 We found that if we could reduce a sequence and get a graphical sequence, then the original sequence was also graphical. But what if the resulting sequence isn't graphical? For example, if we have the sequence, we'll reduce the sequence. If we remove a vertex of degree 6, we reduce the degrees of the six remaining vertices by 1. If we remove a vertex of degree 3, then three of the remaining vertices have their degrees reduced by 1. If we remove a vertex of degree 2, we'd decrease the degree of two vertices by 1, but we can't do that. Since the last sequence isn't graphical, our theorem doesn't apply, because it only applies to our sequences graphical. So part of the problem is we don't actually know which vertices are connected. Well, remember, we can assume anything we want as long as we make it explicit, so suppose our degree sequence is, and it's graphical, and the first vertex is connected to the k vertices corresponding to the next k degrees. Then we can delete it, and our new graph will have the following degree sequence. The first k terms will be reduced by 1, while the rest stay the same. Since this is the actual degree sequence of the new graph, it will also be graphical. There's just one problem. Not all graphs are connected this way. So the graph shown has degree sequence, but the vertex of degree 4 is not connected to vertices of degrees 3, 3, 2, and 1, and in fact, if we remove this vertex of degree 4, we'd get a graph with degree sequence. To implement our algorithm, we want the vertex with the highest degree to be connected to the other vertices of highest degree. In other words, the sum of the degrees of the adjacent vertices should be as great as possible. So consider the degrees of the vertices connected to the first vertex, which we'll call vertex 0. For the degree sum to be as great as possible, we have to use the first k vertices corresponding to the degrees of vertices 1 through k. So to have a lower degree sum, we need to switch out at least one. Say we switch out vertex i with vertex j. Now, if the two degrees are equal, the degree sum is still the greatest possible. So we'll assume di is strictly greater than dj. In other words, we've replaced the connection with one vertex with a connection to a vertex of lower degree. The important thing here is this means vertex i is not connected to vertex 0, but vertex j is. Since the degree of vertex i is greater than the degree of vertex j, then there's at least one vertex k adjacent to i that's not adjacent to j. So we'll make some changes in our graph. We'll remove the link from vertex 0 and j. We'll join j and k. And now we'll remove the link between i and k and join i and our vertex 0. And here we're leaving dotted lines where the edges used to be. So notice that the degree of vertex 0 didn't change because it lost the link to j, but gained the link to i. The degree of vertex j didn't change since it lost the link to 0, but it gained the link to k. The degree of vertex k didn't change since it lost the link to i, but gained the link to j. And the degree of i didn't change since it lost the link to k, but gained the link to 0. And since the only changes were made among the vertices i, j, k, and 0, this means that all of the vertices have the same degrees as before, so our degree sequence is the same as before, except now the degree sum of the vertices adjacent to 0 is greater, because remember the degree of i was greater than the degree of j. And if it's not the greatest possible, lather, rinse, repeat. So this may be a little hard to visualize, so let's see if we can do this. So here's our graph where our vertex of degree 4 is not joined to the vertices of degrees 3, 3, 2, and 1. Let's modify it. So we note the vertex of degree 4 is currently joined to vertices of degree 3, 3, 1, and 1, so we need to switch one of these ones to a 2. So we need to break the link to a degree 1 vertex and form a link to the degree 2 vertex. Now the thing to notice is we now have two edges that don't have endpoints. To reconnect the edges, we need to break one of the edges from the vertex of degree 2. Now rejoin the edges, from the degree 2 vertex to the degree 4 vertex, and from the degree 1 vertex to the freed edge, and every edge is properly secured by two vertices. Note that the degrees of the vertices haven't changed, but now the vertex of degree 4 is connected to vertices of degrees 3, 3, 2, and 1.