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Jun 21, 2016 at 12:34 vote accept Jake B
Jun 21, 2016 at 13:11
Jun 21, 2016 at 11:40 comment added adrianN In this paper the authors examine a similar problem with a slightly different oracle and get $\Theta(n \log \log n)$ in the randomized setting. I believe that one can also solve this problem quicker than $\Theta(n\log n)$.
Jun 21, 2016 at 11:17 comment added Yuval Filmus There is an easy $\Omega(n)$ lower bound since you're recovering $\Omega(n\log n)$ bits of information, and you get $O(\log n)$ bits from each query. Your queries only give you one bit of information, which is why you need $\Omega(n\log n)$ of them.
Jun 21, 2016 at 9:10 comment added adrianN Does anybody know a lower bound? I have a feeling that you might shave of some loglogs by asking about more men in parallel.
Jun 21, 2016 at 9:09 history answered adrianN CC BY-SA 3.0