 Suppose we have a connected graph with n vertices and n minus 1 edges. We found that in this case every edge is a bridge. But we also found that if we have n vertices and n edges, we must contain a cycle. So if we add one edge, we'll get a cycle. And we know that no edge in a cycle can be a bridge. Now, in discrete math, we always ask the question, how many? And so this leads to two questions. How many bridges can we convert into non-bridges? And how many cycles do we get by adding one edge? So if we take any path and join the end to the beginning, we get a cycle. So if every edge in the path was a bridge, we've just converted all these edges into non-bridges. Consequently, we can convert the most bridges by joining the endpoints of the longest path in our graph. Now, suppose we've added one edge, so our graph now has n vertices and n edges. How many cycles could it have? Since the only new edge is the added edge, then any cycle must include this edge. So suppose we have two cycles, c1 and c2, that use the edge. Since both cycles use the edge, they might parallel each other for a while, but if they're different, at one point they have to separate. However, this means that there is another cycle. Namely, the cycle that uses all of the edges except for the common edges. Since a cycle doesn't use the new edge, it would have been in the original graph, which had no cycles. Consequently, if g is a connected graph with n vertices and n minus one edges, adding one edge introduces exactly one cycle. Now, a picture is worth a thousand words, and at 24 frames per second a video is worth a lot more, but we should prove this formally. Now, it's not that bad. Remember that proof is a way of reviewing math and as a way of doing mathematical research. So, suppose we add an edge. If two cycles in our new graph share the edge and possibly some other edges, then we can write the cycles as where here c2 follows the first cycle up to a point and then breaks off, but eventually gets back to its starting point. But now we can create a new cycle. Let's start with a two-cycle split. We'll follow one cycle until it gets to the place where they rejoin, then follow the other cycle backwards to our last common point. And this is a cycle that doesn't use the edge we introduced, so it would have been in our original graph, which is impossible. Now, it's very tempting to reason as follows. If we add one edge to a graph with n vertices and n minus one edges, we get a graph with n vertices and n edges and, by what we just did, one cycle. So, does this mean that connected graphs with n vertices and n edges contain one cycle? Maybe not. Now, we might reason as follows. If we remove one edge, we get a graph with n vertices and n minus one edges, so if we add back in the edge, we'll get exactly one cycle. Or will we? The problem is that our theorem only applies if our graph is connected, so if the edge we remove is a bridge, we can't guarantee its restoration will produce a cycle. And so the natural question to ask is, can a connected graph with n vertices and n edges have a bridge? While we could see what happens if our graph has a bridge, let's focus on the number of cycles the graph has. Suppose a connected graph with n edges and n vertices has a cycle of length p, which includes p vertices and p edges. Suppose it has a second cycle. There are two possibilities. The second cycle is disjoint and shares no vertices with the first, or the second cycle intersects the first cycle. They have some vertices in common. Let's consider these two cases separately. So suppose a connected graph g with n vertices and n edges has two cycles. Let the first cycle have p vertices and p edges. Suppose the second cycle is completely disjoint with q vertices and q edges. Since we're assuming that we have a connected graph, there must be a path joining a point on each cycle that does not include any points on either cycle. If this path passes through r points, it must include r plus 1 edges. So we've used p plus q plus r plus 1 edges and p plus q plus r vertices. Where, remember, a point is a vertex is a node. Now, since we have n edges, this sum p plus q plus r plus 1 has to be less than or equal to n, so p plus q plus r has to be less than the number of vertices. And this means there are additional vertices to be connected, but each vertex connected requires at least one edge, and we've already used more edges than vertices. So we'll run out of edges before connecting the graph. What about the other case? Suppose a connected graph g with n vertices and n edges has two cycles. Let the first cycle have p vertices and p edges. And suppose the second cycle intersects the first. And we can again consider the part of the second cycle that is disjoint from the first. Suppose this passes through q vertices. It must use q plus 1 edges. And so we've used p plus q plus 1 edges to pass through p plus q vertices. And as before, to join the remaining vertices of the graph requires more edges than we have left. And putting our results together, this proves a connected graph with n vertices and n edges contains exactly one cycle. Or does it? Since you can't actually have a cycle on a graph with 1 or 2 vertices, we do require that we have at least 3 vertices. Thank you.