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:
This diagram clearly shows a cycle 0 -> 2 -> 0.
Input:n = 4, e = 3
0 -> 1, 0 -> 2, 1 -> 2, 2 -> 3
Output:No
Explanation:
This diagram clearly shows no cycle.
Solution
Approach: Depth First Traversal can be used to detect 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 (selfloop) or one of its ancestor in the tree produced by DFS. In the following graph, there are 3 back edges, marked with cross sign. It can be observed that these 3 back edges indicate 3 cycles present in the graph.
For a disconnected graph, we get the DFS forest as output. To detect cycle, we can check for cycle in individual trees by checking back edges.
In the previous post, we have discussed a solution that stores visited vertices in a separate array which stores vertices of the current recursion call stack.
In this post, a different solution is discussed. The solution is from CLRS book. The idea is to do DFS of a given graph and while doing traversal, assign one of the below three colours to every vertex.
WHITE : Vertex is not processed yet. Initially, all vertices are WHITE.
GRAY: Vertex is being processed (DFS for this vertex has started, but not finished which means that all descendants (in DFS tree) of this vertex are not processed yet (or this vertex is in the function call stack)
BLACK : Vertex and all its descendants are processed. While doing DFS, if an edge is encountered from current vertex to a GRAY vertex, then this edge is back edge and hence there is a cycle.
Algorithm:
- Create a recursive function that takes the edge and color array (this can be also kept as a global variable)
- Mark the current node as GREY.
- Traverse all the adjacent nodes and if any node is marked GREY then return true as a loop is bound to exist.
- If any adjacent vertex is WHITE then call the recursive function for that node. If the function returns true. Return true.
- If no adjacent node is grey or has not returned true then mark the current Node as BLACK and return false.
Implementation:
C++
// A DFS based approach to find if there is a cycle// in a directed graph. This approach strictly follows// the algorithm given in CLRS book.#include <bits/stdc++.h>using namespace std;enum Color {WHITE, GRAY, BLACK};// Graph class represents a directed graph using// adjacency list representationclass Graph{ int V; // No. of vertices list<int>* adj; // adjacency lists // DFS traversal of the vertices reachable from v bool DFSUtil(int v, int color[]);public: Graph(int V); // Constructor // function to add an edge to graph void addEdge(int v, int w); bool isCyclic();};// ConstructorGraph::Graph(int V){ this->V = V; adj = new list<int>[V];}// Utility function to add an edgevoid Graph::addEdge(int v, int w){ adj[v].push_back(w); // Add w to v's list.}// Recursive function to find if there is back edge// in DFS subtree tree rooted with 'u'bool Graph::DFSUtil(int u, int color[]){ // GRAY : This vertex is being processed (DFS // for this vertex has started, but not // ended (or this vertex is in function // call stack) color[u] = GRAY; // Iterate through all adjacent vertices list<int>::iterator i; for (i = adj[u].begin(); i != adj[u].end(); ++i) { int v = *i; // An adjacent of u // If there is if (color[v] == GRAY) return true; // If v is not processed and there is a back // edge in subtree rooted with v if (color[v] == WHITE && DFSUtil(v, color)) return true; } // Mark this vertex as processed color[u] = BLACK; return false;}// Returns true if there is a cycle in graphbool Graph::isCyclic(){ // Initialize color of all vertices as WHITE int *color = new int[V]; for (int i = 0; i < V; i++) color[i] = WHITE; // Do a DFS traversal beginning with all // vertices for (int i = 0; i < V; i++) if (color[i] == WHITE) if (DFSUtil(i, color) == true) return true; return false;}// Driver code to test aboveint 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
import java.io.*;import java.util.*;class GFG { // A DFS based approach to find if there is a cycle // in a directed graph. This approach strictly follows // the algorithm given in CLRS book. static int WHITE = 0, GRAY = 1, BLACK = 2; // Graph class represents a directed graph using // adjacency list representation static class Graph { int V; LinkedList<Integer>[] adjList; // Constructor Graph(int ver) { V = ver; adjList = new LinkedList[V]; for (int i = 0; i < V; i++) adjList[i] = new LinkedList<>(); } } // Utility function to add an edge static void addEdge(Graph g, int u, int v) { g.adjList[u].add(v); // Add v to u's list. } // Recursive function to find if there is back edge // in DFS subtree tree rooted with 'u' static boolean DFSUtil(Graph g, int u, int[] color) { // GRAY : This vertex is being processed (DFS // for this vertex has started, but not // ended (or this vertex is in function // call stack) color[u] = GRAY; // Iterate through all adjacent vertices for (Integer in : g.adjList[u]) { // If there is if (color[in] == GRAY) return true; // If v is not processed and there is a back // edge in subtree rooted with v if (color[in] == WHITE && DFSUtil(g, in, color) == true) return true; } // Mark this vertex as processed color[u] = BLACK; return false; } // Returns true if there is a cycle in graph static boolean isCyclic(Graph g) { // Initialize color of all vertices as WHITE int[] color = new int[g.V]; for (int i = 0; i < g.V; i++) { color[i] = WHITE; } // Do a DFS traversal beginning with all // vertices for (int i = 0; i < g.V; i++) { if (color[i] == WHITE) { if(DFSUtil(g, i, color) == true) return true; } } return false; } // Driver code to test above public static void main(String args[]) { // Create a graph given in the above diagram Graph g = new Graph(4); addEdge(g, 0, 1); addEdge(g, 0, 2); addEdge(g, 1, 2); addEdge(g, 2, 0); addEdge(g, 2, 3); addEdge(g, 3, 3); if (isCyclic(g)) System.out.println("Graph contains cycle"); else System.out.println("Graph doesn't contain cycle"); }}// This code is contributed by rachana soma |
