 We used our hysterum to create an algorithm to make a graph Hamiltonian, but depending on which add series we added, we obtained different graphs, and we still didn't have a way to decide if G was Hamiltonian to begin with. So a useful strategy for creating mathematics, and crossing the street, look both ways. In our algorithm, we added edges to vertices where the degree sum was less than n. What if we added edges to vertices where the degree sum was at least n? Now why would we do that? Remember, it's the journey, not the destination. We'll try it, see what happens, and see if we can do anything with it. Remember we obtained our hysterum from the contrapositive of, note that our vertices u and v must be non-adjacent since the addition of the edge uv makes the graph Hamiltonian. And so our contrapositive is. And so notice that here we are in this position of adding an edge uv between two vertices whose degree sum is at least n. Intuitively, it's easier to see if a graph with many edges is Hamiltonian. So if we add an edge to G and find it's Hamiltonian, this theorem gives us a way to decide if G was Hamiltonian. So for example, let's say we wanted to decide whether a graph was Hamiltonian. We could look for a Hamiltonian cycle. But if the graph isn't Hamiltonian to begin with, we'd waste a lot of time. So it's best to see if we can determine if the graph is Hamiltonian first. So since the degree of E is 4 and the degree of A is also 4, their sum is at least as great as the number of vertices. So we could add edge Ae. With the added edge, we could find a Hamilton cycle more easily. Or we could add another edge. Since the degree of A is 5 and the degree of F is 3, we could add edge Af. Since the degree of A is now 6 and the degree of D is 2, we could add edge Ad. And at this point, we can get a Hamilton cycle. So now what if we remove the edge as we added? And remember, since our graph with the edge Ad is Hamiltonian, then removing the edge Ad produces another Hamiltonian graph because the degree of sum is at least as great as the number of vertices. But since this graph is Hamiltonian, removing the edge Af produces another Hamiltonian graph because the degree of sum is still at least as great as the number of vertices. But since this graph is Hamiltonian, removing the edge Ae produces yet another Hamiltonian graph, And so we know the original graph was Hamiltonian. This motivates the following definition. Let G be a graph with N vertices. The closure of G is the result of repeatedly joining non-adjacent vertices whose degree sum is at least N until no more edges can be added. So, for example, let's try to find the closure of this graph. There are eight vertices, so if two non-adjacent vertices have degree sum eight or more, we'll join them. We might begin with vertex C, which has degree six. The only non-adjacent vertex is D, but the degree sum is only seven, so we don't join them. Vertex A has degree five. The non-adjacent vertices are... So we can't join AF because the sum of the degrees won't be eight, but we can join AG. So let's do that. What's important to recognize here is that the degree of A has increased, so now A has degree six, and so the degree of A plus the degree of F is at least as great as the number of vertices, and so now we should join AF. Vertex B has degree three, but all non-adjacent vertices all have degree less than five, so it can't be joined to any of them. And a similar argument for E, F, G, and H, and so this is the closure of the graph.