计算平方根的Bakhshali逼近

Bakshali逼近是一种找到数字平方根的逼近的数学方法。它等效于巴比伦方法的两次迭代。
算法：

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)

下面是上述步骤的实现。
执行：

C++

//This program gives result approximated to 5 decimal places.
#include 
 
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(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  "<

---

Java

// 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.

Python3

# 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

C#

// 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.

PHP

 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
?>

Javascript

 
// javascript program gives result approximated
// to 5 decimal places.{
function sqroot(s)
{
 
    // This will be the nearest perfect square to s
    var pSq = 0;
     
    // This is the sqrt of pSq
    var N = 0;
 
    // Find the nearest perfect square to s
    for (i = parseInt(s); i > 0; i--)
    {
        for (j = 1; j < i; j++)
        {
            if (j*j == i)
            {
                pSq = i;
                N = j;
                break;
            }
        }
        if (pSq > 0)
            break;
    }
     
    // calculate d   
    var d = s - pSq;    
     
    // calculate P
    var P = (d/(2.0*N));
     
    // calculate A
    var A = N+P;
      
    // calculate sqrt(S).
    var sqrt_of_s = A-((P*P)/(2.0*A));
    return sqrt_of_s;
}
     
// Driver program
var num = 9.2345;
var sqroot_of_num = sqroot(num);
document.write("Square root of "+num+" = "
                 + (Math.round(sqroot_of_num*100000.0)/100000.0).toFixed(6));
 
// This code is contributed by Princi Singh

输出 ：

Square root of 9.2345 = 3.03883

插图：

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

重要事项：

• 用于查找平方根的近似值。
• 需要需要计算其平方根的数字的最接近完美平方的值。
• 浮点数比整数更有效，因为它可以近似。