N-th root of a number
Last Updated :
23 Mar, 2023
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
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++
#include <bits/stdc++.h>
using namespace std;
double nthRoot( int A, int N)
{
double xPre = rand () % 10;
double eps = 1e-3;
double delX = INT_MAX;
double xK;
while (delX > eps)
{
xK = ((N - 1.0) * xPre +
( double )A/ pow (xPre, N-1)) / ( double )N;
delX = abs (xK - xPre);
xPre = xK;
}
return xK;
}
int main()
{
int N = 4;
int A = 81;
double nthRootValue = nthRoot(A, N);
cout << "Nth root is " << nthRootValue << endl;
return 0;
}
|
Java
class GFG
{
static double nthRoot( int A, int N)
{
double xPre = Math.random() % 10 ;
double eps = 0.001 ;
double delX = 2147483647 ;
double xK = 0.0 ;
while (delX > eps)
{
xK = ((N - 1.0 ) * xPre +
( double )A / Math.pow(xPre, N - 1 )) / ( double )N;
delX = Math.abs(xK - xPre);
xPre = xK;
}
return xK;
}
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 );
}
}
|
Python3
import math
import random
def nthRoot(A,N):
xPre = random.randint( 1 , 101 ) % 10
eps = 0.001
delX = 2147483647
xK = 0.0
while (delX > eps):
xK = ((N - 1.0 ) * xPre +
A / pow (xPre, N - 1 )) / N
delX = abs (xK - xPre)
xPre = xK;
return xK
N = 4
A = 81
nthRootValue = nthRoot(A, N)
print ( "Nth root is " , nthRootValue)
|
C#
using System;
class GFG
{
static double nthRoot( int A, int N)
{
Random rand = new Random();
double xPre = rand.Next(10);;
double eps = 0.001;
double delX = 2147483647;
double xK = 0.0;
while (delX > eps)
{
xK = ((N - 1.0) * xPre +
( double )A / Math.Pow(xPre, N - 1)) / ( double )N;
delX = Math.Abs(xK - xPre);
xPre = xK;
}
return xK;
}
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);
}
}
|
PHP
<?php
function nthRoot( $A , $N )
{
$xPre = rand() % 10;
$eps = 0.001;
$delX = PHP_INT_MAX;
$xK ;
while ( $delX > $eps )
{
$xK = ((int)( $N - 1.0) *
$xPre + $A /
(int)pow( $xPre ,
$N - 1)) / $N ;
$delX = abs ( $xK - $xPre );
$xPre = $xK ;
}
return floor ( $xK );
}
$N = 4;
$A = 81;
$nthRootValue = nthRoot( $A , $N );
echo "Nth root is " ,
$nthRootValue , "\n" ;
?>
|
Javascript
<script>
function nthRoot(A, N)
{
let xPre = Math.random() % 10;
let eps = 0.001;
let delX = 2147483647;
let xK = 0.0;
while (delX > eps)
{
xK = ((N - 1.0) * xPre +
A / Math.pow(xPre, N - 1)) / N;
delX = Math.abs(xK - xPre);
xPre = xK;
}
return xK;
}
let N = 4;
let A = 81;
let nthRootValue = nthRoot(A, N);
document.write( "Nth root is " +Math.round(nthRootValue*1000.0)/1000.0);
</script>
|
Time Complexity: O(log(eps)), where eps is the desired accuracy.
Space Complexity: O(1)
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