A *disjoint-set data structure* is a data structure that keeps track of a set of elements partitioned into a number of disjoint (non-overlapping) subsets. A *union-find algorithm* is an algorithm that performs two useful operations on such a data structure:

* Find:* Determine which subset a particular element is in. This can be used for determining if two elements are in the same subset.

* Union:* Join two subsets into a single subset.

In this post, we will discuss an application of Disjoint Set Data Structure. The application is to check whether a given graph contains a cycle or not.

*Union-Find Algorithm* can be used to check whether an undirected graph contains cycle or not. Note that we have discussed an algorithm to detect cycle. This is another method based on *Union-Find*. This method assumes that graph doesn’t contain any self-loops.

We can keeps track of the subsets in a 1D array, lets call it parent[].

Let us consider the following graph:

For each edge, make subsets using both the vertices of the edge. If both the vertices are in the same subset, a cycle is found.

Initially, all slots of parent array are initialized to -1 (means there is only one item in every subset).

0 1 2 -1 -1 -1

Now process all edges one by one.

*Edge 0-1:* Find the subsets in which vertices 0 and 1 are. Since they are in different subsets, we take the union of them. For taking the union, either make node 0 as parent of node 1 or vice-versa.

0 1 2 <----- 1 is made parent of 0 (1 is now representative of subset {0, 1}) 1 -1 -1

*Edge 1-2:* 1 is in subset 1 and 2 is in subset 2. So, take union.

0 1 2 <----- 2 is made parent of 1 (2 is now representative of subset {0, 1, 2}) 1 2 -1

*Edge 0-2:* 0 is in subset 2 and 2 is also in subset 2. Hence, including this edge forms a cycle.

How subset of 0 is same as 2?

0->1->2 // 1 is parent of 0 and 2 is parent of 1

Based on the above explanation, below are implementations:

## C/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 graph struct Edge { int src, dest; }; // a structure to represent a graph struct 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 edges struct 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 i int find(int parent[], int i) { if (parent[i] == -1) return i; return find(parent, parent[i]); } // A utility function to do union of two subsets void Union(int parent[], int x, int y) { int xset = find(parent, x); int yset = find(parent, y); parent[xset] = yset; } // The main function to check whether a given graph contains // cycle or not int isCycle( struct Graph* graph ) { // Allocate memory for creating V subsets int *parent = (int*) malloc( graph->V * sizeof(int) ); // Initialize all subsets as single element sets memset(parent, -1, sizeof(int) * 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) return 1; Union(parent, x, y); } return 0; } // Driver program to test above functions int main() { /* Let us create 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 graph import java.util.*; import java.lang.*; import java.io.*; 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] == -1) return i; return find(parent, parent[i]); } // A utility function to do union of two subsets void Union(int parent[], int x, int y) { int xset = find(parent, x); int yset = find(parent, y); parent[xset] = yset; } // 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]=-1; // 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 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" ); } }

## Python

# Python Program for union-find algorithm to detect cycle in a undirected graph # we have one egde for any two vertex i.e 1-2 is either 1-2 or 2-1 but not both from collections import defaultdict #This class represents a undirected graph using adjacency list representation class 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] == -1: return i if parent[i]!= -1: return self.find_parent(parent,parent[i]) # A utility function to do union of two subsets def union(self,parent,x,y): x_set = self.find_parent(parent, x) y_set = self.find_parent(parent, y) parent[x_set] = y_set # 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 = [-1]*(self.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 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 diagram g = 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

Output:

graph contains cycle

Note that the implementation of *union()* and *find()* is naive and takes O(n) time in worst case. These methods can be improved to O(Logn) using *Union by Rank or Height*. We will soon be discussing *Union by Rank* in a separate post.

**Related Articles : **

Union-Find Algorithm | Set 2 (Union By Rank and Path Compression)

Disjoint Set Data Structures (Java Implementation)

Greedy Algorithms | Set 2 (Kruskalâ€™s Minimum Spanning Tree Algorithm)

Job Sequencing Problem | Set 2 (Using Disjoint Set)

This article is compiled by Aashish Barnwal and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.