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N-th root of a number

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Given two numbers N and A, find N-th root of A. In mathematics, Nth root of a number A is a real number that gives A, when we raise it to integer power N. These roots are used in Number Theory and other advanced branches of mathematics. 
Refer Wiki page for more information. 
Examples: 

Input : A = 81
        N = 4
Output : 3 
3^4 = 81
Recommended Practice

As this problem involves a real valued function A^(1/N) we can solve this using Newton’s method, which starts with an initial guess and iteratively shift towards the result. 
We can derive a relation between two consecutive values of iteration using Newton’s method as follows, 

according to newton’s method
x(K+1) = x(K) – f(x) / f’(x)        
here    f(x)  = x^(N) – A
so    f’(x) = N*x^(N - 1)
and     x(K) denoted the value of x at Kth iteration
putting the values and simplifying we get,
x(K + 1) = (1 / N) * ((N - 1) * x(K) + A / x(K) ^ (N - 1))

Using above relation, we can solve the given problem. In below code we iterate over values of x, until difference between two consecutive values of x become lower than desired accuracy.
Below is the implementation of above approach: 

C++

// C++ program to calculate Nth root of a number
#include <bits/stdc++.h>
using namespace std;
 
//  method returns Nth power of A
double nthRoot(int A, int N)
{
    // initially guessing a random number between
    // 0 and 9
    double xPre = rand() % 10;
 
    //  smaller eps, denotes more accuracy
    double eps = 1e-3;
 
    // initializing difference between two
    // roots by INT_MAX
    double delX = INT_MAX;
 
    //  xK denotes current value of x
    double xK;
 
    //  loop until we reach desired accuracy
    while (delX > eps)
    {
        //  calculating current value from previous
        // value by newton's method
        xK = ((N - 1.0) * xPre +
              (double)A/pow(xPre, N-1)) / (double)N;
        delX = abs(xK - xPre);
        xPre = xK;
    }
 
    return xK;
}
 
//    Driver code to test above methods
int main()
{
    int N = 4;
    int A = 81;
 
    double nthRootValue = nthRoot(A, N);
    cout << "Nth root is " << nthRootValue << endl;
 
    /*
        double Acalc = pow(nthRootValue, N);
        cout << "Error in difference of powers "
             << abs(A - Acalc) << endl;
    */
 
    return 0;
}

                    

Java

// Java program to calculate Nth root of a number
class GFG
{
     
    // method returns Nth power of A
    static double nthRoot(int A, int N)
    {
         
        // initially guessing a random number between
        // 0 and 9
        double xPre = Math.random() % 10;
     
        // smaller eps, denotes more accuracy
        double eps = 0.001;
     
        // initializing difference between two
        // roots by INT_MAX
        double delX = 2147483647;
     
        // xK denotes current value of x
        double xK = 0.0;
     
        // loop until we reach desired accuracy
        while (delX > eps)
        {
            // calculating current value from previous
            // value by newton's method
            xK = ((N - 1.0) * xPre +
            (double)A / Math.pow(xPre, N - 1)) / (double)N;
            delX = Math.abs(xK - xPre);
            xPre = xK;
        }
     
        return xK;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int N = 4;
        int A = 81;
     
        double nthRootValue = nthRoot(A, N);
        System.out.println("Nth root is "
        + Math.round(nthRootValue*1000.0)/1000.0);
     
        /*
            double Acalc = pow(nthRootValue, N);
            cout << "Error in difference of powers "
                << abs(A - Acalc) << endl;
        */
    }
}
 
// This code is contributed by Anant Agarwal.

