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In the previous slide it showed that alpha + beta < pi yields higher Delaunay. However, this image shows a higher Delaunay is created because the valence of each vertex is 6. Do we prioritize the valence of any triangle mesh to be 6 over the alpha+beta<pi because the flipped edge no longer follows this rule? Is there one general rule that ALWAYS creates the best mesh? mdsavage

"FACT: average valence of any triangle mesh is 6"

This statement is not true; in a tetrahedron, for example, every vertex has valence 3. In fact, this statement isn't true of any surface of genus 0 (i.e., any surface that can be deformed into a sphere): any triangulation on such a surface would have 3V - 6 edges, so the average valence would be 2E/V = (6V - 12) / V < 6. That said, the average valence does approach 6 as the number of vertices increases. keenan

@dvanmali No, there is no rule that always produces the best mesh, because the notion of "best" depends on what the mesh will be used for. Here we're presenting two different notions of local "goodness," namely (i) satisfying the Delaunay condition at each edge or (ii) exhibiting regular degree at each vertex. The Delaunay condition can provide very good behavior in contexts where one wants to solve equations on surfaces (e.g., physical simulation or various geometry processing tasks); regular vertex degree can be valuable for things like mitigating artifacts in subdivision schemes. In general, these two critera may not be compatible, i.e., the Delaunay triangulation may have many vertices of irregular degree, and a triangulation of regular vertex degree may have many non-Delaunay edges. There is no free lunch. keenan

@mdsavage Yes, you are absolutely right. The "FACT" is that the average vertex degree approaches 6 as the number of elements approaches infinity. This could be spelled out more clearly on the slide.