This is a java program to find topological sort of DAG. In computer science, a topological sort (sometimes abbreviated topsort or toposort) or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. For instance, the vertices of the graph may represent tasks to be performed, and the edges may represent constraints that one task must be performed before another; in this application, a topological ordering is just a valid sequence for the tasks. A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). Any DAG has at least one topological ordering, and algorithms are known for constructing a topological ordering of any DAG in linear time.
Here is the source code of the Java Program to Apply DFS to Perform the Topological Sorting of a Directed Acyclic Graph. The Java program is successfully compiled and run on a Windows system. The program output is also shown below.
package com.sanfoundry.graph;
import java.util.InputMismatchException;
import java.util.Scanner;
import java.util.Stack;
public class DigraphTopologicalSortingDFS
{
private Stack<Integer> stack;
public DigraphTopologicalSortingDFS()
{
stack = new Stack<Integer>();
}
public int[] topological(int adjacency_matrix[][], int source)
throws NullPointerException
{
int number_of_nodes = adjacency_matrix[source].length - 1;
int[] topological_sort = new int[number_of_nodes + 1];
int pos = 1;
int j;
int visited[] = new int[number_of_nodes + 1];
int element = source;
int i = source;
visited[source] = 1;
stack.push(source);
while (!stack.isEmpty())
{
element = stack.peek();
while (i <= number_of_nodes)
{
if (adjacency_matrix[element][i] == 1 && visited[i] == 1)
{
if (stack.contains(i))
{
System.out.println("TOPOLOGICAL SORT NOT POSSIBLE");
return null;
}
}
if (adjacency_matrix[element][i] == 1 && visited[i] == 0)
{
stack.push(i);
visited[i] = 1;
element = i;
i = 1;
continue;
}
i++;
}
j = stack.pop();
topological_sort[pos++] = j;
i = ++j;
}
return topological_sort;
}
public static void main(String... arg)
{
int number_no_nodes, source;
Scanner scanner = null;
int topological_sort[] = null;
try
{
System.out.println("Enter the number of nodes in the graph");
scanner = new Scanner(System.in);
number_no_nodes = scanner.nextInt();
int adjacency_matrix[][] = new int[number_no_nodes + 1][number_no_nodes + 1];
System.out.println("Enter the adjacency matrix");
for (int i = 1; i <= number_no_nodes; i++)
for (int j = 1; j <= number_no_nodes; j++)
adjacency_matrix[i][j] = scanner.nextInt();
System.out.println("Enter the source for the graph");
source = scanner.nextInt();
System.out
.println("The Topological sort for the graph is given by ");
DigraphTopologicalSortingDFS toposort = new DigraphTopologicalSortingDFS();
topological_sort = toposort.topological(adjacency_matrix, source);
for (int i = topological_sort.length - 1; i > 0; i--)
{
if (topological_sort[i] != 0)
System.out.print(topological_sort[i] + "\t");
}
}
catch (InputMismatchException inputMismatch)
{
System.out.println("Wrong Input format");
}
catch (NullPointerException nullPointer)
{
}
scanner.close();
}
}
Output:
Enter the number of nodes in the graph 6 Enter the adjacency matrix 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 1 0 0 Enter the source for the graph 1 The Topological sort for the graph is given by TOPOLOGICAL SORT NOT POSSIBLE Enter the number of nodes in the graph 6 Enter the adjacency matrix 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 Enter the source for the graph 1 The Topological sort for the graph is given by 1 2 4 5 6 3
Sanfoundry Global Education & Learning Series – 1000 Java Programs.
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