Python Program to Determine Whether a Given Number is Even or Odd Recursively

When determining whether a number is even or odd, we typically use the modulo operator. However, we can also solve this problem recursively by repeatedly subtracting 2 until we reach a base case.

Recursion breaks down the problem into smaller subproblems. Here, we reduce the number by 2 in each recursive call until we reach 0 (even) or 1 (odd).

How the Recursive Approach Works

The algorithm works by:

• Base case: If the number is less than 2 (0 or 1), check if it's divisible by 2
• Recursive case: Subtract 2 from the number and call the function again

Example

Below is a demonstration of the recursive approach ?

def check_odd_even(my_num):
    if (my_num < 2):
        return (my_num % 2 == 0)
    return (check_odd_even(my_num - 2))

# Test with different numbers
test_numbers = [48, 35, 0, 1, 7, 12]

for number in test_numbers:
    if check_odd_even(number):
        print(f"{number} is even")
    else:
        print(f"{number} is odd")

48 is even
35 is odd
0 is even
1 is odd
7 is odd
12 is even

Step-by-Step Execution

Let's trace how the function works with number 5:

check_odd_even(5)
  ? check_odd_even(3)  # 5 - 2 = 3
    ? check_odd_even(1)  # 3 - 2 = 1
      ? 1 % 2 == 0  # Base case: False (1 is odd)
    ? False
  ? False
? False (5 is odd)

Alternative Implementation

Here's another recursive approach using positive and negative checks ?

def is_even_recursive(n):
    # Handle negative numbers
    if n < 0:
        n = -n
    
    # Base cases
    if n == 0:
        return True
    elif n == 1:
        return False
    else:
        return is_even_recursive(n - 2)

# Test the function
numbers = [8, 15, -6, -3, 0]
for num in numbers:
    result = "even" if is_even_recursive(num) else "odd"
    print(f"{num} is {result}")

8 is even
15 is odd
-6 is even
-3 is odd
0 is even

Explanation

• The function check_odd_even takes a number as parameter
• If the number is less than 2, it checks the remainder when divided by 2
• For numbers ? 2, the function calls itself with the number decremented by 2
• This continues until we reach the base case (0 or 1)
• The final result determines if the original number is even or odd

Conclusion

Recursive even/odd checking works by repeatedly subtracting 2 until reaching a base case. While less efficient than the modulo operator, it demonstrates how recursion can solve mathematical problems by breaking them into smaller subproblems.