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.
- Increment count of visited nodes by 1.
- Decrease in-degree by 1 for all its neighboring nodes.
- 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)
C++
// 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 graphclass Graph { int V; // No. of vertices' // Pointer to an array containing adjacency list 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 // 1 For src Node int cnt = 1; // 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); //while we are pushing elements to the queue we will incrementing the cnt cnt++; } } // Check if there was a cycle if (cnt != V) return true; else return false;}// Driver program to test above functionsint 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;} |
Java
// Java program to check if there is a cycle in // directed graph using BFS.import java.io.*;import java.util.*;class GFG{ // Class to represent a graph static class Graph { int V; // No. of vertices' // Pointer to an array containing adjacency list Vector<Integer>[] adj; @SuppressWarnings("unchecked") Graph(int V) { // Constructor this.V = V; this.adj = new Vector[V]; for (int i = 0; i < V; i++) adj[i] = new Vector<>(); } // function to add an edge to graph void addEdge(int u, int v) { adj[u].add(v); } // Returns true if there is a cycle in the graph // else false. // This function returns true if there is a cycle // in directed graph, else returns false. boolean isCycle() { // Create a vector to store indegrees of all // vertices. Initialize all indegrees as 0. int[] in_degree = new int[this.V]; Arrays.fill(in_degree, 0); // Traverse adjacency lists to fill indegrees of // vertices. This step takes O(V+E) time for (int u = 0; u < V; u++) { for (int v : adj[u]) in_degree[v]++; } // Create an queue and enqueue all vertices with // indegree 0 Queue<Integer> q = new LinkedList<Integer>(); for (int i = 0; i < V; i++) if (in_degree[i] == 0) q.add(i); // Initialize count of visited vertices int cnt = 0; // Create a vector to store result (A topological // ordering of the vertices) Vector<Integer> top_order = new Vector<>(); // One by one dequeue vertices from queue and enqueue // adjacents if indegree of adjacent becomes 0 while (!q.isEmpty()) { // Extract front of queue (or perform dequeue) // and add it to topological order int u = q.poll(); top_order.add(u); // Iterate through all its neighbouring nodes // of dequeued node u and decrease their in-degree // by 1 for (int itr : adj[u]) if (--in_degree[itr] == 0) q.add(itr); cnt++; } // Check if there was a cycle if (cnt != this.V) return true; else return false; } } // Driver Code public static void main(String[] args) { // Create a graph given in the above diagram Graph g = new Graph(6); g.addEdge(0, 1); g.addEdge(1, 2); g.addEdge(2, 0); g.addEdge(3, 4); g.addEdge(4, 5); if (g.isCycle()) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by// sanjeev2552 |
Python3
# A Python3 program to check if there is a cycle in # directed graph using BFS. import mathimport sysfrom collections import defaultdict# Class to represent a graph class Graph: def __init__(self,vertices): self.graph=defaultdict(list) self.V=vertices # No. of vertices' # function to add an edge to graph def addEdge(self,u,v): self.graph[u].append(v)# This function returns true if there is a cycle # in directed graph, else returns false. def isCycleExist(n,graph): # Create a vector to store indegrees of all # vertices. Initialize all indegrees as 0. in_degree=[0]*n # Traverse adjacency lists to fill indegrees of # vertices. This step takes O(V+E) time for i in range(n): for j in graph[i]: in_degree[j]+=1 # Create an queue and enqueue all vertices with # indegree 0 queue=[] for i in range(len(in_degree)): if in_degree[i]==0: queue.append(i) # Initialize count of visited vertices cnt=0 # One by one dequeue vertices from queue and enqueue # adjacents if indegree of adjacent becomes 0 while(queue): # Extract front of queue (or perform dequeue) # and add it to topological order nu=queue.pop(0) # Iterate through all its neighbouring nodes # of dequeued node u and decrease their in-degree # by 1 for v in graph[nu]: in_degree[v]-=1 # If in-degree becomes zero, add it to queue if in_degree[v]==0: queue.append(v) cnt+=1 # Check if there was a cycle if cnt==n: return False else: return True # Driver program to test above functions if __name__=='__main__': # Create a graph given in the above diagram g=Graph(6) g.addEdge(0,1) g.addEdge(1,2) g.addEdge(2,0) g.addEdge(3,4) g.addEdge(4,5) if isCycleExist(g.V,g.graph): print("Yes") else: print("No")# This Code is Contributed by Vikash Kumar 37 |
C#
