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 0 3 1″. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no in-coming edges).
A DFS based solution to find a topological sort has already been discussed.
In this article we will see another way to find the linear ordering of vertices in a directed acyclic graph (DAG). The approach is based on the below fact :
A DAG G has at least one vertex with in-degree 0 and one vertex with out-degree 0.
Proof: There’s a simple proof to the above fact is that a DAG does not contain a cycle which means that all paths will be of finite length. Now let S be the longest path from u(source) to v(destination). Since S is the longest path there can be no incoming edge to u and no outgoing edge from v, if this situation had occurred then S would not have been the longest path
=> indegree(u) = 0 and outdegree(v) = 0
Steps involved in finding the topological ordering of a DAG:
Step-1: Compute in-degree (number of incoming edges) for each of the vertex present in the DAG and initialize the count of visited nodes as 0.
Step-2: Pick all the vertices with in-degree as 0 and add them into a queue (Enqueue operation)
Step-3: Remove a vertex from the queue (Dequeue operation) and then.
- Increment count of visited nodes by 1.
- Decrease in-degree by 1 for all its neighboring nodes.
- If in-degree of a neighboring nodes is reduced to zero, then add it to the queue.
Step 5: Repeat Step 3 until the queue is empty.
Step 5: If count of visited nodes is not equal to the number of nodes in the graph then the topological sort is not possible for the given graph.
How to find in-degree of each node?
There are 2 ways to calculate in-degree of every vertex:
Take an in-degree array which will keep track of
1) Traverse the array of edges and simply increase the counter of the destination node by 1.
for each node in Nodes indegree[node] = 0; for each edge(src,dest) in Edges indegree[dest]++
Time Complexity: O(V+E)
2) Traverse the list for every node and then increment the in-degree of all the nodes connected to it by 1.
for each node in Nodes If (list[node].size()!=0) then for each dest in list indegree[dest]++;
Time Complexity: The outer for loop will be executed V number of times and the inner for loop will be executed E number of times, Thus overall time complexity is O(V+E).
The overall time complexity of the algorithm is O(V+E)
Below is C++ implementation of above algorithm. The implementation uses method 2 discussed above for finding indegrees.
Following is a Topological Sort 4 5 2 0 3 1
This article is contributed by Chirag Agarwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
- Topological Sorting
- Lexicographically Smallest Topological Ordering
- All Topological Sorts of a Directed Acyclic Graph
- Topological Sort of a graph using departure time of vertex
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- Floyd Warshall Algorithm | DP-16
- Prim's algorithm using priority_queue in STL
- KMP Algorithm for Pattern Searching
- Relabel-to-front Algorithm
- Jump Pointer Algorithm
- Hierholzer's Algorithm for directed graph
- Graph Coloring | Set 2 (Greedy Algorithm)
- Dinic's algorithm for Maximum Flow
- Dijkstra’s shortest path algorithm using set in STL
- BFS using vectors & queue as per the algorithm of CLRS