Square root of a number without using sqrt() function

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 <bits/stdc++.h> 
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 

Output:

1.73205

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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