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Nov 12, 2011 at 20:30 comment added Daniel Scocco @kevin cline, I mean to verify the cyclyes analyzing the adjacency matrix only, so this would be n^2 (you basically need to traverse the matrix n x n (where n is the number of vertices). If you use a better algorithm then yeah the complexity will probably be O(V+E).
Nov 12, 2011 at 15:43 comment added kevin cline @Daniel: Definitely not n^2. More like O(n + e).
Nov 12, 2011 at 13:04 vote accept Daniel Scocco
Nov 12, 2011 at 13:00 comment added Daniel Scocco Gotcha. And yeah to verify those on large/complex graphs the cost would be (n^2) if I am not wrong (you basically would need to traverse the adjacency matrix). But for simple ones it might be a good idea. Thanks for the answer.
Nov 12, 2011 at 12:56 history answered thiton CC BY-SA 3.0