Disjoint Set (Or Union-Find) | Set 1 (Detect Cycle in an Undirected Graph)

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 the 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 the graph doesn’t contain any self-loops.
We can keep track of the subsets in a 1D array, let’s call it parent[].

Let us consider the following graph:
cycle-in-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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Based on the above explanation, below are implementations:

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 graph  
class Edge  
{  
    public: 
    int src, dest;  
};  
  
// a structure to represent a graph  
class 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 edges  
Graph* 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 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);  
    if(xset != yset) 
    {  
        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 * 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 code 
int 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 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); 
    if(xset!=yset){ 
       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 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 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.

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.

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

C

Java

Python

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