Python Program for Detect Cycle in a Directed Graph

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 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):
    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"
    print "Graph has no cycle"
# Thanks to Divyanshu Mehta for contributing this code



Graph has a cycle

Please refer complete article on Detect Cycle in a Directed Graph for more details!

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