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The Voronoi cell $V$ determines the lattice $\Lambda$. Consider an ($n-1$)-face $F$ of $V$. The hyperplane of the face $F$ is the perpendicular bisector of the segment between $O$ and a lattice point $p$. Then $V+p$ is the neighboring Voronoi cell that shares face $F$. Marching through the Voronoi cells we can find all cells and thus all lattice points at ...


2

Here is a sketch of a common algorithm. While basic, there are some obvious ways to improve it. For concreteness sake, suppose we wish to plot the contour $f(x,y)=0$, where $$f(x,y) = x^3 - x y + y^4.$$ Let us start with a rectangular grid in the plane. Something like so: This particular grid is $10\times10$ and lives in the square ...


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Three points are collinear if the determinant |x1 y1 1| |x2 y2 1| = x1*y2 + x2*y3 + x3*y1 - x1*y3 - x2*y1 - x3*y2 |x3 y3 1| is zero. That's equivalent to stating that the area of the triangle formed by these three is zero, since the area would be half the value of this determinant by Heron's formula. The determinant of three vectors can also be expressed ...



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