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Square root of a number without using sqrt() function
• Difficulty Level : Medium
• Last Updated : 22 Apr, 2021

Given a number N, the task is to find the square root of N without using sqrt() function.

Examples:

Input: N = 25
Output: 5

Input: N = 3
Output: 1.73205

Input: N = 2.5
Output: 1.58114

Approach:

• Start iterating from i = 1. If i * i = n, then print i as n is a perfect square whose square root is i.
• Else find the smallest i for which i * i is strictly greater than n.
• Now we know square root of n lies in the interval i – 1 and i and we can use Binary Search algorithm to find the square root.
• Find mid of i – 1 and i and compare mid * mid with n, with precision upto 5 decimal places.
1. If mid * mid = n then return mid.
2. If mid * mid < n then recur for the second half.
3. If mid * mid > n then recur for the first half.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Recursive function that returns square root``// of a number with precision upto 5 decimal places``double` `Square(``double` `n, ``double` `i, ``double` `j)``{``    ``double` `mid = (i + j) / 2;``    ``double` `mul = mid * mid;` `    ``// If mid itself is the square root,``    ``// return mid``    ``if` `((mul == n) || (``abs``(mul - n) < 0.00001))``        ``return` `mid;` `    ``// If mul is less than n, recur second half``    ``else` `if` `(mul < n)``        ``return` `Square(n, mid, j);` `    ``// Else recur first half``    ``else``        ``return` `Square(n, i, mid);``}` `// Function to find the square root of n``void` `findSqrt(``double` `n)``{``    ``double` `i = 1;` `    ``// While the square root is not found``    ``bool` `found = ``false``;``    ``while` `(!found) {` `        ``// If n is a perfect square``        ``if` `(i * i == n) {``            ``cout << fixed << setprecision(0) << i;``            ``found = ``true``;``        ``}``        ``else` `if` `(i * i > n) {` `            ``// Square root will lie in the``            ``// interval i-1 and i``            ``double` `res = Square(n, i - 1, i);``            ``cout << fixed << setprecision(5) << res;``            ``found = ``true``;``        ``}``        ``i++;``    ``}``}` `// Driver code``int` `main()``{``    ``double` `n = 3;` `    ``findSqrt(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Recursive function that returns``// square root of a number with``// precision upto 5 decimal places``static` `double` `Square(``double` `n,``                     ``double` `i, ``double` `j)``{``    ``double` `mid = (i + j) / ``2``;``    ``double` `mul = mid * mid;` `    ``// If mid itself is the square root,``    ``// return mid``    ``if` `((mul == n) ||  ``        ``(Math.abs(mul - n) < ``0.00001``))``        ``return` `mid;` `    ``// If mul is less than n,``    ``// recur second half``    ``else` `if` `(mul < n)``        ``return` `Square(n, mid, j);` `    ``// Else recur first half``    ``else``        ``return` `Square(n, i, mid);``}` `// Function to find the square root of n``static` `void` `findSqrt(``double` `n)``{``    ``double` `i = ``1``;` `    ``// While the square root is not found``    ``boolean` `found = ``false``;``    ``while` `(!found)``    ``{` `        ``// If n is a perfect square``        ``if` `(i * i == n)``        ``{``            ``System.out.println(i);``            ``found = ``true``;``        ``}``        ` `        ``else` `if` `(i * i > n)``        ``{` `            ``// Square root will lie in the``            ``// interval i-1 and i``            ``double` `res = Square(n, i - ``1``, i);``            ``System.out.printf(``"%.5f"``, res);``            ``found = ``true``;``        ``}``        ``i++;``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``double` `n = ``3``;` `    ``findSqrt(n);``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the approach``import` `math` `# Recursive function that returns square root``# of a number with precision upto 5 decimal places``def` `Square(n, i, j):` `    ``mid ``=` `(i ``+` `j) ``/` `2``;``    ``mul ``=` `mid ``*` `mid;` `    ``# If mid itself is the square root,``    ``# return mid``    ``if` `((mul ``=``=` `n) ``or` `(``abs``(mul ``-` `n) < ``0.00001``)):``        ``return` `mid;` `    ``# If mul is less than n, recur second half``    ``elif` `(mul < n):``        ``return` `Square(n, mid, j);` `    ``# Else recur first half``    ``else``:``        ``return` `Square(n, i, mid);` `# Function to find the square root of n``def` `findSqrt(n):``    ``i ``=` `1``;` `    ``# While the square root is not found``    ``found ``=` `False``;``    ``while` `(found ``=``=` `False``):` `        ``# If n is a perfect square``        ``if` `(i ``*` `i ``=``=` `n):``            ``print``(i);``            ``found ``=` `True``;``        ` `        ``elif` `(i ``*` `i > n):` `            ``# Square root will lie in the``            ``# interval i-1 and i``            ``res ``=` `Square(n, i ``-` `1``, i);``            ``print` `(``"{0:.5f}"``.``format``(res))``            ``found ``=` `True``        ``i ``+``=` `1``;` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `3``;` `    ``findSqrt(n);` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Recursive function that returns``// square root of a number with``// precision upto 5 decimal places``static` `double` `Square(``double` `n,``                     ``double` `i, ``double` `j)``{``    ``double` `mid = (i + j) / 2;``    ``double` `mul = mid * mid;` `    ``// If mid itself is the square root,``    ``// return mid``    ``if` `((mul == n) ||``        ``(Math.Abs(mul - n) < 0.00001))``        ``return` `mid;` `    ``// If mul is less than n,``    ``// recur second half``    ``else` `if` `(mul < n)``        ``return` `Square(n, mid, j);` `    ``// Else recur first half``    ``else``        ``return` `Square(n, i, mid);``}` `// Function to find the square root of n``static` `void` `findSqrt(``double` `n)``{``    ``double` `i = 1;` `    ``// While the square root is not found``    ``Boolean found = ``false``;``    ``while` `(!found)``    ``{` `        ``// If n is a perfect square``        ``if` `(i * i == n)``        ``{``            ``Console.WriteLine(i);``            ``found = ``true``;``        ``}``        ` `        ``else` `if` `(i * i > n)``        ``{` `            ``// Square root will lie in the``            ``// interval i-1 and i``            ``double` `res = Square(n, i - 1, i);``            ``Console.Write(``"{0:F5}"``, res);``            ``found = ``true``;``        ``}``        ``i++;``    ``}``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``double` `n = 3;` `    ``findSqrt(n);``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`1.73205`

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