Bakhshali Approximation for computing square roots

Bakshali Approximation is a mathematical method of finding an approximation to a square root of a number. It is equivalent to two iterations of Babylonian Method.

Algorithm:

To calculate sqrt(S).

Step 1: Calculate nearest perfect square to S i.e (N2).

Step 2: Calculate d = S - (N2)

Step 3: Calculate P = d/(2*N)

Step 4: Calculate A = N + P

Step 5: Sqrt(S) will be nearly equal to A - (P2/2*A)

Below is the implementation of above steps.

Implementation:

C++

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//This program gives result approximated to 5 decimal places.
#include <iostream>
  
float sqroot(float s)
{
    int pSq = 0; //This will be the nearest perfect square to s
    int N = 0; //This is the sqrt of pSq
  
    // Find the nearest perfect square to s
    for (int i = static_cast<int>(s); i > 0; i--)
    {
        for (int j = 1; j < i; j++)
        {
            if (j*j == i)
            {
                pSq = i;
                N = j;
                break;
            }
        }
        if (pSq > 0)
            break;
    }
  
    float d = s - pSq;     //calculate d
    float P = d/(2.0*N);  //calculate P
    float A = N+P;     //calculate A
    float sqrt_of_s = A-((P*P)/(2.0*A));   //calculate sqrt(S).
    return sqrt_of_s;
}
  
// Driver program to test above function
int main()
{
    float num = 9.2345;
    float sqroot_of_num = sqroot(num);
    std::cout << "Square root of  "<<num<<" = "<<sqroot_of_num;
    return 0;
}

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Java

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// Java program gives result approximated
// to 5 decimal places.
  
class GFG 
{
    static float sqroot(float s)
    {
        // This will be the nearest perfect square to s
        int pSq = 0;
          
        //This is the sqrt of pSq
        int N = 0
      
        // Find the nearest perfect square to s
        for (int i = (int)(s); i > 0; i--)
        {
            for (int j = 1; j < i; j++)
            {
                if (j*j == i)
                {
                    pSq = i;
                    N = j;
                    break;
                }
            }
            if (pSq > 0)
                break;
        }
          
        // calculate d    
        float d = s - pSq;     
          
        // calculate P
        float P = d/(2.0f*N); 
          
        // calculate A
        float A = N+P; 
           
        // calculate sqrt(S).
        float sqrt_of_s = A-((P*P)/(2.0f*A));
        return sqrt_of_s;
    }
      
    // Driver program 
    public static void main (String[] args) 
    {
        float num = 9.2345f;
        float sqroot_of_num = sqroot(num);
        System.out.print("Square root of "+num+" = "
                         + Math.round(sqroot_of_num*100000.0)/100000.0);
    }
}
  
// This code is contributed by Anant Agarwal.

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Python3

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# This Python3 program gives result
# approximated to 5 decimal places.
def sqroot(s):
  
    # This will be the nearest
    # perfect square to s
    pSq = 0
      
    # This is the sqrt of pSq
    N = 0
  
    # Find the nearest 
    # perfect square to s
    for i in range(int(s), 0, -1):
        for j in range(1, i):
            if (j * j == i):
                pSq = i;
                N = j;
                break;
      
        if (pSq > 0):
            break;
  
    d = s - pSq;     # calculate d
    P = d / (2.0 * N); # calculate P
    A = N + P; # calculate A
      
    # calculate sqrt(S).
    sqrt_of_s = A - ((P * P) / (2.0 * A)); 
    return sqrt_of_s;
  
# Driver Code
num = 9.2345;
sqroot_of_num = sqroot(num);
print("Square root of ", num, "="
      round((sqroot_of_num * 100000.0) / 100000.0, 5));
  
# This code is contributed by mits

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C#

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// C# program gives result approximated
// to 5 decimal places.
using System;
  
class GFG {
      
    static float sqroot(float s)
    {
          
        // This will be the nearest 
        // perfect square to s
        int pSq = 0;
          
        //This is the sqrt of pSq
        int N = 0; 
      
        // Find the nearest perfect square to s
        for (int i = (int)(s); i > 0; i--)
        {
            for (int j = 1; j < i; j++)
            {
                if (j * j == i)
                {
                    pSq = i;
                    N = j;
                    break;
                }
            }
              
            if (pSq > 0)
                break;
        }
          
        // calculate d 
        float d = s - pSq;     
          
        // calculate P
        float P = d / (2.0f * N); 
          
        // calculate A
        float A = N + P; 
          
        // calculate sqrt(S).
        float sqrt_of_s = A-((P * P) / (2.0f * A));
        return sqrt_of_s;
    }
      
    // Driver Code
    public static void Main () 
    {
        float num = 9.2345f;
        float sqroot_of_num = sqroot(num);
        Console.Write("Square root of "+num+" = "
                       Math.Round(sqroot_of_num * 100000.0) / 
                                                  100000.0);
    }
}
  
// This code is contributed by Nitin Mittal.

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PHP

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<?php
// This PHP program gives result
// approximated to 5 decimal places.
function sqroot($s)
{
    // This will be the nearest
    // perfect square to s
    $pSq = 0; 
      
    //This is the sqrt of pSq
    $N = 0; 
  
    // Find the nearest 
    // perfect square to s
    for ($i = intval($s); $i > 0; $i--)
    {
        for ($j = 1; $j < $i; $j++)
        {
            if ($j * $j == $i)
            {
                $pSq = $i;
                $N = $j;
                break;
            }
        }
        if ($pSq > 0)
            break;
    }
  
    $d = $s - $pSq;     //calculate d
    $P = $d / (2.0 * $N); //calculate P
    $A = $N + $P;     //calculate A
      
    //calculate sqrt(S).
    $sqrt_of_s = $A - (($P * $P) / 
                       (2.0 * $A)); 
    return $sqrt_of_s;
}
  
// Driver Code
$num = 9.2345;
$sqroot_of_num = sqroot($num);
echo "Square root of ". $num ." = "
              round(($sqroot_of_num
                          100000.0) / 
                        100000.0, 5);
  
// This code is contributed by Sam007
?>

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Output :



Square root of 9.2345 = 3.03883

Illustration:

find sqrt(9.2345)

S = 9.2345

N = 3
 
d = 9.2345 – (3^2) = 0.2345

P = 0.2345/(2*3) = 0.0391

A = 3 + 0.0391  = 3.0391

therefore, sqrt(9.2345) = 3.0391 – (0.0391^2/(2*0.0391)) = 3.0388

Important Points:

  • Used to find approximation to a square root.
  • Requires the value of nearest perfect square of the number whose square root is needed to be calculated.
  • More efficient for floating point numbers than integers as it finds approximation.

Reference:
https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Bakhshali_approximation

This article is contributed by Harsh Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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