Python Program for Topological Sorting

23 Feb, 2022
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 3 1 0”. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no in-coming edges). 
Topological sorting can be implemented recursively and non-recursively. First, we show the clearer recursive version, then provide the non-recursive version with analysis. 
Python3
#Python program to print topological sorting of a DAGfrom collections import defaultdict #Class to represent a graphclass 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)     # A recursive function used by topologicalSort    def topologicalSortUtil(self,v,visited,stack):         # Mark the current node as visited.        visited[v] = True         # Recur for all the vertices adjacent to this vertex        for i in self.graph[v]:            if visited[i] == False:                self.topologicalSortUtil(i,visited,stack)         # Push current vertex to stack which stores result        stack.insert(0,v)     # The function to do Topological Sort. It uses recursive    # topologicalSortUtil()    def topologicalSort(self):        # Mark all the vertices as not visited        visited = [False]*self.V        stack =[]         # Call the recursive helper function to store Topological        # Sort starting from all vertices one by one        for i in range(self.V):            if visited[i] == False:                self.topologicalSortUtil(i,visited,stack)         # Print contents of stack        print (stack) 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 of the given graph
5 4 2 3 1 0
 
The way topological sorting is solved is by processing a node after all of its children are processed. Each time a node is processed, it is pushed into a stack in order to save the final result. This non-recursive solution builds on the same concept of DFS with a little tweak which can be understood above and in this article. However, unlike the recursive solution, which saves the order of the nodes in the stack after all the neighboring elements have been pushed to the program stack, this solution replaces the program stack with a working stack. If a node has a neighbor that has not been visited, the current node and the neighbor are pushed to the working stack to be processed until there are no more neighbors available to be visited.After all the nodes have been visited, what remains is the final result which is found by printing the stack result in reverse.
Python3
#Python program to print topological sorting of a DAGfrom collections import defaultdict #Class to represent a graphclass 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)     # neighbors lazy generator given key    def neighbor_gen(self,v):        for k in self.graph[v]:            yield k         # non recursive topological sort    def nonRecursiveTopologicalSortUtil(self, v, visited,stack):                 # working stack contains key and the corresponding current generator        working_stack = [(v,self.neighbor_gen(v))]                 while len(working_stack) > 0:            # get last element in stack            v, gen = working_stack[-1]            visited[v] = True                         # delete it from stack            working_stack.pop()                         # run through neighbor generator until its empty            continue_flag = True            while continue_flag:                next_neighbor = next(gen,None)                                 # if generator has returned all neighbors                if next_neighbor is None:                    continue_flag = False                    # Save current key into the result stack                    stack.append(v)                    continue                                 # if new neighbor push current key and neighbor into stack                if not(visited[next_neighbor]):                    working_stack.append((v,gen))                    working_stack.append((next_neighbor,self.neighbor_gen(next_neighbor)))                    continue_flag = False                 # The function to do Topological Sort.    def nonRecursiveTopologicalSort(self):        # Mark all the vertices as not visited        visited = [False]*self.V                 # result stack        stack = []         # Call the helper function to store Topological        # Sort starting from all vertices one by one        for i in range(self.V):            if not(visited[i]):                self.nonRecursiveTopologicalSortUtil(i, visited,stack)         # Print contents of the stack in reverse        print(stack[::-1]) 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("The following is a Topological Sort of the given graph")g.nonRecursiveTopologicalSort()# The following code was based of Neelam Yadav's code and is modified by Suhail Alnahari
Output: 
Following is a Topological Sort of the given graph
5 4 2 3 1 0
Time Complexity: O(V + E): The above algorithm is simply DFS with a working stack and a result stack. Unlike the recursive solution, recursion depth is not an issue here.
Auxiliary space: O(V): The extra space is needed for the 2 stacks used.
Please refer complete article on Topological Sorting for more details. 
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