Detect Cycle in a Directed Graph

Given the root of a Directed graph, The task is to check whether the graph contains a cycle or not. 

Examples:

Input: N = 4, E = 6

Example of graph

Output: Yes
Explanation: The diagram clearly shows a cycle 0 -> 2 -> 0

Input: N = 4, E = 4

Output: No
Explanation: The diagram clearly shows no cycle

Approach:

The problem can be solved based on the following idea:

To find cycle in a directed graph we can use the Depth First Traversal (DFS) technique. It is based on the idea that there is a cycle in a graph only if there is a back edge [i.e., a node points to one of its ancestors] present in the graph.

To detect a back edge, we need to keep track of the nodes visited till now and the nodes that are in the current recursion stack [i.e., the current path that we are visiting]. If during recursion, we reach a node that is already in the recursion stack, there is a cycle present in the graph.

Note: If the graph is disconnected then get the DFS forest and check for a cycle in individual trees by checking back edges.

Follow the below steps to Implement the idea:

• Create a recursive dfs function that has the following parameters – current vertex, visited array, and recursion stack.
• Mark the current node as visited and also mark the index in the recursion stack.
• Iterate a loop for all the vertices and for each vertex, call the recursive function if it is not yet visited (This step is done to make sure that if there is a forest of graphs, we are checking each forest):
  • In each recursion call, Find all the adjacent vertices of the current vertex which are not visited:
    • If an adjacent vertex is already marked in the recursion stack then return true.
    • Otherwise, call the recursive function for that adjacent vertex.
  • While returning from the recursion call, unmark the current node from the recursion stack, to represent that the current node is no longer a part of the path being traced.
• If any of the functions returns true, stop the future function calls and return true as the answer.

Illustration:

Consider the following graph:

Example of a Directed Graph

Consider we start the iteration from vertex 0.

• Initially, 0 will be marked in both the visited[] and recStack[] array as it is a part of the current path.

Vertex 0 is visited

• Now 0 has two adjacent vertices 1 and 2. Let us consider traversal to the vertex 1. So 1 will be marked in both visited[] and recStack[].

Vertex 1 is visited

• Vertex 1 has only one adjacent vertex. Call the recursive function for 2 and mark it in visited[] and recStack[].

Vertex 2 is visited

• Vertex 2 also has two adjacent vertices. 
  • Vertex 0 is visited and already marked in the recStack[]. So if 0 is checked first, we will get the answer that there is a cycle present.
  • On the other hand, if vertex 3 is checked first, then 3 will be marked in visited[] and recStack[].

Vertex 3 is visited

• While returning from the recursion call for 3, it will be unmarked from recStack[] as it is now not a part of the path currently being traced.

Vertex 3 is unmarked from recStack[]

• Now we have only one option to check, vertex 0, which is already marked in recStack[]. 

So, we can conclude that a cycle exists. We can also find the cycle if we have traversed to vertex 2 from 0 itself in this same way.

Below is the implementation of the above approach:

C++

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

class Graph {
    // No. of vertices
    int V;

    // Pointer to an array containing adjacency lists
    list<int>* adj;

    // Used by isCyclic()
    bool isCyclicUtil(int v, bool visited[], bool* rs);

public:
    Graph(int V);
    void addEdge(int v, int w);
    bool isCyclic();
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}

void Graph::addEdge(int v, int w)
{
    // Add w to v’s list.
    adj[v].push_back(w);
}

// DFS function to find if a cycle exists
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;
        }
    }

    // Remove the vertex from recursion stack
    recStack[v] = false;
    return false;
}

// Returns true if the graph contains a cycle, else false
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 (!visited[i]
            && isCyclicUtil(i, visited, recStack))
            return true;

    return false;
}

// Driver code
int main()
{
    // Create a graph
    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);

    // Function call
    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<>());
    }

    // Function to check if cycle exists
    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.
    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);

        // Function call
        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.

Python3

# 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 popped 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 + 1)
        recStack = [False] * (self.V + 1)
        for node in range(self.V):
            if visited[node] == False:
                if self.isCyclicUtil(node, visited, recStack) == True:
                    return True
        return False


# Driver code
if __name__ == '__main__':
    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 contains cycle")
    else:
        print("Graph doesn't contain 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>());
    }

    // Function to check if cycle exists
    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
    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);

        // Function call
        if (graph.isCyclic())
            Console.WriteLine("Graph contains cycle");
        else
            Console.WriteLine("Graph doesn't "
                              + "contain cycle");
    }
}

// This code contributed by Rajput-Ji

Javascript

// A JavaScript Program to detect cycle in a graph

let V;
let adj=[];
function Graph(v)
{
    V=v;
    for (let i = 0; i < V; i++)
        adj.push([]);
}

// Function to check if cycle exists
function isCyclicUtil(i,visited,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;
        let children = adj[i];

        for (let c=0;c< children.length;c++)
            if (isCyclicUtil(children, visited, recStack))
                return true;

        recStack[i] = false;

        return false;
}

function addEdge(source,dest)
{
    adj .push(dest);
}

// Returns true if the graph contains a
// cycle, else false.
function isCyclic()
{
    // Mark all the vertices as not visited and
        // not part of recursion stack
        let visited = new Array(V);
        let recStack = new Array(V);
        for(let i=0;i<V;i++)
        {
            visited[i]=false;
            recStack[i]=false;
        }


        // Call the recursive helper function to
        // detect cycle in different DFS trees
        for (let i = 0; i < V; i++)
            if (isCyclicUtil(i, visited, recStack))
                return true;

        return false;
}

// Driver code
Graph(4);
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 2);
addEdge(2, 0);
addEdge(2, 3);
addEdge(3, 3);

if(isCyclic())
    console.log("Graph contains cycle");
else
    console.log("Graph doesn't "
                   + "contain cycle");


// This code is contributed by patel2127

Output

Graph contains cycle

Time Complexity: O(V + E), the Time Complexity of this method is the same as the time complexity of DFS traversal which is O(V+E).
Auxiliary Space: 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