Kahn’s algorithm for Topological Sorting

Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological Sorting for a graph is not possible if the graph is not a DAG.

For example, a topological sorting of the following graph is “5 4 2 3 1 0?. There can be more than one topological sorting for a graph. For example, another topological sorting of the following graph is “4 5 2 0 3 1″. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no in-coming edges).

graph

A DFS based solution to find a topological sort has already been discussed.

In this article we will see another way to find the linear ordering of vertices in a directed acyclic graph (DAG). The approach is based on the below fact :

A DAG G has at least one vertex with in-degree 0 and one vertex with out-degree 0.
Proof: There’s a simple proof to the above fact is that a DAG does not contain a cycle which means that all paths will be of finite length. Now let S be the longest path from u(source) to v(destination). Since S is the longest path there can be no incoming edge to u and no outgoing edge from v, if this situation had occurred then S would not have been the longest path
=> indegree(u) = 0 and outdegree(v) = 0

Algorithm:
Steps involved in finding the topological ordering of a DAG:

Step-1: Compute in-degree (number of incoming edges) for each of the vertex present in the DAG and initialize the count of visited nodes as 0.

Step-2: Pick all the vertices with in-degree as 0 and add them into a queue (Enqueue operation)

Step-3: Remove a vertex from the queue (Dequeue operation) and then.

  1. Increment count of visited nodes by 1.
  2. Decrease in-degree by 1 for all its neighboring nodes.
  3. If in-degree of a neighboring nodes is reduced to zero, then add it to the queue.

Step 5: Repeat Step 3 until the queue is empty.

Step 5: If count of visited nodes is not equal to the number of nodes in the graph then the topological sort is not possible for the given graph.

How to find in-degree of each node?
There are 2 ways to calculate in-degree of every vertex:
Take an in-degree array which will keep track of
1) Traverse the array of edges and simply increase the counter of the destination node by 1.

for each node in Nodes
	indegree[node] = 0;
for each edge(src,dest) in Edges
	indegree[dest]++

Time Complexity: O(V+E)

2) Traverse the list for every node and then increment the in-degree of all the nodes connected to it by 1.

	for each node in Nodes
		If (list[node].size()!=0) then
		for each dest in list
			indegree[dest]++;

Time Complexity: The outer for loop will be executed V number of times and the inner for loop will be executed E number of times, Thus overall time complexity is O(V+E).

The overall time complexity of the algorithm is O(V+E)

Below is C++ implementation of above algorithm. The implementation uses method 2 discussed above for finding indegrees.

C++

// A C++ program to print topological sorting of a graph
// using indegrees.
#include<bits/stdc++.h>
using namespace std;

// Class to represent a graph
class Graph
{
    int V;    // No. of vertices'

    // Pointer to an array containing adjacency listsList
    list<int> *adj;

public:
    Graph(int V);   // Constructor

    // function to add an edge to graph
    void addEdge(int u, int v);

    // prints a Topological Sort of the complete graph
    void topologicalSort();
};

Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}

void Graph::addEdge(int u, int v)
{
    adj[u].push_back(v);
}

// The function to do Topological Sort.
void Graph::topologicalSort()
{
    // Create a vector to store indegrees of all
    // vertices. Initialize all indegrees as 0.
    vector<int> in_degree(V, 0);

    // Traverse adjacency lists to fill indegrees of
    // vertices.  This step takes O(V+E) time
    for (int u=0; u<V; u++)
    {
        list<int>::iterator itr;
        for (itr = adj[u].begin(); itr != adj[u].end(); itr++)
             in_degree[*itr]++;
    }

    // Create an queue and enqueue all vertices with
    // indegree 0
    queue<int> q;
    for (int i = 0; i < V; i++)
        if (in_degree[i] == 0)
            q.push(i);

    // Initialize count of visited vertices
    int cnt = 0;

    // Create a vector to store result (A topological
    // ordering of the vertices)
    vector <int> top_order;

    // One by one dequeue vertices from queue and enqueue
    // adjacents if indegree of adjacent becomes 0
    while (!q.empty())
    {
        // Extract front of queue (or perform dequeue)
        // and add it to topological order
        int u = q.front();
        q.pop();
        top_order.push_back(u);

        // Iterate through all its neighbouring nodes
        // of dequeued node u and decrease their in-degree
        // by 1
        list<int>::iterator itr;
        for (itr = adj[u].begin(); itr != adj[u].end(); itr++)

            // If in-degree becomes zero, add it to queue
            if (--in_degree[*itr] == 0)
                q.push(*itr);

        cnt++;
    }

    // Check if there was a cycle
    if (cnt != V)
    {
        cout << "There exists a cycle in the graph\n";
        return;
    }

    // Print topological order
    for (int i=0; i<top_order.size(); i++)
        cout << top_order[i] << " ";
    cout << endl;
}

// Driver program to test above functions
int main()
{
    // Create a graph given in the above diagram
    Graph g(6);
    g.addEdge(5, 2);
    g.addEdge(5, 0);
    g.addEdge(4, 0);
    g.addEdge(4, 1);
    g.addEdge(2, 3);
    g.addEdge(3, 1);

    cout << "Following is a Topological Sort of\n";
    g.topologicalSort();

    return 0;
}

Java

// A Java program to print topological sorting of a graph
// using indegrees
import java.util.*;

