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Jun 20, 2020 at 19:17 comment added Sandeep Silwal A similar reasoning with the bound of $n^{4/3}$ also appears in the paper 'On Separating Points by Lines' by Har-Peled and Jones
Jun 17, 2020 at 13:56 comment added John Jiang Very nice proof. I initially read the question wrong, and thought the grid was assumed uniform. But rather it can be actively chosen. A couple of points that I had to struggle with: the first estimate can be seen by realizing the binomial factor is 1 + o(1). For the constant 5, note removing a point can promote at most 4 grids to being good (2 on the same row and 2 on the same column), and adding a point can promote at most 1 grid, namely the grid where the point lands. Lastly for part b, two points are not on the same x or y cord with probably 1. How likely is it to get sharp estimate for c?
Jun 10, 2020 at 16:07 vote accept Nikita Kalinin
Jun 10, 2020 at 14:05 history edited Yuval Peres CC BY-SA 4.0
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Jun 10, 2020 at 11:19 history edited Yuval Peres CC BY-SA 4.0
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Jun 10, 2020 at 9:07 comment added user21820 Do you think the constant $k$ for which $p_n(k·n^{4/3}) → r ∈ (0,1)$ as $n→∞$ is rational or irrational? And what do you think it is?
Jun 10, 2020 at 3:45 history answered Yuval Peres CC BY-SA 4.0