*How to find m points on a supersphere x _{1}^{2} + x_{2}^{2} + ... + x_{n}^{2} = 1 in n-dimensional space
such that the minimum distance between any two of these points is as large as possible?*

I first tried the following pascal program:

program maxmin;

label 1;

var i,j,i1,i2,m,n,a,b:integer; max,min,som,s:real; p:array[1..20,1..8] of real;

begin

writeln('dimensie? (n, max 8)'); readln(n);

writeln('aantal punten? (m, max 20)'); readln(m);

max:=0;

while true do

begin

1: min:=100;

for j:=1 to n-1 do p[1,j]:=0; p[1,n]:=1;

for i:=2 to m do

begin

repeat

som:=0;

for j:=1 to n-1 do

begin

p[i,j]:=2*random-1;

som:=som+p[i,j]*p[i,j]

end

until (som<=1);

p[i,n]:=sqrt(1-som); if random>0.5 then p[i,n]:=-p[i,n];

end;

for i1:=1 to m do for i2:=i1+1 to m do

begin

s:=0;

for j:=1 to n do s:=s+(p[i1,j]-p[i2,j])*(p[i1,j]-p[i2,j]);

if s<=min then begin a:=i1; b:=i2; min:=s end;

if min<=max then goto 1;

end;

max:=min; writeln;

writeln(sqrt(max):13:10);

writeln(a:3,b:3);

for i:=1 to m do

begin

for j:=1 to n do write(p[i,j]:8:5); writeln

end; writeln

end

end.

For instance, this program gives a configuration of 10 points on a sphere in 3-dimensional space such that (roughly) the minimal distance between any two of them (0.9524) is 87% of
the theoretical maximum (1.0914) within a few hours.

Next, by varying the coordinates of the two closest points in this configuration with steps of 0.01, and repeating this procedure, we get a new configuration of 10 points on the sphere such that
(roughly) the minimal distance between any two of them (0.9778) is 90% of the theoretical maximum (1.0914).

Next, fixing the coordinates of six points in this configuration at (0,0,1) and (cos(θ)cos(2kπ/5),cos(θ)sin(2kπ/5),sin(θ)) with equal mutual distances,
and choosing the other four points at random like above, we get a new configuration of 10 points on the sphere such that
(roughly) the minimal distance between any two of them (1.025) is 94% of the theoretical maximum (1.0914).

Next, by varying the coordinates of the two closest points in this configuration with steps of 0.001, and repeating this procedure, we get a new configuration of 10 points on the sphere such that
(roughly) the minimal distance between any two of them (1.0258) is 94% of the theoretical maximum (1.0914).

This configuration is (roughly) the following:

0.00000 0.00000 1.00000

0.89443 0.00000 0.44721

0.27639 0.85065 0.44721

-0.72361 0.52573 0.44721

-0.72361 -0.52573 0.44721

0.27639 -0.85065 0.44721

-0.78118 -0.20289 -0.59042

-0.26384 0.70751 -0.65561

0.80143 0.24997 -0.54335

0.32744 -0.73356 -0.59554

We can find coordinates for the ten points in Danzer's optimal configuration as follows:

O1(0,-cos(θ_{1}),-sin(θ_{1}))

O2(0,cos(θ_{1}),-sin(θ_{1}))

T1(cos(α)cos(θ_{2}),-sin(α)cos(θ_{2}),-sin(θ_{2}))

T2(cos(α)cos(θ_{2}),sin(α)cos(θ_{2}),-sin(θ_{2}))

T3(-cos(α)cos(θ_{2}),sin(α)cos(θ_{2}),-sin(θ_{2}))

T4(-cos(α)cos(θ_{2}),-sin(α)cos(θ_{2}),-sin(θ_{2}))

U1(0,-cos(θ_{3}),sin(θ_{3}))

U2(0,cos(θ_{3}),sin(θ_{3}))

V1(cos(θ_{4}),0,sin(θ_{4}))

V2(-cos(θ_{4}),0,sin(θ_{4}))

with θ_{1} and θ_{4} 'about' π/3, with θ_{2} and θ_{3} 'about' π/6, and with α 'about' π/4.

By requiring O1O2 = O1T1 = T1T2 = T1U1 = U1V1 and O1U1 = V1V2, we find in degrees rounded to 2 decimals after the decimal point

α = 33.63, θ_{1} = 56.61, θ_{2} = 6.28, θ_{3} = 34.15, θ_{4} = 44.62 .

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