Project Euler #2 - Even Fibonacci numbers

I had a blast reading through the many creative and varied approaches to solving Problem #1. There's something very fun and enlightening about seeing the same problem solved in different ways:

Project Euler #1 - Multiples of 3 and 5

Peter Kim Frank ・ Jun 11 '18

#projecteuler #challenge

Not sure if this will become a regular thing, but I wanted to keep this rolling and present Problem #2. Again, this is from the wonderful Project Euler.

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.

Look forward to seeing solutions in your language of choice!

Comments

[Massimo Artizzu]

Math to the rescue again!
To compute the n-th Fibonacci number, you can use the Binet formula:

where φ is the Golden Ratio, (1 + √5)/2. Also, it's given that f(0) = 0 and f(1) = 1, instead of having 1 and 2 as the first two Fibonacci's numbers.

In order to know which Fibonacci number is the highest below 4 million, we can solve by n... which isn't really easy. So, instead, we get to obtain a (pretty good) estimate by ignoring the second term (-1/φ)^n, which quickly approaches to 0, and solve:

In fact, f(33) = 3524578 and f(34) = 5702887.
Now, it's time to sum. First of all, let's notice that the even terms in the Fibonacci sequence happen once every three: f(0) = 0, f(3) = 2, f(6) = 8, f(9) = 34 and so on. So, we just have to sum every f(3 k) for k from 0 to 11. Using the known formula:

and using a = φ, h = 3 and n = 11, we have:

So, more in general, using JavaScript (for example):

const SQ5 = 5 ** .5;
const PHI = (1 + SQ5) / 2;
function allEvenFibonacciUpTo(limit) {
  const highestIndexBelowLimit = Math.floor(Math.log(limit * SQ5) / Math.log(PHI));
  // I love expressive variable names, but the formula could get too long!
  const n = Math.floor(highestIndexBelowLimit / 3);
  return ((PHI ** (3 * n + 3) - 1) / (PHI ** 3 - 1)
    - ((1 - PHI) ** (3 * n + 3) - 1) / ((1 - PHI) ** 3 - 1)) / SQ5;
}

console.log(allEvenFibonacciUpTo(4e6)); // 4613731.999999999

Well... I guess we'll have to deal with some rounding problems 😅
But another O(1) solution, yay! 🎉

(By the way, -1/φ = 1 - φ if you were confused.)

[Kushan Joshi]

Ah nice to see maths, I came up with the same thing however with a different approach of using generator functions.

Let us forget about fibonacci and think of the even fibonacci as a new series. (I probably think there would be a mathematical way of deducing it, but I just used brute force to find it out)

series = 2, 8, 34, 144, 610, 2584...
34 = 8*4 + 2,
144 = 34*4 + 8,
610 = 144*4 + 34;
Number = 4 * (Previous Number) +  Previous Previous Number

So the recurrence relation for this series would be

F_n= 4F_n-1 + F_n-2.
To find out the Sum of n elements in this series (let's call this S_n), we can rewrite the recurrence relation like this:
4F_n-1 = F_n - F_n-2
and we can move all n's by 1:
4F_n = F_n+1 - F_n-1

Now we can use the same logic to finder other n's:
4F_n = F_n+1 - F_n-1
4F_n-1 = F_n - F_n-2
4F_n-2 = F_n-1 - F_n-3
4F_n-3 = F_n-2 - F_n-3
...
...
4F₄ = F₅ - F₃
4F₃ = F₄ - F₂
4F₂ = F₃ - F₁
4F₁ = 4F₁

For folks not familiar with this technique, we are simply doing to cancel out similar terms when we add all equations. (Note all items in pair are canceled out since x - x = 0).

4F_n = F_n+1 - F_n-1
4F_n-1 = F_n - F_n-2
4F_n-2 = F_n-1 - F_n-3
4F_n-3 = F_n-2 - F_n-3
...
...
4F₄ = F₅ - F₃
4F₃ = F₄ - F₂
4F₂ = F₃ - F₁
4F₁ = 4F₁

4F_n + 4F_n-1 + 4F_n-2 + ... + 4F₂ + 4F₁ = F_n+1 + F_n - F₂ + 3F₁

If you notice the left hand side is essentially equivalent to
4S_n

Now we can finally write the relation between sum and series as:

S_n= ( F_n+1 + F_n - F₂ + 3F₁ ) / 4

The point is this avoids the for loop for summing all the values. Go ahead and try substituting values and you will find this sums it up. You will have to use F(1) = 2 and F(2) = 8.

Now that we have removed one for loop, how do we get a direct formula for getting the F_n.

This involves a bit more complicated math(for adventurous folks visit austinrochford.com/posts/2013-11-0...).
In short the closed form for this recurrence relation is

And if you compute the roots and follow along the url I posted above, you will end up with this

This gives us a function to compute F_n .

Now we can combine our summing and this formula to get a simple O(1) arithmetic function to get the sum. Side note: Running a for loop will take much less brain time :P

[Massimo Artizzu]

I applaud your solution, fellow math lover! 👏

[Nested Software]

These are really great solutions. I think it would be worthwhile for you to publish them as standalone articles. One quibble: I don't think this is O(1) since arithmetic operations are not constant time as a function of input, although I guess as long as we’re sticking with floating point numbers O(1) is probably valid.

[edA‑qa mort‑ora‑y]

You're correct that addition is O(log N) for a number N, or O(log n) where n is the number of digits. The power function x^n also has a O(log N) time complexity if N is an integer (I presume it's also linear on number of significant bits).

Given fixed sized types on a computer though I believe most of these become constant time, as the inputs have a fixed upper limit on bits. It probably involves an unrolled loop, or fixed iterations on a CPU.

