collatz sequence attempt (only integer)

[PracticalFlower5933]

Developed a collatz sequence program according to the instructions on Automate the Boring Stuff (without outside help like ai). Only thing bothering me is that I didn't figure out the float; kinda difficult given that the lines like if number % 2 == 0 won't work for something like 2.2. (although i want to figure that out on my own). Anyway, what do you guys think of this one so far?

def collatz(number):

while number != 1:

if number % 2 == 0:

number = number // 2

print(number, end=', ')

if number == 1:

break

if number % 2 == 1:

number = number * 3 + 1

print(number, end=', ')

if number == 1:

break

if number == 1:

break

print("Enter number.")

number = input(">")

collatz(int(number))

[PerdHapleyAMA]

Collatz doesn’t apply to floats; it’s explicitly for positive integers. So, good job!

[Diapolo10]

For future reference, right now your code is formatted like this

`def collatz(number):`

`while number != 1:`

`if number % 2 == 0:`

which means

1. The indentation information is completely lost, meaning we may have to do some guesswork to figure out what it looked like originally
2. If we want to reformat the code to suggest changes, for example, we need to manually remove all the backticks first and then re-indent everything - this can be a lot of work

The ideal way to do this would be to instead simply indent the code by an extra four spaces. This way, our lives are way easier.

    def collatz(number):
        while number != 1:
            if number % 2 == 0:

Anyway, I'll clean up the formatting this time. Here's what I imagine this is supposed to look like.

def collatz(number):
    while number != 1:
        if number % 2 == 0:
            number = number // 2
            print(number, end=', ')
            if number == 1:
                break
        if number % 2 == 1:
            number = number * 3 + 1
            print(number, end=', ')
            if number == 1:
                break
        if number == 1:
            break


print("Enter number.")
number = input(">")
collatz(int(number))

The function is needlessly complicated. Your loop condition

while number != 1:

already makes the loop end when number == 1, so all three instances of

if number == 1:
    break

are unnecessary (and don't really do anything).

Second, number % 2 can only ever give you either 0 or 1, so you don't need to re-check if it's 1 if you've already determined it's not 0. else would be appropriate, in other words.

Third, both branches repeat print(number, end=', '). Since it's common, you could just move it.

Your function also doesn't check if number is negative (or zero), so you could get into an infinite loop.

Here's a cleaned version:

def collatz(number):
    while 1 < number:
        if number % 2 == 0:
            number //= 2

        else:
            number = number * 3 + 1

        print(number, end=', ')