Finding Bridges Online

We are given an undirected graph. A bridge is an edge whose removal makes the graph disconnected (or, more precisely, increases the number of connected components). Our task is to find all the bridges in the given graph.

Informally this task can be put as follows: we have to find all the "important" roads on the given road map, i.e. such roads that the removal of any of them will lead to some cities being unreachable from others.

There is already the article Finding Bridges in $O(N+M)$ which solves this task with a Depth First Search traversal. This algorithm will be much more complicated, but it has one big advantage: the algorithm described in this article works online, which means that the input graph doesn't have to be known in advance. The edges are added once at a time, and after each addition the algorithm recounts all the bridges in the current graph. In other words the algorithm is designed to work efficiently on a dynamic, changing graph.

More rigorously the statement of the problem is as follows: Initially the graph is empty and consists of $n$ vertices. Then we receive pairs of vertices $(a, b)$, which denote an edge added to the graph. After each received edge, i.e. after adding each edge, output the current number of bridges in the graph.

It is also possible to maintain a list of all bridges as well as explicitly support the 2-edge-connected components.

The algorithm described below works in $O(n \log n + m)$ time, where $m$ is the number of edges. The algorithm is based on the data structure Disjoint Set Union. However the implementation in this article takes $O(n \log n + m \log n)$ time, because it uses the simplified version of the DSU without Union by Rank.


First let's define a $k$-edge-connected component: it is a connected component that remains connected whenever you remove fewer than $k$ edges.

It is very easy to see, that the bridges partition the graph into 2-edge-connected components. If we compress each of those 2-edge-connected components into vertices and only leave the bridges as edges in the compressed graph, then we obtain an acyclic graph, i.e. a forest.

The algorithm described below maintains this forest explicitly as well as the 2-edge-connected components.

It is clear that initially, when the graph is empty, it contains $n$ 2-edge-connected components, which by themselves are not connect.

When adding the next edge $(a, b)$ there can occur three situations:

Consequently the whole task is reduced to the effective implementation of all these operations over the forest of 2-edge-connected components.

Data Structures for storing the forest

The only data structure that we need is Disjoint Set Union. In fact we will make two copies of this structure: one will be to maintain the connected components, the other to maintain the 2-edge-connected components. And in addition we store the structure of the trees in the forest of 2-edge-connected components via pointers: Each 2-edge-connected component will store the index par[] of its ancestor in the tree.

We will now consistently disassemble every operation that we need to learn to implement:


Here is the final implementation of the whole algorithm.

As mentioned before, for the sake of simplicity the DSU of the 2-edge-connected components is written without Union by rank, therefore the resulting complexity will be $O(\log n)$ on average.

Also in this implementation the bridges themselves are not stored, only their count bridges. However it will not be difficult to create a set of all bridges.

Initially you call the function init(), which initializes the two DSUs (creating a separate set for each vertex, and setting the size equal to one), and sets the ancestors par.

The main function is add_edge(a, b), which processes and adds a new edge.

vector<int> par, dsu_2ecc, dsu_cc, dsu_cc_size;
int bridges;
int lca_iteration;
vector<int> last_visit;

void init(int n) {
    lca_iteration = 0;
    last_visit.assign(n, 0);
    for (int i=0; i<n; ++i) {
        dsu_2ecc[i] = i;
        dsu_cc[i] = i;
        dsu_cc_size[i] = 1;
        par[i] = -1;
    bridges = 0;

int find_2ecc(int v) {
    if (v == -1)
        return -1;
    return dsu_2ecc[v] == v ? v : dsu_2ecc[v] = find_2ecc(dsu_2ecc[v]);

int find_cc(int v) {
    v = find_2ecc(v);
    return dsu_cc[v] == v ? v : dsu_cc[v] = find_cc(dsu_cc[v]);

void make_root(int v) {
    v = find_2ecc(v);
    int root = v;
    int child = -1;
    while (v != -1) {
        int p = find_2ecc(par[v]);
        par[v] = child;
        dsu_cc[v] = root;
        child = v;
        v = p;
    dsu_cc_size[root] = dsu_cc_size[child];

void merge_path (int a, int b) {
    vector<int> path_a, path_b;
    int lca = -1;
    while (lca == -1) {
        if (a != -1) {
            a = find_2ecc(a);
            if (last_visit[a] == lca_iteration){
                lca = a;
            last_visit[a] = lca_iteration;
            a = par[a];
        if (b != -1) {
            b = find_2ecc(b);
            if (last_visit[b] == lca_iteration){
                lca = b;
            last_visit[b] = lca_iteration;
            b = par[b];


    for (int v : path_a) {
        dsu_2ecc[v] = lca;
        if (v == lca)
    for (int v : path_b) {
        dsu_2ecc[v] = lca;
        if (v == lca)

void add_edge(int a, int b) {
    a = find_2ecc(a);
    b = find_2ecc(b);
    if (a == b)

    int ca = find_cc(a);
    int cb = find_cc(b);

    if (ca != cb) {
        if (dsu_cc_size[ca] > dsu_cc_size[cb]) {
            swap(a, b);
            swap(ca, cb);
        par[a] = dsu_cc[a] = b;
        dsu_cc_size[cb] += dsu_cc_size[a];
    } else {
        merge_path(a, b);

The DSU for the 2-edge-connected components is stored in the vector dsu_2ecc, and the function returning the representative is find_2ecc(v). This function is used many times in the rest of the code, since after the compression of several vertices into one all these vertices cease to exist, and instead only the leader has the correct ancestor par in the forest of 2-edge-connected components.

The DSU for the connected components is stored in the vector dsu_cc, and there is also an additional vector dsu_cc_size to store the component sizes. The function find_cc(v) returns the leader of the connectivity component (which is actually the root of the tree).

The re-rooting of a tree make_root(v) works as descibed above: if traverses from the vertex $v$ via the ancestors to the root vertex, each time redirecting the ancestor par in the opposite direction. The link to the representative of the connected component dsu_cc is also updated, so that it points to the new root vertex. After re-rooting we have to assign the new root the correct size of the connected component. Also we have to be careful that we call find_2ecc() to get the representatives of the 2-edge-connected component, rather than some other vertex that have already been compressed.

The cycle finding and compression function merge_path(a, b) is also implemented as descibed above. It searches for the LCA of $a$ and $b$ be rising these nodes in parallel, until we meet a vertex for the second time. For efficiency purposes we choose a unique identifier for each LCA finding call, and mark the traversed vertices with it. This works in $O(1)$, while other approaches like using $set$ perform worse. The passed paths are stored in the vectors path_a and path_b, and we use them to walk through them a second time up to the LCA, thereby obtaining all vertices of the cycle. All the vertices of the cycle get compressed by attaching them to the LCA, hence the average complexity is $O(\log n)$ (since we don't use Union by rank). All the edges we pass have been bridges, so we subtract 1 for each edge in the cycle.

Finally the query function add_edge(a, b) determines the connected components in which the vertices $a$ and $b$ lie. If they lie in different connectivity components, then a smaller tree is re-rooted and then attached to the larger tree. Otherwise if the vertices $a$ and $b$ lie in one tree, but in different 2-edge-connected components, then the function merge_path(a, b) is called, which will detect the cycle and compress it into one 2-edge-connected component.