Closest Pair of Points

Source: GeeksforGeeks:

We are given an array of n points in the plane, and the problem is to find out the closest pair of points in the array. This problem arises in a number of applications.

For example, in air-traffic control, you may want to monitor planes that come too close together, since this may indicate a possible collision.

Solution & Analysis

1)We sort all points according to x coordinates.

2)Divide all points in two halves.

3)Recursively find the smallest distances in both subarrays.

4)Take the minimum of two smallest distances. Let the minimum be d.

5)Create an array strip[] that stores all points which are at most d distance away from the middle line dividing the two sets.

6)Find the smallest distance in strip[].

7)Return the minimum of d and the smallest distance calculated in above step 6.

The great thing about the above approach is, if the array strip[] is sorted according to y coordinate, then we can find the smallest distance in strip[] in O(n) time. In the implementation discussed in previous post, strip[] was explicitly sorted in every recursive call that made the time complexity O(n (Logn)^2), assuming that the sorting step takes O(nLogn) time.

In this post, we discuss an implementation where the time complexity is O(nLogn). The idea is to presort all points according to y coordinates. Let the sorted array be Py[]. When we make recursive calls, we need to divide points of Py[] also according to the vertical line. We can do that by simply processing every point and comparing its x coordinate with x coordinate of middle line.

Following is C++ implementation of O(nLogn) approach.

// A divide and conquer program in C++ to find the smallest distance from a 
// given set of points. 

#include <iostream> 
#include <float.h> 
#include <stdlib.h> 
#include <math.h> 
using namespace std; 

// A structure to represent a Point in 2D plane 
struct Point 
    int x, y; 

/* Following two functions are needed for library function qsort(). 
Refer: */

// Needed to sort array of points according to X coordinate 
int compareX(const void* a, const void* b) 
    Point *p1 = (Point *)a, *p2 = (Point *)b; 
    return (p1->x - p2->x); 
// Needed to sort array of points according to Y coordinate 
int compareY(const void* a, const void* b) 
    Point *p1 = (Point *)a, *p2 = (Point *)b; 
    return (p1->y - p2->y); 

// A utility function to find the distance between two points 
float dist(Point p1, Point p2) 
    return sqrt( (p1.x - p2.x)*(p1.x - p2.x) + 
                (p1.y - p2.y)*(p1.y - p2.y) 

// A Brute Force method to return the smallest distance between two points 
// in P[] of size n 
float bruteForce(Point P[], int n) 
    float min = FLT_MAX; 
    for (int i = 0; i < n; ++i) 
        for (int j = i+1; j < n; ++j) 
            if (dist(P[i], P[j]) < min) 
                min = dist(P[i], P[j]); 
    return min; 

// A utility function to find minimum of two float values 
float min(float x, float y) 
    return (x < y)? x : y; 

// A utility function to find the distance beween the closest points of 
// strip of given size. All points in strip[] are sorted accordint to 
// y coordinate. They all have an upper bound on minimum distance as d. 
// Note that this method seems to be a O(n^2) method, but it's a O(n) 
// method as the inner loop runs at most 6 times 
float stripClosest(Point strip[], int size, float d) 
    float min = d; // Initialize the minimum distance as d 

    // Pick all points one by one and try the next points till the difference 
    // between y coordinates is smaller than d. 
    // This is a proven fact that this loop runs at most 6 times 
    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; 

// A recursive function to find the smallest distance. The array Px contains 
// all points sorted according to x coordinates and Py contains all points 
// sorted according to y coordinates 
float closestUtil(Point Px[], Point Py[], int n) 
    // If there are 2 or 3 points, then use brute force 
    if (n <= 3) 
        return bruteForce(Px, n); 

    // Find the middle point 
    int mid = n/2; 
    Point midPoint = Px[mid]; 

    // Divide points in y sorted array around the vertical line. 
    // Assumption: All x coordinates are distinct. 
    Point Pyl[mid+1]; // y sorted points on left of vertical line 
    Point Pyr[n-mid-1]; // y sorted points on right of vertical line 
    int li = 0, ri = 0; // indexes of left and right subarrays 
    for (int i = 0; i < n; i++) 
    if (Py[i].x <= midPoint.x) 
        Pyl[li++] = Py[i]; 
        Pyr[ri++] = Py[i]; 

    // Consider the vertical line passing through the middle point 
    // calculate the smallest distance dl on left of middle point and 
    // dr on right side 
    float dl = closestUtil(Px, Pyl, mid); 
    float dr = closestUtil(Px + mid, Pyr, n-mid); 

    // Find the smaller of two distances 
    float d = min(dl, dr); 

    // Build an array strip[] that contains points close (closer than d) 
    // to the line passing through the middle point 
    Point strip[n]; 
    int j = 0; 
    for (int i = 0; i < n; i++) 
        if (abs(Py[i].x - midPoint.x) < d) 
            strip[j] = Py[i], j++; 

    // Find the closest points in strip. Return the minimum of d and closest 
    // distance is strip[] 
    return min(d, stripClosest(strip, j, d) ); 

// The main functin that finds the smallest distance 
// This method mainly uses closestUtil() 
float closest(Point P[], int n) 
    Point Px[n]; 
    Point Py[n]; 
    for (int i = 0; i < n; i++) 
        Px[i] = P[i]; 
        Py[i] = P[i]; 

    qsort(Px, n, sizeof(Point), compareX); 
    qsort(Py, n, sizeof(Point), compareY); 

    // Use recursive function closestUtil() to find the smallest distance 
    return closestUtil(Px, Py, n); 

// Driver program to test above functions 
int main() 
    Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}}; 
    int n = sizeof(P) / sizeof(P[0]); 
    cout << "The smallest distance is " << closest(P, n); 
    return 0; 


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