⭐ Must Read: web//https://www.geeksforgeeks.org/detect cycle in a graph - สรุปดราม่าล่าสุด

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.

Example,

Input: n = 4, e = 6
0 -> 1, 0 -> 2, 1 -> 2, 2 -> 0, 2 -> 3, 3 -> 3
Output: Yes
Explanation:
Diagram:

The diagram clearly shows a cycle 0 -> 2 -> 0


Input:n = 4, e = 3
0 -> 1, 0 -> 2, 1 -> 2, 2 -> 3
Output:No
Explanation:
Diagram:

The diagram clearly shows no cycle

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Solution using Depth First Search or DFS

Approach: Depth First Traversal can be used to detect a cycle in a Graph. DFS for a connected graph produces a tree. There is a cycle in a graph only if there is a back edge present in the graph. A back edge is an edge that is from a node to itself (self-loop) or one of its ancestors in the tree produced by DFS. In the following graph, there are 3 back edges, marked with a cross sign. We can observe that these 3 back edges indicate 3 cycles present in the graph.

For a disconnected graph, Get the DFS forest as output. To detect cycle, check for a cycle in individual trees by checking back edges.

To detect a back edge, keep track of vertices currently in the recursion stack of function for DFS traversal. If a vertex is reached that is already in the recursion stack, then there is a cycle in the tree. The edge that connects the current vertex to the vertex in the recursion stack is a back edge. Use recStack[] array to keep track of vertices in the recursion stack.

Dry run of the above approach:
Algorithm:
1. Create the graph using the given number of edges and vertices.
2. Create a recursive function that initializes the current index or vertex, visited, and recursion stack.
3. Mark the current node as visited and also mark the index in recursion stack.
4. Find all the vertices which are not visited and are adjacent to the current node. Recursively call the function for those vertices, If the recursive function returns true, return true.
5. If the adjacent vertices are already marked in the recursion stack then return true.
6. Create a wrapper class, that calls the recursive function for all the vertices and if any function returns true return true. Else if for all vertices the function returns false return false.

Implementation:

C++

// A C++ Program to detect cycle in a graph 
#include<bits/stdc++.h> 
  
using namespace std; 
  
class Graph 
{ 
    int V;    // No. of vertices 
    list<int> *adj;    // Pointer to an array containing adjacency lists 
    bool isCyclicUtil(int v, bool visited[], bool *rs);  // used by isCyclic() 
public: 
    Graph(int V);   // Constructor 
    void addEdge(int v, int w);   // to add an edge to graph 
    bool isCyclic();    // returns true if there is a cycle in this graph 
}; 
  
Graph::Graph(int V) 
{ 
    this->V = V; 
    adj = new list<int>[V]; 
} 
  
void Graph::addEdge(int v, int w) 
{ 
    adj[v].push_back(w); // Add w to v’s list. 
} 
  
// This function is a variation of DFSUtil() in https://www.geeksforgeeks.org/archives/18212 
bool Graph::isCyclicUtil(int v, bool visited[], bool *recStack) 
{ 
    if(visited[v] == false) 
    { 
        // Mark the current node as visited and part of recursion stack 
        visited[v] = true; 
        recStack[v] = true; 
  
        // Recur for all the vertices adjacent to this vertex 
        list<int>::iterator i; 
        for(i = adj[v].begin(); i != adj[v].end(); ++i) 
        { 
            if ( !visited[*i] && isCyclicUtil(*i, visited, recStack) ) 
                return true; 
            else if (recStack[*i]) 
                return true; 
        } 
  
    } 
    recStack[v] = false;  // remove the vertex from recursion stack 
    return false; 
} 
  
// Returns true if the graph contains a cycle, else false. 
// This function is a variation of DFS() in https://www.geeksforgeeks.org/archives/18212 
bool Graph::isCyclic() 
{ 
    // Mark all the vertices as not visited and not part of recursion 
    // stack 
    bool *visited = new bool[V]; 
    bool *recStack = new bool[V]; 
    for(int i = 0; i < V; i++) 
    { 
        visited[i] = false; 
        recStack[i] = false; 
    } 
  
