 We'd like to find sufficient conditions for a graph to be Hamiltonian. One possibility is to find the contrapositive of a conditional of the form if G is not Hamiltonian then, but how is G not Hamiltonian? A common strategy is to consider extreme cases. In this case, suppose G is almost Hamiltonian. We can interpret this as meaning that adding one edge would make it Hamiltonian. So remember, concrete doesn't hurt. Let's try a specific example and see if we can gain some insights. So suppose adding an edge, UV, would make the graph Hamiltonian, which means there must be some cycle that includes all vertices of our graph. Now consider the graph without this edge. There's still a path from U to V that passes through all vertices. So we'll assume that U is connected to some vertices and V is also connected to some vertices. But if they're connected to adjacent vertices, we can take a shortcut and get a Hamilton circuit. What's a Hamilton circuit, you ask? Remember that every term in graph theory usually has several variants depending on who's using it. And in this case, a circuit is also a cycle. And which term somebody uses is a litmus test of what field they're in, or in my case, what I read most recently. So if the vertices are connected to adjacent vertices, we can get a Hamilton circuit from U to the vertex adjacent to V, the shortcut to V, from V to the vertex adjacent to U, then take the shortcut to U. So if our graph isn't Hamiltonian, this can't happen. Consequently, there must be some limit on how many vertices U and V can be adjacent to. So let's see how many vertices we can join to U and V without producing a Hamiltonian graph. So let's analyze some possibilities. Suppose U is connected to all the other vertices. If V is connected to any other vertex, we'll get a Hamilton circuit. And since, by assumption, we don't actually have a Hamilton circuit, this can't happen. So we note that we have seven vertices, and the degree of U is five. And the only way we can avoid having a Hamilton circuit is to make the degree of V one. So the sum of the degrees of U and V is six. Contrapositively, if the sum of the degrees of U and V is more than six, we'd have a Hamilton cycle. Circuit, whatever we call it. Or would we? So remember, we assume that U was connected to every vertex except V. If U was connected to fewer vertices, how many vertices could V be connected to without creating a Hamilton cycle? So let's try a different case. So suppose U is connected to fewer vertices like this. In the figure, if V is connected to U one, we'd get a Hamilton cycle. So it can't be connected to U one. By the same logic, it can't be connected to U four. But it can be connected to U three. Or U two. But not both. So we note the degree of U is three, and the degree of V is at most two. And so the sum of the degrees is five. So what if U was connected to three other vertices? Then V would create a Hamilton cycle if it was connected to, so V can only connect to U three. In this case, U has degree four, V has degree two, so the sum of the degrees of U and V is six. Let's summarize our observations. We consider to graph G with seven vertices, where the existence of an edge U V would have created a Hamilton cycle, but the omission of the edge makes the graph non-Hamiltonian. We found three examples where the sum of the degrees of U and V was at most one less than the number of vertices. And this suggested conjecture. Suppose G is a non-Hamiltonian graph with N vertices, and suppose adding an edge creates a Hamilton cycle. Then the sum of the degrees is less than or equal to N minus one. Can we prove it? Well, let's find out.