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Find maximum and minimum distance between magnets
  • Difficulty Level : Basic
  • Last Updated : 29 May, 2018

Given coordinates of two pivot points (x0, y0) & (x1, y1) in coordinates plane. Along with each pivot, two different magnets are tied with the help of a string of length r1 and r2 respectively. Find the distance between both magnets when they repelling each other and when they are attracting each other.

Examples :

Input : x1=0, y1=0, x2=5, y2=0, r1=2, r2=2
Output : Distance while repulsion = 9, Distance while attraction = 1

Input : x1=0, y1=0, x2=5, y2=0, r1=3, r2=3
Output : Distance while repulsion = 11, Distance while attraction = 0

As we all know about the properties of magnet that they repel each other when they are facing each other with the same pole and attract each other when they are facing each other with opposite pole. Also, the force of attraction, as well as repulsion, always work in a straight line.



We have two pivots points on coordinates, so distance between these points are D = ((x1-x2)2 +(y1-y2)2 )1/2.
Also, we can conclude that distance between magnet is maximum while repulsion and that too should be the distance between pivots + sum of the length of both strings.
In case of attraction we have two cases to take care of:
Either the minimum distance is the distance between pivots – the sum of the length of both strings
Or minimum distance should be zero in case if the sum of the length of strings is greater than the distance between pivot points.

Illustration with help of diagram:

C++




// C++ program for max and min distance
#include <bits/stdc++.h>
using namespace std;
  
// Function for finding distance between pivots
int pivotDis(int x0, int y0, int x1, int y1)
{
    return sqrt((x1 - x0) * (x1 - x0) + 
                (y1 - y0) * (y1 - y0));
}
  
// Function for minimum distance
int minDis(int D, int r1, int r2)
{
    return max((D - r1 - r2), 0);
}
  
// Function for maximum distance
int maxDis(int D, int r1, int r2)
{
    return D + r1 + r2;
}
  
// Drivers code
int main()
{
    int x0 = 0, y0 = 0, x1 = 8, y1 = 0, r1 = 4, r2 = 5;
    int D = pivotDis(x0, y0, x1, y1);
    cout << "Distance while repulsion = " << maxDis(D, r1, r2);
    cout << "\nDistance while attraction = " << minDis(D, r1, r2);
    return 0;
}

Java




// Java program for max
// and min distance
import java.io.*;
  
class GFG 
{
      
// Function for finding 
// distance between pivots
static int pivotDis(int x0, int y0, 
                    int x1, int y1)
{
    return (int)Math.sqrt((x1 - x0) * 
                          (x1 - x0) + 
                          (y1 - y0) * 
                          (y1 - y0));
}
  
// Function for
// minimum distance
static int minDis(int D, int r1, int r2)
{
    return Math.max((D - r1 - r2), 0);
}
  
// Function for
// maximum distance
static int maxDis(int D, int r1, int r2)
{
    return D + r1 + r2;
}
  
// Driver Code
public static void main (String[] args) 
{
int x0 = 0, y0 = 0, x1 = 8
    y1 = 0, r1 = 4, r2 = 5;
int D = pivotDis(x0, y0, x1, y1);
System.out.print( "Distance while "
                     "repulsion = "
                  maxDis(D, r1, r2));
System.out.print("\nDistance while "
                     "attraction = "
                   minDis(D, r1, r2));
}
}
  
// This code is contributed by anuj_67.

Python3




# Python 3 program for max and min 
# distance
import math
  
# Function for finding distance between
# pivots
def pivotDis(x0, y0, x1, y1):
  
    return math.sqrt((x1 - x0) * (x1 - x0)
                  + (y1 - y0) * (y1 - y0))
  
# Function for minimum distance
def minDis( D, r1, r2):
  
    return max((D - r1 - r2), 0)
  
# Function for maximum distance
def maxDis( D, r1, r2):
  
    return D + r1 + r2
  
# Drivers code
x0 = 0
y0 = 0
x1 = 8
y1 = 0
r1 = 4
r2 = 5
D = pivotDis(x0, y0, x1, y1)
print("Distance while repulsion = ",
               int(maxDis(D, r1, r2)))
                 
print("Distance while attraction = ",
                    minDis(D, r1, r2))
  
# This code is contributed by Smitha

C#




// C# program for max and min distance
using System;
  
class GFG {
      
    // Function for finding 
    // distance between pivots
    static int pivotDis(int x0, int y0, 
                        int x1, int y1)
    {
        return (int)Math.Sqrt((x1 - x0) * 
                              (x1 - x0) + 
                              (y1 - y0) * 
                              (y1 - y0));
    }
      
    // Function for
    // minimum distance
    static int minDis(int D, int r1, int r2)
    {
        return Math.Max((D - r1 - r2), 0);
    }
      
    // Function for
    // maximum distance
    static int maxDis(int D, int r1, int r2)
    {
        return D + r1 + r2;
    }
      
    // Driver Code
    public static void Main () 
    {
    int x0 = 0, y0 = 0, x1 = 8, 
        y1 = 0, r1 = 4, r2 = 5;
    int D = pivotDis(x0, y0, x1, y1);
      
    Console.WriteLine( "Distance while "
                          "repulsion = "
                       maxDis(D, r1, r2));
    Console.WriteLine("Distance while "
                        "attraction = "
                      minDis(D, r1, r2));
    }
}
  
// This code is contributed by anuj_67.

PHP




<?php
// PHP program for max and 
// min distance
  
// Function for finding
// distance between pivots
function pivotDis($x0, $y0
                  $x1, $y1)
{
    return sqrt(($x1 - $x0) *
                ($x1 - $x0) + 
                ($y1 - $y0) * 
                ($y1 - $y0));
}
  
// Function for minimum distance
function minDis( $D, $r1, $r2)
{
    return max(($D - $r1 - $r2), 0);
}
  
// Function for maximum distance
function maxDis( $D, $r1, $r2)
{
    return $D + $r1 + $r2;
}
  
    // Driver code
    $x0 = 0; $y0 = 0;
    $x1 = 8; $y1 = 0;
    $r1 = 4; $r2 = 5;
    $D = pivotDis($x0, $y0,
                  $x1, $y1);
    echo "Distance while repulsion = " 
              , maxDis($D, $r1, $r2);
    echo "\nDistance while attraction = "
               , minDis($D, $r1, $r2);
  
// This code is contributed by anuj_67.
?>
Output:
Distance while repulsion = 17
Distance while attraction = 0

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