    // Call the recursive helper function to detect cycle in different 
    // DFS trees 
    for(int i = 0; i < V; i++) 
        if (isCyclicUtil(i, visited, recStack)) 
            return true; 
  
    return false; 
} 
  
int main() 
{ 
    // Create a graph given in the above diagram 
    Graph g(4); 
    g.addEdge(0, 1); 
    g.addEdge(0, 2); 
    g.addEdge(1, 2); 
    g.addEdge(2, 0); 
    g.addEdge(2, 3); 
    g.addEdge(3, 3); 
  
    if(g.isCyclic()) 
        cout << "Graph contains cycle"; 
    else
        cout << "Graph doesn't contain cycle"; 
    return 0; 
} 

Java

// A Java Program to detect cycle in a graph 
import java.util.ArrayList; 
import java.util.LinkedList; 
import java.util.List; 
  
class Graph { 
      
    private final int V; 
    private final List<List<Integer>> adj; 
  
    public Graph(int V)  
    { 
        this.V = V; 
        adj = new ArrayList<>(V); 
          
        for (int i = 0; i < V; i++) 
            adj.add(new LinkedList<>()); 
    } 
      
    // This function is a variation of DFSUtil() in  
    // https://www.geeksforgeeks.org/archives/18212 
    private boolean isCyclicUtil(int i, boolean[] visited, 
                                      boolean[] recStack)  
    { 
          
        // Mark the current node as visited and 
        // part of recursion stack 
        if (recStack[i]) 
            return true; 
  
        if (visited[i]) 
            return false; 
              
        visited[i] = true; 
  
        recStack[i] = true; 
        List<Integer> children = adj.get(i); 
          
        for (Integer c: children) 
            if (isCyclicUtil(c, visited, recStack)) 
                return true; 
                  
        recStack[i] = false; 
  
        return false; 
    } 
  
    private void addEdge(int source, int dest) { 
        adj.get(source).add(dest); 
    } 
  
    // Returns true if the graph contains a  
    // cycle, else false. 
    // This function is a variation of DFS() in  
    // https://www.geeksforgeeks.org/archives/18212 
    private boolean isCyclic()  
    { 
          
        // Mark all the vertices as not visited and 
        // not part of recursion stack 
        boolean[] visited = new boolean[V]; 
        boolean[] recStack = new boolean[V]; 
          
          
        // Call the recursive helper function to 
        // detect cycle in different DFS trees 
        for (int i = 0; i < V; i++) 
            if (isCyclicUtil(i, visited, recStack)) 
                return true; 
  
        return false; 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        Graph graph = new Graph(4); 
        graph.addEdge(0, 1); 
        graph.addEdge(0, 2); 
        graph.addEdge(1, 2); 
        graph.addEdge(2, 0); 
        graph.addEdge(2, 3); 
        graph.addEdge(3, 3); 
          
        if(graph.isCyclic()) 
            System.out.println("Graph contains cycle"); 
        else
            System.out.println("Graph doesn't "
                                    + "contain cycle"); 
    } 
} 
  
// This code is contributed by Sagar Shah. 

Python

# Python program to detect cycle  
# in a graph 
  
from collections import defaultdict 
  
class Graph(): 
    def __init__(self,vertices): 
        self.graph = defaultdict(list) 
        self.V = vertices 
  
    def addEdge(self,u,v): 
        self.graph[u].append(v) 
  
    def isCyclicUtil(self, v, visited, recStack): 
  
        # Mark current node as visited and  
        # adds to recursion stack 
        visited[v] = True
        recStack[v] = True
  
