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I know that it can be done in O(V+E) using topological sorting. But I think it can also be done in the same complexity using BFS.

Dominique Fortin
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Aditya Joshee
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  • Can you elaborate more what do you mean by (in the most optimal way)? If what you mean is the complexity then yes. A topological sort is the application of DFS that is why it has a complexity of O(V+E). I think it is more natural to ask "can we do a topological sort using BFS?". This question has already been asked. – Abdulhakeem Apr 25 '17 at 03:58
  • What I meant was can we just use BFS instead of visiting the nodes in topological order. – Aditya Joshee Apr 25 '17 at 09:59

2 Answers2

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The topological sort is nothing except that it is one DFS run. Then as each vertex is finished you put it onto the front of a linked list. It has a running time of O(V+E). As I said before, I think it is more natural here to ask "can we do a topological sort using BFS?". This has already been asked and answered.

If what you mean is mere BFS, then I would say no. Have a look to this example. A topological sort is a linear ordering of all its vertices such that if graph G contains an edge (u, v), then u appears before v in the ordering. You can see in that example there is an edge from c to d but d appears before c. I think you can update BFS to find the topological sort but again it has the same complexity as running DFS. I would recommend you to have a look at this post Using BFS for topological sort

Glorfindel
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Abdulhakeem
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  • Let's forget about topological sort for some time. If I use breadth first traversal for a DAG, starting with a source node s, then I can find the shortest distance to all other nodes in O(V+E) time by maintaining a distance array (initialized with infinity) and updating it when we get a value smaller than the currently stored value for a node. So coming back to my original question, isn't this approach of same complexity as using a topological order for traversing nodes in a DAG to find shortest paths? – Aditya Joshee Apr 25 '17 at 17:32
  • I also have a same question, why can't we simply use BFS and find the shortest distance from the source vertex to all other vertices in a DAG? – user3260035 Jun 06 '21 at 05:08
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Using simple BFS is not optimal because if the node, say(A), is already updated from INT_MAX to some x and pushed into the queue, and we found a way to reach A with some value less than x, then we update it and again push into the queue now here we can reduce the time complexity slightly if A is already in the queue, we dont push it ,but what if A is not in the queue but already processed once then we have to push A again and all the adjacent nodes have to get processed again.

To avoid this repetation we travel in topologically sorted order so we can avoid this kind of looping.

Joshith Murthy
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