Given a set of $n$ two-dimensional points in the plane $$\{ (x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)\}$$ and a real number $M$, I want to transform this set of points into a graph with the points as vertices with an edge between points $A$ and $B$ if the distance between $A$ and $B$ is at most $M$.

What is a fastest algorithm that can perform this task?

  • $\begingroup$ One thing to note is that if $M$ is very very large, then you need to create all m ~ n² pairs, so you cannot guarantee that you don't need to perform n² operations. Hence, if you're constructing the actual graph, a linear time algorithm (in the number of points) is not possible. $\endgroup$
    – Pål GD
    Jan 15 '20 at 16:16

As you can imagine, evaluating every pair of points is very expansive ($O(N^2)$ for $N$ points).

If your set of point is sufficiently dense with respect to $M$, a simple solution is to use a grid. Build a grid of $M \times M$ cells, then for each cell, compute the list of points inside (doing one loop on points $O(N)$). Finally you just have to compare every point with every other point in the same cell or in one of the 8 surrounding it.

If you set of points is very dense you even can use a second grid of $\frac{M}{\sqrt{2}} \times \frac{M}{\sqrt{2}}$ cells so the points in the same cells have necessarly pair-wise distance lower than $M$. This may save some computation.

Note that in the case you have lot of empty cells (small density), it may generate memory issue. Imagine for instance 2 pairs of points very distant to each other (with respect to $M$). It may require a tremendeous grid, just to process 4 points... In this case, the basic "all-pair" comparison would be largely more efficient. k-d tree may be a solution to this problem. Or you can either identify the empty zones to process the points subset by subset.

  • 2
    $\begingroup$ What is the suggested algorithm and what is its complexity? $\endgroup$
    – Pål GD
    Jan 15 '20 at 16:12
  • $\begingroup$ @Pål_GD Basically, I suggest the $M \times M$ cell grid algorithm. And of course, you cannot guarantee that it will do better than $O(N^2)$ or even worse in the low density case. The remain of my answer try to treat extreme cases. $\endgroup$
    – Optidad
    Jan 16 '20 at 10:32

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