Square Root without square root.
snippet in c

Square Root without square root.

user4109

#include<stdio.h>

void main()
{
    int number;

    float temp, sqrt;

    printf("Provide the number: \n");

    scanf("%d", &number);

    // store the half of the given number e.g from 256 => 128
    sqrt = number / 2;
    temp = 0;

    // Iterate until sqrt is different of temp, that is updated on the loop
    while(sqrt != temp){
        // initially 0, is updated with the initial value of 128
        // (on second iteration = 65)
        // and so on
        temp = sqrt;

        // Then, replace values (256 / 128 + 128 ) / 2 = 65
        // (on second iteration 34.46923076923077)
        // and so on
        sqrt = ( number/temp + temp) / 2;
    }

    printf("The square root of '%d' is '%f'", number, sqrt);
}

Square Root without square root.

user4142

# 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 

Square Root without square root.

user1529

// 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); 
} 
}