An iterative algorithm for Fibonacci numbers

You are returning a value within a loop, so the function is exiting before the value of y ever gets to be any more than 1.

If I may suggest something shorter, and much more pythonful:

def fibs(n):                                                                                                 
    fibs = [0, 1, 1]                                                                                           
    for f in range(2, n):                                                                                      
        fibs.append(fibs[-1] + fibs[-2])                                                                         
    return fibs[n]

This will do exactly the same thing as your algorithm, but instead of creating three temporary variables, it just adds them into a list, and returns the nth fibonacci number by index.

The problem is that your return y is within the loop of your function. So after the first iteration, it will already stop and return the first value: 1. Except when n is 0, in which case the function is made to return 0 itself, and in case n is 1, when the for loop will not iterate even once, and no return is being execute (hence the None return value).

To fix this, just move the return y outside of the loop.

Alternative implementation

Following KebertX’s example, here is a solution I would personally make in Python. Of course, if you were to process many Fibonacci values, you might even want to combine those two solutions and create a cache for the numbers.

def f(n):
    a, b = 0, 1
    for i in range(0, n):
        a, b = b, a + b
    return a