Given a number N, the task is to find the square root of N without using sqrt() function.
Examples:
Input: N = 25
Output: 5Input: N = 3
Output: 1.73205Input: 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.
- If mid * mid = n then return mid.
- If mid * mid < n then recur for the second half.
- 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; } |
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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 |
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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 |
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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 |
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Output:
1.73205
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