Bridges in a connected graph

Question

I have a programming task(not homework.) where I have to find the bridges in a graph. I worked on it a bit myself, but could not come up with anything satisfactory. So i googled it , I did find something but I am unable to understand the algorithm as it is presented. Could someone please take a look at this code and give me an explanation.?

public Bridge(Graph G) {
    low = new int[G.V()];
    pre = new int[G.V()];
    for (int v = 0; v < G.V(); v++) low[v] = -1;
    for (int v = 0; v < G.V(); v++) pre[v] = -1;

    for (int v = 0; v < G.V(); v++)
        if (pre[v] == -1)
            dfs(G, v, v);
}

public int components() { return bridges + 1; }

private void dfs(Graph G, int u, int v) {
    pre[v] = cnt++;
    low[v] = pre[v];
    for (int w : G.adj(v)) {
        if (pre[w] == -1) {
            dfs(G, v, w);
            low[v] = Math.min(low[v], low[w]);
            if (low[w] == pre[w]) {
                StdOut.println(v + "-" + w + " is a bridge");
                bridges++;
            }
        }

        // update low number - ignore reverse of edge leading to v
        else if (w != u)
            low[v] = Math.min(low[v], pre[w]);
    }
}

Answer 1

Def: Bridge is an edge, when removed, will disconnect the graph (or increase the number of connected components by 1).

One observation regarding bridges in graph; none of the edges that belong to a loop can be a bridge. So in a graph such as A--B--C--A, removing any of the edge A--B, B--C and C--A will not disconnect the graph. But, for an undirected graph, the edge A--B implies B--A; and this edge could still be a bridge, where the only loop it is in is A--B--A. So, we should consider only those loops formed by a back edge. This is where the parent information you've passed in the function argument helps. It will help you to not use the loops such as A--B--A.

Now to identify the back edge (or the loop), A--B--C--A we use the low and pre arrays. The array pre is like the visited array in the dfs algorithm; but instead of just flagging that the vertex as visited, we identify each vertex with a different number (according to its position in the dfs tree). The low array helps to identify if there is a loop. The low array identifies the lowest numbered (from pre array) vertex that the current vertex can reach.

Lets work through this graph A--B--C--D--B.

Starting at A

dfs:   ^                 ^                 ^                 ^              ^
pre:   0 -1 -1 -1 -1  0--1 -1 -1  1  0--1--2 -1  1  0--1--2--3  1  0--1--2--3--1
graph: A--B--C--D--B  A--B--C--D--B  A--B--C--D--B  A--B--C--D--B  A--B--C--D--B
low:   0 -1 -1 -1 -1  0--1 -1 -1  1  0--1--2 -1  1  0--1--2--3  1  0--1--2--3->1

At this point, you've encountered a cycle/loop in graph. In your code if (pre[w] == -1) will be false this time. So, you'll enter the else part. The if statement there is checking if B is the parent vertex of D. It is not, so D will absorb B's pre value into low. Continuing the example,

dfs:            ^
pre:   0--1--2--3
graph: A--B--C--D
low:   0--1--2--1   

This low value of D propagates back to C through the code low[v] = Math.min(low[v], low[w]);.

dfs:         ^           ^           ^
pre:   0--1--2--3--1  0--1--2--3--1  0--1--2--3--1
graph: A--B--C--D--B  A--B--C--D--B  A--B--C--D--B
low:   0--1--1--1--1  0--1--1--1--1  0--1--1--1--1

Now, that the cycle/loop is identified, we note that the vertex A is not part of the loop. So, you print out A--B as a bridge. The code low['B'] == pre['B'] means an edge to B will be a bridge. This is because, the lowest vertex we can reach from B is B itself.

Hope this explanation helps.

Answer 2

Not a new answer, but I needed this in Python. Here's a translation of the algorithm for an undirected NetworkX Graph object G:

def bridge_dfs(G,u,v,cnt,low,pre,bridges):
    cnt    += 1
    pre[v]  = cnt
    low[v]  = pre[v]

    for w in nx.neighbors(G,v):
        if (pre[w] == -1):
            bridge_dfs(G,v,w,cnt,low,pre,bridges)

            low[v] = min(low[v], low[w])
            if (low[w] == pre[w]):
                bridges.append((v,w))

        elif (w != u):
            low[v] = min(low[v], pre[w])

def get_bridges(G):
    bridges = []
    cnt     = 0
    low     = {n:-1 for n in G.nodes()}
    pre     = low.copy()

    for n in G.nodes():
         bridge_dfs(G, n, n, cnt, low, pre, bridges)

    return bridges # <- List of (node-node) tuples for all bridges in G

Be careful of Python's recursion depth limiter for large graphs...

Answer 3

Not a new answer, but I needed this for the JVM/Kotlin. Here's a translation that relies upon com.google.common.graph.Graph.

/**
 * [T] The type of key held in the [graph].
 */
private class BridgeComputer<T>(private val graph: ImmutableGraph<T>) {
    /**
     * Counter.
     */
    private var count = 0
    /**
     * `low[v]` = Lowest preorder of any vertex connected to `v`.
     */
    private val low: MutableMap<T, Int> =
        graph.nodes().map { it to -1 }.toMap(mutableMapOf())
    /**
     * `pre[v]` = Order in which [depthFirstSearch] examines `v`.
     */
    private val pre: MutableMap<T, Int> =
        graph.nodes().map { it to -1 }.toMap(mutableMapOf())

    private val foundBridges = mutableSetOf<Pair<T, T>>()

    init {
        graph.nodes().forEach { v ->
            // DO NOT PRE-FILTER!
            if (pre[v] == -1) {
                depthFirstSearch(v, v)
            }
        }
    }

    private fun depthFirstSearch(u: T, v: T) {
        pre[v] = count++
        low[v] = checkNotNull(pre[v]) { "pre[v]" }
        graph.adjacentNodes(v).forEach { w ->
            if (pre[w] == -1) {
                depthFirstSearch(v, w)
                low[v] =
                    Math.min(checkNotNull(low[v]) { "low[v]" }, checkNotNull(low[w]) { "low[w]" })
                if (low[w] == pre[w]) {
                    println("$v - $w is a bridge")
                    foundBridges += (v to w)
                }
            } else if (w != u) {
                low[v] =
                    Math.min(checkNotNull(low[v]) { "low[v]" }, checkNotNull(pre[w]) { "pre[w]" })
            }
        }
    }

    /**
     * Holds the computed bridges.
     */
    fun bridges() = ImmutableSet.copyOf(foundBridges)!!
}

Hopefully this makes someone's life easier.

Answer 4

Lets say you are given an edge (c,d) and you have to find if it is a bridge or not
There are several methods to solve this problem but lets concentrate on one.

• Starting from c, you have to do a BFS.
• If there is an edge c-d then don't visit it.
• Keep a track of vertex by making a boolean visited.

In the end, if you found that d is visited, this means, by removing c-d we can still visit d from the source c, hence c-d is not a bridge. Here is the short implementation of the above :

int isBridge(int V, ArrayList<ArrayList<Integer>> adj,int c,int d)
{
    Queue<Integer> q = new LinkedList<>();
    boolean visited[] = new boolean[V];
    ArrayList<Integer> ls = new ArrayList<>();
    
    q.add(c);
    
    while(!q.isEmpty()) {
        
        Integer v = q.remove();
        
        if(visited[v])
            continue;

        visited[v] = true;
        ls.add(v);
        
        for(Integer e: adj.get(v)) {
            
            if(visited[e] || (c == v && d == e))
                continue;
            q.add(e);
        }
    }
    
    if(visited[d] == true)
        return 0;
    return 1;
    
}