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May 8, 2022 at 22:37 comment added D.W. Please credit the original source of all copied material: cs.stackexchange.com/help/referencing
May 8, 2022 at 13:55 history edited Metropola Official CC BY-SA 4.0
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May 8, 2022 at 12:01 comment added Yuval Filmus The worst-case complexity is precisely $n$. Ask for the numbers $1,\ldots,n$. If $k$ is equal to one of them, you're done, and otherwise, $k = 0$. You can show that $n$ queries are necessary using an adversary argument: after $n-1$ questions answered negatively, there are still at least two possible values of $k$ which are consistent with everything known so far.
May 8, 2022 at 11:31 history edited Metropola Official CC BY-SA 4.0
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S May 8, 2022 at 11:23 review First questions
May 8, 2022 at 15:27
S May 8, 2022 at 11:23 history asked Metropola Official CC BY-SA 4.0