Making a collatz program automate the boring stuff

Question

I\'m trying to write a Collatz program using the guidelines from a project found at the end of chapter 3 of Automate the Boring Stuff with Python. I\'m using python 3.

Answer 1

Nuncjo got the solution that works. I tweaked it a little to add try and except statements for error handling.

def collatz(number):
    if number % 2 == 0:
        print(number // 2)
        return number // 2

    elif number % 2 == 1:
        result = 3 * number + 1
        print(result)
        return result

try:
    n = input("Enter number: ")
    while n > 0 and n!= 1:
        n = collatz(int(n))
except ValueError:
    print('whoops, type an integer, bro.')

Answer 2

I managed to get it right without using any return statements and nested a while loop inside the function.

number=int(input('Enter number:\n'))

def collatz(number):

    while number !=1:
        if number% 2 == 0:
            number= number//2
            print(number)

        else:
           number=  3 * number + 1
           print(number)    


collatz(number)

Answer 3

def collatz(number):

    if number % 2 == 0:
        print(number // 2)
        return number // 2

    elif number % 2 == 1:
        result = 3 * number + 1
        print(result)
        return result

n = input("Give me a number: ")
while n != 1:
    n = collatz(int(n))

Output:

Give me a number: 3
10
5
16
8
4
2
1

Give me a number: 11
34
17
52
26
13
40
20
10
5
16
8
4
2
1

Answer 4

My Code

def collatz(number):
    while number != 1:
        if number % 2 == 0:
            print(number // 2)
            number = number // 2
        elif number % 2 == 1:
            print(number * 3 + 1)
            number =  number *3 + 1

try:
    print ('Enter the number to Collatz:')
    collatz(int(input()))
except ValueError:
    print('Enter a valid integer')

Answer 5

def collatz(number):
    while number != 1:
        if number %2==0:
            number = number//2
            yield number
        elif number %2 ==1:
            number=number*3 +1
            yield number

while True:
    try:
        for n in collatz(int(input('Enter number:'))):
            print(n)
        break
    except ValueError:
        print('Please enter an integer')

The extra while True loop will help the program to continue functioning after the user inputs a non-integer.

Answer 6

Your collatz() function should print & return only the next value. (It ends when it returns.)

The while loop should not be inside the collatz() function.

You've also got inconsistent variable names (n, number, nnumber), and some important code is commented out.