Generator for the collatz conjecture sequence

Question

I tried to write this code as concisely as possible. Is this the best way to do it?

def collatz(n):
 """
 Generator for collatz sequence beginning with n
 >>> list(collatz(10))
 [5, 16, 8, 4, 2, 1]
 """
 while n != 1:
 n = n / 2 if n % 2 == 0 else 3*n + 1
 yield int(n)

Answer 1

The only improvement I see here is to divide n by 2 using // (since we are dealing with Python 3.x) and to remove the explicit conversion to int (int(n)):

while n != 1:
 n = n // 2 if n % 2 == 0 else 3*n + 1
 yield n

Also, I suggest you put a single space before and after the multiplication operator in 3*n, so that it becomes 3 * n.

Hope that helps.

Answer 2

I would use while n > 1, just in case someone tries to call this function with n=0 or even a negative n.

• Dair: Along those lines OP should probably mention in the docstring that n is a positive integer.

Answer 3

I expect collatz(1) to produce 1. Rather, it generates an empty sequence. So, in my opinion, the function should immediately yield n before the loop.

• Vermillion: You're right, that was an oversight