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    Can you explain how DFS will clearly identify that cycle does not exist in your example.I agree that the cycle does not exist in provided example.But if we go from A->B and then A->C->B we will find that B was already visited and its parent is A not C..and i read that DFS will detect the cycle by comparing the parent of already visited element with current node from which direction we are checking at this moment.am i getting DFS wrong or what? Commented Oct 20, 2017 at 14:44
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    All you've shown here is that this particular implementation doesn't work, not that it's impossible with BFS. In fact, it is possible, although it takes more work and space. Commented Nov 17, 2017 at 17:45
  • @Prune: All the threads(i think) here are trying to prove that bfs wont work for detecting cycles. If you know how to counter prove you should give the proof. Simply saying that efforts are greater wont suffice Commented Nov 19, 2017 at 6:47
  • Since the algorithm is given in the linked postings, I don't feel it's appropriate to repeat the outline here. Commented Nov 20, 2017 at 16:11
  • I couldnt find any linked postings, hence asked for the same. I agree with your point about bfs capability and have just thought about the implementation. Thanks for the tip :) Commented Nov 23, 2017 at 9:37