Detect Cycle in a Directed Graph using BFS

Given a directed graph, check whether the graph contains a cycle or not. Your function should return true if the given graph contains at least one cycle, else return false. For example, the following graph contains two cycles 0->1->2->3->0 and 2->4->2, so your function must return true.

We have discussed a DFS based solution to detect cycle in a directed graph. In this post, BFS based solution is discussed.



The idea is to simply use Kahn’s algorithm for Topological Sorting

Steps involved in detecting cycle in a directed graph using BFS.

Step-1: Compute in-degree (number of incoming edges) for each of the vertex present in the graph and initialize the count of visited nodes as 0.

Step-2: Pick all the vertices with in-degree as 0 and add them into a queue (Enqueue operation)



Step-3: Remove a vertex from the queue (Dequeue operation) and then.

  1. Increment count of visited nodes by 1.
  2. Decrease in-degree by 1 for all its neighboring nodes.
  3. If in-degree of a neighboring nodes is reduced to zero, then add it to the queue.

Step 4: Repeat Step 3 until the queue is empty.

Step 5: If count of visited nodes is not equal to the number of nodes in the graph has cycle, otherwise not.

How to find in-degree of each node?
There are 2 ways to calculate in-degree of every vertex:
Take an in-degree array which will keep track of
1) Traverse the array of edges and simply increase the counter of the destination node by 1.

for each node in Nodes
    indegree[node] = 0;
for each edge(src,dest) in Edges
    indegree[dest]++

Time Complexity: O(V+E)

2) Traverse the list for every node and then increment the in-degree of all the nodes connected to it by 1.

    for each node in Nodes
        If (list[node].size()!=0) then
        for each dest in list
            indegree[dest]++;

Time Complexity: The outer for loop will be executed V number of times and the inner for loop will be executed E number of times, Thus overall time complexity is O(V+E).

The overall time complexity of the algorithm is O(V+E)

filter_none

edit
close

play_arrow

link
brightness_4
code

// A C++ program to check if there is a cycle in 
// directed graph using BFS.
#include <bits/stdc++.h>
using namespace std;
  
// Class to represent a graph
class Graph {
    int V; // No. of vertices'
  
    // Pointer to an array containing adjacency lisr
    list<int>* adj;
  
public:
    Graph(int V); // Constructor
  
    // function to add an edge to graph
    void addEdge(int u, int v);
  
    // Returns true if there is a cycle in the graph
    // else false.
    bool isCycle();
};
  
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
  
void Graph::addEdge(int u, int v)
{
    adj[u].push_back(v);
}
  
// This function returns true if there is a cycle
// in directed graph, else returns false.
bool Graph::isCycle()
{
    // Create a vector to store indegrees of all
    // vertices. Initialize all indegrees as 0.
    vector<int> in_degree(V, 0);
  
    // Traverse adjacency lists to fill indegrees of
    // vertices. This step takes O(V+E) time
    for (int u = 0; u < V; u++) {
        for (auto v : adj[u])
            in_degree[v]++;
    }
  
    // Create an queue and enqueue all vertices with
    // indegree 0
    queue<int> q;
    for (int i = 0; i < V; i++)
        if (in_degree[i] == 0)
            q.push(i);
  
    // Initialize count of visited vertices
    int cnt = 0;
  
    // Create a vector to store result (A topological
    // ordering of the vertices)
    vector<int> top_order;
  
    // One by one dequeue vertices from queue and enqueue
    // adjacents if indegree of adjacent becomes 0
    while (!q.empty()) {
  
        // Extract front of queue (or perform dequeue)
        // and add it to topological order
        int u = q.front();
        q.pop();
        top_order.push_back(u);
  
        // Iterate through all its neighbouring nodes
        // of dequeued node u and decrease their in-degree
        // by 1
        list<int>::iterator itr;
        for (itr = adj[u].begin(); itr != adj[u].end(); itr++)
  
            // If in-degree becomes zero, add it to queue
            if (--in_degree[*itr] == 0)
                q.push(*itr);
  
        cnt++;
    }
  
    // Check if there was a cycle
    if (cnt != V) 
        return true;
    else
        return false;
}
  
// Driver program to test above functions
int main()
{
    // Create a graph given in the above diagram
    Graph g(6);
    g.addEdge(0, 1);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(3, 4);
    g.addEdge(4, 5);
  
    if (g.isCycle())
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

chevron_right


Output:

Yes

Time Complexity : O(V+E)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : MilanToriya