TopologicalSort.md

Topological Sort

Topological Sort of a directed graph (a graph with unidirectional edges) is a linear ordering of its vertices such that for every directed edge (U, V) from vertex U to vertex V, U comes before V in the ordering.

Example 1:

Input: Vertices=4, Edges=[3, 2], [3, 0], [2, 0], [2, 1]
Output: Following are the two valid topological sorts for the given graph:
1) 3, 2, 0, 1
2) 3, 2, 1, 0

Solution:

👉 Here are a few fundamental concepts related to topological sort:

Source: Any node that has no incoming edge and has only outgoing edges is called a source.

Sink: Any node that has only incoming edges and no outgoing edge is called a sink.

So, we can say that a topological ordering starts with one of the sources and ends at one of the sinks.

A topological ordering is possible only when the graph has no directed cycles, i.e. if the graph is a Directed Acyclic Graph (DAG). If the graph has a cycle, some vertices will have cyclic dependencies which makes it impossible to find a linear ordering among vertices.

👉 Algorithm:

To find the topological sort of a graph we can traverse the graph in a Breadth First Search (BFS) way. We will start with all the sources, 
and in a stepwise fashion, save all sources to a sorted list. We will then remove all sources and their edges from the graph. 
After the removal of the edges, we will have new sources, so we will repeat the above process until all vertices are visited.

Implementation of above algorithm:

This is how we can implement this algorithm:

a. Initialization:

We will store the graph in Adjacency Lists, which means each parent vertex will have a list containing all of its children. We will do this using a HashMap where the ‘key’ will be the parent vertex number and the value will be a List containing children vertices. To find the sources, we will keep a HashMap to count the in-degrees i.e., count of incoming edges of each vertex. Any vertex with ‘0’ in-degree will be a source.

b. Build the graph and find in-degrees of all vertices:

We will build the graph from the input and populate the in-degrees HashMap.

c. Find all sources

All vertices with ‘0’ in-degrees will be our sources and we will store them in a Queue.

d. Sort

For each source, do the following things: Add it to the sorted list. Get all of its children from the graph. Decrement the in-degree of each child by 1. If a child’s in-degree becomes ‘0’, add it to the sources Queue. Repeat step 1, until the source Queue is empty.

from collections import deque


def topological_sort(vertices, edges):
  sortedOrder = []
  if vertices <= 0:
    return sortedOrder

  # a. Initialize the graph
  inDegree = {i: 0 for i in range(vertices)}  # count of incoming edges
  graph = {i: [] for i in range(vertices)}  # adjacency list graph

  # b. Build the graph
  for edge in edges:
    parent, child = edge[0], edge[1]
    graph[parent].append(child)  # put the child into it's parent's list
    inDegree[child] += 1  # increment child's inDegree

  # c. Find all sources i.e., all vertices with 0 in-degrees
  sources = deque()
  for key in inDegree:
    if inDegree[key] == 0:
      sources.append(key)

  # d. For each source, add it to the sortedOrder and subtract one from all of its children's in-degrees
  # if a child's in-degree becomes zero, add it to the sources queue
  while sources:
    vertex = sources.popleft()
    sortedOrder.append(vertex)
    for child in graph[vertex]:  # get the node's children to decrement their in-degrees
      inDegree[child] -= 1
      if inDegree[child] == 0:
        sources.append(child)

  # topological sort is not possible as the graph has a cycle
  if len(sortedOrder) != vertices:
    return []

  return sortedOrder


def main():
  print("Topological sort: " +
        str(topological_sort(4, [[3, 2], [3, 0], [2, 0], [2, 1]])))
  print("Topological sort: " +
        str(topological_sort(5, [[4, 2], [4, 3], [2, 0], [2, 1], [3, 1]])))
  print("Topological sort: " +
        str(topological_sort(7, [[6, 4], [6, 2], [5, 3], [5, 4], [3, 0], [3, 1], [3, 2], [4, 1]])))


main()

Time Complexity

In step ‘d’, each vertex will become a source only once and each edge will be accessed and removed once. Therefore, the time complexity of the above algorithm will be O(V+E), where ‘V’ is the total number of vertices and ‘E’ is the total number of edges in the graph.

Space Complexity

The space complexity will be O(V+E), since we are storing all of the edges for each vertex in an adjacency list.

Similar Problems

Problem 1: Find if a given Directed Graph has a cycle in it or not.

Solution: If we can’t determine the topological ordering of all the vertices of a directed graph, the graph has a cycle in it. This was also referred to in the above code:

if (sortedOrder.size() != vertices) // topological sort is not possible as the graph has a cycle
  return new ArrayList<>();