// C# program to check if there is a cycle in // directed graph using BFS.using System;using System.Collections.Generic;class GFG{ // Class to represent a graphpublic class Graph{ // No. of vertices' public int V; // Pointer to an array containing // adjacency list public List<int>[] adj; public Graph(int V) { // Constructor this.V = V; this.adj = new List<int>[V]; for (int i = 0; i < V; i++) adj[i] = new List<int>(); } // Function to add an edge to graph public void addEdge(int u, int v) { adj[u].Add(v); } // Returns true if there is a cycle in the // graph else false. // This function returns true if there is // a cycle in directed graph, else returns // false. public bool isCycle() { // Create a vector to store indegrees of all // vertices. Initialize all indegrees as 0. int[] in_degree = new int[this.V]; // Traverse adjacency lists to fill indegrees // of vertices. This step takes O(V+E) time for(int u = 0; u < V; u++) { foreach(int v in adj[u]) in_degree[v]++; } // Create an queue and enqueue all // vertices with indegree 0 Queue<int> q = new Queue<int>(); for(int i = 0; i < V; i++) if (in_degree[i] == 0) q.Enqueue(i); // Initialize count of visited vertices int cnt = 0; // Create a vector to store result // (A topological ordering of the // vertices) List<int> top_order = new List<int>(); // One by one dequeue vertices from // queue and enqueue adjacents if // indegree of adjacent becomes 0 while (q.Count != 0) { // Extract front of queue (or perform // dequeue) and add it to topological // order int u = q.Peek(); q.Dequeue(); top_order.Add(u); // Iterate through all its neighbouring // nodes of dequeued node u and decrease // their in-degree by 1 foreach(int itr in adj[u]) if (--in_degree[itr] == 0) q.Enqueue(itr); cnt++; } // Check if there was a cycle if (cnt != this.V) return true; else return false; }}// Driver Codepublic static void Main(String[] args) { // Create a graph given in the above diagram Graph g = new Graph(6); g.addEdge(0, 1); g.addEdge(1, 2); g.addEdge(2, 0); g.addEdge(3, 4); g.addEdge(4, 5); if (g.isCycle()) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by Princi Singh |
Javascript
<script>// JavaScript program to check if there is a cycle in // directed graph using BFS.// Class to represent a graph// No. of vertices'var V = 0; // Pointer to an array containing// adjacency listvar adj ;function initialize(v){ // Constructor V = v; adj = Array.from(Array(V), ()=>Array(V));}// Function to add an edge to graphfunction addEdge(u, v){ adj[u].push(v);}// Returns true if there is a cycle in the// graph else false.// This function returns true if there is // a cycle in directed graph, else returns// false.function isCycle() { // Create a vector to store indegrees of all // vertices. Initialize all indegrees as 0. var in_degree = Array(V).fill(0); // Traverse adjacency lists to fill indegrees // of vertices. This step takes O(V+E) time for(var u = 0; u < V; u++) { for(var v of adj[u]) in_degree[v]++; } // Create an queue and enqueue all // vertices with indegree 0 var q = []; for(var i = 0; i < V; i++) if (in_degree[i] == 0) q.push(i); // Initialize count of visited vertices var cnt = 0; // Create a vector to store result // (A topological ordering of the // vertices) var top_order = []; // One by one dequeue vertices from // queue and enqueue adjacents if // indegree of adjacent becomes 0 while (q.length != 0) { // Extract front of queue (or perform // dequeue) and add it to topological // order var u = q[0]; q.shift(); top_order.push(u); // Iterate through all its neighbouring // nodes of dequeued node u and decrease // their in-degree by 1 for(var itr of adj[u]) if (--in_degree[itr] == 0) q.push(itr); cnt++; } // Check if there was a cycle if (cnt != V) return true; else return false;}// Create a graph given in the above diagraminitialize(6)addEdge(0, 1);addEdge(1, 2);addEdge(2, 0);addEdge(3, 4);addEdge(4, 5);if (isCycle()) document.write("Yes");else document.write("No");</script> |
Output:
Yes
Time Complexity: O(V+E)
Auxiliary Space: O(V)
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