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Bakhshali Approximation for computing square roots
• Last Updated : 22 Apr, 2021

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++

 `//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``<``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  "``<

## 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

 ``

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.