 So, remember a graph is connected if there is a path between any two vertices. We might be able to disconnect the graph if we remove an edge or vertex. This could be desirable, or we might want to avoid it. So maybe we can disrupt a trafficking ring by blocking a road, or maybe we can disrupt internet service by cutting a link. A bridge is an edge whose removal breaks a graph into disconnected subgraphs. Some of the graph shown, let's find any bridges. And it should be clear that if we remove this edge between A and G, the graph splits apart into two pieces, and so this edge is a bridge. But if you have a complex graph, it can be difficult to find bridges if they even exist. So let's introduce a useful strategy in math and in life. Look both ways. While we want to find bridges, let's find edges that aren't bridges. Now let's think about this. If an edge between two vertices isn't a bridge, you can detour around it. So here we have an edge whose removal won't disconnect the graph. But if we choose not to use it, we can take a detour. Recalculating. And then you can use the bridge to get back to your starting point. Consequently, this edge is part of a cycle. So suppose you have a cycle in a connected graph. Consider any edge that is part of the cycle. If we remove the edge, a path still exists between the vertices because we can use the cycle backwards. So the two vertices are still connected. But what about the other points? So consider two other points, i, j, and the graph. Again, since the graph was originally connected, there's a path between them. So if we remove that bridge, either the path does not include the removed edge, in which case even if we remove the bridge, the path still exists, or the path includes the removed edge. But in that case, we can take the path up to the bridge. Recalculating. Take the detour and continue on. So a path still exists and the graph is connected. So while I'm waiting for my offer from Pixar for these quality animations, let's present a formal proof. So suppose an edge E1 is in some cycle. Even if we remove E1, a path still exists between vertices V1 and V0. And since edges are non-directional, there's also a path from V0 to V1 that avoids the edge E1. Now by assumption, our original graph was connected, so there's a path between any other vertices Vi Vj. If this path didn't include E1, it still exists, even if we remove this edge. But suppose the path included the edge E1. By the proceeding, there's a path from V0 to V1 that avoids the edge, giving us the path, so the graph is still connected. And consequently, if an edge is in a cycle, it cannot be a bridge. And again, part of the goal is not to learn mathematics, but to learn how to create mathematics. And so here's a key strategy. Prove as many things as you can. If they happen to be what you want to prove, even better. Now to get a statement to prove, remember that any conditional, if A, then B, has three related conditionals. The inverse, if not A, then not B. The converse, if B, then A. And the counter positive, if not B, then not A. Now the counter positive has the same truth value as the original, so it can be regarded as a corollary. The others need to be proved if they're true. So we prove that if an edge is in a cycle, then it is not a bridge. So what if an edge is not in a cycle? That's how the inverse would start. Or what if an edge is not a bridge? That's how we'd start the converse. Well, let's take a look. So let's start with the inverse. Suppose an edge in a connected graph is not in any cycle. So if we remove it, there could be no path between the two vertices. Because if there's still a path, we can then join the edge to the path and get a cycle. But if there's no path between the vertices, then the graph is no longer connected. And consequently, if an edge is not in any cycle, then it is a bridge. We could also consider the converse of the theorem. If an edge is not a bridge, then it is a cycle. So suppose an edge in a connected graph is not a bridge. Then its removal will leave the graph connected. Consequently, there must be an alternate path between the two vertices. And if we append the removed edge, we form a cycle. Consequently, if an edge in a connected graph is not a bridge, then it is part of a cycle. And notice that we've proved two statements. If an edge is part of a cycle, then it's not a bridge. And also, if an edge is not a bridge, it's part of a cycle. So because this is a statement and its converse, we can incorporate them into a single statement in a connected graph. An edge is part of a cycle if and only if it is not a bridge. However, it's probably better to state this as in a connected graph, if an edge is part of a cycle, then it is not a bridge and conversely. Now, part of the reason that it's useful to do this is this leads to an algorithm. So an algorithm is a method of solving a problem that has a finite number of steps. We say that the algorithm terminates and always yields a solution. Computer scientists are interested in algorithms and mathematician should be more interested than they are. So this connection between bridges and cycles suggests an algorithm for finding them. Find all the cycles in a graph. Any edge that is not on a cycle is a bridge. Since a graph can only have a finite number of cycles, the algorithm terminates. And since every edge not on a cycle is a bridge, the algorithm always returns a solution. Now, it is worth pointing out the algorithm might not be practical more about this later. Until then, let's consider this graph. Now, any edge on a cycle can't be a bridge, so let's find a few cycles. And we'll color the edges as we find them to be part of a cycle. For example, we have this cycle that goes from one to two to eight back to one. And we can find a few more. Now, some edges will appear on more than one cycle, so we'll change colors to make these a little bit more obvious. Now, while we haven't necessarily found all the cycles, it's a little suspicious that the edge between one and nine never seems to be on a cycle. And so we might suspect that this is actually a bridge. Remember, we can move the points as long as we maintain the edges, and we find that if we move that point nine, this edge between one and nine is obviously a bridge.