Automate the Boring Stuff Practice Problem: Collatz Sequence--Running the sequence on the number 1 without going into an infinite loop

Total programming newbie here, trying to learn python with Automate the Boring Stuff online. I've looked at all the other questions on this problem and haven't found a solution that does what I want, so I am asking it as a new question.

I've been working on a problem from chapter 3 of ATBS, and at the end there is a practice problem to create a program that will repeatedly execute the Collatz Sequence on any positive integer that a user enters, until it reduces to the number 1. The Collatz sequence is an unsolved puzzle in math where any starting number will always end up reducing to 1, but nobody can prove why.

Anyway, in the Collatz sequence, if the number is an even number, you divide by 2. If the number is odd, you multiply by 3 and then add 1. You do the same thing to the resulting output. So, if someone entered 5, that would first become 16 (i.e. 3*5 +1). The 16 then becomes 8 (i.e. 16 / 2), the 8 becomes 4, the 4 becomes 2, and finally the 2 becomes 1.

The number that you start with has to be a non-zero whole number. Negative numbers don't work, zero doesn't work, and non-integers don't work.

I've written code that works almost perfectly, but it has one problem. If the user enters 1, it just treats the sequence as complete instead of treating it like any other odd number and doing the sequence (i.e. it won't do this: 1*3 +1 becomes 4, then 4 becomes 2, then 2 becomes 1). I realize this might be technically correct. But it still bothers me that for the life of me I can't figure out how to make it work the other way.

If anyone has a solution for this, I would love to see it.

Here's my code:

def collatz(myNum):
    if myNum % 2 == 0 and myNum > 0:  # Collatz operation for even numbers
        print(myNum // 2)
        return myNum // 2
    elif myNum % 2 != 0 and myNum > 0:  # Collatz operation for odd numbers
        print(3 * myNum + 1)
        return 3 * myNum + 1

while True:
    try:
        myNum = int(input('Please enter a positive integer other than one: '))
        incomplete = True  # flag variable to track completion
        while myNum  > 1: # repeats until reduces to 1 and constrains input to positive numbers
            myNum = collatz(myNum)
            if myNum == 1:
                incomplete = False  # set flag to indicate completion
        if not incomplete:
            quit = input('Please enter q to quit or any other key to play again: ')
            if quit == 'q': #allows user to exit or try with a different number
                break
    except ValueError: #to deal with non-integers
        print('Integers only please.')

If I change "while myNum > 1" to while myNum > 0, then every time it run, it enters an infinite loop and won't stop once it reduces to 1. I can get it to either stop at the appropriate time, or I can get it to execute the sequence on the number 1, but I can't figure out how to get it to do both.

So first of all, let me recap your situation.

• Your code works
• You are not happy

Let me refer to the famous saying "If it works, don't fix it".

But I understand this is your way of learning, so let me help out.
The way I see it, our issue here is that the collatz function is never called because the while myNum > 1 condition is never validated when myNum == 1.

To avoid this, you can simply use while incomplete instead. Since you've declared incomplete as True, beforehand and you set it to False once the Collatz iteration reaches 1. it should work perfectly.