Python Solution for Project Euler #2 (Fibonacci Sums)

Question

I'm a fairly new programmer (just started yesterday!). I decided to tackle Project Euler #2 today, as I did #1 yesterday without many problems. I came up with what seems to me to be a working solution, but I feel like I did it in an exceedingly ugly way. Could anyone suggest improvements to my code and/or logic?

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.

fib = 1
fib2 = 2
temp = 0
total = 0

while temp <=4000000:
    temp = fib2
    if temp % 2 == 0:
        total += temp
    temp = fib + fib2
    fib = fib2
    fib2 = temp

print total

I just sort of used my temp variable as a holding ground for the results, and then juggled some variables around to make it do what I wanted... Seems like quite a mess. But it does result in 4613732, which seems to be correct!

Answer 1

I would use separately Fibonacci generator

def even_fib(limit):
    a, b = 0, 1
    while a < limit:
        if not a % 2:         
            yield a
        a, b = b, a + b

And sum function to calculate sum of items

print sum(even_fib(4000000))

Answer 2

I'm also a fellow newbie but the way I would approach it is to first create a separate function for calculating the value for a given nth term in the Fibonacci sequence. Here's my solution:

def fibonacci(n):
    if n == 1:
        return 1
    elif n == 2:
        return 1
    else:
        return fibonacci(n-2) + fibonacci(n-1)

total = 0
i = 1
while fibonacci(i) <= 4000000:
    if fibonacci(i) % 2 == 0:
        total += fibonacci(i)
    i += 1
print(total)

Answer 3

First of all, I'd like to tell you that the fibonnaci series starts at 1 and the second number is 1, because 1+0=1, so a fibonacci series starts like this: [1, 1, 2, 3, 5, 8...]

I know you can say it's kind of cheatting but anyways: Fibonacci series follows a pattern, because to get an even number out of a sum you'll need 2 even or 2 odd numbers(3+3=6, 6 is even; 3+5=8; 2+8=10). Starting at 1 (odd) it follows this pattern: [odd, odd, even, odd, odd, even...].

Cheating or not this is my solution:

def generate_fib(max_value):
    series = [1, 1]
    while True:
        next_last = series[-1] + series[-2]
        if next_last > max_value:
            break
        series.append(next_last)
    return series

series = generate_fib(4000000)
evens = [series[x] for x in range(2, len(series), 3)]
print(sum(evens))

Answer 4

@San4ez gave you a good solution, I would use the Single Responsability principle and define:

def fibs(limit):
    a, b = 0, 1
    while a < limit:
        yield a
        a, b = b, a + b

print(sum(i for i in fib(4000000) if i % 2 == 0))

so that the fibs function does not check evenness