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May 9, 2011 at 20:42 vote accept user1855
May 9, 2011 at 20:42 vote accept user1855
May 9, 2011 at 20:42
Apr 28, 2011 at 15:36 comment added Charles Right, using Bareiss' algorithm space needed is at most 0.5 n lg n + O(n) due to a theorem of Hadamard. This means that each operation takes time $O(n(\log n)^{2+\varepsilon})$ giving overall time complexity $O(n^4(\log n)^{2+\varepsilon}).$
Apr 28, 2011 at 13:14 history edited Emil Jeřábek CC BY-SA 3.0
clarify and include refs
Apr 28, 2011 at 11:13 history answered Emil Jeřábek CC BY-SA 3.0