Project Euler Problem 2 and Fibonacci numbers

We can't talk about Fibonaacci numbers in a programming blog without at least touching on the idea of recursion. Of course, it's not a terribly efficient way to calculate Fibonacci numbers, it still warrants attention. That said, what is a Fibonacci number? Well, more accurately, it's the Fibonacci sequence and it starts with [0, 1]. So, we'll notate that by saying F₀ = 0 and F₁ = 1. Each subsequent number in the series is the sum of the previous two numbers.

F₂ = F₁ + F₀

1 = 0 + 1

And more generally:

F_n = F_{n - 1} + F_{n - 2}

The first 10 numbers in the series looks like this:

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34]

With that overly simple explanation, on to question #2!

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Notice that the Project Euler question starts with 1, 2. That's fine, it isn't going to affect us here. The first thing we'll want to do is calculate Fibonacci numbers, I'm almost certain of that. Since the sequence is defined as a self referential sum of the previous 2 numbers, we could create a function like this:

def fib(n):
    """Recursive Fibonacci function.  
    Return the nth Fibonacci number.
    """
    if n == 0:
        return 0
    elif n == 1:
        return 1
    else:
        return(fib(n - 1) + fib(n - 2))

Now, you can call that function with any number and get the nth Fibonacci number. Sadly, this is terribly slow, and it's not really what we need to solve the problem. I just put it here to show you an example of recursion. If you don't know what recursion is, then you don't know what recursion is. So, hopefully, now you know. Maybe we really want a generator function, that will just continuously yield subsequent fibonacci numbers. Something like this:

def fib_gen():
    """Fibonacci generator function.  
    Just keep yielding Fibonacci numbers.
    """
    a, b = 0, 1
    while 1:
        yield a
        a, b = b, a + b

Or maybe we can just brute force the whole thing? This is how I actually solved it without really thinking too hard at all:

second_previous = 0
previous = 1
fib_num = 0
total = 0

while fib_num < 4000000:
    fib_num = previous + second_previous
    if fib_num < 4000000:
        if fib_num % 2 == 0:
            total = total + fib_num
        second_previous = previous
        previous = fib_num
print(total)

It prints out the total that problem 2 is asking for. That solution runs very fast for me. There is one other interesting property, and that is that every third Fibonacci number is even.