I am looking for a triangulation of an $n$-dimensional simplex such that all sub-simplices are of comparable size, and are "as close as possible" to a regular simplex : the latter property could be formalized as "the minimum $n$-dimensional angle is bounded away from $0$, uniformly in the size of the triangulation". I have read that the Delaunay triangulation had this kind of properties, but I did not find quantitative results about it, like lower bounds on the minimum angle.

Thank you!

  • $\begingroup$ Well for starters, what is bad for your purposes about the barycentric subdivision? $\endgroup$ Commented Oct 20, 2021 at 12:25
  • 3
    $\begingroup$ I think that in the barycentric subdivision, the triangles become flatter and flatter as the subdivision becomes thinner and thinner. This is typically the kind of property that I would like to avoid. $\endgroup$
    – Bruno
    Commented Oct 20, 2021 at 12:30

1 Answer 1


You are looking for the edgewise subdivision:
Edelsbrunner, H.; Grayson, D. R., Edgewise subdivision of a simplex, Discrete Comput. Geom. 24, No. 4, 707-719 (2000). ZBL0968.51016.

The basic idea is to slice your simplex k times along each coordinate hyperplane. Then you get some small uniform shapes, which are not simplices except at the corners (or in dimension 2). So then you slice each one of those shapes up in the same ad hoc way. If you do the details right, IIRC you can even get it so that every edge length is either $1/k$ or $\sqrt{2}/k$. Probably you can use that to figure out the minimal angle.

Another way, with worse constants but which is simpler to describe completely, is this: Split your $n$-simplex into $n+1$ (distorted) cubes with vertices at the barycenter, then subdivide the cubes cubically, and then split the little cubes back into simplices.

  • $\begingroup$ Thank you so much, this is exactly what I was looking for! $\endgroup$
    – Bruno
    Commented Oct 27, 2021 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.