Determine whether a universal sink exists in a directed graph

Determine whether a universal sink exists in a directed graph. A universal sink is a vertex which has no edge emanating from it, and all other vertices have an edge towards the sink.

Input : 
v1 -> v2 (implies vertex 1 is connected to vertex 2)
v3 -> v2
v4 -> v2
v5 -> v2
v6 -> v2                        
Output :
Sink found at vertex 2

Input : 
v1 -> v6
v2 -> v3
v2 -> v4
v4 -> v3
v5 -> v3
Output :
No Sink

We try to eliminate n – 1 non-sink vertices in O(n) time and check the remaining vertex for the sink property.
To eliminate vertices, we check whether a particular index (A[i][j]) in the adjacency matrix is a 1 or a 0. If it is a 0, it means that the vertex corresponding to index j cannot be a sink. If the index is a 1, it means the vertex corresponding to i cannot be a sink. We keep increasing i and j in this fashion until either i or j exceeds the number of vertices.
Using this method allows us to carry out the universal sink test for only one vertex instead of all n vertices. Suppose we are left with only vertex i.
We now check for whether row i has only 0s and whether row j as only 1s except for A[i][i], which will be 0.

Illustration :



v1 -> v2 
v3 -> v2
v4 -> v2
v5 -> v2
v6 -> v2                     
We can visualize the adjacency matrix for 
the above as follows:
0 1 0 0 0 0
0 0 0 0 0 0
0 1 0 0 0 0
0 1 0 0 0 0
0 1 0 0 0 0 

We observe that vertex 2 does not have any emanating edge, and that every other vertex has an edge in vertex 2. At A[0][0] (A[i][j]), we encounter a 0, so we increment j and next
look at A[0][1]. Here we encounter a 1. So we have to increment i by 1. A[1][1] is 0, so we keep increasing j. We notice that A[1][2], A[1][3].. etc are all 0, so j will exceed the
number of vertices (6 in this example). We now check row i and column i for the sink property. Row i must be completely 0, and column i must be completely 1 except for the index A[i][i]

Adjacency Matrix

Adjacency Matrix

Second Example:

v1 -> v6
v2 -> v3
v2 -> v4
v4 -> v3
v5 -> v3
We can visualize the adjacency matrix
for the above as follows:
0 0 0 0 0 1
0 0 1 1 0 0
0 0 0 0 0 0
0 0 1 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0

In this example, we observer that in row 1, every element is 0 except for the last column. So we will increment j until we reach the 1. When we reach 1, we increment i as long as
the value of A[i][j] is 0. If i exceeds the number of vertices, it is not possible to have a sink, and in this case, i will exceed the number of vertices.

Adjacency Matrix

Adjacency Matrix

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// Java program to find whether a universal sink
// exists in a directed graph
import java.io.*;
import java.util.*;
  
class Graph
{
    int vertices;
    int[][] adjacency_matrix;
  
    // constructor to initialize number of vertices and
    // size of adjacency matrix
    public graph(int vertices)
    {
        this.vertices = vertices;
        adjacency_matrix = new int[vertices][vertices];
    }
  
    public void insert(int source, int destination)
    {
        // make adjacency_matrix[i][j] = 1 if there is
        // an edge from i to j
        adjacency_matrix[destination-1] = 1;
    }
  
    public boolean issink(int i)
    {
        for (int j = 0 ; j < vertices ; j++)
        {
            // if any element in the row i is 1, it means
            // that there is an edge emanating from the
            // vertex, which means it cannot be a sink
            if (adjacency_matrix[i][j] == 1)
                return false;
  
            // if any element other than i in the column
            // i is 0, it means that there is no edge from
            // that vertex to the vertex we are testing
            // and hence it cannot be a sink
            if (adjacency_matrix[j][i] == 0 && j != i)
                return false;
        }
        //if none of the checks fails, return true
        return true;
    }
  
    // we will eliminate n-1 non sink vertices so that
    // we have to check for only one vertex instead of
    // all n vertices
    public int eliminate()
    {
        int i = 0, j = 0;
        while (i < vertices && j < vertices)
        {
            // If the index is 1, increment the row we are
            // checking by 1
            // else increment the column
            if (adjacency_matrix[i][j] == 1)
                i = i + 1;
            else
                j = j + 1;
  
        }
  
        // If i exceeds the number of vertices, it
        // means that there is no valid vertex in
        // the given vertices that can be a sink
        if (i > vertices)
            return -1;
        else if (!issink(i))
            return -1;
        else return i;
    }
}
  
public class Sink
{
    public static void main(String[] args)throws IOException
    {
        int number_of_vertices = 6;
        int number_of_edges = 5;
        graph g = new graph(number_of_vertices);
        /*
        //input set 1
        g.insert(1, 6);
        g.insert(2, 6);
        g.insert(3, 6);
        g.insert(4, 6);
        g.insert(5, 6);
        */
        //input set 2
        g.insert(1, 6);
        g.insert(2, 3);
        g.insert(2, 4);
        g.insert(4, 3);
        g.insert(5, 3);
  
        int vertex = g.eliminate();
  
        // returns 0 based indexing of vertex. returns
        // -1 if no sink exits.
        // returns the vertex number-1 if sink is found
        if (vertex >= 0)
            System.out.println("Sink found at vertex "
                                     + (vertex + 1));
        else
            System.out.println("No Sink");
    }
}

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Output:

input set 1: 
Sink found at vertex 6
input set 2:
No Sink

This program eliminates non-sink vertices in O(n) complexity and checks for the sink property in O(n) complexity.

You may also try The Celebrity Problem, which is an application of this concept

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