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

*chevron_right*

*filter_none*

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

This article is contributed by **Deepak Srivatsav**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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