This method will not work. My counterexample consists of 5 points,
4 red, and 1 blue, shown below. I also show that for the lower-left
red point, it has two GG neighbors that are also red, so it will be
discarded. The blue point is not a GG neighbor.
Now I show that for the upper-left red point, it has 3 GG neighbors
that are red, and the blue point is not a GG neighbor, so the upper-left
red point will also be discarded. By symmetry the same applies for the
lower-right red point. Trivially, we can see that no other points are
So our set XG consists of the 2 red points and 1 blue point at
the top of the following image. Let us query the original lower-left red
point, now shown black.
We see that the blue point is the closest, so the red point will be
classified as blue, incorrectly.