Calculating n-th real root using binary search
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.
- 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- - Otherwise, we take low = 1 and high = x.
- 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. - Declare a variable guess and initialize it to guess=(low+high)/2.
- Run a loop such that:
- if the absolute error of our guess is more than epsilon then do:
- if guessn > x, then high=guess
- else low=guess
- 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++
#include <bits/stdc++.h>
using namespace std;
void findNthRoot( double x, int n)
{
double low, high;
if (x >= 0 and x <= 1)
{
low = x;
high = 1;
}
else
{
low = 1;
high = x;
}
double epsilon = 0.00000001;
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;
}
int main()
{
double x = 5;
int n = 2;
findNthRoot(x, n);
}
|
Java
class GFG
{
static void findNthRoot( double x, int n)
{
double low, high;
if (x >= 0 && x <= 1 )
{
low = x;
high = 1 ;
}
else
{
low = 1 ;
high = x;
}
double epsilon = 0.00000001 ;
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);
}
public static void main(String[] args)
{
double x = 5 ;
int n = 2 ;
findNthRoot(x, n);
}
}
|
Python3
def findNthRoot(x, n):
x = float (x)
n = int (n)
if (x > = 0 and x < = 1 ):
low = x
high = 1
else :
low = 1
high = x
epsilon = 0.00000001
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)
x = 5
n = 2
findNthRoot(x, n)
|
C#
using System;
public class GFG {
static void findNthRoot( double x, int n)
{
double low, high;
if (x >= 0 && x <= 1)
{
low = x;
high = 1;
}
else
{
low = 1;
high = x;
}
double epsilon = 0.00000001;
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);
}
static public void Main()
{
double x = 5;
int n = 2;
findNthRoot(x, n);
}
}
|
Javascript
<script>
function findNthRoot(x, n)
{
let low, high;
if (x >= 0 && x <= 1)
{
low = x;
high = 1;
}
else
{
low = 1;
high = x;
}
let epsilon = 0.00000001;
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>
|
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.
Last Updated :
28 Mar, 2022
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