Closest Pair of Points using Divide and Conquer algorithm - GeeksforGeeks

25 Apr, 2022
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. Recall the following formula for distance between two points p and q.The Brute force solution is O(n^2), compute the distance between each pair and return the smallest. We can calculate the smallest distance in O(nLogn) time using Divide and Conquer strategy. In this post, a O(n x (Logn)^2) approach is discussed. We will be discussing a O(nLogn) approach in a separate post.
Algorithm Following are the detailed steps of a O(n (Logn)^2) algorithm. Input: An array of n points P[] Output: The smallest distance between two points in the given array.As a pre-processing step, the input array is sorted according to x coordinates.1) Find the middle point in the sorted array, we can take P[n/2] as middle point. 2) Divide the given array in two halves. The first subarray contains points from P[0] to P[n/2]. The second subarray contains points from P[n/2+1] to P[n-1].3) Recursively find the smallest distances in both subarrays. Let the distances be dl and dr. Find the minimum of dl and dr. Let the minimum be d. 
4) From the above 3 steps, we have an upper bound d of minimum distance. Now we need to consider the pairs such that one point in pair is from the left half and the other is from the right half. Consider the vertical line passing through P[n/2] and find all points whose x coordinate is closer than d to the middle vertical line. Build an array strip[] of all such points. 
5) Sort the array strip[] according to y coordinates. This step is O(nLogn). It can be optimized to O(n) by recursively sorting and merging. 6) Find the smallest distance in strip[]. This is tricky. From the first look, it seems to be a O(n^2) step, but it is actually O(n). It can be proved geometrically that for every point in the strip, we only need to check at most 7 points after it (note that strip is sorted according to Y coordinate). See this for more analysis.7) Finally return the minimum of d and distance calculated in the above step (step 6)
Implementation Following is the implementation of the above algorithm.  
C++
C
Java
Python3
// A divide and conquer program in C++// to find the smallest distance from a// given set of points. #include <bits/stdc++.h>using namespace std; // A structure to represent a Point in 2D planeclass Point{    public:    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 coordinateint 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 coordinateint 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 pointsfloat 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 nfloat 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 valuesfloat min(float x, float y){    return (x < y)? x : y;}  // A utility function to find the// distance between the closest points of// strip of 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 timesfloat stripClosest(Point strip[], int size, float d){    float min = d; // Initialize the minimum distance as d     qsort(strip, size, sizeof(Point), compareY);     // 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 P contains// all points sorted according to x coordinatefloat closestUtil(Point P[], int n){    // If there are 2 or 3 points, then use brute force    if (n <= 3)        return bruteForce(P, n);     // Find the middle point    int mid = n/2;    Point midPoint = P[mid];     // 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(P, mid);    float dr = closestUtil(P + mid, 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(P[i].x - midPoint.x) < d)            strip[j] = P[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 function that finds the smallest distance// This method mainly uses closestUtil()float closest(Point P[], int n){    qsort(P, n, sizeof(Point), compareX);     // Use recursive function closestUtil()    // to find the smallest distance    return closestUtil(P, n);} // Driver codeint 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;} // This is code is contributed by rathbhupendra
