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Jul 25, 2023 at 8:37 history edited yash vinayvanshi CC BY-SA 4.0
added 20 characters in body
Jul 24, 2023 at 18:53 history became hot network question
Jul 24, 2023 at 13:33 comment added Peter Taylor @FedorPetrov, correct.
S Jul 24, 2023 at 13:29 history suggested ARA CC BY-SA 4.0
fixed grammar
Jul 24, 2023 at 11:33 comment added Fedor Petrov @PeterTaylor you probably mean $n$ copies of 2
Jul 24, 2023 at 11:30 answer added Fedor Petrov timeline score: 5
Jul 24, 2023 at 11:19 comment added Peter Taylor The bound isn't going to be much lower than $O(2^n)$ because if $n$ is even and $S$ contains $n$ copies of $\frac n2$ you get $\binom{n}{n/2} \approx \frac{2^{n-1/2}}{\sqrt{\pi n}}$
Jul 24, 2023 at 11:02 review Suggested edits
S Jul 24, 2023 at 13:29
S Jul 24, 2023 at 10:51 review First questions
Jul 24, 2023 at 10:52
S Jul 24, 2023 at 10:51 history asked yash vinayvanshi CC BY-SA 4.0