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Assume that $y$ is not in the polyhedron (it is easy to check whether it is, and we know that the distance is zero in that case). If $y$ is outside then the closest point will be on the surface of the polyhedron. So I came up with the following (horrible) algorithm, which will give you an upper bound. Let $y^0=y$. Find distance of point $y^n$ to all planes $...


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