For the "bad" triangles with 3 around the vertex, what would happen if we applied the edge flip function as implemented in the tutorial to the lower left edge? It seems like the flipped edge would be 'outside' the two triangles the edge connects. I guess in general do we have to check if the two triangles form a convex quad before we apply the function?

keenan

@Log Right, excellent observation. An edge cannot be flipped if it is contained in a pair of triangles that forms a nonconvex quad. Serendipitously, the angle sum $\alpha + \beta$ for such an edge (as defined on this slide) can never be greater than $\pi$! In other words: a "nonconvex edge" is already Delaunay. So if your goal is to make a mesh Delaunay by flipping, you will never run into this case...

adam

One thing I realize I'm a little unclear on after the quiz - is the valence of 6 something that we try to enforce on the mesh, or is it something that occurs as the number of triangles approaches infinity? If the latter, why does an infinite number of triangles approximate hexagons?

nrauen

I think that in the best case scenario we would like them to be hexagons as the degrees would fit around a 60 degree expectations, but if you think of the number of triangles approaching infinity there is a higher likely hood of points being not on the edge. Thus 360 degrees of coverage must be accounted for which leads to a valence of 6 on average as each angle accounts for on average 60 degrees.

keenan

@adam You can prove that the average vertex degree approaches 6 as the number of mesh elements approaches infinity, but this of course doesn't mean you can't have very high (or low) degree vertices on a very fine mesh. This statement is really just about the connectivity, by the way: you can still have crazy angles all over the place (far from 60) even on a really high-res triangle mesh.

cou

On the quiz I was also confused as to why the number of vertices in the ratio was 1 in the 1:3:2 V:E:F ratio? I was out of town today interviewing so I'm not sure if you explained this in lecture.

keenan

@cou I just sent you the solutions via email (please don't redistribute, as we may want to reuse these next year!)

For the "bad" triangles with 3 around the vertex, what would happen if we applied the edge flip function as implemented in the tutorial to the lower left edge? It seems like the flipped edge would be 'outside' the two triangles the edge connects. I guess in general do we have to check if the two triangles form a convex quad before we apply the function?

@Log Right, excellent observation. An edge cannot be flipped if it is contained in a pair of triangles that forms a nonconvex quad. Serendipitously, the angle sum $\alpha + \beta$ for such an edge (as defined on this slide) can never be greater than $\pi$! In other words: a "nonconvex edge" is already Delaunay. So if your goal is to make a mesh Delaunay by flipping, you will never run into this case...

One thing I realize I'm a little unclear on after the quiz - is the valence of 6 something that we try to enforce on the mesh, or is it something that occurs as the number of triangles approaches infinity? If the latter, why does an infinite number of triangles approximate hexagons?

I think that in the best case scenario we would like them to be hexagons as the degrees would fit around a 60 degree expectations, but if you think of the number of triangles approaching infinity there is a higher likely hood of points being not on the edge. Thus 360 degrees of coverage must be accounted for which leads to a valence of 6 on average as each angle accounts for on average 60 degrees.

@adam You can prove that the

averagevertex degree approaches 6 as the number of mesh elements approaches infinity, but this of course doesn't mean you can't have very high (or low) degree vertices on a very fine mesh. This statement is really just about the connectivity, by the way: you can still have crazy angles all over the place (far from 60) even on a really high-res triangle mesh.On the quiz I was also confused as to why the number of vertices in the ratio was 1 in the 1:3:2 V:E:F ratio? I was out of town today interviewing so I'm not sure if you explained this in lecture.

@cou I just sent you the solutions via email (please don't redistribute, as we may want to reuse these next year!)