Maximum distance between two points in coordinate plane using Rotating Caliper's Method

Maximum distance between two points in coordinate plane using Rotating Caliper’s Method

Difficulty Level : Expert
Last Updated : 07 Dec, 2020

Prerequisites: Graham Scan’s Convex Hull, Orientation.
Given a set of N points in a coordinates plane, the task is to find the maximum distance between any two points in the given set of planes.

Examples:

Input: n = 4, Points: (0, 3), (3, 0), (0, 0), (1, 1)
Output: Maximum Distance = 4.24264
Explanation:
Points having maximum distance between them are (0, 3) and (3, 0)

Input: n = 5, Points: (4, 0), (0, 2), (-1, -7), (1, 10), (2, -3)
Output: Maximum Distance = 17.11724
Explanation:
Points having maximum distance between them are (-1, 7) and (1, 10)

Naive Approach: The naive idea is to try every possible pair of points from the given set and calculate the distances between each of them and print the maximum distance among all the pairs.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Function calculates distance 
// between two points 
long dist(pair<long, long> p1,
          pair<long, long> p2) 
{ 
    long x0 = p1.first - p2.first; 
    long y0 = p1.second - p2.second; 
    return x0 * x0 + y0 * y0; 
} 
 
// Function to find the maximum 
// distance between any two points 
double maxDist(pair<long, long> p[], int n) 
{ 
    double Max = 0; 
 
    // Iterate over all possible pairs 
    for(int i = 0; i < n; i++)
    { 
        for(int j = i + 1; j < n; j++)
        { 
             
            // Update max 
            Max = max(Max, (double)dist(p[i], 
                                        p[j])); 
        } 
    } 
 
    // Return actual distance 
    return sqrt(Max); 
} 
 
// Driver code   
int main()
{
     
    // Number of points 
    int n = 5; 
 
    pair<long, long> p[n]; 
 
    // Given points 
    p[0].first = 4, p[0].second = 0; 
    p[1].first = 0, p[1].second = 2; 
    p[2].first = -1, p[2].second = -7; 
    p[3].first = 1, p[3].second = 10;
    p[4].first = 2, p[4].second = -3; 
 
    // Function call 
    cout << fixed << setprecision(14) 
         << maxDist(p, n) <<endl; 
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

import java.awt.*;
import java.util.ArrayList;
 
public class Main {
 
    // Function calculates distance
    // between two points
    static long dist(Point p1, Point p2)
    {
        long x0 = p1.x - p2.x;
        long y0 = p1.y - p2.y;
        return x0 * x0 + y0 * y0;
    }
 
    // Function to find the maximum
    // distance between any two points
    static double maxDist(Point p[])
    {
        int n = p.length;
        double max = 0;
 
        // Iterate over all possible pairs
        for (int i = 0; i < n; i++) {
 
            for (int j = i + 1; j < n; j++) {
 
                // Update max
                max = Math.max(max,
                               dist(p[i],
                                    p[j]));
            }
        }
 
        // Return actual distance
        return Math.sqrt(max);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Number of points
        int n = 5;
 
        Point p[] = new Point[n];
 
        // Given points
        p[0] = new Point(4, 0);
        p[1] = new Point(0, 2);
        p[2] = new Point(-1, -7);
        p[3] = new Point(1, 10);
        p[4] = new Point(2, -3);
 
        // Function Call
        System.out.println(maxDist(p));
    }
}

Python3

from math import sqrt
 
# Function calculates distance
# between two points
def dist(p1, p2):
     
    x0 = p1[0] - p2[0]
    y0 = p1[1] - p2[1]
    return x0 * x0 + y0 * y0
 
# Function to find the maximum
# distance between any two points
def maxDist(p):
 
    n = len(p)
    maxm = 0
 
    # Iterate over all possible pairs
    for i in range(n):
        for j in range(i + 1, n):
             
            # Update maxm
            maxm = max(maxm, dist(p[i], p[j]))
 
    # Return actual distance
    return sqrt(maxm)
     
# Driver Code
if __name__ == '__main__':
     
    # Number of points
    n = 5
 
    p = []
     
    # Given points
    p.append([4, 0])
    p.append([0, 2])
    p.append([-1, -7])
    p.append([1, 10])
    p.append([2, -3])
 
    # Function Call
    print(maxDist(p))
 
# This code is contributed by mohit kumar 29

C#

using System;
class GFG {
     
    // Function calculates distance
    // between two points
    static long dist(Tuple<int, int> p1, Tuple<int, int> p2)
    {
        long x0 = p1.Item1 - p2.Item1;
        long y0 = p1.Item2 - p2.Item2;
        return x0 * x0 + y0 * y0;
    }
  
