Python Program to Calculate Sum of Nth Power using Recursion

Python Program to Calculate Sum of Nth Power using Recursion

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Recursion:

Recursion is the process by which a function calls itself directly or indirectly, and the associated function is known as a recursive function. Certain issues can be addressed fairly easily using a recursive approach. Towers of Hanoi (TOH), Inorder /Preorder/Postorder Tree Traversals, DFS of Graph, and other analogous issues are examples.

Find the number of ways a given integer can be represented as the sum of the Nth power of the unique, natural integers using this program. For example, if X = 100 and N = 2, we must identify all square combinations that add up to 100. The potential solutions are (10,2), (8^2+6^2), (1^2+3^2+4^2+5^2+7^2). As a result, the total number of solutions is three.

Examples:

Example1:

Input:

Given number = 25

Given N =2

Output:

The possible solutions count = 2

Example2:

Input:

Given number = 100
Given N =2

Output:

The possible solutions count = 3

Python Program to Calculate Sum of Nth Power using Recursion

Below are the ways to write a program that calculates the sum of Nth  Power using Recursion.

Method #1: Using Recursion (Static Input)

Approach:

  • Give the number and N as static input and store them in two separate variables.
  • Create a function countNPower() function which returns the total number of possible solutions count.
  • Pass the given number and N,1 as arguments to the countNPower() function.
  • Check to see if the number equals 1 power N, if so, there is only one potential/possible solution.
  • There is no solution if the number is smaller than 1 power of N.
  • If number is more than 1 power N, then countNPower(value, N, num+1)+countNPower(number, N, num+1) is returned.
  • The first countNPowercall includes the 1 power N value, but the second one excludes it.
  • Print the Result.
  • The Exit of the Program.

Below is the implementation:

# Create a function countNPower() function which
# returns the total number of possible solutions count.


def countNPower(Number, n, numb):
    resultval = Number-pow(numb, n)
    # There is no solution if the number is smaller than 1 power of N.
    if resultval < 0:
        return 0
    # Check to see if the number equals 1 power N,
    # if so, there is only one potential/possible solution.
    elif resultval == 0:
        return 1
    # If number is more than 1 power N, then
    # countNPower(value, N, num+1)+countNPower(number, N, num+1) is returned.
    # The first countNPowercall includes the 1 power N value,
    # but the second one excludes it.
    else:
        return countNPower(resultval, n, numb+1)+countNPower(Number, n, numb+1)


# Give the number and N as static input and store them in two separate variables.
Numb = 25
n = 2
# Pass the given number and N,1 as arguments to the countNPower() function.
print('The possible solutions count = ', countNPower(Numb, n, 1))

Output:

The possible solutions count =  2

Method #2: Using Recursion (User Input)

Approach:

  • Give the number and N as user input using map(),split(), and input() functions.
  • Store them in two separate variables.
  • Create a function countNPower() function which returns the total number of possible solutions count.
  • Pass the given number and N,1 as arguments to the countNPower() function.
  • Check to see if the number equals 1 power N, if so, there is only one potential/possible solution.
  • There is no solution if the number is smaller than 1 power of N.
  • If number is more than 1 power N, then countNPower(value, N, num+1)+countNPower(number, N, num+1) is returned.
  • The first countNPowercall includes the 1 power N value, but the second one excludes it.
  • Print the Result.
  • The Exit of the Program.

Below is the implementation:

# Create a function countNPower() function which
# returns the total number of possible solutions count.


def countNPower(Number, n, numb):
    resultval = Number-pow(numb, n)
    # There is no solution if the number is smaller than 1 power of N.
    if resultval < 0:
        return 0
    # Check to see if the number equals 1 power N,
    # if so, there is only one potential/possible solution.
    elif resultval == 0:
        return 1
    # If number is more than 1 power N, then
    # countNPower(value, N, num+1)+countNPower(number, N, num+1) is returned.
    # The first countNPowercall includes the 1 power N value,
    # but the second one excludes it.
    else:
        return countNPower(resultval, n, numb+1)+countNPower(Number, n, numb+1)


# Give the number and N as user input using map(),split(), and input() functions.
# Store them in two separate variables.
Numb, n = map(int, input(
    'Enter some random number and N separated by spaces = ').split())
# Pass the given number and N,1 as arguments to the countNPower() function.
print('The possible solutions count = ', countNPower(Numb, n, 1))

Output:

Enter some random number and N separated by spaces = 100 2
The possible solutions count = 3

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