Distances in many dimensions

160 random points in a cube. The chart shows every pairwise distance, divided by the average distance.

Dimensions2

Farthest ÷ nearest neighbor40.8×

Spread of distances±48% of average

Challenges

0 of 2 complete

Find the fewest dimensions at which a point's farthest neighbor is less than twice as far away as its nearest (the ratio reads below 2×). (not yet)
Squeeze the spread of distances below ±5% of the average. (not yet)

Try this

Compare 2, 10, 100, and 1,000 dimensions. Where does the farthest-to-nearest ratio drop below 2?

Distance concentration in a few lines of codePython with NumPy

import numpy as np

rng = np.random.default_rng(0)
n = 160  # random points in the unit cube, as in the demo

for d in (2, 10, 100, 1000):
    x = rng.random((n, d))
    # All pairwise distances, from |a - b|^2 = |a|^2 + |b|^2 - 2 a.b
    sq = (x**2).sum(axis=1)
    dist = np.sqrt(np.maximum(sq[:, None] + sq[None, :] - 2 * x @ x.T, 0))
    np.fill_diagonal(dist, np.nan)  # ignore each point's distance to itself
    ratio = np.mean(np.nanmax(dist, axis=1) / np.nanmin(dist, axis=1))
    pairs = dist[np.triu_indices(n, k=1)]
    spread = pairs.std() / pairs.mean()
    print(f"{d:>4} dims: farthest / nearest = {ratio:6.2f}, spread = ±{spread:.0%}")

   2 dims: farthest / nearest =  43.91, spread = ±48%
  10 dims: farthest / nearest =   2.89, spread = ±19%
 100 dims: farthest / nearest =   1.35, spread = ±6%
1000 dims: farthest / nearest =   1.10, spread = ±2%