    // Function to find the maximum
    // distance between any two points
    static double maxDist(Tuple<int, int>[] p)
    {
        int n = p.Length;
        double max = 0;
  
        // Iterate over all possible pairs
        for (int i = 0; i < n; i++) {
  
            for (int j = i + 1; j < n; j++) {
  
                // Update max
                max = Math.Max(max, dist(p[i],p[j]));
            }
        }
  
        // Return actual distance
        return Math.Sqrt(max);
    }
 
  // Driver code
  static void Main() {
      
        // Given points
        Tuple<int, int>[] p =
        {
            Tuple.Create(4, 0),
            Tuple.Create(0, 2),
            Tuple.Create(-1, -7),
            Tuple.Create(1, 10),
            Tuple.Create(2, -3),
        };
       
        // Function Call
        Console.WriteLine(maxDist(p));
  }
}
 
// This code is contributed by divyesh072019

Output:

17.11724276862369

Time Complexity: O(N²), where N is the total number of points.
Auxiliary Space: O(1)

Efficient Approach: The above naive approach can be optimized using Rotating Caliper’s Method.

Rotating Calipers is a method for solving a number of problems from the field of computational geometry. It resembles the idea of rotating an adjustable caliper around the outside of a polygon’s convex hull. Originally, this method was invented to compute the diameter of convex polygons. It can also be used to compute the minimum and maximum distance between two convex polygons, the intersection of convex polygons, the maximum distance between two points in a polygon, and many things more.

To implement the above method we will use the concept of the Convex Hull. Before we begin a further discussion about the optimal approach, we need to know about the following:

Unsigned Area Of Triangle: If we are given three points P1(x1, y1), P2(x2, y2) and P3(x3, y3) then

$\left ( \frac {(x2 - x1)*(y3 - y2) - (x3 - x2)*(y2 - y1)}{2} \right )$

is the signed area of triangle. If the area is positive then three points are in the clockwise order, Else they are in anti-clockwise order and if the area equals to zero then, the points are co-linear. If we take absolute value, then this will represent the unsigned area of the triangle. Here, unsigned basically means area without direction i.e., we just need the relative absolute value of the area. Therefore, we can remove (1/2) from the formula. Hence,

Relative Area of Triangle = abs((x2-x1)*(y3-y2)-(x3-x2)*(y2-y1))

Antipodal Points: It is those points which are diametrically opposite to each other. But for us, antipodal points are those which are farthest from each other in the convex polygon. If we choose one point from the given set, then this point can only achieve it’s maximum distance if and only if we can find it’s antipodal point from the given set.

Below are the steps:

Two points having maximum distance must lie on the boundary of the convex polygon formed from the given set. Therefore, use Graham Scan’s convex hull method to arrange points in counter-clockwise order.
We have N points, Initially start from point P1 and include those points from set of given points such that area of region always increases by including any points from the set.
Starting from point P1, Choose K = 2 and increment K while area(PN, P1, PK) is increasing and stop before it starts decreasing. Now the current point PK might be the antipodal point for P1. Similarly, find antipodal point for p2 by finding area(P1, P2, PK) and incrementing K form where we previously stopped and so on.
Keep updating the maximum distance for each antipodal points occurs in the above steps as the distance between intial point and point by including area was maximum.

Below is the implementation of the above approach:

Java

// Java Program for the above approach
import java.awt.*;
import java.util.*;
import java.util.Map.Entry;
 
public class Main {
 
    // Function to detect the orientation
    static int orientation(Point p,
                           Point q,
                           Point r)
    {
        int x = area(p, q, r);
 
        // If area > 0 then
        // points are clockwise
        if (x > 0) {
            return 1;
        }
 
        // If area<0 then
        // points are counterclockwise
        if (x < 0) {
            return -1;
        }
 
        // If area is 0 then p, q
        // and r are co-linear
        return 0;
    }
 
    // Function to find the area
    static int area(Point p, Point q, Point r)
    {
        // 2*(area of triangle)
        return ((p.y - q.y) * (q.x - r.x)
                - (q.y - r.y) * (p.x - q.x));
    }
 
    // Function to find the absolute Area
    static int absArea(Point p,
                       Point q, Point r)
    {
        // Unsigned area
        // 2*(area of triangle)
        return Math.abs(area(p, q, r));
    }
 
