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VC dimension of the class of convex polygons with $ k $ vertices is known to be $ 2k + 1$.

For the general case I was able to derive a bound of the type $ O(k^2log(k)) $ (probably can be easily improved to O(klog(k)), this is not the issue). However, I found some hits (e.g. https://cs.nyu.edu/~mohri/ml/ml08/sol2.pdf) that VC dimension for this case is in fact also $2k + 1$. Any idea if this is correct?

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  • $\begingroup$ Which of Mohri’s arguments gives you indication that it’s 2k+1? I don’t see it. $\endgroup$
    – Aryeh
    Apr 24 at 8:42
  • $\begingroup$ Indeed, there's no argument in dat document, just a statement. $\endgroup$ Apr 30 at 15:44

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Assuming that the $k$-gon is simple (i.e., does not intersect itself and has no holes) and $k>3$, the two-ears theorem, https://en.wikipedia.org/wiki/Two_ears_theorem implies that it can be triangulated -- i.e., expressed as a union of $k$ or fewer triangles. Now triangles in the plane have VC-dim 7, and, by Lemma 3.2.3 of Blumer et al. (1989), $k$-fold unions of triangles have VC-dim at most $$ 14k\log(3k)=O(k\log k). $$ I still don't see how you can get $2k+1$.

A. Blumer, A. Ehrenfeucht, D. Haussler, and M. K. Warmuth. Learnability and the Vapnik-Chervonenkis dimension. J. Assoc. Comput. Mach., 36(4):929–965, 1989. ISSN 0004-5411.

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  • $\begingroup$ Thank you, Aryeh! I used the same lemma, but with a "convex partition by segments" - cs.jhu.edu/~misha/Spring16/05.pdf . I still have an intuition (which can very well be wrong) that it might be 2k +1. My reasoning is like this: a) For any n points, even for general polygons, the best case (maximum number of label assignments ) is still achieved for points that are all vertexes of the convex hull. In that case, the fact that non-convex polygons can be used is not useful- the VC dimension is the same as in the case of convex polygons. However, I was not able to actually show this. $\endgroup$ Apr 30 at 15:38
  • $\begingroup$ Interesting, worth thinking about! $\endgroup$
    – Aryeh
    Apr 30 at 18:45

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