 Remember, a bridge is an edge whose removal disconnects a connected graph. We can remove an edge if we eliminate either endpoint, but we might be able to do even more by removal of a vertex. So we introduce the following idea. Let G be a connected graph, vertex V is a cut point if it's removal, and removal of all incident edges disconnects the graph. So every bridge gives two cut points. Or does it? We'd have to think about that a little more. Bridges at least give us a starting point because we know how to find bridges, but some other points might also disconnect the graph. Can we identify them? It might seem to be an easy task, but maybe not. In fact, this is a tiny graph. A real graph might have thousands of vertices and thousands of edges. So let's see if we can characterize the cut points. Suppose V is a cut point of a connected graph. Then its removal will separate G into two subgraphs, G1 and G2, which are not connected. So if we take a point in G1 and G2, since the original graph was connected, there had to have been a path from U to W, but since the only change is the removal of the cut point V, then all paths between the two points had to pass through V. Giving us the following result, if V is a cut point, then there are points U and W for which all paths between the two points pass through V. Unfortunately, this only tells us what happens if we already know V is a cut point. So consider the converse. Suppose G is a connected graph, so any two points U and W have a path between them. If all paths pass through V, then removing V means there is no path between U and W, which separates the graph. As this proves a converse, suppose all paths between U and W pass through V, then V is a cut point. And this means we could find cut points by finding all paths between all points, but maybe there's an easier way. We'll return to this later. In the meantime, let's introduce some other ideas. A graph is non-separable if there are no cut points. A block in a graph is a maximal non-separable subgraph. In other words, we can't introduce more vertices and still be non-separable. So let's see what happens. Suppose we have a block. Consider any two points U and V in the block. There's a path, because it's connected, from U to V. Since the block is non-separable, there are no cut points, so there must be an alternate path from V to U, so the two together form a cycle containing U and V. Or do they? So remember, if you don't find the flaws in your work, someone else will. So it's important to look very carefully at all the work that you produce. Our conclusion relies on assuming that our two paths have no points in common. But can we guarantee this? Suppose we have a point on our first path, then, since we have no cut points, there has to be at least one path that does not pass through this point. But it might pass through another point on our first path. If that happens, then we don't have a cycle containing U and V, because remember the points of a cycle have to be distinct. So let's take another approach. Suppose instead we consider all points U that share a cycle with our chosen point. Now if V is in U, then they're on a common cycle. So suppose V is not in U. So let W be the point in U closest to V. Now since W is in U, there's a cycle that passes through U and W. So let P1 be the part of the cycle from U to W, and P2 the part of the cycle from W back to U. And since the two form a cycle, the vertices in the two paths have to be distinct. Now consider the geodesic from W to V. Since W is the point in U closest to V, the geodesic can't include any points in U. And since G is connected, there has to be a path from U to V as well. In general, we don't know what this path is going to look like, and it might intersect the cycle many times. So let U' be the last point on this path that is on the cycle containing U and W, and it really doesn't matter whether it's on P1 or P2, so we'll assume that it's on P1. The remaining part of the path from U' to V will either pass through a point W' on the geodesic from W to V, or it will not pass through a point on the geodesic. Now suppose the path from U' to V passes through points on the geodesic. Let W' be the point on the geodesic closest to W. Then we can take the path from U to U' in this first part of the cycle, the path from U' to W', which by assumption includes no vertices in the cycle, the path from W' back to W, and the path from W' back to U, and this produces a cycle, including W', which can't happen. Because remember no point on this geodesic can have a cycle with U, so the path can't pass through a point on the geodesic. But if the path from U' to V can't pass through a point on the geodesic, we can concern the cycle. From U to U' from U' to V, from V to W, and from W to U, so U and V are in a common cycle. Consequently, suppose G is non-several. Then any two points U and V are on a common cycle. So again a useful way of learning how to create mathematics is to follow the consequences of any result. So now suppose G is a graph where any two points U and V are on a common cycle. What can we conclude? We'll take a look at that next.