        # Recur for all neighbours 
        # if any neighbour is visited and in  
        # recStack then graph is cyclic 
        for neighbour in self.graph[v]: 
            if visited[neighbour] == False: 
                if self.isCyclicUtil(neighbour, visited, recStack) == True: 
                    return True
            elif recStack[neighbour] == True: 
                return True
  
        # The node needs to be poped from  
        # recursion stack before function ends 
        recStack[v] = False
        return False
  
    # Returns true if graph is cyclic else false 
    def isCyclic(self): 
        visited = [False] * self.V 
        recStack = [False] * self.V 
        for node in range(self.V): 
            if visited[node] == False: 
                if self.isCyclicUtil(node,visited,recStack) == True: 
                    return True
        return False
  
g = Graph(4) 
g.addEdge(0, 1) 
g.addEdge(0, 2) 
g.addEdge(1, 2) 
g.addEdge(2, 0) 
g.addEdge(2, 3) 
g.addEdge(3, 3) 
if g.isCyclic() == 1: 
    print "Graph has a cycle"
else: 
    print "Graph has no cycle"
  
# Thanks to Divyanshu Mehta for contributing this code 

C#

// A C# Program to detect cycle in a graph  
using System; 
using System.Collections.Generic; 
  
public class Graph {  
      
    private readonly int V;  
    private readonly List<List<int>> adj;  
  
    public Graph(int V)  
    {  
        this.V = V;  
        adj = new List<List<int>>(V);  
          
        for (int i = 0; i < V; i++)  
            adj.Add(new List<int>());  
    }  
      
    // This function is a variation of DFSUtil() in  
    // https://www.geeksforgeeks.org/archives/18212  
    private bool isCyclicUtil(int i, bool[] visited,  
                                    bool[] recStack)  
    {  
          
        // Mark the current node as visited and  
        // part of recursion stack  
        if (recStack[i])  
            return true;  
  
        if (visited[i])  
            return false;  
              
        visited[i] = true;  
  
        recStack[i] = true;  
        List<int> children = adj[i];  
          
        foreach (int c in children)  
            if (isCyclicUtil(c, visited, recStack))  
                return true;  
                  
        recStack[i] = false;  
  
        return false;  
    }  
  
    private void addEdge(int sou, int dest) {  
        adj[sou].Add(dest);  
    }  
  
    // Returns true if the graph contains a  
    // cycle, else false.  
    // This function is a variation of DFS() in  
    // https://www.geeksforgeeks.org/archives/18212  
    private bool isCyclic()  
    {  
          
        // Mark all the vertices as not visited and  
        // not part of recursion stack  
        bool[] visited = new bool[V];  
        bool[] recStack = new bool[V];  
          
          
        // Call the recursive helper function to  
        // detect cycle in different DFS trees  
        for (int i = 0; i < V; i++)  
            if (isCyclicUtil(i, visited, recStack))  
                return true;  
  
        return false;  
    }  
  
    // Driver code  
    public static void Main(String[] args)  
    {  
        Graph graph = new Graph(4);  
        graph.addEdge(0, 1);  
        graph.addEdge(0, 2);  
        graph.addEdge(1, 2);  
        graph.addEdge(2, 0);  
        graph.addEdge(2, 3);  
        graph.addEdge(3, 3);  
          
        if(graph.isCyclic())  
            Console.WriteLine("Graph contains cycle");  
        else
            Console.WriteLine("Graph doesn't "
                                    + "contain cycle");  
    }  
}  
  
// This code contributed by Rajput-Ji 

Output:

Graph contains cycle

Complexity Analysis:
- Time Complexity: O(V+E).
  Time Complexity of this method is same as time complexity of DFS traversal which is O(V+E).
- Space Complexity: O(V).
  To store the visited and recursion stack O(V) space is needed.

In the below article, another O(V + E) method is discussed :
Detect Cycle in a direct graph using colors

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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My Personal Notes

Oct	NOV	Dec
	17
2019	2020	2021

Detect Cycle in a Directed Graph

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python

C#

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