Find cubic root of a number

3.7

Given a number n, find the cube root of n.

Examples:

Input:  n = 3
Output: Cubic Root is 1.442250

Input: n = 8
Output: Cubic Root is 2.000000

We can use binary search. First we define error e. Let us say 0.0000001 in our case. The main steps of our algorithm for calculating the cubic root of a number n are:

  1. Initialize start = 0 and end = n
  2. Calculate mid = (start + end)/2
  3. Check if the absolute value of (n – mid*mid*mid) < e. If this condition holds true then mid is our answer so return mid.
  4. If (mid*mid*mid)>n then set end=mid
  5. If (mid*mid*mid)

Below is the implementation of above idea.

C++

// C++ program to find cubic root of a number
// using Binary Search
#include <bits/stdc++.h>
using namespace std;

// Returns the absolute value of n-mid*mid*mid
double diff(double n,double mid)
{
    if (n > (mid*mid*mid))
        return (n-(mid*mid*mid));
    else
        return ((mid*mid*mid) - n);
}

// Returns cube root of a no n
double cubicRoot(double n)
{
    // Set start and end for binary search
    double start = 0, end = n;

    // Set precision
    double e = 0.0000001;

    while (true)
    {
        double mid = (start + end)/2;
        double error = diff(n, mid);

        // If error is less than e then mid is
        // our answer so return mid
        if (error <= e)
            return mid;

        // If mid*mid*mid is greater than n set
        // end = mid
        if ((mid*mid*mid) > n)
            end = mid;

        // If mid*mid*mid is less than n set
        // start = mid
        else
            start = mid;
    }
}

// Driver code
int main()
{
    double n = 3;
    printf("Cubic root of %lf is %lf\n",
           n, cubicRoot(n));
    return 0;
}

Java

// Java program to find cubic root of a number
// using Binary Search
import java.io.*;

class GFG 
{
    // Returns the absolute value of n-mid*mid*mid
    static double diff(double n,double mid)
    {
        if (n > (mid*mid*mid))
            return (n-(mid*mid*mid));
        else
            return ((mid*mid*mid) - n);
    }
    
    // Returns cube root of a no n
    static double cubicRoot(double n)
    {
        // Set start and end for binary search
        double start = 0, end = n;
 
        // Set precision
        double e = 0.0000001;
 
        while (true)
        {
            double mid = (start + end)/2;
            double error = diff(n, mid);
 
            // If error is less than e then mid is
            // our answer so return mid
            if (error <= e)
                return mid;
 
            // If mid*mid*mid is greater than n set
            // end = mid
            if ((mid*mid*mid) > n)
                end = mid;
 
            // If mid*mid*mid is less than n set
            // start = mid
            else
                start = mid;
        }
    }
    
    // Driver program to test above function
    public static void main (String[] args) 
    {
        double n = 3;
        System.out.println("Cube root of "+n+" is "+cubicRoot(n));
    }
}

// This code is contributed by Pramod Kumar

Python3

# Python 3 program to find cubic root 
# of a number using Binary Search

# Returns the absolute value of 
# n-mid*mid*mid
def diff(n, mid) :
    if (n > (mid * mid * mid)) :
        return (n - (mid * mid * mid))
    else :
        return ((mid * mid * mid) - n)
        
# Returns cube root of a no n
def cubicRoot(n) :
    
    # Set start and end for binary 
    # search
    start = 0
    end = n
    
    # Set precision
    e = 0.0000001
    while (True) :
        
        mid = (start + end) / 2
        error = diff(n, mid)

        # If error is less than e 
        # then mid is our answer
        # so return mid
        if (error <= e) :
            return mid
            
        # If mid*mid*mid is greater
        # than n set end = mid
        if ((mid * mid * mid) > n) :
            end = mid
            
        # If mid*mid*mid is less 
        # than n set start = mid
        else :
            start = mid
            
# Driver code
n = 3
print("Cubic root of", n, "is", 
      round(cubicRoot(n),6))


# This code is contributed by Nikita Tiwari.


Output:

Cubic root of 3.000000 is 1.442250

Time Complexity : O(Log n)

This article is contributed by Madhur Modi .If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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