Recursion

def steps(number):
    if number <= 0:
        raise ValueError("Only positive integers are allowed")
    if number == 1:
        return 0
    number = number / 2 if number % 2 == 0 else number * 3 + 1
    return 1 + steps(number)

This approach uses recursion to solve the challenge. Recursion is less common as a strategy in Python than in other "fully-functional" programming languages such as Elixir, Haskell, and Clojure. While it can be powerful, it can also be trickier to implement than looping constructs.

This approach starts with checking if number <= 0 and raising a ValueError if it is. Next, we return zero if number == 1. This is the base case.

We then assign number to the same conditional expression as seen in the ternary operator approach. Finally, we return one plus the result of calling steps(), with the updated number value. This is the recursive case.

Solving this exercise in this way removes the need for a counter variable and the creation of a loop. If number is not equal to one, we call 1 + steps(number). Then steps() can execute the same code again with new values. This makes a long chain (or stack) of 1 + steps(number) — until number == 1 and the code adds zero and exits. That translates to something like: 1 + 1 + 1 + 1 + 0.

Python doesn't have tail call optimization, so the stack of 1 + steps(number) will continue to grow until the base case triggers resolution, or the code reaches the recursion limit.