal to three times the number of faces. Meanwhile, an edge can only be on at most two faces, so each edge can contribute at most two to this sum, and so the sum of the number of edges along each face has to be less than or equal to twice the number of edges. Consequently, three times the number of faces must be less than or equal to twice the number of edges, and so the number of faces is less than or equal to two-thirds the number of edges. And we can use this to rewrite Euler's formula without the number of faces. So substituting it to Euler's formula, we'll replace the number of faces with something greater and simplify. So if g is planar, the number of edges is less than or equal to 3v minus 6. Now since we often don't know the graph as planar, we'll take the counter positive. If e is not less than or equal to 3v minus 6, then g is not planar. Or in other words, if e is greater than 3v minus 6, g is not planar. So let's go back to K5. K5 has five vertices and ten edges. Our theorem has an inequality between the number of edges and vertices, and we check, and we find that it is not planar. How about something like K33, the complete bipartite graph between two sets of three vertices a piece? K33 has six vertices and nine edges, and we check, and since the number of edges is not greater than 3v minus 6, we know the K33 is not necessarily planar. Our theorem only tells us if a graph is not planar. So it's conceivable the K33 is planar, but there doesn't seem to be any way to place the vertices so the edges don't cross. Can we prove it possible? So one other thing to notice is that every face of a planar graph is also a cycle. But since K33 is bipartite, all cycles are even, and what this means is that every face has to use at least four edges. And so if we go back to our earlier analysis, four times the number of faces must be less than or equal to the sum of the edges, and we still have the sum less than or equal to twice the number of edges, which gives us, or the number of faces is less than or equal to half the number of edges. So again, replacing this into Euler's formula gives us, and so if G is a planar graph with E edges and V vertices, if every face has at least four edges, then E is less than or equal to 2v minus 4. Since we generally don't know if the graph is planar, we'll use the contrapositive, let G be a graph with E edges and V vertices where all cycles include at least four edges. If E is greater than 2v minus 4, then G is not planar. So returning to K33, we have nine edges and six vertices, and every cycle has at least four edges. So we check and we find, and so K33 can't be planar. Since neither K5 nor K33 are planar, this gives us a starting point for identifying non-planar graphs. It should be clear that every subgraph of a planar graph is planar. Contrapositively, if G has a non-planar subgraph, G is non-planar. So if K5 or K33 is a subgraph of G, then G is not planar.