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Context, I have a geometric algorithm that is sensitive to collinear points and receives as input a list of points in 2D generated randomly. Suppose that I have a Boolean function nonCollinear(x,y,z) that given three points returns true if they are non collinear and false otherwise. Is there an efficient algorithm to generate a list of random points such that no 3 points in the list are collinear? The points coordinates x y are integers, the amount of points is N and the grid size is RxC, I know that there is a restriction between N and R C, so I am fine if the algorithm relies in some assumption (e.g RxC is way bigger than N)

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  • $\begingroup$ Do you need some restrictions on the distribution generated? Some fixed shapes (e.g., points along the edge of an approximate circle; in general, the vertices of any convex shape) can be easily generated to have no three points collinear $\endgroup$ – Curtis F Jul 20 '18 at 23:11
  • $\begingroup$ Ideally I would want the points closer to uniformly distributed $\endgroup$ – dpalma Jul 20 '18 at 23:33
  • $\begingroup$ However any easy algorithm will do if it is random in some sense. $\endgroup$ – dpalma Jul 20 '18 at 23:43
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For simplicity , assume the grid is a square $N \times N$ grid and $N$ is a prime.

Its easy to see that from each row we can pick $\leq 2$ points only , so the maximum number of points we can chose is $2N$.

Now consider the set of points $\{(i,i^2\ mod\ n)\ |\ 0\leq i \leq n-1 \}$.

For any set of 3 points to be collinear (Lets call them $(x_1,y_1),(x_2,y_2),(x_3,y_3) , x_1 < x_2 < x_3$ ) we must have $\frac{y_3 - y_2}{x_3 - x_2} = \frac{y_2 - y_1}{x_2 - x_1}$ , putting in $y_i = x_i^2\ mod\ n$ , we obtain $x_3 + x_2 \equiv x_2 + x_1 \mod n$ and this is only possible when $x_1 = x_3$.

Hence we have a set of $N$ points where no 3 points are collinear and it is the best we can obtain upto a constant factor.

Since this is a deterministic construction , to get a random family of points , instead of choosing $i^2$ , we can take any random quadratic polynomial $p$ and take the set of points $\{(i , p(i)) \ mod\ n \ |\ 0 \leq i \leq n-1 \}$.

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