Bridges and Articulation Points

Articulation Points Is a node u of a graph such as if you remove u from the graph then the number of components increases 3 components 1 component

Articulation Points u is an articulation point if u isn't a root and there exists one of its subtrees that has no back edges for an ancestor of u u is root and has more than 2 children p u Right Wrong The second and third subtree all have back edges to p (an ancestor of u), while the first one hasn't, therefore u is an articulation point (notice that if you remove it, the first subtree will be disconnected from the rest of the graph).

Bridges Is an edge uv of a graph such as if you remove uv from the graph then the number of components increases 2 components 1 component

Bridges uv is a bridge if and only if it doesn't belong to a cycle. Proof uv is a bridge => doesn't belong to a cycle uv doesn't belong to a cycle => uv is a bridge u v u v uv is not in a cycle so if uv is removed there is no other way to get from u to v, so you are cutting the graph into 2 components (one with u and another with v) If you remove uv there is another path that takes from u to v (the graph is still connected), so uv is not a bridge

Bridges The definition on the previous slide gives us that: uv is a bridge ⇔ there is no back edge from a descendant of u to an ancestor of u uv is a bridge => there is no back edge... uv doesn't have a back edge... => uv is a bridge u u v v Assume that uv is not a bridge, then once it’s removed it must have a path between u and v, but we are assuming that there is no back edge, contradiction. Assume that there is a back edge, then uv is part of a cycle and by the previous slide this is a contradiction

Interesting fact If you remove a bridge of a graph the number of components increases by exactly one but if you remove an articulation point the number of components can increase by more than one.

Naïve Implementation (not optimal) Find a vertex or an edge that when removed will disconnect the UNDIRECTED graph Calculate the number of connected components Try, one-by-one, removing each edge/vertex Recalculate number of CCs If more CCs, then it’s an articulation point or bridge.

A better approach Compute DFS, but also store two values per node: d: the order in which the node was reached during DFS Low: smallest d reachable from current subtree At a vertex, the Low value is the lowest of: Your own value (obviously, you can reach yourself) Any back edge The lowest returned from a recursive call on all children Notice: it does NOT include the parent If a neighbor vertex has a Low value greater or equal to your own d, then you’re at an articulation point Root node is special: it is an articulation point iff there are 2 (or more) children not part of the same branch (i.e. if during DFS, it has to start recursive calls on at least 2) If Low(u) > d(v), then u-v is a bridge Likewise, if Low(v) > d(u), then u-v is a bridge

How to implement? d[v] = time instant that you enter a node low[v] = the smallest d that v can reach through its descendants (including v itself) Example: d[a] = 1 d[b] = 2 d[c] = 3 d[d] = 8 d[e] = 4 d[f] = 7 d[g] = 5 d[h] = 6 low[a] = 1 low[b] = 1 low[c] = 1 low[d] = 8 low[e] = 3 low[f] = 2 low[g] = 3 low[h] = 6 u is an articulation point if low[v] >= d[u] for some children v uv is a bridge if low[v] > d[u] for some children of v Notice that low[e] = 3 and not 1 Notice that a has 2 children (b and c), but they are connected, so only 1 is a “child” in the DFS search tree a b c d e f g h

d Low 1 2 d is when visited Low is lowest seen in that subgraph 7 3 6 9 4 3 8 5 3

d Low 1 2 Articulation point if neighbor has a low value >= your d 7 3 6 9 4 3 8 5 3

d Low 1 2 Bridge if Low(u) > d(v) for some edge u-v 7 3 6 9 4 3 8 5 3