Print all possible paths in a DAG from vertex whose indegree is 0

Given a Directed Acyclic Graph (DAG), having N vertices and M edges, The task is to print all path starting from vertex whose in-degree is zero.

Indegree of a vertex is the total number of incoming edges to a vertex.

Example:

Input: N = 6, edges[] = {{5, 0}, {5, 2}, {4, 0}, {4, 1}, {2, 3}, {3, 1}}
Output: All possible paths:
4 0
4 1
5 0
5 2 3 1
Explaination:
The given graph can be represented as:

There are two vertices whose indegree are zero i.e vertex 5 and 4, after exploring these vertices we got the fallowing path:
4 -> 0
4 -> 1
5 -> 0
5 -> 2 -> 3 -> 1
Input: N = 6, edges[] = {{0, 5}}
Output: All possible paths:
0 5
Explaination:
There will be only one possible path in the graph.

Approach:

Create a boolean array indeg0 which store a true value for all those vertex whose indegree is zero.
Apply DFS on all those vertex whose indegree is 0.
Print all path starting from a vertex whose indegree is 0 to a vertex whose outdegrees are zero.

Illustration:
For the above graph:
indeg0[] = {False, False, False, False, True, True}
Since indeg[4] = True, so appling DFS on vertex 4 and printing all path terminating to 0 outdegree vertex are as follow:
4 -> 0
4 -> 1
Also indeg[5] = True, so appling DFS on vertex 5 and printing all path terminating to 0 outdegree vertex are as follow:
5 -> 0
5 -> 2 -> 3 -> 1

Here is the implementation of the above approach:

Python3

# Python program to print all 
# possible paths in a DAG 
  
# Recursive function to print all paths 
def dfs(s): 
    # Append the node in path 
    # and set visited 
    path.append(s) 
    visited[s] = True
  
    # Path started with a node 
    # having in-degree 0 and 
    # current node has out-degree 0, 
    # print current path 
    if outdeg0[s] and indeg0[path[0]]: 
        print(*path) 
  
    # Recursive call to print all paths 
    for node in adj[s]: 
        if not visited[node]: 
            dfs(node) 
  
    # Remove node from path 
    # and set unvisited 
    path.pop() 
    visited[s] = False
  
  
def print_all_paths(n): 
    for i in range(n): 
  
        # for each node with in-degree 0 
        # print all possible paths 
        if indeg0[i] and adj[i]: 
            path = [] 
            visited = [False] * (n + 1) 
            dfs(i) 
  
  
# Driver code 
from collections import defaultdict 
  
n = 6
# set all nodes unvisited 
visited = [False] * (n + 1) 
path = [] 
  
# edges = (a, b): a -> b 
edges = [(5, 0), (5, 2), (2, 3), 
         (4, 0), (4, 1), (3, 1)] 
  
# adjacency list for nodes 
adj = defaultdict(list) 
  
# indeg0 and outdeg0 arrays 
indeg0 = [True]*n 
outdeg0 = [True]*n 
  
for edge in edges: 
    u, v = edge[0], edge[1] 
    # u -> v 
    adj[u].append(v) 
  
    # set indeg0[v] <- false 
    indeg0[v] = False
  
    # set outdeg0[u] <- false 
    outdeg0[u] = False
  
print('All possible paths:') 
print_all_paths(n) 

Output:

All possible paths:
4 0
4 1
5 0
5 2 3 1

Time Complexity: O (N + E)²
Auxiliary Space: O (N)

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Python3

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