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 $\mathbb R^d$.
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 higher time complexity - since it runs over "extra simplexes").
Thanks!
