Challenge 3: A Fibonacci Generator — Possible Solution
====================================================================

def fibonacci(n):
    a, b = 0, 1
    count = 0
    while count < n:
        yield a
        a, b = b, a + b
        count += 1

for num in fibonacci(8):
    print(num)

Output:
0
1
1
2
3
5
8
13

WHY THIS WORKS AS AN ANSWER
------------------------------
def fibonacci(n): containing a yield statement reuses this chapter's
own count_up_to(n) structure exactly — the presence of yield anywhere
in the function body is what makes calling fibonacci(8) return a
generator object rather than immediately running the whole function.

a, b = 0, 1 reuses tuple-unpacking assignment from Course 1 to track
the two numbers needed to compute the next Fibonacci value. yield a
pauses the function and hands back the current value, exactly like
count_up_to's yield i — and just like that example, the function's
local state (a, b, and count) is preserved automatically between
calls, which is the entire reason a generator can produce a running
sequence like this at all; a plain variable inside a regular function
would reset every time the function was called again.

a, b = b, a + b reuses simultaneous tuple assignment to advance both
values in one step: b becomes the new a, and a + b becomes the new b
— the standard Fibonacci recurrence, computed without needing a
temporary variable.

The for loop over fibonacci(8) reuses the chapter's own for num in
count_up_to(5): pattern — each iteration resumes the generator right
after its last yield, continuing until count reaches n (8), at which
point the while loop's condition becomes false, the function reaches
its natural end, and the generator raises StopIteration internally —
which the for loop handles automatically, ending the loop cleanly.