    // Function to find the distance
    static int dist(Point p1, Point p2)
    {
        // squared-distance b/w
        // p1 and p2 for precision
        return ((p1.x - p2.x) * (p1.x - p2.x)
                + (p1.y - p2.y) * (p1.y - p2.y));
    }
 
    // Function to implement Convex Hull
    // Approach
    static ArrayList<Point>
    convexHull(Point points[])
    {
        int n = points.length;
 
        Point min = new Point(Integer.MAX_VALUE,
                              Integer.MAX_VALUE);
 
        // Choose point having min.
        // y-coordinate and if two points
        // have same y-coordinate choose
        // the one with minimum x-coordinate
        int ind = -1;
 
        // Iterate Points[]
        for (int i = 0; i < n; i++) {
            if (min.y > points[i].y) {
                min.y = points[i].y;
                min.x = points[i].x;
                ind = i;
            }
            else if (min.y == points[i].y
                     && min.x > points[i].x) {
                min.x = points[i].x;
                ind = i;
            }
        }
        points[ind] = points[0];
        points[0] = min;
 
        // Sort points which have
        // minimum polar angle wrt
        // Point min
        Arrays.sort(points, 1, n,
                    new Comparator<Point>() {
 
                        @Override
                        public int compare(Point o1,
                                           Point o2)
                        {
 
                            int o = orientation(min, o1, o2);
 
                            // If points are co-linear
                            // choose the one having smaller
                            // distance with min first.
                            if (o == 0) {
                                return dist(o1, min)
                                    - dist(o2, min);
                            }
 
                            // If clockwise then swap
                            if (o == 1) {
                                return 1;
                            }
 
                            // If anticlockwise then
                            // don't swap
                            return -1;
                        }
                    });
 
        Stack<Point> st = new Stack<>();
 
        // First hull point
        st.push(points[0]);
 
        int k;
        for (k = 1; k < n - 1; k++) {
            if (orientation(points[0],
                            points[k],
                            points[k + 1])
                != 0)
                break;
        }
 
        // Second hull point
        st.push(points[k]);
 
        for (int i = k + 1; i < n; i++) {
            Point top = st.pop();
 
            while (orientation(st.peek(),
                               top,
                               points[i])
                   >= 0) {
                top = st.pop();
            }
 
            st.push(top);
            st.push(points[i]);
        }
 
        ArrayList<Point> hull
            = new ArrayList<>();
 
        // Iterate stack and add node to hull
        for (int i = 0; i < st.size(); i++) {
            hull.add(st.get(i));
        }
        return hull;
    }
 
    // Function to find the maximum
    // distance between any two points
    // from a set of given points
    static double
    rotatingCaliper(Point points[])
    {
        // Takes O(n)
        ArrayList<Point> convexHull
            = convexHull(points);
        int n = convexHull.size();
 
        // Convex hull point in counter-
        // clockwise order
        Point hull[] = new Point[n];
        n = 0;
 
        while (n < convexHull.size()) {
            hull[n] = convexHull.get(n++);
        }
 
        // Base Cases
        if (n == 1)
            return 0;
        if (n == 2)
            return Math.sqrt(dist(hull[0], hull[1]));
        int k = 1;
 
        // Find the farthest vertex
        // from hull[0] and hull[n-1]
        while (absArea(hull[n - 1],
                       hull[0],
                       hull[(k + 1) % n])
               > absArea(hull[n - 1],
                         hull[0],
                         hull[k])) {
            k++;
        }
 
        double res = 0;
 
        // Check points from 0 to k
        for (int i = 0, j = k;
             i <= k && j < n; i++) {
            res = Math.max(res,
                           Math.sqrt((double)dist(hull[i],
                                                  hull[j])));
 
            while (j < n
                   && absArea(hull[i],
                              hull[(i + 1) % n],
                              hull[(j + 1) % n])
                          > absArea(hull[i],
                                    hull[(i + 1) % n],
                                    hull[j])) {
 
                // Update res
                res = Math.max(
                    res,
                    Math.sqrt(dist(hull[i],
                                   hull[(j + 1) % n])));
                j++;
            }
        }
 
        // Return the result distance
        return res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Total points
        int n = 5;
        Point p[] = new Point[n];
 
        // Given Points
        p[0] = new Point(4, 0);
        p[1] = new Point(0, 2);
        p[2] = new Point(-1, -7);
        p[3] = new Point(1, 10);
        p[4] = new Point(2, -3);
 
        // Function Call
        System.out.println(rotatingCaliper(p));
    }
}

Output:

17.11724276862369

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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