Given an undirected graph, the task is to check if the graph contains a cycle or not, using DSU.
Examples:
Input: The following is the graph
Output: Yes
Explanation: There is a cycle of vertices {0, 1, 2}.
We already have discussed an algorithm to detect cycle in directed graph. Here Union-Find Algorithm can be used to check whether an undirected graph contains cycle or not. The idea is that,
Initially create subsets containing only a single node which are the parent of itself. Now while traversing through the edges, if the two end nodes of the edge belongs to the same set then they form a cycle. Otherwise, perform union to merge the subsets together.
Note: This method assumes that the graph doesn’t contain any self-loops.
Illustration:
Follow the below illustration for a better understanding
Let us consider the following graph:
Use an array to keep track of the subsets and which nodes belong to that subset. Let the array be parent[].
Initially, all slots of parent array are initialized to hold the same values as the node.
parent[] = {0, 1, 2}. Also when the value of the node and its parent are same, that is the root of that subset of nodes.
Now process all edges one by one.
Edge 0-1:
=> Find the subsets in which vertices 0 and 1 are.
=> 0 and 1 belongs to subset 0 and 1.
=> Since they are in different subsets, take the union of them.
=> For taking the union, either make node 0 as parent of node 1 or vice-versa.
=> 1 is made parent of 0 (1 is now representative of subset {0, 1})
=> parent[] = {1, 1, 2}Edge 1-2:
=> 1 is in subset 1 and 2 is in subset 2.
=> Since they are in different subsets, take union.
=> Make 2 as parent of 1. (2 is now representative of subset {0, 1, 2})
=> parent[] = {1, 2, 2}Edge 0-2:
=> 0 is in subset 2 and 2 is also in subset 2.
=> Because 1 is parent of 0 and 2 is parent of 1. So 0 also belongs to subset 2
=> Hence, including this edge forms a cycle.Therefore, the above graph contains a cycle.
Follow the below steps to implement the idea:
- Initially create a parent[] array to keep track of the subsets.
- Traverse through all the edges:
- Check to which subset each of the nodes belong to by finding the parent[] array till the node and the parent are the same.
- If the two nodes belong to the same subset then they belong to a cycle.
- Otherwise, perform union operation on those two subsets.
- If no cycle is found, return false.
Below is the implementation of the above approach.
C++
// A union-find algorithm to detect cycle in a graph#include <bits/stdc++.h>using namespace std;// a structure to represent an edge in graphclass Edge {public: int src, dest;};// a structure to represent a graphclass Graph {public: // V-> Number of vertices, E-> Number of edges int V, E; // graph is represented as an array of edges Edge* edge;};// Creates a graph with V vertices and E edgesGraph* createGraph(int V, int E){ Graph* graph = new Graph(); graph->V = V; graph->E = E; graph->edge = new Edge[graph->E * sizeof(Edge)]; return graph;}// A utility function to find the subset of an element iint find(int parent[], int i){ if (parent[i] == i) return i; return find(parent, parent[i]);}// A utility function to do union of two subsetsvoid Union(int parent[], int x, int y) { parent[x] = y; }// The main function to check whether a given graph contains// cycle or notint isCycle(Graph* graph){ // Allocate memory for creating V subsets int* parent = new int[graph->V]; // Initialize all subsets as single element sets for(int i = 0; i < graph->V; i++) { parent[i] = i; } // Iterate through all edges of graph, find subset of // both vertices of every edge, if both subsets are // same, then there is cycle in graph. for (int i = 0; i < graph->E; ++i) { int x = find(parent, graph->edge[i].src); int y = find(parent, graph->edge[i].dest); if (x == y) return 1; Union(parent, x, y); } return 0;}// Driver codeint main(){ /* Let us create the following graph 0 | \ | \ 1---2 */ int V = 3, E = 3; Graph* graph = createGraph(V, E); // add edge 0-1 graph->edge[0].src = 0; graph->edge[0].dest = 1; // add edge 1-2 graph->edge[1].src = 1; graph->edge[1].dest = 2; // add edge 0-2 graph->edge[2].src = 0; graph->edge[2].dest = 2; if (isCycle(graph)) cout << "Graph contains cycle"; else cout << "Graph doesn't contain cycle"; return 0;}// This code is contributed by rathbhupendra |
C
