# What is the best approximate of points on a sphere?

I have a unit radius sphere with a set $S$ of $n$ points on it. How can I find a map $f:S\to \mathbb{R}^4$ which minimizes $$\sum_{x,y\in S} \bigg( d_{\text{geodesic}} (x,y)^{2} - d(f(x),f(y))^{2}\bigg)^2$$

$d_{\text{geodesic}}(x,y)$ is the length of the shortest path on the sphere from $x$ to $y$, and $d(f(x),f(y))$ is the Euclidean distance in $\mathbb{R}^4$.$\hspace{-0.03 in}\big)$

• ... append a 0 to their coordinate tuples. $\;$ – user57159 Nov 15 '15 at 9:01
• No, the original distances are geometric not Euclidean! I am not sure about the name maybe geographical distance! – remo Nov 15 '15 at 10:17
• Do you have in mind a particular measure of how well they're preserved? $\:$ When I did something like that, I used sum-of-squares-of-logs-of-ratios-of-distances (over the choice of pairs of distinct points), but I've also seen a measure which is equivalent to replacing that "sum" with "max" (where the original metric is a discrete metric). $\;\;\;\;$ – user57159 Nov 15 '15 at 18:46
• Thank you for your answer. I need SSE. It seems there are multiple solutions with minimum error, the final solution depends on the initial values for the new dimension. I programmed it in GAMS and tried different solvers to minimize the squared errors, but I was unsuccessful. – remo Nov 15 '15 at 19:19
• Are the "squared errors" just $\big(\hspace{-0.02 in}d_{\hspace{.03 in}0\hspace{-0.02 in}}(x,\hspace{-0.03 in}y)\hspace{-0.03 in}-\hspace{-0.03 in}d_{\hspace{.02 in}1\hspace{-0.02 in}}(x,\hspace{-0.03 in}y)\hspace{-0.04 in}\big)^2\hspace{-0.04 in}$, or a type of relative error? $\;$ – user57159 Nov 15 '15 at 19:30