Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Basics (2015) Letter Game App Even or Odd Loop

Ian Cole
Ian Cole
454 Points
Posted by Ian Cole
Ian Cole
454 Points

Task 1 suddenly not passing on last step even though Task 1 is literally just importing the random library??????

Exactly what it says on the tin. I get to the last step of this challenge and suddenly Task 1 stops passing even though Task 1 is just importing the random library. I cannot express how frustrated I am in words without breaking community guidelines. Is it something with my code?

even.py 
import random
def even_odd(num):
    # If % 2 is 0, the number is even.
    # Since 0 is falsey, we have to invert it with not.
    return not num % 2
start = 5
while start not False:
    numb = random.randint(1, 99)
    if numb % 2 == 0:
        print("{} is even".format(numb))
    else:
        print("{} is odd".format(numb))
    start -= 0

3 Answers

Kip Yin
Kip Yin
4,847 Points

Task 1 is not passing because you have a SyntaxError in your while loop: start not False. You may either do while start, which breaks the loop as soon as start == 0, or do while start != 0. The former way is preferred in Python.

Also, I noticed that you have start -= 0 at the end. This is not decrementing start. Typo?

Kip Yin
Kip Yin
4,847 Points

Otherwise, your code is good to go.

Ian Cole
Ian Cole
454 Points
Ian Cole
Ian Cole
454 Points

Thanks for pointing that out, knowing me I would have never noticed. Now it's saying that there is too many prints as the error... Go figure, almost done with a course and the only part I have trouble with is the last part

Kip Yin
Kip Yin
4,847 Points

I changed the while-loop condition to start and changed the last line to start -= 1 and nothing else. Passed the challenge without a problem.

If you also changed the same things as I did, your code should look exactly like this:

import random
def even_odd(num):
    # If % 2 is 0, the number is even.
    # Since 0 is falsey, we have to invert it with not.
    return not num % 2
start = 5
while start:
    numb = random.randint(1, 99)
    if numb % 2 == 0:
        print("{} is even".format(numb))
    else:
        print("{} is odd".format(numb))
    start -= 1
Ian Cole
Ian Cole
454 Points
Ian Cole
Ian Cole
454 Points

Agh, I had an already existing typo made after I had put in the code for the discussion. Thanks a ton for your help!

Kip Yin
Kip Yin
4,847 Points

No problem! You can click "Best Answer" to mark this question as solved.

Gerald Wells
Gerald Wells
12,763 Points
Gerald Wells
Gerald Wells
12,763 Points

True and False are actually have a base type of Int, here is a little trick to exploit that and reduce your overall lines of code.

start = 5

while start > 0:
    rand = random.randint(1,99)
    print("{} is {}".format(rand, (True,False)[int(even_odd(rand))]))
    start -= 1
Kip Yin
Kip Yin
4,847 Points

It's interesting that your answer is even valid since it should be ("even", "odd") instead of (True, False).

However, if (True, False) is valid, why not just

while start:
    rand = random.randint(1, 99)
    print("{} is {}".format(rand, even_odd(rand))
    start -= 1

Amazingly, this also works:

while start:
    print("{0} is {0}".format(random.randint(1, 99)))
    start -= 1

And this:

while start:
    print("{} is".format(random.randint(1, 99)))
    start -= 1

This might be the shortest code possible for this problem.

Gerald Wells
Gerald Wells
12,763 Points

True and False base values are 1, 0 respectivley. So by looking for a not true value those values are inversed and being that the predefined values for odd and even now map to the interger values as defined in the function its logical that my solution is valid.