Python Program for Detect Cycle in a Directed Graph

Last Updated : 19 Apr, 2023

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 three cycles 0->2->0, 0->1->2->0 and 3->3, so your function must return true.

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

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 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 
        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 