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Apr 3, 2020 at 10:12 vote accept user153694
Mar 17, 2020 at 20:14 answer added Seva timeline score: 2
Mar 16, 2020 at 20:56 comment added Seva The answer is somewhere between $1/2^{n-1}$ (attained for the set $\{2^{i+1-n}\colon 0\le i\le n-1\}$) and $n/(2^n-1)$ (we have $2^n$ subset sums residing in the interval $[0,n]$). So, the fight can be for a logarithmic factor only.
Mar 16, 2020 at 16:40 review First posts
Mar 16, 2020 at 17:01
Mar 16, 2020 at 16:35 history asked user153694 CC BY-SA 4.0