Given two positive integers N and P. The task is to find the Nth root of P.
Examples:
Input: P = 1234321, N = 2
Output: 1111
Explanation: square root of 1234321 is 1111.Input: P = 123456785, N = 20
Output: 2.53849
Approach: There are various ways to solve the given problem. Here the below algorithm is based on Mathematical Concept called Bisection Method for finding roots. To find the N-th power root of a given number P we will form an equation is formed in x as ( xp – P = 0 ) and the target is to find the positive root of this equation using the Bisection Method.
How Bisection Method works?
Take an interval (a, b) such that its already known that the root is existing in that interval. After this find the mid of the interval and examine the value of function and it’s derivative at x = mid.
- If the value of function is 0 that means root is found
- If the value of the function is positive and its derivative is negative that means the root is lying in the right half.
- If the value of the function is positive and its derivative is positive that means the root is lying in the left half.
Below is the implementation of the above approach:
C++14
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function that returns the value// of the function at a given value of xdouble f(double x, int p, double num){ return pow(x, p) - num;}// calculating the value// of the differential of the functiondouble f_prime(double x, int p){ return p * pow(x, p - 1);}// The function that returns// the root of given numberdouble root(double num, int p){ // Defining range // on which answer can be found double left = -num; double right = num; double x; while (true) { // finding mid value x = (left + right) / 2.0; double value = f(x, p, num); double prime = f_prime(x, p); if (value * prime <= 0) left = x; else right = x; if (value < 0.000001 && value >= 0) { return x; } }}// Driver codeint main(){ double P = 1234321; int N = 2; double ans = root(P, N); cout << ans;} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function that returns the value // of the function at a given value of x static double f(double x, int p, double num) { return Math.pow(x, p) - num; } // calculating the value // of the differential of the function static double f_prime(double x, int p) { return p * Math.pow(x, p - 1); } // The function that returns // the root of given number static double root(double num, int p) { // Defining range // on which answer can be found double left = -num; double right = num; double x; while (true) { // finding mid value x = (left + right) / 2.0; double value = f(x, p, num); double prime = f_prime(x, p); if (value * prime <= 0) left = x; else right = x; if (value < 0.000001 && value >= 0) { return x; } } } // Driver code public static void main(String args[]) { double P = 1234321; int N = 2; double ans = root(P, N); System.out.print(ans); }}// This code is contributed by ihritik |
Python3
# python program for above approach# Function that returns the value# of the function at a given value of xdef f(x, p, num): return pow(x, p) - num# calculating the value# of the differential of the functiondef f_prime(x, p): return p * pow(x, p - 1)# The function that returns# the root of given numberdef root(num, p): # Defining range # on which answer can be found left = -num right = num x = 0 while (True): # finding mid value x = (left + right) / 2.0 value = f(x, p, num) prime = f_prime(x, p) if (value * prime <= 0): left = x else: right = x if (value < 0.000001 and value >= 0): return x# Driver codeif __name__ == "__main__": P = 1234321 N = 2 ans = root(P, N) print(ans)# This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;public class GFG { // Function that returns the value // of the function at a given value of x static double f(double x, int p, double num) { return Math.Pow(x, p) - num; } // calculating the value // of the differential of the function static double f_prime(double x, int p) { return p * Math.Pow(x, p - 1); } // The function that returns // the root of given number static double root(double num, int p) { // Defining range // on which answer can be found double left = -num; double right = num; double x; while (true) { // finding mid value x = (left + right) / 2.0; double value = f(x, p, num); double prime = f_prime(x, p); if (value * prime <= 0) left = x; else right = x; if (value < 0.000001 && value >= 0) { return x; } } } // Driver code public static void Main(string []args) { double P = 1234321; int N = 2; double ans = root(P, N); Console.Write(ans); }}// This code is contributed by AnkThon |
Javascript
<script> // JavaScript Program to implement // the above approach // Function that returns the value // of the function at a given value of x function f(x, p, num) { return Math.pow(x, p) - num; } // calculating the value // of the differential of the function function f_prime(x, p) { return p * Math.pow(x, p - 1); } // The function that returns // the root of given number function root(num, p) { // Defining range // on which answer can be found let left = -num; let right = num; let x; while (true) { // finding mid value x = (left + right) / 2.0; let value = f(x, p, num); let prime = f_prime(x, p); if (value * prime <= 0) left = x; else right = x; if (value < 0.000001 && value >= 0) { return x; } } } // Driver code let P = 1234321; let N = 2; let ans = Math.floor(root(P, N)); document.write(ans);// This code is contributed by Potta Lokesh </script> |
Output
1111
Time Complexity: O(log P).
Auxiliary Space: O(1).
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