Automate The Boring Stuff With Python - Collatz Sequence

user_13538552 0
Viewed 686

I'm very new to any sort of coding, currently using python 3.3. I've managed to run the Collatz Sequence accurately in python with the following:

while True: # The main game loop.
number = int(input('Enter number:\n'))  
def collatz(number):
    while number !=1:
        if number % 2==0: #even numbers
            number=number//2
            print(number)
        elif number % 2!=0: #odd numbers
            number=number*3+1
            print(number)
collatz(number)

However, I'm unsure of how and where to add a ValueError strong, for when the user enters a non-integer, something like the following:

except ValueError:
    print('Only integers accepted.')

I'm very new to python, so if any answers could have a little explanation I'd be very appreciative. Thanks

collatz python
5 Answers

Put it at the very very top. Parameter constraints should always happen as soon as possible, so that you don't waste time running code you're just going to error out of.

def progress(percentage):
    if percentage < 0 or percentage > 100:
        raise ValueError
    #  logic
user_100016880

I assumed that you're referring to Exception Handling, Validation part must be done in the beginning.

    while True: # The main game loop.
        
        try:
            number = int(input('Enter number:\n'))
   
        except ValueError:
            print("Only integers accepted! Please try again ...")
        else:
            collatz(number)
#output:
#
#Enter number:
#abc
#Only integers accepted! Please try again ...
#Enter number:
#5
#16
#8
#4
#2
#1
#Enter number:

But program will continue looping, termination conditions needed.

user_83880570 
number = None

while number != int():
    try:
        number = int(input("Enter number: "))
        break
    except:
        print("Enter a valid Number")

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

collatz(number)
user_144174730 
def collatz():

    try:
        number = int(input("Enter number: "))

        while True:
            if number == 1 or number == 0:
                break

            elif number % 2 == 0:
                number = number // 2
                print(number)

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

    except:
        print("Enter in a valid number")


collatz()
user_163615240

Try this:

def collatz(number):
    if number % 2 == 0:
        print(number // 2)
        return number // 2
    else:
        print(3 * number + 1)
        return 3 * number + 1

while True:
    try:
        number = int(input("Enter a Number or type 0 to exit: "))
        if number == 0:
            break
        while True:
            if number != 1:
                number = collatz(number)
            else:
                break
    except ValueError:
        print("Enter an integer number")
user_163282030
Related