Topological sorting for **D**irected **A**cyclic **G**raph (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.

Given a DAG, print all topological sorts of the graph.

For example, consider the below graph.All topological sorts of the given graph are: 4 5 0 2 3 1 4 5 2 0 3 1 4 5 2 3 0 1 4 5 2 3 1 0 5 2 3 4 0 1 5 2 3 4 1 0 5 2 4 0 3 1 5 2 4 3 0 1 5 2 4 3 1 0 5 4 0 2 3 1 5 4 2 0 3 1 5 4 2 3 0 1 5 4 2 3 1 0

In a Directed acyclic graph many a times we can have vertices which are unrelated to each other because of which we can order them in many ways. These various topological sorting is important in many cases, for example if some relative weight is also available between the vertices, which is to minimize then we need to take care of relative ordering as well as their relative weight, which creates the need of checking through all possible topological ordering.

We can go through all possible ordering via backtracking , the algorithm step are as follows :

- Initialize all vertices as unvisited.
- Now choose vertex which is unvisited and has zero indegree and decrease indegree of all those vertices by 1 (corresponding to removing edges) now add this vertex to result and call the recursive function again and backtrack.
- After returning from function reset values of visited, result and indegree for enumeration of other possibilities.

Below is C++ implementation of above steps.

// C++ program to print all topological sorts of a graph #include <bits/stdc++.h> using namespace std; class Graph { int V; // No. of vertices // Pointer to an array containing adjacency list list<int> *adj; // Vector to store indegree of vertices vector<int> indegree; // A function used by alltopologicalSort void alltopologicalSortUtil(vector<int>& res, bool visited[]); public: Graph(int V); // Constructor // function to add an edge to graph void addEdge(int v, int w); // Prints all Topological Sorts void alltopologicalSort(); }; // Constructor of graph Graph::Graph(int V) { this->V = V; adj = new list<int>[V]; // Initialising all indegree with 0 for (int i = 0; i < V; i++) indegree.push_back(0); } // Utility function to add edge void Graph::addEdge(int v, int w) { adj[v].push_back(w); // Add w to v's list. // increasing inner degree of w by 1 indegree[w]++; } // Main recursive function to print all possible // topological sorts void Graph::alltopologicalSortUtil(vector<int>& res, bool visited[]) { // To indicate whether all topological are found // or not bool flag = false; for (int i = 0; i < V; i++) { // If indegree is 0 and not yet visited then // only choose that vertex if (indegree[i] == 0 && !visited[i]) { // reducing indegree of adjacent vertices list<int>:: iterator j; for (j = adj[i].begin(); j != adj[i].end(); j++) indegree[*j]--; // including in result res.push_back(i); visited[i] = true; alltopologicalSortUtil(res, visited); // resetting visited, res and indegree for // backtracking visited[i] = false; res.erase(res.end() - 1); for (j = adj[i].begin(); j != adj[i].end(); j++) indegree[*j]++; flag = true; } } // We reach here if all vertices are visited. // So we print the solution here if (!flag) { for (int i = 0; i < res.size(); i++) cout << res[i] << " "; cout << endl; } } // The function does all Topological Sort. // It uses recursive alltopologicalSortUtil() void Graph::alltopologicalSort() { // Mark all the vertices as not visited bool *visited = new bool[V]; for (int i = 0; i < V; i++) visited[i] = false; vector<int> res; alltopologicalSortUtil(res, visited); } // Driver program to test above functions int main() { // Create a graph given in the above diagram Graph g(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); cout << "All Topological sorts\n"; g.alltopologicalSort(); return 0; }

Output :

All Topological sorts 4 5 0 2 3 1 4 5 2 0 3 1 4 5 2 3 0 1 4 5 2 3 1 0 5 2 3 4 0 1 5 2 3 4 1 0 5 2 4 0 3 1 5 2 4 3 0 1 5 2 4 3 1 0 5 4 0 2 3 1 5 4 2 0 3 1 5 4 2 3 0 1 5 4 2 3 1 0

This articles is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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