Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1.
Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1.
def collatz(number):
if number % 2 == 0:
print(number//2)
else:
result = 3*number + 1
print(result)
try:
n = input("Give me a number: ")
while n != 1:
n = collatz(int(n))
except ValueError:
print('Type a number please!')message:
Traceback (most recent call last):
File "C:\Python36\kodovi\coll.py", line 10, in <module>
n = collatz(int(n))
TypeError: int() argument must be a string, a bytes-like object or a number, not 'NoneType'
Not sure why am I receiving this message.
Is this your full code? How are you running it? Are you providing any input to it? If so, please provide it.
your collatz function is not returning anything so you should change it to something like this:
def collatz(number):
if number % 2 == 0:
return number//2
else:
return 3*number + 1and then print the result inside your while loop.
Hello,
the main problem is that your function doesn't return any value, like mlieqo says. Also you have two points more to revise:
1- input function returns a 'char'. Then you are comparing in your loop in its first iteration a char with a int. '1' is not the same that 1. You solve it to cast the input.
2- You are not controlling negative values. Then, a -1, -5, -100... values are allowed (maybe "while n>1")
Maybe anything like this:
def collatz(number):
if number % 2 == 0:
result = number//2
return result
else:
result = 3*number + 1
return result
try:
n = int(input("Give me a number: "))
while n != 1:
print(n)
n = collatz(n)
except ValueError:
print('Type a number please!')... Sorry for my low English level...
(Apr-17-2018, 11:10 PM)micseydel Wrote: [ -> ]Is this your full code? How are you running it? Are you providing any input to it? If so, please provide it.
After I run the file in command prompt ( like any other code ) it asks me for a number and after I add it it gives me that error..
(Apr-18-2018, 09:29 AM)mlieqo Wrote: [ -> ]your collatz function is not returning anything so you should change it to something like this:
def collatz(number):
if number % 2 == 0:
return number//2
else:
return 3*number + 1and then print the result inside your while loop.
def collatz(number):
if number % 2 == 0:
print(number//2)
return number//2
else:
result = 3*number + 1
print(result)
return result
try:
n = input("Give me a number: ")
while n != 1:
n = collatz(int(n))
except ValueError:
print('Type a number please!')This code works fine after I add return to if/else statement.
vaison, this edit solved the issue with negative integers.
while n != 1:
n = collatz(abs(int(n)))
Hello,
I think that in your first iteration you're still comparing a char with an integer. Try to introduce '1' and verify if the execution doesn't enter to the loop. I think that you must do the cast to integer in the input() function and not in the collatz function.
try:
n = int(input("Give me a number: "))
while n != 1:
print(n)
n = collatz(abs(n))
except ValueError:
print('Type a number please!')What do you think?
Hi,
When I input 1 it gives:
4
2
1
I don't know what exactly to think :)
Hello,
Do you want that when you input 1, the loop condition doesn't acomplish?
while n != 1:
Then, if you input 1, there should not be any print at the output. Not?
I said that you are comparing a char (input()) with an integer 1. Then you can cast to int the returned value of the input() function.
My English level is not good. I explained myself well?
Aha, now I get what you mean. Yes, you are correct.
The final code:
def collatz(number):
if number % 2 == 0:
print(number//2)
return number//2
else:
result = 3*number + 1
print(result)
return result
try:
n = int(input("Give me a number: "))
while n != 1:
print(n)
n = collatz(abs(n))
except ValueError:
print('Type a number please!')
Truman, in your final code, you are now printing the results twice, first in your collatz function and then in your while loop. So now if I input f.e. number 3 output is:
Output:
3
10
10
5
5
16
16
8
8
4
4
2
2
1
You should remove print from your collatz function, and probably add one more condition for printing number 1.
def collatz(number):
if number % 2 == 0:
return number//2
else:
result = 3*number + 1
return result
try:
n = int(input("Give me a number: "))
while n != 1:
print(n)
n = collatz(abs(n))
if n == 1:
print(n)
except ValueError:
print('Type a number please!')Also one more thing to consider -> I dont think user should be able to input negative number, from what I read on wikipedia -
Quote:The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined as follows: start with any positive integer n.
Actually in this case, if statement from my comment above
if n == 1:
should be replaced with else(same result, but nicer).