Closest Pair of Points using Divide and Conquer algorithm | A divide and conquer program in C ++ to find the smallest distance from a given set of points . ; A structure to represent a Point in 2D plane ; Needed to sort array of points according to X coordinate ; Needed to sort array of points according to Y coordinate ; A utility function to find the distance between two points ; A Brute Force method to return the smallest distance between two points in P [ ] of size n ; A utility function to find minimum of two float values ; 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 ; 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 ; A recursive function to find the smallest distance . The array P contains all points sorted according to x coordinate ; If there are 2 or 3 points , then use brute force ; Find the middle point ; Consider the vertical line passing through the middle point calculate the smallest distance dl on left of middle point and dr on right side ; Find the smaller of two distances ; Build an array strip [ ] that contains points close ( closer than d ) to the line passing through the middle point ; Find the closest points in strip . Return the minimum of d and closest distance is strip [ ] ; The main function that finds the smallest distance This method mainly uses closestUtil ( ) ; Use recursive function closestUtil ( ) to find the smallest distance ; Driver code
#include <bits/stdc++.h> NEW_LINE using namespace std ; class Point { public : int x , y ; } ; int compareX ( const void * a , const void * b ) { Point * p1 = ( Point * ) a , * p2 = ( Point * ) b ; return ( p1 -> x - p2 -> x ) ; } int compareY ( const void * a , const void * b ) { Point * p1 = ( Point * ) a , * p2 = ( Point * ) b ; return ( p1 -> y - p2 -> y ) ; } float dist ( Point p1 , Point p2 ) { return sqrt ( ( p1 . x - p2 . x ) * ( p1 . x - p2 . x ) + ( p1 . y - p2 . y ) * ( p1 . y - p2 . y ) ) ; } 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 ; } float min ( float x , float y ) { return ( x < y ) ? x : y ; } float stripClosest ( Point strip [ ] , int size , float d ) { float min = d ; qsort ( strip , size , sizeof ( Point ) , compareY ) ; 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 closestUtil ( Point P [ ] , int n ) { if ( n <= 3 ) return bruteForce ( P , n ) ; int mid = n / 2 ; Point midPoint = P [ mid ] ; float dl = closestUtil ( P , mid ) ; float dr = closestUtil ( P + mid , n - mid ) ; float d = min ( dl , dr ) ; Point strip [ n ] ; int j = 0 ; for ( int i = 0 ; i < n ; i ++ ) if ( abs ( P [ i ] . x - midPoint . x ) < d ) strip [ j ] = P [ i ] , j ++ ; return min ( d , stripClosest ( strip , j , d ) ) ; } float closest ( Point P [ ] , int n ) { qsort ( P , n , sizeof ( Point ) , compareX ) ; return closestUtil ( P , n ) ; } 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 ; }