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]

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.

`// 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"` `); ` ` ` `} ` `} ` |

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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