top of page


Pubblico·10 membri
Wyatt Collins
Wyatt Collins

No One Left To Lie To: The Triangulations Of Wi...

V,I worked with a Narcissistic woman for a couple weeks. She was terrible and loved making me feel inferior every time we were left alone. Coincidentally, she would never say/do anything bad to me when other people were around. After suffering from her for a couple days I went to talk to the manager and guess what happened? I was fired two days later: all my co-workers believed her lies, never did anything to change her ways, and thought I was delusional.

No One Left to Lie To: The Triangulations of Wi...

I encountered the same problem when i was implementing Bowyer-Watson algorithm described here: I couldn't find anything helpful on internet and even asked at my university, but with no result. After a while I came up with a solution.I started with discovery that for the problem to disappear, the vertices of bounding triangle should ideally lie at infinity, which is not practical. So what does triangles circumcircle look like if triangle has one or two vertices at infinity? It is just line going through the other points. So testing if point lies in triangles circumcircle changes to testing if point lies left or right of line.

If it's sharing two vertices: check if point lies to the left/right of line defined by these two vertices but shifted to the third point. In other words: you take only the slope vector from line between these shared vertices and shift it so that the line passes through the third point. two vertices in infinity

I see now that the term "arching away" wasn't precise enough. What we need is that the $n-1$ points lie on a (non-circular) arc that is everywhere concave as seen from the $n$-th point, that is, that the lines connecting the $n$-th point to the arc lie everywhere on the concave side of the arc and nowhere inside its convex hull. For instance, if the $n$-th point is $(0,0)$, the $n-1$ points could lie on the curve $y=\mathrm e^x$ to the left of the point $(1,\mathrm e^1)$. At that point, the line from $(0,0)$ to the curve is tangent to the curve; for points to the left, the line lies on the concave side of the curve, whereas for points to the right the line crosses the curve and ends on its convex side. 041b061a72


Ti diamo il benvenuto nel gruppo! Qui puoi entrare in contat...
bottom of page