// A union-find algorithm to detect cycle in a graph#include <stdio.h>#include <stdlib.h>#include <string.h> // a structure to represent an edge in graphstruct Edge { int src, dest;}; // a structure to represent a graphstruct Graph { // V-> Number of vertices, E-> Number of edges int V, E; // graph is represented as an array of edges struct Edge* edge;}; // Creates a graph with V vertices and E edgesstruct Graph* createGraph(int V, int E){ struct Graph* graph = (struct Graph*)malloc(sizeof(struct Graph)); graph->V = V; graph->E = E; graph->edge = (struct Edge*)malloc( graph->E * sizeof(struct Edge)); return graph;} // A utility function to find the subset of an element iint find(int parent[], int i){ if (parent[i] == -1) return i; return find(parent, parent[i]);} // A utility function to do union of two subsetsvoid Union(int parent[], int x, int y) { parent[y] = x; } // The main function to check whether a given graph contains// cycle or notint isCycle(struct Graph* graph){ // Allocate memory for creating V subsets int* parent = (int*)malloc(graph->V); // Initialize all subsets as single element sets memset(parent, -1, sizeof(graph->V)); // Iterate through all edges of graph, find subset of // both vertices of every edge, if both subsets are // same, then there is cycle in graph. for (int i = 0; i < graph->E; ++i) { int x = find(parent, graph->edge[i].src); int y = find(parent, graph->edge[i].dest); if (x == y && (x!=-1 && y!=-1)) return 1; Union(parent, x,y); } return 0;} // Driver program to test above functionsint main(){ /* Let us create the following graph 0 | \ | \ 1---2 */ int V = 3, E = 3; struct Graph* graph = createGraph(V, E); // // add edge 0-1 graph->edge[0].src = 0; graph->edge[0].dest = 1; // add edge 1-2 graph->edge[1].src = 1; graph->edge[1].dest = 2; //add edge 0-2 graph->edge[2].src = 0; graph->edge[2].dest = 2; if (isCycle(graph)) printf("Graph contains cycle"); else printf("Graph doesn't contain cycle"); return 0;} |
Java
// Java Program for union-find algorithm to detect cycle in// a graphimport java.io.*;import java.lang.*;import java.util.*;public class Graph { int V, E; // V-> no. of vertices & E->no.of edges Edge edge[]; // /collection of all edges class Edge { int src, dest; }; // Creates a graph with V vertices and E edges Graph(int v, int e) { V = v; E = e; edge = new Edge[E]; for (int i = 0; i < e; ++i) edge[i] = new Edge(); } // A utility function to find the subset of an element i int find(int parent[], int i) { if (parent[i] == i) return i; return find(parent, parent[i]); } // A utility function to do union of two subsets void Union(int parent[], int x, int y) { parent[x] = y; } // The main function to check whether a given graph // contains cycle or not int isCycle(Graph graph) { // Allocate memory for creating V subsets int parent[] = new int[graph.V]; // Initialize all subsets as single element sets for (int i = 0; i < graph.V; ++i) parent[i] = i; // Iterate through all edges of graph, find subset // of both vertices of every edge, if both subsets // are same, then there is cycle in graph. for (int i = 0; i < graph.E; ++i) { int x = graph.find(parent, graph.edge[i].src); int y = graph.find(parent, graph.edge[i].dest); if (x == y) return 1; graph.Union(parent, x, y); } return 0; } // Driver Method public static void main(String[] args) { /* Let us create the following graph 0 | \ | \ 1---2 */ int V = 3, E = 3; Graph graph = new Graph(V, E); // add edge 0-1 graph.edge[0].src = 0; graph.edge[0].dest = 1; // add edge 1-2 graph.edge[1].src = 1; graph.edge[1].dest = 2; // add edge 0-2 graph.edge[2].src = 0; graph.edge[2].dest = 2; if (graph.isCycle(graph) == 1) System.out.println("Graph contains cycle"); else System.out.println( "Graph doesn't contain cycle"); }} |
Python3
# Python Program for union-find algorithm# to detect cycle in a undirected graph# we have one edge for any two vertex# i.e 1-2 is either 1-2 or 2-1 but not bothfrom collections import defaultdict# This class represents a undirected graph# using adjacency list representationclass Graph: def __init__(self, vertices): self.V = vertices # No. of vertices self.graph = defaultdict(list) # default dictionary to store graph # function to add an edge to graph def addEdge(self, u, v): self.graph[u].append(v) # A utility function to find the subset of an element i def find_parent(self, parent, i): if parent[i] == i: return i if parent[i] != i: return self.find_parent(parent, parent[i]) # A utility function to do union of two subsets def union(self, parent, x, y): parent[x] = y # The main function to check whether a given graph # contains cycle or not def isCyclic(self): # Allocate memory for creating V subsets and # Initialize all subsets as single element sets parent = [0]*(self.V) for i in range(self.V): parent[i] = i # Iterate through all edges of graph, find subset of both # vertices of every edge, if both subsets are same, then # there is cycle in graph. for i in self.graph: for j in self.graph[i]: x = self.find_parent(parent, i) y = self.find_parent(parent, j) if x == y: return True self.union(parent, x, y)# Create a graph given in the above diagramg = Graph(3)g.addEdge(0, 1)g.addEdge(1, 2)g.addEdge(2, 0)if g.isCyclic(): print("Graph contains cycle")else: print("Graph does not contain cycle ")# This code is contributed by Neelam Yadav |
C#