// A divide and conquer program in C/C++ to find the smallest distance from a// given set of points. #include <stdio.h>#include <float.h>#include <stdlib.h>#include <math.h> // A structure to represent a Point in 2D planestruct 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 coordinateint 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 coordinateint 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 pointsfloat 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 nfloat 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 valuesfloat 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 timesfloat stripClosest(Point strip[], int size, float d){    float min = d;  // Initialize the minimum distance as d     qsort(strip, size, sizeof(Point), compareY);     // 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 P contains// all points sorted according to x coordinatefloat closestUtil(Point P[], int n){    // If there are 2 or 3 points, then use brute force    if (n <= 3)        return bruteForce(P, n);     // Find the middle point    int mid = n/2;    Point midPoint = P[mid];     // 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(P, mid);    float dr = closestUtil(P + mid, 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(P[i].x - midPoint.x) < d)            strip[j] = P[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 function that finds the smallest distance// This method mainly uses closestUtil()float closest(Point P[], int n){    qsort(P, n, sizeof(Point), compareX);     // Use recursive function closestUtil() to find the smallest distance    return closestUtil(P, n);} // Driver program to test above functionsint main(){    Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};    int n = sizeof(P) / sizeof(P[0]);    printf("The smallest distance is %f ", closest(P, n));    return 0;}
import java.text.DecimalFormat;import java.util.Arrays;import java.util.Comparator; // A divide and conquer program in Java// to find the smallest distance from a// given set of points. // A structure to represent a Point in 2D planeclass Point {  public int x;  public int y;   Point(int x, int y) {    this.x = x;    this.y = y;  }   // A utility function to find the  // distance between two points  public static float dist(Point p1, Point p2) {    return (float) Math.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  public static float bruteForce(Point[] P, int n) {    float min = Float.MAX_VALUE;    float currMin = 0;     for (int i = 0; i < n; ++i) {      for (int j = i + 1; j < n; ++j) {        currMin = dist(P[i], P[j]);        if (currMin < min) {          min = currMin;        }      }    }    return min;  }   // A utility function to find the  // distance between the closest points of  // strip of 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  public static float stripClosest(Point[] strip, int size, float d) {    float min = d; // Initialize the minimum distance as d     Arrays.sort(strip, 0, size, new PointYComparator());     // 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 P contains  // all points sorted according to x coordinate  public static float closestUtil(Point[] P,                                  int startIndex,                                  int endIndex)  {         // If there are 2 or 3 points, then use brute force    if ((endIndex - startIndex) <= 3) {      return bruteForce(P, endIndex);    }     // Find the middle point    int mid = startIndex + (endIndex - startIndex) / 2;    Point midPoint = P[mid];     // 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(P, startIndex, mid);    float dr = closestUtil(P, mid, endIndex);     // Find the smaller of two distances    float d = Math.min(dl, dr);     // Build an array strip[] that contains    // points close (closer than d)    // to the line passing through the middle point    Point[] strip = new Point[endIndex];    int j = 0;    for (int i = 0; i < endIndex; i++) {      if (Math.abs(P[i].x - midPoint.x) < d) {        strip[j] = P[i];        j++;      }    }     // Find the closest points in strip.    // Return the minimum of d and closest    // distance is strip[]    return Math.min(d, stripClosest(strip, j, d));  }   // The main function that finds the smallest distance  // This method mainly uses closestUtil()  public static float closest(Point[] P, int n) {    Arrays.sort(P, 0, n, new PointXComparator());     // Use recursive function closestUtil()    // to find the smallest distance    return closestUtil(P, 0, n);  } } // A structure to represent a Point in 2D planeclass PointXComparator implements Comparator<Point> {   // Needed to sort array of points  // according to X coordinate  @Override  public int compare(Point pointA, Point pointB) {    return Integer.compare(pointA.x, pointB.x);  } } class PointYComparator implements Comparator<Point> {   // Needed to sort array of points  // according to Y coordinate  @Override  public int compare(Point pointA, Point pointB) {    return Integer.compare(pointA.y, pointB.y);  } } public class ClosestPoint {   // Driver code  public static void main(String[] args) {    Point[] P = new Point[]{      new Point(2, 3),      new Point(12, 30),      new Point(40, 50),      new Point(5, 1),      new Point(12, 10),      new Point(3, 4)       };     DecimalFormat df = new DecimalFormat("#.######");    System.out.println("The smallest distance is " +                       df.format(Point.closest(P, P.length)));  } } // This code is contributed by sanjay sharma 1
