 Suppose G has N vertices of varying degree. We can order the vertices by degree, putting them in what's called non-decreasing order because there could be a quality. This produces the degree sequence of the graph. For example, suppose we want to produce the degree sequence of the graph shown. So we see this node has degree 3. This node has degree 3. There's a 2, a 4, and two 1s. Given a graph, we can find its degree sequence. But can we go backwards? Can we produce a graph from a given degree sequence? We say that a sequence where each term is greater than or equal to the next is graphical if a graph exists with the given degree sequence. For example, we might see if this sequence is graphical. So remember, this sequence gives us the degrees of every vertex in the graph. So if this graph exists, this would be a graph with 6 vertices. Unfortunately, there's a lot of them. And we can't just rely on a graph falling out of the sky and hitting us on the head. Well, actually, how about this one? And in fact, this is the graph that produced this degree sequence. But what if we hadn't already seen this graph? Could we construct a graph with a particular degree sequence? For example, how about this one? Such a graph would have to have four points, one of which is connected to 12 others, which is impossible, so no such graph can exist. Now, while this analysis might work for small graphs, it will be much harder if there are more vertices. And even if a graph with a given degree sequence exists, we might still not know how to connect it. So let's investigate. So first, we know that in a graph, the number of vertices with odd degree must be even. So we might take a sequence, and we can immediately decide since there are three vertices of odd degree, no graph exists with this degree sequence. While the Handshake Theorem tells us what a graph doesn't exist, we can't use it to say whether a graph could exist. For example, this is not graphical, even though there are an even number of vertices with odd degree. To proceed, let's consider a simpler problem. By that, let's talk about a smaller graph. So suppose we remove one vertex from a graph. If the vertex has degree D, then D of the remaining vertices will have their degrees reduced by one. Now, Lather rins repeat since every removal reduces the number of terms by one. Eventually, we'll either get a sequence that is obviously graphical or obviously non-graphical. So let's see if this sequence is graphical. We'll eliminate one of the degree one vertices. Since this vertex was connected to one other, we'll reduce another degree by one. But which one? The problem is that when we eliminate a vertex, we can only reduce the degrees of its adjacent vertices. But we can't tell which vertices are adjacent from the degree sequence alone. So what can we do? If you build it. So instead of removing vertices, let's add vertices. That way we can choose which vertices increase their degree. So suppose we have a graphical sequence. If we add a vertex with degree K, then K of the terms will increase by one. Since the degree sequence should be non-decreasing, let's assume K is greater than the first term and add one to the first K terms, giving us, since we began with a graphical sequence and produced a graph, the new sequence is also graphical, giving us an important result. Suppose we have a sequence that's graphical. If K is greater than the first term, then we can prepend the sequence with K and add one to the first K terms to get another graphical sequence. And this suggests we can reduce a sequence by starting with the largest term. So, for example, we might try our sequence again. We'll eliminate the degree for vertex and reduce the first four degrees by one to give us. Now keep in mind that this is a new graph, and because we want to find the degree sequence, we would put the degrees of the vertices in order. And as long as they're all there, it doesn't make a difference. So we can rearrange this degree sequence as... Next, we'll eliminate the degree to vertex and reduce the first two terms by one, which we can rearrange as... Now, at this point, we have a graph with four vertices, two of which are isolated, and two of which have degree one. Let's try to produce a graph with this degree sequence. So let's put down four vertices. Two of the vertices have degree zero, so there will be no edge incident on those vertices. And two of the vertices have degree one, and we can achieve that by linking them. Now, let's see if we can work our way backwards to the original graph. Remember, we obtained this sequence by reducing and rearranging. What this means is that we removed a degree to vertex that was linked to two vertices whose degrees became one and zero. So working backwards, we'll want to add a vertex linked to a vertex of degree one and a vertex of degree zero. So we'll link over to this vertex with degree one and up to this vertex with degree zero. Let's try to take another step back. So remember, we got this sequence by reducing. And again, we removed a vertex of degree four attached to vertices of degree three, three, two, and one, leaving them with degrees two, two, one, and zero. So we'll add a vertex and link it to vertices of degree two, two, one, and zero to get our original degree sequence and our graph. While this algorithm terminates, it doesn't necessarily give us an answer. If the final sequence is graphical, then the original sequence is also graphical. But if the final degree sequence isn't graphical, it's conceivable that the original was graphical. And remember, as long as the process might return undecided, it can't be called an algorithm. So let's look at this further.