While loop does not break though condition is met (Collatz Sequence)

Question

Note: I was trying to figure out why my while statement did not evaluate to False when the integer was, so I don't believe this is a duplicate

Doing an automating the boring stuff exercise in python where the program receives an input and reduces it to 1 using the following algorithm.

#even / 2
#odd * 3 + 1


def collatz():
    print("Enter number:")
    number = input()
    try:
        data = int(number) # Input Validation

        while int(data) != 1:
            if data % 2 == 0: #Number is even
                data = int(data/2)
                print(data)
            if data % 2 == 1: # Number is odd
                data = int(3*data+1)
                print(data)

    except:
        print("Please input a valid value")
        collatz()

collatz()

Instead of the while loop breaking, when the number is reduced to 1. The loop continues and it multiplies 1 by 3 and adds 1(Just as normal odd number). Btw many int conversions were done as I thought it may have returned a floating point.

So please tell me where are the unncessary int conversions and how to break it using a while statement. Any cleanup to code is appreciated

Joe Iddon · Accepted Answer · 2017-12-30T10:02:24.957

There are a few things that you should neaten up to make your code work properly and just generally better:

Rename data to n, this isn't going to make much of a difference, but I would say it just makes more sense.
There is no need to do endless conversions to int, n only needs to be converted from a string to an integer once.
Don't put your entire code in a function then call it once, instead have the main algorithm in a smaller function which you then can call from the main body of your code or even create another function to call the algorithm which handles the inputting side of things.
Use the integer division operator (//) instead of the floating point divider (/), since the number will be even, you can be sure that the decimal place will be 0.
You don't need to check if n % 2 is 0 and then on the next line check if it n % 2 is 1, you can just use an if ... else clause.

And that's about it! Here's what it looks like:

#even / 2
#odd * 3 + 1

def collatz(n):
    while n != 1:
        if n % 2 == 0: #Number is even
            n = n // 2
        else:
            n = n * 3 + 1
        print(n)

number = input("Enter number:")

try:
    number = int(number)
    collatz(number)
except ValueError:
    print("Please input a valid value")

And a test shows it works (input of 24):

Thank you for all the tips and improvements! If I were to leave all the code in one function, why did my while loop not break when n or in my case data was 1? — KOOTS HOOTS, Dec 30 '17 at 10:12
@KOOTSHOOTS Got it! It's because you are doing two `if` statements, so i the case where `data` was `2`, it would pass the first test and divide by `2` to get `1`, but then the second `if` would see it was now odd and keep it going by multiplying by `3` and adding `1`. So that is why my function is working, because I used an `else`. **:)** Please accept then! — Joe Iddon, Dec 30 '17 at 11:19

While loop does not break though condition is met (Collatz Sequence)

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