*Written by Stefanie Zbinden.* *Translated by Stefanie Zbinden.*

Contents

We define bridges and articulation point and come up with an algorithm that computes them in linear time.

As many graph algorithms, this is a clever application of DFS. If you are not familiar with dfs, you might consider reading the dfs article first.

In this article, all graphs are simple and undirected.

## DFS preorder, postorder and other properties

We will first have a look at dfs and some properties to use this later in order to find an efficient algorithm to find articulation points and bridges (we will explain later what they are).

For simplicity, we will assume our graph is connected and observe what happens when we run a dfs starting from some node $r$. The dfs code we use for this is the following:

void dfs(vector<vector<int>>& graph, vector<int>& visited, int node) { visited[node] = 1; for (int next : graph[node]) if (visited[next] == 0) dfs(graph, visited, next); visited[node] = 2; }

**Remark:** $graph$ is the adjacency list of our graph and $visited$ is the visited array, which is set to 0 at the beginning.

At each step of the dfs, the nodes will have one of the following states:

- The node has
**not been called**by the dfs yet. In this case, visited is equal to 0. We will say such a node is a*white node*. - The node has been called by the dfs but the call has
**not returned yet**. In this case, the dfs is still processing the children of that node and visited is equal to 1. We will say that such a node is a*grey node*. - The node has been called by the dfs and the call has returned. In this case, all the neighbors of the node have been called and the node will
**not be called anymore**. Visited will be equal to 2 and we will say that such a node is*black*.

Also we can create a **rooted tree** called the **dfs tree** from the order in which the nodes are called. The root will be the node with which we initially call the dfs (in our case this is $r$). If when processing node $a$ in the dfs we call node $b$ then $a$ will be the parent of $b$ in the dfs tree (so there will be an edge from $a$ to $b$). As we can only call neighbors from the original graph during the dfs our tree will be a subgraph of our original graph.

Now we might ask ourselves (or our rubber mice) why this tree is interesting. Using it we can classify the edges of our original graph into three categories (here, we make a distinction between the edge $(u, v)$ and $(v, u)$ for nodes $u$ and $v$).

**Tree edges:**These are edges that belong to the dfs tree (so edges from a node to one of its children in the dfs tree).**Back edges:**Edges that go from a node to one of its ancestors.**Forward edges:**Edges that go from a node to one of its successors that is not its child.

**Why are these all the possible edges?** Assume there are nodes $a$ and $b$ such that there exists an edge between them but none is the ancestor of the other and assume $a$ is called before $b$ during the dfs, then this implies that we try to go to $b$ but somehow it is not possible anymore. This implies that $b$ has been visited while $a$ was grey. But all nodes that have been called while $a$ was grey will be the children of $a$, their children or the children of their children and so on. So $a$ is an ancestor of $b$.

**Disclaimer:** This only works so nicely because we assumed that our graph is **undirected**. In the directed case there are more cases that can appear.

What also might be interesting is the order in which the nodes are called. This order is called **dfs-preorder** and is useful for many algorithms.

We can modify our algorithm to calculate this during the dfs as follows:

int counter=0; void dfs(vector<vector<int>>& graph, vector<int>& visited, vector<int>& preorder, int node) { visited[node] = 1; preorder[node] = counter; counter++; for (int next : graph[node]) if (visited[next] == 0) dfs(graph, visited, next); visited[node] = 2; }

There is a nice synergy between the dfs tree and the dfs preorder: the ancestor of each node has to be called before each node, so its preorder index is smaller.

**Remark:** The same can be done with the times when we return. This is then called the **dfs-postorder**.

## Articulation points and bridges

### Definition

- An
**articulation point**is a node whose removal increases the number of connected components in the graph. - Similarly, a
**bridge**is an edge whose removal increases the number of connected components in the graph. - A
**leaf**is a node with only one edge (in other words a node whose degree is one).

Note that by removing a node from the graph we also remove all its edges. Furthermore when our graph was connected being an articulation point or bridge means that the removal will disconnect the graph.

**Remark:** From now on, we will mark all articulation points green and all bridges and leaves blue.

Here are some examples:

What you can observe is that each bridge has only endpoints that are articulation points or leaves. The question is whether this is a coincidence or is always the case. In other words: Do there exist graphs which have a bridge whose endpoint is neither a leaf nor an articulation point? Does there exist a graph with two articulation points that are endpoint of an edge which is not a bridge? Is every articulation point endpoint of some bridge?

**Do there exist graphs which have a bridge whose endpoint is neither a leaf nor an articulation point? No.**
Assume we have a bridge. Removing it will lead to an increase of components in the graph, splitting a component $C$ into two components $C1$ and $C2$. When we remove an endpoint of the bridge instead of the bridge itself, then the same will happen, however if the endpoint was its own connected component (this happens exactly if it was a leaf) this component vanishes.

**Does there exist a graph with two articulation points that are endpoint of an edge which is not a bridge? Yes.**

An example for this is the following graph:

There is an edge from $b$ to $d$ which are both articulation points but the edge is not a bridge.

**Is every articulation point endpoint of some bridge?** **No, not necessarily**.

We can again look at a counterexample:

Both $(a, b, c)$ and $(a, d, e)$ form a cycle, so removing any edge does not disconnect the graph. However, removing $a$ will disconnect the graph (there will be the components $(b, c)$ and $(d, e)$ left and thus $a$ is an articulation point.

### Brute force

**How can we find all the articulation points and bridges in a graph?** What we could do is for each node and each edge remove it once from the graph and count the number of connected components compared to the number of connected component in the original graph.

As we can find the number of components in a graph using dfs in $O(V+E)$ we can find all the articulation points in $O(V\cdot(V+E))$ and the bridges in $O(E\cdot(V+E))$.

### Efficient algorithm - Tarjan

We will give a short overview of the algorithm here , more information can be found e.g. here: Codeforces or CP Algorithms

The main idea of the algorithm is to calculate the preorder and introduce an additional notion, the $low$-values. $low[v]$ denote the smallest preorder index one can reach from the vertex $v$ by a path using only tree edges and at most one back edge.

A vertex is an articulation point if it fullfills one of the following conditions:

- It is the troot and has at least two incident tree edges,
- it is not the root and has a child $w$ in the dfs tree such that $low[w] \ge pre[v]$.

An edge $e = \{u,v\}$ with $v$ being a DFS child of $u$ is a bridge if and only if $low[v]>pre[u]$.

Implementation in C++:

vector<vector<int>> graph; vector<int> pre, low; // -1 unvisited int timer; void dfs(int v, int parent = -1) { low[v] = pre[v] = timer; timer++; int children = 0; for(int w: graph[v]) { if(w == parent) continue; // skip parent edge if(pre[w] != -1){ // forward & back edges low[v] = min(low[v], pre[w]); continue; } // tree edges dfs(w, v); low[v] = min(low[v], low[w]); if (pre[v] < low[w]){ // {v, w} is a bridge } if (pre[v] <= low[w] && parent != -1){ // v is an articulation point } children++; } if(parent == -1 && children > 1){ // v is root and has at least two children // v is an articulation point } } void find_cutpoints(){ int n = graph.size(); pre.assign(n, -1); low.assign(n, -1); timer = 0; for(int i = 0; i < n; i++){ if(pre[i] == -1) dfs(i); } }

**Remark** If your graph is not connected, don’t forget to start the dfs from multiple points and adjust the root node (as done when doing normal dfs).