How to make closest possible to $n$ equidistant points in $\mathbb{R}^3$?

Question

Given $n$ points in $\mathbb{R}^3$, the most exactly equidistant points we can have is $n=4$. I was wondering for larger cases how to construct the closest possible situation to having all points equidistant. I was thinking of maybe having one point be the center of a sphere, and then the other points being on the surface of the sphere and being approximately mutually equidistant from each other. I'm not sure if this is the right idea, but even if it is I'm not sure exactly how to execute it. Any help would be appreciated.

Answer 1

I do not think, that I understand your problem yet but I'm going to give it a try. But first, let us verify a bunch of stuff. Let $P^{(1)},\dots,P^{(n)} \in \mathbb{R}^3$

• What do you mean with distant? Assuming it would be the standard metric in $\mathbb{R}^3$, then let denote it by $$ \left\lVert P^{(i)} - P^{(j)}\right\rVert = \sqrt{x^{(i)}x^{(j)} + y^{(i)}y^{(j)}+ z^{(i)}z^{(j)}} \qquad \text{ where } x,y,z \in \mathbb{R} \qquad \text{ for } i,j = 1,\dots, n$$
• What do you mean by equidistant? Does it mean - for some constant $c \in \mathbb{R}$ $$ \left\lVert P^{(i)} - P^{(j)}\right\rVert = c \qquad \forall i,j = 1, \dots n $$ If this is the case. Then with $n = 2$ the points will make a segment, $n =3$ is a quilateral triangle and $n = 4$ is a triangular pyramid with all equilateral triangle. We can think of a way to generalize this for the case $\mathbb{R}^d$ by elaborating about the construction of equaliteral triangle but not with the sphere. The reason we can't use a sphere is, because the diameter is always twice the radius, thus any two points lying opposite to each other always have a greater distance than that towards their neighbor. Instead one can guesses, that the figure as the result of the complete graph from all these points, may be shaped by a bunch of equiliteral triangles.

Answer 2

Probably the simplest approach is to place your points in a hexagonal close packed array with outer envelope about a sphere. It is like the centers of a bunch of tightly packed spheres. Each point is the same distance from its neighbors and roughly speaking the maximum distance between any two points is the diameter of the sphere. I suspect this is about the best you can do if your criterion is the ratio of maximum to minimum distance. For small numbers of points there will be perturbations. Packing problems are hard.

Answer 3

One sensible way to interpret your question is given $n$ points in the plane, what is the maximum number of pairs of points with distance between $d-\varepsilon$ and $d+\varepsilon$ between them? For $\varepsilon =0$ this is some discrete geometry problem that I would not know how to solve. But for $\varepsilon > 0$, I present the following possibly tight construction. Simply place four tight groups of points of size $n/4$ on the four vertices of a regular tetrahedron. Then $3/4$ of the pairs have the desired distance.