#plan an efficient finnish bus stop on a sphere [tag:code-challenge][tag:test-battery] Apparently respecting personal space is very important at [finnish bus stops](https://twitter.com/vilsepi/status/795254546763313152?lang=it). Now given some "minimum-personal-space-angle" \$\vartheta\$, your job is designing a bus stop on a sphere for as many people as possible respecting the "minimum-personal-space-angle". Let us rephrase this a little bit more clearly: Let \$S^2 = \{x \in \mathbb R^3 \mid \Vert x \Vert_2 =1 \}\$ be the unit sphere in \$\mathbb R^3\$. Given the angle \$\vartheta \in (0,\pi)\$ you should find a set \$U \subset S^2\$ such that all pairs of vectors \$x,y \in U\$ (\$x \neq y\$) are at least an angle of \$\vartheta\$ apart, that is \$x \cdot y \leqslant \cos \vartheta\$. And this set \$U\$ should be as large as possible - but this does not mean that your program needs to find the largest possible \$U\$ (this is a hard unsolved problem), but it should try to make it as large as possible as this will be part of the score. Let us define \$a_\vartheta = \vert U \vert\$ as the number of vectors your program found for \$\vartheta\$. The score \$s\$ of your submission will be $$ s=\frac{1}{N}\sum_{n=1}^N a_{\vartheta_n} w_n$$ where \$\vartheta_n = 1/n\$, \$w_n = 1/n^2\$. And you can choose \$N \in \mathbb N\$ as large as you want. Inspired by [this reddit thread](https://www.reddit.com/r/math/comments/e3go8b/question_maximum_number_of_vectors_in_rn_that_are/). #META: I think the choice of \$\vartheta_n\$ and \$w_n\$ needs some fine tuning to make the challenge interesting. My thoughts so far: The idea is that \$a_{\vartheta} \leqslant c \frac{1}{\vartheta^2}\$ since every vector on the sphere needs a circle of a radius that is at least \$\vartheta/2\$, so the area of such a circle is about \$\pi (\vartheta/2)^2\$ which means we can fit at most \$\frac{4\pi}{\pi (\vartheta/2)^2} = \frac{1}{\vartheta^2}\$ (just as a rough estimate). So I think with current choice of \$\vartheta_n\$ and \$w_n\$ the score should be bounded. But I fear that with the current choice of these sequences the greatest score will be achieved by a relatively simple solution where someone just chooses \$N=1,2\$ or so. Can we alleviate this by adding a factor of \$\log n\$ to \$w_n\$? Unfortunately I think this would incentivise using very large \$N\$. ----- **EDIT**: Instead of using \$\log n\$ I think using a *bounded* increasing sequence would work. E.g. \$(1-1/n)\$ so \$w_n = \frac{1}{n^2}(1- \frac{1}{n})\$. ----- If you have any thoughts or ideas, please share!