# Why Kosaraju's algorithm cannot use original graph by increasing finish time? [duplicate]

This question already has an answer here:

I saw this post, and it explains about why kosaraju's algorithm includes transpose graph $G^T$.

But, I totally didn't understand why cannot use increasing order of original graph. Kosaraju's algorithm uses decreasing order of transpose graph.

Is there an any counter example about this? I'm really confusing.

## marked as duplicate by Apass.Jack, Evil, David Richerby, Yuval Filmus, Discrete lizard♦Feb 8 at 18:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 1. Something is wrong with your link. 2. Have you tried to construct a counterexample? Try enumerating a bunch of graphs with 3 or 4 or 5 vertices. – D.W. May 23 '18 at 15:32
• I edited my link and I tried to find counterexample by drawing the graph, but I cannot find... – molamola May 23 '18 at 23:03
• Check cs.stackexchange.com/q/47298/755, cs.stackexchange.com/q/60503/755, cs.stackexchange.com/q/48625/755. Do any of those help? – D.W. May 23 '18 at 23:09
• Thank you for your answer, but I have an exam 2 hours later, so I will check later. – molamola May 23 '18 at 23:11
• Possible duplicate of Kosaraju's Algorithm-Strongly connected components, which contains the same answer below by K. Ali and whose accepted answer contains a counterexample. – Apass.Jack Feb 8 at 8:06