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Given are a number of control points $c_i$ and a point $p$ (all in cartesian 3D space). $i$ may be anywhere between 8 and roughly 200. The point $p$ can be written as a convex combination of $c_i$: \begin{equation} p = \sum_i w_i c_i \end{equation} given the constraints that $\sum w_i = 1$ and $0 \leq w_i \leq 1$ (for all $i$).

I would like to find a set of weights $w_i$. Since I suppose there are infinitely many solutions to this, I would like to find an intuitive solution (whatever that means, exactly) - maybe one that keeps the weights as evenly distributed as possible.

How would I find those weights?

Daerst
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  • build the convex hull of the point and tetrahedrisize once. – Asinomás Jun 16 '17 at 15:34
  • @TacNayn When I have built the convex hull of $w_i$ and tetrahedralized it (so that each tetrahedron has $p$ as one corner, I suppose), how do I then find the weights I am looking for? – Daerst Jun 19 '17 at 18:03
  • no, you don't do it so that every tetrahedron has p as a corner, you do it so that the corners are among the $c_i$ – Asinomás Jun 19 '17 at 19:22
  • Once you have tetrahedrized properly and you have a point $p$, the first step is to find inside of which of the tetrahdera point $p$ is . Suppose that it is inside the tetrahedra $d_1,d_2,d_3,d_4$. – Asinomás Jun 19 '17 at 19:26
  • notice that $d_2-d_1,d_3-d_1,d_4-d_1$ forms a basis for $\mathbb R^n$, express $p-d_1$ in this basis, and then ad $d_1$ to the found expression. This expression is convex and gives you $p$. – Asinomás Jun 19 '17 at 19:37
  • This way, only the four points $d_1, d_2, d_3, d_4$ will have a weight $\gt 0$, right? As stated in the question, I would prefer a solution that distributes the weights evenly among the points $c_i$. – Daerst Jun 20 '17 at 08:07
  • oh sorry.${}{}{}{}{}{}{}{}$ – Asinomás Jun 20 '17 at 20:47
  • For others stumbling upon this: I ended up using Mean Value Coordinates [Ju et al. 2005]. There are various alternatives, e.g. Harmonic Coordinates [Joshi et al. 2007] and Green Coordinates [Lipman et al. 2008] that may work better, depending on the specific requirements. – Daerst Jul 06 '17 at 06:57

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