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Jul 4, 2020 at 0:01 history edited Fedor Petrov CC BY-SA 4.0
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Jul 3, 2020 at 23:54 comment added H A Helfgott OK, I think this works, and disproves that the absolute sum is bounded by the number of minimal sets of $\mathscr{S}$. At the same time, the set $X$ is pretty large (it can be of size $2^{|\mathbf{P}|}$ or close to that), so one can still have a useful bound for this problem (see my answer) and not for the other one.
Jul 3, 2020 at 23:43 history edited Fedor Petrov CC BY-SA 4.0
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Jul 3, 2020 at 22:35 history answered Fedor Petrov CC BY-SA 4.0