Topological sort python

Question

I coded a solution for DFS non-recursive, but i can't modify it to make a topological sort:

def dfs(graph,start):
    path = []
    stack = [start]    
    while stack != []: 
        v = stack.pop()
        if v not in path: path.append(v)
        for w in reversed(graph[v]): 
            if w not in path and not w in stack:
                stack.append(w) 
    return path

Any ideas how to modify it?

With the recursive version i can easy have the sorting:

def dfs_rec(graph,start,path):
    path = path + [start]
    for edge in graph[start]: 
        if edge not in path:
            path = dfs_rec(graph, edge,path)
    print start
    return path

Input:

>>> graph = {
        1: [2, 3],
        2: [4, 5, 6],
        3: [4,6],
        4: [5,6],
        5: [6],
        6: []
    }
>>> dfs_rec(graph,1,[])
6
5
4
2
3
1
[1, 2, 4, 5, 6, 3]
>>> dfs(graph,1)
[1, 2, 4, 5, 6, 3]
>>> graph = {
        1: [3],
        3: [5,6],
        5: [4],
        4: [7],
        7: [],
        6: []
    }
>>> print dfs_rec(graph,1,[])
7
4
5
6
3
1
[1, 3, 5, 4, 7, 6]
>>> print dfs(graph,1)
[1, 3, 5, 4, 7, 6]

so i need to get this ordering in the non-recursive also.

Non-recursive solution:

I think that this also could be the solution, mark me if i am wrong.

def dfs(graph,start):
    path = []
    stack = [start]
    label = len(graph)
    result = {}  
    while stack != []:
        #this for loop could be done in other ways also
        for element in stack:
            if element not in result:
                result[element] = label
                label = label - 1

        v = stack.pop()
        if v not in path: path.append(v)
        for w in reversed(graph[v]): 
            if w not in path and not w in stack:
                stack.append(w) 

    result = {v:k for k, v in result.items()}
    return path,result

Input:

graph = { 1: [3], 3:[5,6] , 5:[4] , 4:[7], 7:[],6:[]}
print dfs(graph,1) 

Output:

([1, 3, 5, 4, 7, 6], {1: 7, 2: 4, 3: 5, 4: 6, 5: 3, 6: 1})

        1
       / 
      3
     /\
    5  6
   /
  4
 /
7    

Solution 1

FWIW, here is some code I worked up for a non-recursive topological sort.

from collections import defaultdict, namedtuple
from itertools import islice

Results = namedtuple('Results', ['sorted', 'cyclic'])

def topological_sort(dependency_pairs):
    'Sort values subject to dependency constraints'
    num_heads = defaultdict(int)   # num arrows pointing in
    tails = defaultdict(list)      # list of arrows going out
    heads = []                     # unique list of heads in order first seen
    for h, t in dependency_pairs:
        num_heads[t] += 1
        if h in tails:
            tails[h].append(t)
        else:
            tails[h] = [t]
            heads.append(h)

    ordered = [h for h in heads if h not in num_heads]
    for h in ordered:
        for t in tails[h]:
            num_heads[t] -= 1
            if not num_heads[t]:
                ordered.append(t)
    cyclic = [n for n, heads in num_heads.items() if heads]
    return Results(ordered, cyclic)

if __name__ == '__main__':
    print( topological_sort('aa'.split()) )
    print( topological_sort('ah bg cf ch di ed fb fg hd he ib'.split()) )

Solution 2

from collections import defaultdict, deque


class Graph:
    def __init__(self, directed=False, nodes=None, edges=None):
        self.graph = defaultdict(list)
        self.directed = directed
        self.add_nodes(nodes)
        self.add_edges(edges)

    @property
    def nodes(self):
        if not self.directed:
            return list(self.graph.keys())
        elif self.directed:
            nodes = set()
            nodes.update(self.graph.keys())
            for node in self.graph.keys():
                for neighbor in self.graph[node]:
                    nodes.add(neighbor)
            return list(nodes)

    def add_node(self, node):
        if node not in self.nodes:
            self.graph[node] = list()

    def add_nodes(self, nodes):
        if nodes is None:
            return None
        for node in nodes:
            self.add_node(node)

    @property
    def edges(self):
        edges = list()
        for source, neighbors in self.graph.items():
            for neighbor in neighbors:
                edges.append((source, neighbor))
        return edges

    def add_edge(self, edge):
        node1, node2 = edge
        self.graph[node1].append(node2)
        if not self.directed:
            self.graph[node2].append(node1)

    def add_edges(self, edges):
        if edges is None:
            return None
        for edge in edges:
            self.add_edge(edge)

    def topological_util(self, node, visited, label):
        visited[node] = True
        for edge in self.graph[node]:
            if not visited[edge]:
                self.topological_util(edge, visited, label)
        label.appendleft(node)

    def topological_sort(self):
        visited = dict.fromkeys(self.nodes, False)
        # store all nodes in topological order, the index is the position
        label = deque()
        for node in self.nodes:
            if not visited[node]:
                self.topological_util(node, visited, label)
        return label

Comments

[ovgolovin]

Could you give an example of input?

[Eric]

What's the problem here? In both cases, the return value of dfs_rec matches that of dfs. What results do you want instead?

[badc0re]

Yes the result matches, but in dfs_rec when the recursion ends it gives me the (by print start) the topological ordering of the graph, so now i want to make a topological ordering on the non-recursive function (dfs) but i could not succeed in doing it.

[Eric]

So you want both functions to return [7, 4, 5, 6, 3, 1] in the second case?

[badc0re]

[6, 5, 4, 2, 3, 1] and [7, 4, 5, 6, 3, 1] for dfs(graph,1) the non-recursive function. As you can see i already have it for the first function dfs_rec.

[Rob Cowie]

Unless this is an academic exercise don't reinvent the wheel, use NetworkX, it's awesome

[badc0re]

Yeah it is academic exercise, of course i would probably use something that is already coded.