Detect cycle in an undirected graph using BFS

Last Updated : 17 Apr, 2024

Given an undirected graph, how to check if there is a cycle in the graph? For example, the following graph has a cycle of 1-0-2-1.

cycleGraph

We have discussed cycle detection for the directed graph. We have also discussed a union-find algorithm for cycle detection in undirected graphs.. The time complexity of the union-find algorithm is O(ELogV). Like directed graphs, we can use DFS to detect a cycle in an undirected graph in O(V+E) time. We have discussed DFS based solution for cycle detection in an undirected graph.

In this article, the BFS based solution is discussed. The basic idea is to keep track of visited nodes during BFS traversal. If a node is encountered more than once during BFS, it indicates the presence of a cycle in the graph.

Here are the key steps for detecting cycles:

Start BFS traversal from each unvisited node in the graph.
While traversing, mark each visited node.
If a node is encountered that is already marked as visited, it implies the presence of a cycle.
Continue BFS traversal until all nodes are visited or a cycle is detected.

Implementation:

C++

// C++ Program to detect cycle in an undirected graph using BFS
#include <bits/stdc++.h>
using namespace std;

void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}

bool isCyclicConnected(vector<int> adj[], int s, int V, vector<bool>& visited){
    // Create a queue for BFS
    queue<int> q;

    // Enqueue the current node
    q.push(s);

    while (!q.empty()) {
        // Dequeue a vertex from queue and print it
        int v = q.front();
        q.pop();

        if (visited[v] == 1) {
            return true; // Cycle detected
        }

        visited[v] = 1; // Mark as visited

        // Visit adjacent nodes
        for (auto it : adj[v]) {
            if (visited[it] == 0) {
                q.push(it);
            }
        }
    }
    return false;
}

bool isCyclicDisconnected(vector<int> adj[], int V){
    // Mark all the vertices as not visited
    vector<bool> visited(V, false);

    for (int i = 0; i < V; i++) {
        if (!visited[i]
            && isCyclicConnected(adj, i, V, visited))
            return true;
    }
    return false;
}

// Driver program to test methods of graph class
int main(){
    int V = 4;
    vector<int> adj[V];
    addEdge(adj, 0, 1);
    addEdge(adj, 1, 2);
    addEdge(adj, 2, 0);
    addEdge(adj, 2, 3);

    if (isCyclicDisconnected(adj, V))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}
// This code is contributed by Yash Agarwal(yashagarwal2852002)

Java

import java.util.*;

public class Main {
    public static void addEdge(ArrayList<Integer>[] adj, int u, int v) {
        adj[u].add(v);
        adj[v].add(u);
    }

    public static boolean isCyclicConnected(ArrayList<Integer>[] adj, int s, int V, boolean[] visited) {
        Queue<Integer> q = new LinkedList<>();
        q.add(s);

        while (!q.isEmpty()) {
            int v = q.poll();

            if (visited[v]) {
                return true; // Cycle detected
            }

            visited[v] = true; // Mark as visited

            for (int it : adj[v]) {
                if (!visited[it]) {
                    q.add(it);
                }
            }
        }
        return false;
    }

    public static boolean isCyclicDisconnected(ArrayList<Integer>[] adj, int V) {
        boolean[] visited = new boolean[V];

        for (int i = 0; i < V; i++) {
            if (!visited[i] && isCyclicConnected(adj, i, V, visited)) {
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        int V = 4;
        ArrayList<Integer>[] adj = new ArrayList[V];
        for (int i = 0; i < V; i++) {
            adj[i] = new ArrayList<>();
        }
        addEdge(adj, 0, 1);
        addEdge(adj, 1, 2);
        addEdge(adj, 2, 0);
        addEdge(adj, 2, 3);

        if (isCyclicDisconnected(adj, V))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

from collections import deque

def addEdge(adj, u, v):
    adj[u].append(v)
    adj[v].append(u)

def isCyclicConnected(adj, s, V, visited):
    q = deque()
    q.append(s)

    while q:
        v = q.popleft()

        if visited[v]:
            return True  # Cycle detected

        visited[v] = True  # Mark as visited

        for it in adj[v]:
            if not visited[it]:
                q.append(it)

    return False

def isCyclicDisconnected(adj, V):
    visited = [False] * V

    for i in range(V):
        if not visited[i] and isCyclicConnected(adj, i, V, visited):
            return True

    return False

V = 4
adj = [[] for _ in range(V)]
addEdge(adj, 0, 1)
addEdge(adj, 1, 2)
addEdge(adj, 2, 0)
addEdge(adj, 2, 3)

if isCyclicDisconnected(adj, V):
    print("Yes")
else:
    print("No")

using System;
using System.Collections.Generic;

class MainClass {
    public static void addEdge(List<int>[] adj, int u, int v) {
        adj[u].Add(v);
        adj[v].Add(u);
    }

    public static bool isCyclicConnected(List<int>[] adj, int s, int V, bool[] visited) {
        Queue<int> q = new Queue<int>();
        q.Enqueue(s);

        while (q.Count > 0) {
            int v = q.Dequeue();

            if (visited[v]) {
                return true; // Cycle detected
            }

            visited[v] = true; // Mark as visited

            foreach (int it in adj[v]) {
                if (!visited[it]) {
                    q.Enqueue(it);
                }
            }
        }
        return false;
    }

    public static bool isCyclicDisconnected(List<int>[] adj, int V) {
        bool[] visited = new bool[V];

        for (int i = 0; i < V; i++) {
            if (!visited[i] && isCyclicConnected(adj, i, V, visited)) {
                return true;
            }
        }
        return false;
    }

    public static void Main(string[] args) {
        int V = 4;
        List<int>[] adj = new List<int>[V];
        for (int i = 0; i < V; i++) {
            adj[i] = new List<int>();
        }
        addEdge(adj, 0, 1);
        addEdge(adj, 1, 2);
        addEdge(adj, 2, 0);
        addEdge(adj, 2, 3);

        if (isCyclicDisconnected(adj, V))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

Javascript

function addEdge(adj, u, v) {
    adj[u].push(v);
    adj[v].push(u);
}

function isCyclicConnected(adj, s, V, visited) {
    let q = [];
    q.push(s);

    while (q.length > 0) {
        let v = q.shift();

        if (visited[v]) {
            return true; // Cycle detected
        }

        visited[v] = true; // Mark as visited

        for (let it of adj[v]) {
            if (!visited[it]) {
                q.push(it);
            }
        }
    }
    return false;
}

function isCyclicDisconnected(adj, V) {
    let visited = new Array(V).fill(false);

    for (let i = 0; i < V; i++) {
        if (!visited[i] && isCyclicConnected(adj, i, V, visited)) {
            return true;
        }
    }
    return false;
}

let V = 4;
let adj = [...Array(V)].map(() => []);
addEdge(adj, 0, 1);
addEdge(adj, 1, 2);
addEdge(adj, 2, 0);
addEdge(adj, 2, 3);

if (isCyclicDisconnected(adj, V))
    console.log("Yes");
else
    console.log("No");

Output

Yes

Time Complexity: The program does a simple BFS Traversal of the graph and the graph is represented using an adjacency list. So the time complexity is O(V+E)
Space Complexity: O(V) for visited vector.

kartik

Improve

Traversal of a Graph in lexicographical order using BFS

Detect Cycle in a Directed Graph using BFS

BFS in different language

Variations of BFS implementations

Easy problems on BFS

Intermediate problems on BFS

Hard Problems on BFS

Detect cycle in an undirected graph using BFS

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