I understand that we keep a constant δ value between points, and so at best create a grid of 8 points in a 2δ x δ rectangle. However, if we try and maximise the number of delta nearest points on such a grid, then theoretically our selected point will at most be able to reach a maximum of 5, not 7 points. This is because any points in the opposite corner of our rectangle will have sqrt(2δ^2) distance which is greater than δ. Not sure what I am misunderstanding about this, and why a seemingly unnecessary extra 2 points have to be checked when it does not seem possible to me to construct a point that has more than 5 points within δ. Why not just take every points contained within a circle of radius δ around a point and calculate the distance, which would be 5 points maximum.
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1You get an O(n log n) algorithm either way. My guess is that your source either wanted to get the simplest proof possible or to handle the L-infinity metric. – David Eisenstat Sep 27 '22 at 11:55