Coding DFS and BFS Graph Algorithms JAVA

Java Programming – Union-Find Algorithm | Set 1 (Detect Cycle in an Undirected Graph)

Java Programming - Union-Find Algorithm | Set 1 (Detect Cycle in an Undirected Graph) - Graph Algorithms - A disjoint-set data structure is a data.

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.

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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:

Java Programming:

// 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" );
    }
}

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.

About the author

Venkatesan Prabu

Venkatesan Prabu

Wikitechy Founder, Author, International Speaker, and Job Consultant. My role as the CEO of Wikitechy, I help businesses build their next generation digital platforms and help with their product innovation and growth strategy. I'm a frequent speaker at tech conferences and events.

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