Consider a semicircle C with radius r in the plane and a set P of n points lying on or inside C. Can we give an upper bound on the cost of an MST of P as a function of r (independent of n)?
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`math.pi * r` is definitely an upper bound. And it must be linear in `r`. – user58697 Jul 02 '18 at 08:06
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I have edited the question. The points can lie on or inside C. – user2311963 Jul 02 '18 at 11:02
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i don't understand how the cost can be independent of n? As the cost is sum is calculated over n right? or do you mean that on vertice has an upper bound. – Gijs Den Hollander Jul 02 '18 at 13:52
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you mean a euclidean MST right? – Gijs Den Hollander Jul 02 '18 at 13:53
1 Answers
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There is no upper bound independent of n.
Consider, for example, the Koch snowflake: https://en.wikipedia.org/wiki/Koch_snowflake
It fits into a finite area, and the perimeter is an MST, but its length is infinite (or approaches infinity as you iterate the construction).
More generally, you can force an MST to follow any curve you like, however closely you want, by sampling the curve at sufficiently small increments. Since there is no limit on the length of a curve you can stuff into an area, there is no limit on the cost of an MST you can stuff into an area.
Matt Timmermans
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