# A divide and conquer program in Python3# to find the smallest distance from a# given set of points.import mathimport copy# A class to represent a Point in 2D planeclass Point():    def __init__(self, x, y):        self.x = x        self.y = y # A utility function to find the# distance between two pointsdef dist(p1, p2):    return math.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 ndef bruteForce(P, n):    min_val = float('inf')    for i in range(n):        for j in range(i + 1, n):            if dist(P[i], P[j]) < min_val:                min_val = dist(P[i], P[j])     return min_val # A utility function to find the# distance between the closest points of# strip of 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 timesdef stripClosest(strip, size, d):         # Initialize the minimum distance as d    min_val = 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 i in range(size):        j = i + 1        while j < size and (strip[j].y -                            strip[i].y) < min_val:            min_val = dist(strip[i], strip[j])            j += 1     return min_val # A recursive function to find the# smallest distance. The array P contains# all points sorted according to x coordinatedef closestUtil(P, Q, n):         # If there are 2 or 3 points,    # then use brute force    if n <= 3:        return bruteForce(P, n)     # Find the middle point    mid = n // 2    midPoint = P[mid]     #keep a copy of left and right branch    Pl = P[:mid]    Pr = P[mid:]     # Consider the vertical line passing    # through the middle point calculate    # the smallest distance dl on left    # of middle point and dr on right side    dl = closestUtil(Pl, Q, mid)    dr = closestUtil(Pr, Q, n - mid)     # Find the smaller of two distances    d = min(dl, dr)     # Build an array strip[] that contains    # points close (closer than d)    # to the line passing through the middle point    stripP = []    stripQ = []    lr = Pl + Pr    for i in range(n):        if abs(lr[i].x - midPoint.x) < d:            stripP.append(lr[i])        if abs(Q[i].x - midPoint.x) < d:            stripQ.append(Q[i])     stripP.sort(key = lambda point: point.y) #<-- REQUIRED    min_a = min(d, stripClosest(stripP, len(stripP), d))    min_b = min(d, stripClosest(stripQ, len(stripQ), d))              # Find the self.closest points in strip.    # Return the minimum of d and self.closest    # distance is strip[]    return min(min_a,min_b) # The main function that finds# the smallest distance.# This method mainly uses closestUtil()def closest(P, n):    P.sort(key = lambda point: point.x)    Q = copy.deepcopy(P)    Q.sort(key = lambda point: point.y)        # Use recursive function closestUtil()    # to find the smallest distance    return closestUtil(P, Q, n) # Driver codeP = [Point(2, 3), Point(12, 30),     Point(40, 50), Point(5, 1),     Point(12, 10), Point(3, 4)]n = len(P)print("The smallest distance is",                   closest(P, n)) # This code is contributed# by Prateek Gupta (@prateekgupta10)
Output: 
The smallest distance is 1.414214
Time Complexity Let Time complexity of above algorithm be T(n). Let us assume that we use a O(nLogn) sorting algorithm. The above algorithm divides all points in two sets and recursively calls for two sets. After dividing, it finds the strip in O(n) time, sorts the strip in O(nLogn) time and finally finds the closest points in strip in O(n) time. So T(n) can expressed as follows T(n) = 2T(n/2) + O(n) + O(nLogn) + O(n) T(n) = 2T(n/2) + O(nLogn) T(n) = T(n x Logn x Logn)
Notes 1) Time complexity can be improved to O(nLogn) by optimizing step 5 of the above algorithm. We will soon be discussing the optimized solution in a separate post. 2) The code finds smallest distance. It can be easily modified to find the points with the smallest distance. 3) The code uses quick sort which can be O(n^2) in the worst case. To have the upper bound as O(n (Logn)^2), a O(nLogn) sorting algorithm like merge sort or heap sort can be used 
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References: http://www.cs.umd.edu/class/fall2013/cmsc451/Lects/lect10.pdf http://www.youtube.com/watch?v=vS4Zn1a9KUc http://www.youtube.com/watch?v=T3T7T8Ym20M http://en.wikipedia.org/wiki/Closest_pair_of_points_problem 
rathbhupendra
Akanksha_Rai
PrateekGupta10
kjs9
rleecharlie
sumitgumber28
varshagumber28
sanjay sharma 1
Closest Pair of Points
Divide and Conquer
Geometric
Divide and Conquer
Geometric
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