Closest Pair of Points Problem


In this problem, a set of n points are given on the 2D plane. In this problem, we have to find the pair of points, whose distance is minimum.

To solve this problem, we have to divide points into two halves, after that smallest distance between two points is calculated in a recursive way. Using distances from the middle line, the points are separated into some strips. We will find the smallest distance from the strip array. At first two lists are created with data points, one list will hold points which are sorted on x values, another will hold data points, sorted on y values.

The time complexity of this algorithm will be O(n log n).

Input and Output

Input:
A set of different points are given. (2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (3, 4)
Output:
Find the minimum distance from each pair of given points.
Here the minimum distance is: 1.41421 unit

Algorithm

findMinDist(pointsList, n)

Input: Given point list and number of points in the list.

Output −  Finds minimum distance from two points.

Begin
   min := ∞
   for all items i in the pointsList, do
      for j := i+1 to n-1, do
         if distance between pointList[i] and pointList[j] < min, then
            min = distance of pointList[i] and pointList[j]
     done
   done
   return min
End

stripClose(strips, size, dist)

Input − Different points in the strip, number of points, distance from the midline.

Output − Closest distance from two points in a strip.

Begin
   for all items i in the strip, do
      for j := i+1 to size-1 and (y difference of ithand jth points) <min, do
         if distance between strip[i] and strip [j] < min, then
            min = distance of strip [i] and strip [j]
      done
   done
   return min
End

findClosest(xSorted, ySorted, n)

Input − Points sorted on x values, and points sorted on y values, number of points.

Output − Find minimum distance from the total set of points.

Begin
   if n <= 3, then
      call findMinDist(xSorted, n)
      return the result
   mid := n/2
   midpoint := xSorted[mid]
   define two sub lists of points to separate points along vertical line.
   the sub lists are, ySortedLeft and ySortedRight

   leftDist := findClosest(xSorted, ySortedLeft, mid)         //find left distance
   rightDist := findClosest(xSorted, ySortedRight, n - mid)   //find right distance

   dist := minimum of leftDist and rightDist

   make strip of points
   j := 0
   for i := 0 to n-1, do
      if difference of ySorted[i].x and midPoint.x<dist, then
         strip[j] := ySorted[i]
         j := j+1
   done

   close := stripClose(strip, j, dist)
   return minimum of close and dist
End

Example

#include <iostream>
#include<cmath>
#include<algorithm>
using namespace std;

struct point {
   int x, y;
};

intcmpX(point p1, point p2) {    //to sort according to x value
   return (p1.x < p2.x);
}

intcmpY(point p1, point p2) {    //to sort according to y value
   return (p1.y < p2.y);
}

float dist(point p1, point p2) {    //find distance between p1 and p2
   return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
}

float findMinDist(point pts[], int n) {    //find minimum distance between two points in a set
   float min = 9999;
   for (int i = 0; i < n; ++i)
      for (int j = i+1; j < n; ++j)
         if (dist(pts[i], pts[j]) < min)
            min = dist(pts[i], pts[j]);
   return min;
}

float min(float a, float b) {
   return (a < b)? a : b;
}

float stripClose(point strip[], int size, float d) {    //find closest distance of two points in a strip
   float min = d;
   for (int i = 0; i < size; ++i)
      for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
         if (dist(strip[i],strip[j]) < min)
            min = dist(strip[i], strip[j]);
   return min;
}

float findClosest(point xSorted[], point ySorted[], int n){
   if (n <= 3)
      return findMinDist(xSorted, n);
   int mid = n/2;

   point midPoint = xSorted[mid];
   point ySortedLeft[mid+1];     // y sorted points in the left side
   point ySortedRight[n-mid-1];  // y sorted points in the right side
   intleftIndex = 0, rightIndex = 0;

   for (int i = 0; i < n; i++) {       //separate y sorted points to left and right
      if (ySorted[i].x <= midPoint.x)
         ySortedLeft[leftIndex++] = ySorted[i];
      else
         ySortedRight[rightIndex++] = ySorted[i];
   }

   float leftDist = findClosest(xSorted, ySortedLeft, mid);
   float rightDist = findClosest(ySorted + mid, ySortedRight, n-mid);
   float dist = min(leftDist, rightDist);

   point strip[n];      //hold points closer to the vertical line
   int j = 0;

   for (int i = 0; i < n; i++)
      if (abs(ySorted[i].x - midPoint.x) <dist) {
         strip[j] = ySorted[i];
         j++;
      }
   return min(dist, stripClose(strip, j, dist));    //find minimum using dist and closest pair in strip
}

float closestPair(point pts[], int n) {    //find distance of closest pair in a set of points
   point xSorted[n];
   point ySorted[n];

   for (int i = 0; i < n; i++) {
      xSorted[i] = pts[i];
      ySorted[i] = pts[i];
   }

   sort(xSorted, xSorted+n, cmpX);
   sort(ySorted, ySorted+n, cmpY);
   return findClosest(xSorted, ySorted, n);
}

int main() {
   point P[] ={{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
   int n = 6;
   cout<< "The minimum distance is " <<closestPair(P, n);
}

Output

The minimum distance is 1.41421

Samual Sam
Samual Sam

Learning faster. Every day.

Updated on: 16-Jun-2020

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