 In the previous lecture we proved if g is non-separable then any two points u and v are on a common cycle Let's try to form another chain of conditionals So let's start by supposing g is a connected graph where any two points u and v are on a common cycle What if we have a point u and an edge vw? Since any two points are on a common cycle, there is some cycle c that includes u and v Now suppose this cycle also includes w Then the cycle must go from u to v along some path, from v to w along some path, and from w to u along some path Or some permutation of this But since there's an edge that joins v and w we can take a shortcut and produce a cycle that includes the edge Now suppose w is not on the cycle that includes u and v Because the graph is connected there must be a path from u to w that does not pass through v And that's because if every path from u to w passes through v It would be impossible to form a cycle containing u and w So again we have part of our cycle from u to v and the remaining part of the cycle from v back to you and The path from u to w might have points in common with these paths But there's going to be a last common point u prime, which we'll assume is on p2 So let's take the path from u prime to w and it has to have no points in common with the cycle containing u and v Remember u prime was the last point in common with the cycle And so we can form a cycle by taking the path from u to v Taking the edge from v to w Taking the path from w to u prime and then taking the remaining part of our cycle back to you And this gives us a theorem suppose g is a connected graph if every pair of points is on a common cycle Then any point and edge will be on a common cycle To continue our chain. Let's suppose g is connected and any point and edge are on a common cycle What if we have two edges? Suppose our edges are those between vertices u and v and that between vertices r and s Then vertex u and the edge are on a common cycle If v is also on this cycle, we can use the uv shortcut to produce a cycle that includes both edges Otherwise we can find a path from v to r that does not pass through s Here because any point and any edge must be on a common cycle We can just use the part of the cycle that does not include s So again, we're assuming that any point and edge are on a common cycle So there's a cycle that contains u and the edge rs and again We don't know what this cycle looks like and it might cross this path Let v prime be the first point this path has in common with the cycle including u and the edge rs Then we can form the cycle From u to v along the edge From v to v prime going through no points on the cycle And from v prime to the edge sr Back home to you and so suppose g is a connected graph if any point and edge is on a common cycle Then any pair of edges is also on a common cycle Let's take stock so far. We've considered two points a point and an edge and two edges And in each case we have two things What if we have three things? Let's consider two points u v and an edge between two vertices rs So again begin where we end suppose g is a connected graph where every pair of edges is on a common cycle Now since each point can be used as one vertex of an edge we have two edges Which must be on a common cycle So that means any two points are on a common cycle But remember that in a connected graph if any two points are on a common cycle Then any point and edge will be on a common cycle And so there's a cycle containing the point u and the edge rs If v is on this cycle Then there's a path from u to v that uses the edge And if not, there's a cycle containing v and the edge rs And as before if u is on this cycle, we can find a path that uses the edge So suppose u and v are on different cycles We can follow the path from u to v prime the first point that is on the cycle that includes v the edge rs From v prime to the edge along the cycle including v across the edge And then back to v along the path that can't include any vertices between u and v prime And so suppose g is a connected graph if any two edges are on a common cycle Then given two points at an edge there is a path between the two points using the edge And notice we've switched from cycles to paths So what if we have three points? So again begin where we ended Suppose g is a connected graph where given any two points at an edge There is a path between the two points that uses the edge Since any point in a connected graph is incident on some edge We can take any of these points and form an edge There is a path between the points that uses the edge So there's a path that passes through all three points And whichever point is on the edge will be between the other two And so suppose g is a connected graph if given any two points at an edge There is a path between the two points using the edge then given any three points There is a path from one to the other passing through the third Now suppose that in a connected graph g given any three points There is a path between any two passing through the third So given three points u, v, and w we have a path from u to w passing through v So there's a path from u to v that avoids w So in a connected graph if given any three points There is a path between two that passes through the third Then there's a path between two that does not pass through the third But if we can always find a path between two points that bypasses any third point Then we have no cut points And so this takes us home Suppose g is a connected graph if given three points There is a path between two that does not pass through the third Then g is non separable Now our conditional chain means that any one of these statements implies all the others for a connected graph g g is non separable Which tells us that any two points are on a common cycle Which tells us that any point and edge of g are in a common cycle Which tells us that any two edges are on a common cycle Which tells us that given two points in an edge. There's a path between the points using the edge Which tells us that given any three points. There's a path between two that passes through the third Which tells us that there's a path between two that does not pass through the third Which tells us that g is non separable