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1000 alien spaceships meet in a 4-dimensional battlefield. At an agreed time (ignoring relativistic effects on their clocks) every spaceship fires its laser to the spaceship which is closest (assume that all distances between the spaceships are different). What is the maximum number of hits a single spaceship can suffer?

This problem is equivalent to this question: "How many points can you place on the surface of a 4D-sphere, such that all distances between any two points are larger than the radius of the sphere?"

If you consider these 16 points: (0 0 $\pm 1 \, {\pm\phi}$), (0 $\pm 1 \, {\pm\phi}\; 0$), ($\pm 1 \, {\pm\phi}\; 0\ 0$), (${\pm\phi}\; 0\ 0\ {\pm 1}$) with $\phi = \frac{1 + \sqrt{5}}{2}$, you have a lower bound. I have failed to find a solution with more than 16 points. Any idea?

Rüdi Jehn
  • 624
  • I think the number should be higher, because I think it should be at least the kissing number in 4 dimensions, which is 24. Unless I misunderstood the problem. – Immanuel Aug 25 '24 at 12:32
  • Sorry I think I should have said at most the kissing number in 4 dimensions. If the sphere packing which gives you a kissing number of 24, has no touching spheres, then the answer should be exactly 24. Otherwise you could probably rearrange it s.t. you get 23 as an answer. But maybe I got your question wrong. – Immanuel Aug 25 '24 at 12:51
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    How are those 2 questions equivalent? – Calvin Lin Aug 25 '24 at 13:12
  • Assuming that all distances between spaceships are different, you cannot place more than one spaceship on any given hypersphere around your target spaceship. You might place several ships close to a hypersphere, but only one will be exactly on it. – David K Aug 25 '24 at 13:29
  • I was pointed to this page: http://neilsloane.com/packings/. It seems the answer is 22 but I have not yet found an example solution. Note, the distance between the points must be larger than the sphere radius. If it can be equal, the solution should be 24. – Rüdi Jehn Aug 27 '24 at 05:43

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