Until yesterday I looked upon the terms "lim sup" and "lim inf" with suspicion and had no use for the concepts, but yesterday I had a need, so when I looked it up it made sense. With the phi-based method of placing points on a square torus, as N, the number of points increases, there is a lim sup *and* a lim inf of the ratio (minimum distance between lattice points) / (1/sqrt(N)) (the "figure of merit" to maximize). For reference, with a square lattice the ratio is 1. With an equilateral triangle lattice, it's sqrt(2/sqrt(3)), about 1.07457 -- that has to be the best case of any method. The best ratios of the phi method do gradually approach-- that is, their limit superior is-- 1.07457. You might think this method gets more reliable with more points, but no, there's also a limit inferior! It's about 0.8137. As I try larger N's (up to about 2 million now), the record worst case keeps approaching that, so phi is no panacea. I could look at one of the worst solutions and find the dimensions of the cells, but I don't see right away how to find the exact pessimal solution. I haven't found a match to 0.8137 among expressions of the form sqrt(a/sqrt(b)).

From: Yoav Kallus <colonelmustard@gmail.com> Date: 9/11/20, 5:01 PM

I asked a question similar to Steve's "peppering problem" for the circle on MathOverflow a few years ago. My assumption was also that k phi (mod 1), k = 1, 2, ... would be the way to maximize lim inf [N d(N)], where d(N) is the minimum separation between the first N points. This turned out to be wrong, and you can do substantially better (0.721 vs 0.618 for the golden ratio process).

https://mathoverflow.net/questions/275158/sequential-addition-of-points-on-a...

Awesome! Thank you Colonel Mustard, you are definitely in the study. --Steve