//Class to represent a graph
class Graph
{
	int V;// No. of vertices
	
	//An Array of List which contains 
	//references to the Adjacency List of 
	//each vertex
	List <Integer> adj[];
	public Graph(int V)// Constructor
	{
		this.V = V;
		adj = new ArrayList[V];
		for(int i = 0; i < V; i++)
			adj[i]=new ArrayList<Integer>();
	}
	
	// function to add an edge to graph
	public void addEdge(int u,int v)
	{
		adj[u].add(v);
	}
	// prints a Topological Sort of the complete graph	
	public void topologicalSort()
	{
		// Create a array to store indegrees of all
		// vertices. Initialize all indegrees as 0.
		int indegree[] = new int[V];
		
		// Traverse adjacency lists to fill indegrees of
		// vertices. This step takes O(V+E) time		
		for(int i = 0; i < V; i++)
		{
			ArrayList<Integer> temp = (ArrayList<Integer>) adj[i];
			for(int node : temp)
			{
				indegree[node]++;
			}
		}
		
		// Create a queue and enqueue all vertices with
		// indegree 0
		Queue<Integer> q = new LinkedList<Integer>();
		for(int i = 0;i < V; i++)
		{
			if(indegree[i]==0)
				q.add(i);
		}
		
		// Initialize count of visited vertices
		int cnt = 0;
		
		// Create a vector to store result (A topological
		// ordering of the vertices)
		Vector <Integer> topOrder=new Vector<Integer>();
		while(!q.isEmpty())
		{
			// Extract front of queue (or perform dequeue)
			// and add it to topological order
			int u=q.poll();
			topOrder.add(u);
			
			// Iterate through all its neighbouring nodes
			// of dequeued node u and decrease their in-degree
			// by 1
			for(int node : adj[u])
			{
				// If in-degree becomes zero, add it to queue
				if(--indegree[node] == 0)
					q.add(node);
			}
			cnt++;
		}
		
		// Check if there was a cycle		
		if(cnt != V)
		{
			System.out.println("There exists a cycle in the graph");
			return ;
		}
		
		// Print topological order			
		for(int i : topOrder)
		{
			System.out.print(i+" ");
		}
	}
}
// Driver program to test above functions
class Main
{
	public static void main(String args[])
	{
		// Create a graph given in the above diagram
		Graph g=new Graph(6);
		g.addEdge(5, 2);
	    g.addEdge(5, 0);
	    g.addEdge(4, 0);
	    g.addEdge(4, 1);
	    g.addEdge(2, 3);
	    g.addEdge(3, 1);
	    System.out.println("Following is a Topological Sort");
	    g.topologicalSort();

	}
}

Python

# A Python program to print topological sorting of a graph
# using indegrees
from collections import defaultdict

#Class to represent a graph
class Graph:
	def __init__(self,vertices):
		self.graph = defaultdict(list) #dictionary containing adjacency List
		self.V = vertices #No. of vertices

	# function to add an edge to graph
	def addEdge(self,u,v):
		self.graph[u].append(v)


	# The function to do Topological Sort. 
	def topologicalSort(self):
		
		# Create a vector to store indegrees of all
		# vertices. Initialize all indegrees as 0.
		in_degree = [0]*(self.V)
		
		# Traverse adjacency lists to fill indegrees of
   		# vertices.  This step takes O(V+E) time
		for i in self.graph:
			for j in self.graph[i]:
				in_degree[j] += 1

		# Create an queue and enqueue all vertices with
		# indegree 0
		queue = []
		for i in range(self.V):
			if in_degree[i] == 0:
				queue.append(i)

		#Initialize count of visited vertices
		cnt = 0

		# Create a vector to store result (A topological
		# ordering of the vertices)
		top_order = []

		# One by one dequeue vertices from queue and enqueue
		# adjacents if indegree of adjacent becomes 0
		while queue:

			# Extract front of queue (or perform dequeue)
			# and add it to topological order
			u = queue.pop(0)
			top_order.append(u)

			# Iterate through all neighbouring nodes
			# of dequeued node u and decrease their in-degree
			# by 1
			for i in self.graph[u]:
				in_degree[i] -= 1
				# If in-degree becomes zero, add it to queue
				if in_degree[i] == 0:
					queue.append(i)

			cnt += 1

		# Check if there was a cycle
		if cnt != self.V:
			print "There exists a cycle in the graph"
		else :
			#Print topological order
			print top_order


g= Graph(6)
g.addEdge(5, 2);
g.addEdge(5, 0);
g.addEdge(4, 0);
g.addEdge(4, 1);
g.addEdge(2, 3);
g.addEdge(3, 1);

print "Following is a Topological Sort of the given graph"
g.topologicalSort()

#This code is contributed by Neelam Yadav 

Output :

Following is a Topological Sort
4 5 2 0 3 1

This article is contributed by Chirag Agarwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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