Researchers achieve ‘absurdly fast’ algorithm for network flow

Re:Researchers achieve ‘absurdly fast’ algorithm for network... by im3w1l on Thursday June 09, 2022 @01:36AM (#31677901)

I'm trying to come up with an example to check my understanding. n (log n)^k for some constant k would qualify right?

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Re:Researchers achieve ‘absurdly fast’ algorithm for network... by mauricioc on Thursday June 09, 2022 @04:51AM (#31679003)

Yes. (log n)/(n^epsilon) tends to zero for any positive epsilon, and so log n = n^(o(1)). The same holds for (log n)^k.

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