Python3
# Python program to detect cycle in # a directed graphfrom collections import defaultdictclass Graph(): def __init__(self, V): self.V = V self.graph = defaultdict(list) def addEdge(self, u, v): self.graph[u].append(v) def DFSUtil(self, u, color): # GRAY : This vertex is being processed (DFS # for this vertex has started, but not # ended (or this vertex is in function # call stack) color[u] = "GRAY" for v in self.graph[u]: if color[v] == "GRAY": return True if color[v] == "WHITE" and self.DFSUtil(v, color) == True: return True color[u] = "BLACK" return False def isCyclic(self): color = ["WHITE"] * self.V for i in range(self.V): if color[i] == "WHITE": if self.DFSUtil(i, color) == True: return True return False# Driver program to test above functionsg = 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)print ("Graph contains cycle" if g.isCyclic() == True\ else "Graph doesn't contain cycle") # This program is contributed by Divyanshu Mehta |
C#
// A C# program to detect cycle in// an undirected graph using BFS.using System;using System.Collections.Generic;class GFG { // A DFS based approach to find if // there is a cycle in a directed graph. // This approach strictly follows the // algorithm given in CLRS book. static int WHITE = 0, GRAY = 1, BLACK = 2; // Graph class represents a directed graph // using adjacency list representation public class Graph { public int V; public List<int>[] adjList; // Constructor public Graph(int ver) { V = ver; adjList = new List<int>[V]; for (int i = 0; i < V; i++) adjList[i] = new List<int>(); } } // Utility function to add an edge static void addEdge(Graph g, int u, int v) { g.adjList[u].Add(v); // Add v to u's list. } // Recursive function to find if there is back edge // in DFS subtree tree rooted with 'u' static bool DFSUtil(Graph g, int u, int[] color) { // GRAY : This vertex is being processed (DFS // for this vertex has started, but not // ended (or this vertex is in function // call stack) color[u] = GRAY; // Iterate through all adjacent vertices foreach (int iN in g.adjList[u]) { // If there is if (color[iN] == GRAY) return true; // If v is not processed and there is a back // edge in subtree rooted with v if (color[iN] == WHITE && DFSUtil(g, iN, color) == true) return true; } // Mark this vertex as processed color[u] = BLACK; return false; } // Returns true if there is a cycle in graph static bool isCyclic(Graph g) { // Initialize color of all vertices as WHITE int[] color = new int[g.V]; for (int i = 0; i < g.V; i++) { color[i] = WHITE; } // Do a DFS traversal beginning with all // vertices for (int i = 0; i < g.V; i++) { if (color[i] == WHITE) { if(DFSUtil(g, i, color) == true) return true; } } return false; } // Driver Code public static void Main(String []args) { // Create a graph given in the above diagram Graph g = new Graph(4); addEdge(g, 0, 1); addEdge(g, 0, 2); addEdge(g, 1, 2); addEdge(g, 2, 0); addEdge(g, 2, 3); addEdge(g, 3, 3); if (isCyclic(g)) Console.WriteLine("Graph contains cycle"); else Console.WriteLine("Graph doesn't contain cycle"); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// A Javascript program to detect cycle in// an undirected graph using BFS.// A DFS based approach to find if // there is a cycle in a directed graph. // This approach strictly follows the // algorithm given in CLRS book. var WHITE = 0, GRAY = 1, BLACK = 2;// Graph class represents a directed graph // using adjacency list representation class Graph { // Constructor constructor(ver) { this.V = ver; this.adjList = Array.from( Array(this.V), () => Array(this.V)); }}// Utility function to add an edge function addEdge(g, u, v){ // Push v to u's list. g.adjList[u].push(v); }// Recursive function to find if there is back edge // in DFS subtree tree rooted with 'u' function DFSUtil(g, u, color) { // GRAY : This vertex is being processed (DFS // for this vertex has started, but not // ended (or this vertex is in function // call stack) color[u] = GRAY; // Iterate through all adjacent vertices for(var iN of g.adjList[u]) { // If there is if (color[iN] == GRAY) return true; // If v is not processed and there is a back // edge in subtree rooted with v if (color[iN] == WHITE && DFSUtil(g, iN, color) == true) return true; } // Mark this vertex as processed color[u] = BLACK; return false;}// Returns true if there is a cycle in graphfunction isCyclic(g){ // Initialize color of all vertices as WHITE var color = Array(g.V); for(var i = 0; i < g.V; i++) { color[i] = WHITE; } // Do a DFS traversal beginning with all // vertices for(var i = 0; i < g.V; i++) { if (color[i] == WHITE) { if (DFSUtil(g, i, color) == true) return true; } } return false;}// Driver Code// Create a graph given in the above diagramvar g = new Graph(4);addEdge(g, 0, 1);addEdge(g, 0, 2);addEdge(g, 1, 2);addEdge(g, 2, 0);addEdge(g, 2, 3);addEdge(g, 3, 3);if (isCyclic(g)) document.write("Graph contains cycle");else document.write("Graph doesn't contain cycle"); // This code is contributed by rrrtnx</script> |
Output
Graph contains cycle
Complexity Analysis:
- Time complexity: O(V + E), where V is the number of vertices and E is the number of edges in the graph.
- Space Complexity :O(V).
Since an extra color array is needed of size V.
This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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