// C# Program for union-find// algorithm to detect cycle// in a graphusing System;class Graph { // V-> no. of vertices & // E->no.of edges public int V, E; // collection of all edges public Edge[] edge; public class Edge { public int src, dest; }; // Creates a graph with V // vertices and E edges public Graph(int v, int e) { V = v; E = e; edge = new Edge[E]; for (int i = 0; i < e; ++i) edge[i] = new Edge(); } // A utility function to find // the subset of an element i int find(int[] parent, int i) { if (parent[i] == i) return i; return find(parent, parent[i]); } // A utility function to do // union of two subsets void Union(int[] parent, int x, int y) { parent[x] = y; } // The main function to check // whether a given graph // contains cycle or not int isCycle(Graph graph) { // Allocate memory for // creating V subsets int[] parent = new int[graph.V]; // Initialize all subsets as // single element sets for (int i = 0; i < graph.V; ++i) parent[i] = i; // Iterate through all edges of graph, // find subset of both vertices of every // edge, if both subsets are same, then // there is cycle in graph. for (int i = 0; i < graph.E; ++i) { int x = graph.find(parent, graph.edge[i].src); int y = graph.find(parent, graph.edge[i].dest); if (x == y) return 1; graph.Union(parent, x, y); } return 0; } // Driver code public static void Main(String[] args) { /* Let us create the following graph 0 | \ | \ 1---2 */ int V = 3, E = 3; Graph graph = new Graph(V, E); // add edge 0-1 graph.edge[0].src = 0; graph.edge[0].dest = 1; // add edge 1-2 graph.edge[1].src = 1; graph.edge[1].dest = 2; // add edge 0-2 graph.edge[2].src = 0; graph.edge[2].dest = 2; if (graph.isCycle(graph) == 1) Console.WriteLine("Graph contains cycle"); else Console.WriteLine( "Graph doesn't contain cycle"); }}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program for union-find // algorithm to detect cycle // in a graph// V-> no. of vertices & // E->no.of edges var V, E; // Collection of all edgesvar edge; class Edge{ constructor() { this.src = 0; this.dest = 0; }};// Creates a graph with V // vertices and E edgesfunction initialize(v,e){ V = v; E = e; edge = Array.from(Array(E), () => Array());}// A utility function to find // the subset of an element ifunction find(parent, i){ if (parent[i] == i) return i; return find(parent, parent[i]);}// A utility function to do // union of two subsetsfunction Union(parent, x, y){ parent[x] = y;}// The main function to check // whether a given graph// contains cycle or notfunction isCycle(){ // Allocate memory for // creating V subsets var parent = Array(V).fill(0); // Initialize all subsets as // single element sets for(var i = 0; i < V; ++i) parent[i] = i; // Iterate through all edges of graph, // find subset of both vertices of every // edge, if both subsets are same, then // there is cycle in graph. for (var i = 0; i < E; ++i) { var x = find(parent, edge[i].src); var y = find(parent, edge[i].dest); if (x == y) return 1; Union(parent, x, y); } return 0;}// Driver code/* Let us create the following graph 0 | \ | \ 1---2 */var V = 3, E = 3;initialize(V, E);// Add edge 0-1edge[0].src = 0;edge[0].dest = 1;// Add edge 1-2edge[1].src = 1;edge[1].dest = 2;// Add edge 0-2edge[2].src = 0;edge[2].dest = 2;if (isCycle() == 1) document.write("Graph contains cycle");else document.write("Graph doesn't contain cycle"); // This code is contributed by rutvik_56</script> |
Output
Graph contains cycle
The time and space complexity of the given code is as follows:
Time Complexity:
- Creating the graph takes O(V + E) time, where V is the number of vertices and E is the number of edges.
- Finding the subset of an element takes O(log V) time in the worst case, where V is the number of vertices. The worst case occurs when the tree is skewed, and the depth of the tree is V.
- Union of two subsets takes O(1) time.
- The loop iterating through all edges takes O(E) time.
- Therefore, the overall time complexity of the algorithm is O(E log V).
However, in practice, it can be much faster than O(E log V) because the worst-case scenario of finding the subset of an element does not happen often.
Space Complexity:
- The space complexity of creating the graph is O(E).
- The space complexity of creating the parent array is O(V).
- The space complexity of the algorithm is O(max(V,E)) because at any point in time, there can be at most max(V,E) subsets.
- Therefore, the overall space complexity of the algorithm is O(max(V,E)).
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