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.
Python Program for Topological Sorting
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.
Recursive Topological Sorting
#Python program to print topological sorting of a DAG 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)
# 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 |
Following is a Topological Sort of the given graph [5, 4, 2, 3, 1, 0]
Non-recursive Topological Sorting Python
Algorithm:
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 onto 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.
#Python program to print topological sorting of a DAG 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)
# neighbors 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 working_stack:
# get last element from stack
v, gen = working_stack.pop()
visited[v] = True
# run through neighbor generator until it's empty
for next_neighbor in gen:
if not visited[next_neighbor]: # not seen before?
# remember current work
working_stack.append((v,gen))
# restart with new neighbor
working_stack.append((next_neighbor, self .neighbor_gen(next_neighbor)))
break
else :
# no already-visited neighbor (or no more of them)
stack.append(v)
# 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
stack.reverse()
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 ( "The following is a Topological Sort of the given graph" )
g.nonRecursiveTopologicalSort() |
The following is a Topological Sort of the given graph [5, 4, 2, 3, 1, 0]
Complexity Analysis:
- 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.