# Calculating n-th real root using binary search

Last Updated : 28 Mar, 2022

Given two number x and n, find n-th root of x.

Examples:

Input : 5 2
Output : 2.2360679768025875

Input :  x = 5, n = 3
Output : 1.70997594668

In order to calculate nth root of a number, we can use the following procedure.

1. If x lies in the range [0, 1) then we set the lower limit low = x and upper limit high = 1, because for this range of numbers the nth root is always greater than the given number and can never exceed 1.
eg-
2. Otherwise, we take low = 1 and high = x.
3. Declare a variable named epsilon and initialize it for accuracy you need.
Say epsilon=0.01, then we can guarantee that our guess for nth root of the given number will be
correct up to 2 decimal places.
4. Declare a variable guess and initialize it to guess=(low+high)/2.
5. Run a loop such that:
• if the absolute error of our guess is more than epsilon then do:
1. if guessn > x, then high=guess
2. else low=guess
3. Making a new better guess i.e., guess=(low+high)/2.
• If the absolute error of our guess is less than epsilon then exit the loop.

Absolute Error: Absolute Error can be calculated as abs(guessn -x)

## C++

 // C++ Program to find// n-th real root of x#include using namespace std; void findNthRoot(double x, int n){     // Initialize boundary values    double low, high;    if (x >= 0 and x <= 1)    {        low = x;        high = 1;    }    else    {        low = 1;        high = x;    }     // Used for taking approximations    // of the answer    double epsilon = 0.00000001;     // Do binary search    double guess = (low + high) / 2;    while (abs((pow(guess, n)) - x) >= epsilon)     {        if (pow(guess, n) > x)        {            high = guess;        }        else        {            low = guess;        }        guess = (low + high) / 2;    }     cout << fixed << setprecision(16) << guess;} // Driver codeint main(){    double x = 5;    int n = 2;    findNthRoot(x, n);} // This code is contributed// by Subhadeep

## Java

 // Java Program to find n-th real root of xclass GFG {    static void findNthRoot(double x, int n)    {         // Initialize boundary values        double low, high;        if (x >= 0 && x <= 1)         {            low = x;            high = 1;        }        else        {            low = 1;            high = x;        }         // used for taking approximations        // of the answer        double epsilon = 0.00000001;         // Do binary search        double guess = (low + high) / 2;        while (Math.abs((Math.pow(guess, n)) - x)               >= epsilon)         {            if (Math.pow(guess, n) > x)            {                high = guess;            }            else            {                low = guess;            }            guess = (low + high) / 2;        }         System.out.println(guess);    }     // Driver code    public static void main(String[] args)    {        double x = 5;        int n = 2;        findNthRoot(x, n);    }} // This code is contributed// by mits

## Python3

 # Python Program to find n-th real root# of x  def findNthRoot(x, n):     # Initialize boundary values    x = float(x)    n = int(n)    if (x >= 0 and x <= 1):        low = x        high = 1    else:        low = 1        high = x     # used for taking approximations    # of the answer    epsilon = 0.00000001     # Do binary search    guess = (low + high) / 2    while abs(guess ** n - x) >= epsilon:        if guess ** n > x:            high = guess        else:            low = guess        guess = (low + high) / 2    print(guess)  # Driver codex = 5n = 2findNthRoot(x, n)

## C#

 // C# Program to find n-th real root of x using System; public class GFG {    static void findNthRoot(double x, int n)    {         // Initialize boundary values        double low, high;        if (x >= 0 && x <= 1)         {            low = x;            high = 1;        }        else        {            low = 1;            high = x;        }         // used for taking approximations        // of the answer        double epsilon = 0.00000001;         // Do binary search        double guess = (low + high) / 2;        while (Math.Abs((Math.Pow(guess, n)) - x)               >= epsilon)        {            if (Math.Pow(guess, n) > x)             {                high = guess;            }            else            {                low = guess;            }            guess = (low + high) / 2;        }         Console.WriteLine(guess);    }     // Driver code    static public void Main()    {        double x = 5;        int n = 2;        findNthRoot(x, n);    }} // This code is contributed by akt_mit

## Javascript

 

Output
2.2360679768025875

Time Complexity: O( log( x * 10d)*logguess(n) )

Auxiliary Space: O(1)

Here d is the number of decimal places upto which we want the result accurately.

Explanation of first example with epsilon = 0.01

Since taking too small value of epsilon as taken in our program might not be feasible for
explanation because it will increase the number of steps drastically so for the sake of
simplicity we are taking epsilon = 0.01
The above procedure will work as follows:
Say we have to calculate the then x = 5, low = 1, high = 5.Taking epsilon = 0.01First Guess:guess = (1 + 5) / 2 = 3Absolute error = |32 - 5| = 4 > epsilonguess2 = 9 > 5(x) then high = guess --> high = 3Second Guess:guess = (1 + 3) / 2 = 2Absolute error = |22 - 5| = 1 > epsilonguess2 = 4 > 5(x) then low = guess --> low = 2Third Guess:guess = (2 + 3) / 2 = 2.5Absolute error = |2.52 - 5| = 1.25 > epsilonguess2 = 6.25 > 5(x) then high = guess --> high = 2.5and proceeding so on we will get the  correct up to 2 decimal places i.e.,  = 2.23600456We will ignore the digits after 2 decimal places since they may or may not be correct.

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