 Suppose g is a non-Hamiltonian graph with n vertices where adding an edge uv produced a Hamilton circuit. It appeared that in such cases the sum of the degrees of u and v was less than or equal to n minus 1. Let's try to prove this. So suppose g with edge uv would have a Hamilton circuit but g does not. Then there is a path joining u and v. Now suppose the degree of u is k. What can we say about the degree of v? So notice that if v is adjacent to any vertex left of a neighbor of u, we'll be able to complete a Hamilton circuit. This means there are at least k vertices v cannot be adjacent to. So the degree of v is at most n minus 1 minus k. And so the degree of u plus the degree of v is less than or equal to n minus 1. And this proves our theorem. Since our theorem begins with a non-Hamiltonian graph, the contrapositive of a conditional could give us a sufficient condition for a graph to be Hamiltonian. But which conditional? So let's consider our theorem. Our antecedent can be broken into three components. g is a non-Hamiltonian graph with n vertices, and adding an edge creates a Hamilton cycle. And finally we note our traditional disregard for actually counting. Anyway, since we want to conclude that g is Hamiltonian, then we need to make this first part our antecedent, and the other part has to be part of our main premise. And so we need the contrapositive of, which will be, while the contrapositive is true, it's useless. The problem is in order to use it we need to know that g plus the edge uv is Hamiltonian, but that means we need a way to decide if a graph is Hamiltonian. So let's consider a different tactic. So we assume that adding a specific edge uv would create a Hamilton cycle. So let's suppose that adding any edge would create a Hamilton cycle. In other words, let's assume that g is a maximal non-Hamiltonian graph. In that case our degree sum is less than or equal to n minus one for all vertices u and v. So we'll consider the contrapositive of, which will be, again the contrapositive is automatically true and does not require a separate proof. However, it might not be obvious why this is an improvement, so let's consider how g could not be a maximal non-Hamiltonian graph. So if it's not a maximal non-Hamiltonian graph, this means we could add an edge and it still wouldn't be Hamiltonian. But remember that adding any edge to g makes it Hamiltonian, so the only way g couldn't be a maximal non-Hamiltonian graph is if g is already Hamiltonian. Now it still seems we need to know if g plus uv is Hamiltonian. Can we avoid that? Let's consider. Suppose for some graph g it's already true that the vertex i is greater than or equal to n for all vertices u and v. And suppose g is not already Hamiltonian. We can add edges until g' is a maximal non-Hamiltonian graph and then when we add the next edge we get a Hamiltonian graph. But that means that we can use our theorem, which will tell us that g' is Hamiltonian as well, which is a contradiction, so our original graph must be Hamiltonian. And this gives us an important result. Apologies to all Niska fans out there. Urus theorem. Named after Euston Ura, a 20th century Norwegian mathematician. Suppose g is a graph with n vertices. If the degree sum of any two vertices is greater than or equal to n, then g is Hamiltonian. And as a quick example of how we might use this, we might try to find what k regular graphs are Hamiltonian. The k regular graph, remember, all vertices have degree k, so a degree sum will be 2k for all vertices. Consequently, we'll need 2k to be greater than or equal to n, the number of vertices. Or n is less than or equal to 2k, giving us a k regular graph with fewer than 2k vertices is Hamiltonian. For example, cubic, three regular graphs with up to six vertices are definitely Hamiltonian.