Question

I am looking at the wikipedia entry for how to solve this. It lists five steps

1.Sort points along the x-coordinate

2.Split the set of points into two equal-sized subsets by a vertical line x = xmid

3.Solve the problem recursively in the left and right subsets. This will give the left-side and right-side minimal distances dLmin and dRmin respectively.

4.Find the minimal distance dLRmin among the pair of points in which one point lies on the left of the dividing vertical and the second point lies to the right.

5.The final answer is the minimum among dLmin, dRmin, and dLRmin.

The fourth step I am having trouble understanding. How do I choose what point to the left of the line to compare to a point right of the line. I know I am not supposed to compare all points, but I am unclear about how to choose points to compare. Please do not send me a link, I have searched, gone to numerous links, and have not found an explanation that helps me understand step 4.

Thanks

Aaron

Answer 1

The answer to your question was in the next paragraph of the wikipedia article:

It turns out that step 4 may be
accomplished in linear time. Again, a
naive approach would require the
calculation of distances for all
left-right pairs, i.e., in quadratic
time. The key observation is based on
the following sparsity property of the
point set. We already know that the
closest pair of points is no further
apart than dist = min(dLmin,dRmin).
Therefore for each point p of the left
of the dividing line we have to
compare the distances to the points
that lie in the rectangle of
dimensions (dist, 2 * dist) to the
right of the dividing line, as shown
in the figure. And what is more, this
rectangle can contain at most 6 points
with pairwise distances at least
dRmin. Therefore it is sufficient to
compute at most 6n left-right
distances in step 4. The recurrence
relation for the number of steps can
be written as T(n) = 2T(n / 2) + O(n),
which we can solve using the master
theorem to get O(n log n).

I don’t think I can put it much clearer than they already have, but do you have any specific questions about this step of the algorithm?