Skip to content
Related Articles

Related Articles

Calculating n-th real root using binary search
  • Difficulty Level : Medium
  • Last Updated : 06 May, 2021

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- $\sqrt{0.09} = 0.3$.
  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 <bits/stdc++.h>
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 code
int 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 x
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;
        }
 
        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 code
x = 5
n = 2
findNthRoot(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




<script>
 
    // Javascript Program to find n-th
    // real root of x
     
    function findNthRoot(x, n)
    {
  
        // Initialize boundary values
        let low, high;
        if (x >= 0 && x <= 1)
        {
            low = x;
            high = 1;
        }
        else
        {
            low = 1;
            high = x;
        }
  
        // used for taking approximations
        // of the answer
        let epsilon = 0.00000001;
  
        // Do binary search
        let guess = parseInt((low + high) / 2, 10);
        while (Math.abs((Math.pow(guess, n)) - x)
                >= epsilon)
        {
            if (Math.pow(guess, n) > x)
            {
                high = guess;
            }
            else
            {
                low = guess;
            }
            guess = (low + high) / 2;
        }
  
        document.write(guess);
    }
     
    let x = 5;
    let n = 2;
    findNthRoot(x, n);
     
</script>
Output
2.2360679768025875

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 $\sqrt{5}$,then x = 5, low = 1, high = 5.
Taking epsilon = 0.01
First Guess:
guess = (1 + 5) / 2 = 3
Absolute error = |32 - 5| = 4 > epsilon
guess2 = 9 > 5(x) then high = guess --> high = 3
Second Guess:
guess = (1 + 3) / 2 = 2
Absolute error = |22 - 5| = 1 > epsilon
guess2 = 4 > 5(x) then low = guess --> low = 2
Third Guess:
guess = (2 + 3) / 2 = 2.5
Absolute error = |2.52 - 5| = 1.25 > epsilon
guess2 = 6.25 > 5(x) then high = guess --> high = 2.5
and proceeding so on we will get the $\sqrt{5}$ correct up to 2 decimal places i.e., $\sqrt{5}$ = 2.23600456
We will ignore the digits after 2 decimal places since they may or may not be correct.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :