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A monotone polygon is such that it is crossed at most two times by lines perpendicular to a given direction. That means that the polygon can be split in two chains of vertices that are sorted in that direction. Sorted order is beneficial as it can often reduce the complexity of computations. For instance, the convex hull of a monotone polygon can be ...


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This is called polygon offsetting rather than scaling. In the general case, this is much harder than it seems, as different parts of the inflated polygon can overlap each other, letting holes appear, and some edges may completely disappear. The way to handle corners isn't so well defined either. For an exact solution, have a look at Clipper. For a simple ...


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Your isosurface is a sphere, having center at the centroid of the $p_i$ and whose radius $r$ is given by $$ r^2={v\over n}-{1\over n^2}\sum_{i\ne j}(\vec p_i-\vec p_j)^2. $$ That follows from the identity $$ \sum_{i=1}^n(\vec x-\vec p_i)^2= n\left(\vec x -\sum_{i=1}^n{\vec p_i\over n}\right)^2+ {1\over n}\sum_{i\ne j}{(\vec p_i-\vec p_j)^2}. $$ Once $r$ is ...


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I'd test the convex hulls of the control points for overlap. If there is no overlap, then the answer is definitely "no" because the curves stay inside the convex hull. If there is overlap in such a way that the endpoints of the first curve are not inside the convex hull for the second curve, and vice versa, and the direct connections of the endpoints ...


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The convex hull is the smallest convex body that contains the given set. Any other convex body that contains the set will necessarily contain its convex hull. A convex body enclosed in another has both smaller area and smaller perimeter in 2D, or smaller surface area and smaller volume in 3D. So the convex hull is also the smallest enclosing convex body in ...



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