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Theory Clustering is unlikely to work in this case because your red points are separated from each other by the green points. You could use more clusters, but this will require a lot of manual inspection and fiddling. A standard approach to this sort of problem is to use a nonlinear support vector machine. The idea, simply described, is that although your ...


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I don't know if the following is a good idea. But it is an idea, and I hope that it helps. This problem can be recast to "find a function $f: \mathbb{R}^2\to \mathbb{R}$ and $z\in\mathbb{R}$ s.t. $f(x,y)-z \geq 0$ if the point $(x,y)$ is "inside", and negative otherwise. For the basis of $f$ you can pick tensor products of Legendre polynomials ...


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One method could be to apply the (modified) Gram Schmidt process. To do this you would need to choose two vectors $v_1$ and $v_2$ not parallel to $nn$ such that $\operatorname{span}(v_1, v_2, nn) = \mathbb{R}^3$. Then set $nn$ as the first vector, $u_1$, in the Gram Schmidt process and the second and third vectors ($u_2$ and $u_3$) will be two orthogonal ...


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As suggested in Kindlmann the curvature of a surface is defined by the relationship between positional changes in the neighborhood of a point placed on the surface and the change in the surface normal. Given a level-set $\Phi(\mathbf{x})$, we consider that the value of the level-set is positive inside the object, negative outside. Hence, we define the ...


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I ended up implementing the suggestion by @DanielShapero (or at least how I understood it): import numpy as np def distance(a, b): # a & b contains each the first 6 values of direction cosine matrix ax, ay = a # ax is the first vector, ay is the second vector bx, by = b # bx is the first vector, by is the second vector az = np.cross(ax, ay) # ...


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You can find an algorithm to find a vector that is perpendicular to your first vector $\vec{u}$ here . Starting with this couple of vectors (we will call the vector perpendicular to $\vec{u}$ as $\vec{v}$) you can simply do the cross product to find the vector that is perpendicular to both $\vec{w} = \vec{u} \times \vec{v}$ and hence obtain the couple of ...


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