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The Bowyer-Watson algorithm for Delaunay triangulation is known to run in O(n^2) according to the authors, where n is the number of data points in R^d.

In addition, the algorithm (for example, as is written in Bowyer's paper, at stage 5 of the algorithm), runs over all the simplexes, (also runs over simplexes that will be later deleted).

How does it settle with the fact that there are O(n^{d/2}) simplexes in Delaunay triangulation?

One should expect from the algorithm to run in Omega(n^{d/2}), (and maybe even in a worse time complexity - since it runs over "extra simplexes").

Thanks!

Joe Doe
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  • There aren't too many algorithm geeks here. You might do better asking on the CS stack exchange. – Gene Aug 14 '21 at 20:49
  • what is `d` in your question? – jwezorek Aug 18 '21 at 21:31
  • Anyway you can O(n log n) time complexity if you have some way of finding the triangle a given point is in in O(log n) time. See my answer here: https://stackoverflow.com/a/59582271/1413244 – jwezorek Aug 18 '21 at 21:33
  • Thanks for the answer @jwezorek $d$ is the dimension - meaning that the points are in R^d. Also, the fact that we can run the algorithm in O(n log n) doesn't solve the question. – Joe Doe Aug 23 '21 at 23:19

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