c++ Closest Pair of Points | O(nlogn) Implementation

Travinth Dayalaeaswaran answered on July 8, 2022 Popularity 6/10 Helpfulness 1/10

answer c++ Closest Pair of Points | O(nlogn) Implementation

c++ Closest Pair of Points | O(nlogn) Implementation

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// 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: http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
 
// 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) ? (p1->x - p2->x) : (p1->y - p2->y);
}
// 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) ? (p1->y - p2->y) : (p1->x - p2->x);
}
 
// 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 a minimum of two float values
float min(float x, float y)
{
    return (x < y)? x : y;
}
 
 
// A utility function to find the distance between the closest points of
// strip of a given size. All points in strip[] are sorted according 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];   // y sorted points on left of vertical line
    Point Pyr[n-mid];  // 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 || (Py[i].x == midPoint.x && Py[i].y < midPoint.y)) && li<mid)
         Pyl[li++] = Py[i];
      else
         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 stripClosest(strip, j, d);
}
 
// The main function 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;
}

Popularity 6/10 Helpfulness 1/10 Language cpp

Source: www.geeksforgeeks.org

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Contributed on Jul 08 2022

Travinth Dayalaeaswaran

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c++ Closest Pair of Points | O(nlogn) Implementation

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