2D Closest Pair Problem

Question

PROBLEM:
Given an arbitrarily number of data points in 2-d space, this algorithm finds a pair of data points with minimum Euclidean distance from each other.
I read in points from a data file as follows:
100 //number of points
12.5 100.2 //each line contains the x and y coordinates of one data point
0.7 1.5
I know it seems strange to be posting such a topic since it has been addressed several places before. I have found a decent psuedocode that addresses the algorithm:
closestPair of (xP, yP)
               where xP is P(1) .. P(N) sorted by x coordinate, and
                     yP is P(1) .. P(N) sorted by y coordinate (ascending order)
if N ≤ 3 then
  return closest points of xP using brute-force algorithm
else
  xL ← points of xP from 1 to ⌈N/2⌉
  xR ← points of xP from ⌈N/2⌉+1 to N
  xm ← xP(⌈N/2⌉)x
  yL ← { p ∈ yP : px ≤ xm }
  yR ← { p ∈ yP : px > xm }
  (dL, pairL) ← closestPair of (xL, yL)
  (dR, pairR) ← closestPair of (xR, yR)
  (dmin, pairMin) ← (dR, pairR)
  if dL < dR then
    (dmin, pairMin) &#8592; (dL, pairL)
  endif
  yS &#8592; { p &#8712; yP : |xm - px| < dmin }
  nS &#8592; number of points in yS
  (closest, closestPair) &#8592; (dmin, pairMin)
  for i from 1 to nS - 1
    k &#8592; i + 1
    while k &#8804; nS and yS(k)y - yS(i)y < dmin
      if |yS(k) - yS(i)| < closest then
        (closest, closestPair) &#8592; (|yS(k) - yS(i)|, {yS(k), yS(i)})
      endif
      k &#8592; k + 1
    endwhile
  endfor
  return closest, closestPair
endif I don't understand this a 100% and how I would implement this. This has a more mathematical approach and that seems to be my problem when I look online. Is there anymore resources that anyone is aware of. Thanks for any help.

Answer 1

    static double d;
    private static void bruteForceClosestPair (ArrayList<Pair> points) {
        ArrayList<Pair> minPoints = new ArrayList();
        if (points.size() < 2) {
            System.out.println("Only one point!");
        else {
            d = distance (points.get(0).getX(),points.get(0).getY(), points.get(1).getX(), points.get(1).getY());
            minPoints.add(points.get(0));
            minPoints.add(points.get(1));
            for (int i = 0; i < points.size()-1;i++){
                for (int j = i+1;j < points.size(); j++) {
                    System.out.println("Checking "+ points.get(i) +  " and " + points.get(j) );
                    if(distance(points.get(i).getX(), points.get(i).getY(), points.get(j).getX(), points.get(j).getY())< d){
                        d = distance(points.get(i).getX(), points.get(i).getY(), points.get(j).getX(), points.get(j).getY());
                        minPoints.clear();
                        minPoints.add(points.get(i));
                        minPoints.add(points.get(j));
        System.out.println("minimum is " + d);
        System.out.println("My pair points be " + minPoints.toString());
    }I am able to check every point using a brute force attack, but the problem that I am trying to accomplish is finding a divide and conquer/recursive approach to solving this problem. I hope this makes sense. How can I go about this?