                    

Python3

# Python3 program to calculate
# Nth root of a number
import math
import random
 
# method returns Nth power of A
def nthRoot(A,N):
 
    # initially guessing a random number between
    # 0 and 9
    xPre = random.randint(1,101) % 10
  
    #  smaller eps, denotes more accuracy
    eps = 0.001
  
    # initializing difference between two
    # roots by INT_MAX
    delX = 2147483647
  
    #  xK denotes current value of x
    xK=0.0
  
    #  loop until we reach desired accuracy
    while (delX > eps):
 
        # calculating current value from previous
        # value by newton's method
        xK = ((N - 1.0) * xPre +
              A/pow(xPre, N-1)) /N
        delX = abs(xK - xPre)
        xPre = xK;
         
    return xK
 
# Driver code
N = 4
A = 81
nthRootValue = nthRoot(A, N)
 
print("Nth root is ", nthRootValue)
 
## Acalc = pow(nthRootValue, N);
## print("Error in difference of powers ",
##             abs(A - Acalc))
 
# This code is contributed
# by Anant Agarwal.

                    

C#

// C# program to calculate Nth root of a number
using System;
class GFG
{
     
    // method returns Nth power of A
    static double nthRoot(int A, int N)
    {
        Random rand = new Random();
        // initially guessing a random number between
        // 0 and 9
        double xPre = rand.Next(10);;
     
        // smaller eps, denotes more accuracy
        double eps = 0.001;
     
        // initializing difference between two
        // roots by INT_MAX
        double delX = 2147483647;
     
        // xK denotes current value of x
        double xK = 0.0;
     
        // loop until we reach desired accuracy
        while (delX > eps)
        {
            // calculating current value from previous
            // value by newton's method
            xK = ((N - 1.0) * xPre +
            (double)A / Math.Pow(xPre, N - 1)) / (double)N;
            delX = Math.Abs(xK - xPre);
            xPre = xK;
        }
     
        return xK;
    }
     
    // Driver code
    static void Main()
    {
        int N = 4;
        int A = 81;
     
        double nthRootValue = nthRoot(A, N);
        Console.WriteLine("Nth root is "+Math.Round(nthRootValue*1000.0)/1000.0);
    }
}
 
// This code is contributed by mits

                    

PHP

<?php
// PHP program to calculate
// Nth root of a number
 
// method returns
// Nth power of A
function nthRoot($A, $N)
{
    // initially guessing a
    // random number between
    // 0 and 9
    $xPre = rand() % 10;
 
    // smaller eps, denotes
    // more accuracy
    $eps = 0.001;
 
    // initializing difference
    // between two roots by INT_MAX
    $delX = PHP_INT_MAX;
 
    // xK denotes current
    // value of x
    $xK;
 
    // loop until we reach
    // desired accuracy
    while ($delX > $eps)
    {
        // calculating current
        // value from previous
        // value by newton's method
        $xK = ((int)($N - 1.0) *
                     $xPre + $A /
                     (int)pow($xPre,
                              $N - 1)) / $N;
        $delX = abs($xK - $xPre);
        $xPre = $xK;
    }
 
    return floor($xK);
}
 
// Driver code
$N = 4;
$A = 81;
 
$nthRootValue = nthRoot($A, $N);
echo "Nth root is " ,
      $nthRootValue ,"\n";
 
// This code is contributed by akt_mit
?>

                    

Javascript

<script>
 
// Javascript program for the above approach
   
    // method returns Nth power of A
    function nthRoot(A, N)
    {
           
        // initially guessing a random number between
        // 0 and 9
        let xPre = Math.random() % 10;
       
        // smaller eps, denotes more accuracy
        let eps = 0.001;
       
        // initializing difference between two
        // roots by INT_MAX
        let delX = 2147483647;
       
        // xK denotes current value of x
        let xK = 0.0;
       
        // loop until we reach desired accuracy
        while (delX > eps)
        {
            // calculating current value from previous
            // value by newton's method
            xK = ((N - 1.0) * xPre +
            A / Math.pow(xPre, N - 1)) / N;
            delX = Math.abs(xK - xPre);
            xPre = xK;
        }
       
        return xK;
    }
        
 
// Driver Code
     
        let N = 4;
        let A = 81;
       
        let nthRootValue = nthRoot(A, N);
        document.write("Nth root is "+Math.round(nthRootValue*1000.0)/1000.0);
         
</script>

                    

Output
Nth root is 3

Time Complexity: O(log(eps)), where eps is the desired accuracy.
Space Complexity: O(1)




 


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Last Updated : 23 Mar, 2023
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