[Massimo Artizzu]

Yes, indeed. I am in fact using just plain double precision numbers there, and they do take a fixed amount of ticks for common operations like those on a modern CPU.

I wouldn't be so certain if I used JavaScript's BigInt numbers (or rather, I would be certain of the opposite).

Thank you for the explanation!

[Ali Spittel]

OOhh Project Euler problems are my favorite!

"""
Even Fibonacci numbers: Project Euler Problem 2
Solution in Python

https://projecteuler.net/problem=2

Takes 0.4s on my computer
"""
def fibonacci_sequence(n):
    numbers = [0, 1]

    while numbers[-1] < n:
        numbers.append(numbers[-1] + numbers[-2])

    return numbers


def sum_even_fibonacci_numbers(n):
    # O(N) Complexity
    sequence = fibonacci_sequence(n)
    sequence = [n for n in sequence if n % 2 == 0]
    return sum(sequence)

print(sum_even_fibonacci_numbers(4000000))

[Priyansh Jain]

I tried timing your algorithm, with couple other implementations I wrote. One was recursive, one was iterative. Check this out.

# recursive
def recursive_fibonacci(num, prev_num, sum=0):
    if num % 2 == 0:
        sum += num
    if num < 4000000:
        return recursive_fibonacci(num + prev_num, num, sum)
    return sum

# iterative
def iterative_fibonacci(num, second_num):
    sum = 0
    while num < 4000000:
        if num % 2 == 0:
            sum += num
        temp = num
        num += second_num
        second_num = temp
    return sum

# Other
def fibonacci_sequence(n):
    numbers = [0, 1]

    while numbers[-1] < n:
        numbers.append(numbers[-1] + numbers[-2])

    return numbers


def sum_even_fibonacci_numbers(n):
    # O(N) Complexity
    sequence = fibonacci_sequence(n)
    sequence = [n for n in sequence if n % 2 == 0]
    return sum(sequence)

import time

start_time_recursive = time.time()
print(recursive_fibonacci(1, 1))
print("--- Recursive %s seconds ---" % (time.time() - start_time_recursive))

start_time_iterative = time.time()
print(iterative_fibonacci(1, 1))
print("--- Iterative %s seconds ---" % (time.time() - start_time_iterative))

start_time_other = time.time()
print(sum_even_fibonacci_numbers(4000000))
print("--- Other %s seconds ---" % (time.time() - start_time_other))

The output was interesting!

[Phil Nash]

I went for what I thought was an interesting solution with Ruby this time. It uses a lazy enumerable over an infinite range and calculates the sum and the fibonacci numbers until the end condition is met.

def sum_even_fibonacci_numbers_up_to(max)
  range = 1..Float::INFINITY
  sum = 0
  last_two_fibs = [0,1]

  range.lazy.take_while do
    next_fib = last_two_fibs[0] + last_two_fibs[1] 
    sum += next_fib if next_fib % 2 == 0
    last_two_fibs = [last_two_fibs[1], next_fib]
    next_fib < max
  end.force

  sum
end

sum_even_fibonacci_numbers_up_to(4_000_000)
# => 4613732

This uses constant space, so does not compute an array of fibonacci numbers, just holds the latest two and the current sum. It runs in O(n) time and for 4,000,000 took less than 0.2s on my Macbook Pro.

[Ross Henderson]

SQL

I have an answer, but I'm not 100% happy with it. It's not dynamic enough, but I need to do some more digging into Cycles it seems :)

with FIBONACCI (i, FIBO, PREV) as 
(
   select  
      1 i, 0 FIBO, cast(null as number) PREV 
   from 
      DUAL
   union all
   select 
      f1.i + 1  i, f1.FIBO + nvl(f1.PREV,1) FIBO, f1.FIBO PREV
   from 
      FIBONACCI f1
   where 
      f1.i < 35
)

select 
   sum(FIBO) as "Answer"
from 
   FIBONACCI
where 
   mod(FIBO, 2) = 0 
order by i;

-----------
Answer: 4613732

[Josh Cheek]

My original solution (Ruby)

a, b, sum = 1, -1, 0
while a <= 4_000_000
  a, b = a+b, a 
  sum += a if a.even?
end
sum # => 4613732

I also thought it would be fun to try it as a lazy enumerator, though I still like my original solution better.

a, b = 1, -1
loop.lazy
    .map { a, b = a+b, a; a }
    .take_while { |n| n <= 4_000_000 }
    .select(&:even?)
    .sum # => 4613732

[Oskar Larsson]

In Javascript using filter and reduce.

var memo = [0, 1];
function fib (n) {
    if(memo.length-1 < n) {
        memo[n] = fib(n-1)+fib(n-2);
    }
    return memo[n];
}

// creates array with elements from function that takes the index as argument while a given condition holds
function takeWhile (fromFunc, cond, arr=[]) {
    var n = arr.length;
    var val = fromFunc(n);
    if(cond(val)) {
        arr.push(val);
        return takeWhile(fromFunc, cond, arr);
    }
    return arr;
}

var sum = 
    takeWhile(fib, n => n < 4000000)
    .filter(n => n%2==0)
    .reduce((acc, c) => acc+c, 0);

console.log(sum);

Haskell:

fibo a b = a:fibo b (a+b)
isEven = (== 0) . flip mod 2

fibSumEvenUnder n = sum $ filter isEven $ takeWhile (<n) $ fibo 1 2
ans = fibSumEvenUnder 4000000

[Jay]

A simple iterative Rust Solution.
Takes about 0.9 sec, and optimized one takes 0.35 sec

fn fibo(mut a: i64, mut b: i64, max: i64) -> i64 {
    let mut sum = 0;
    while a < max {
        if a % 2 == 0 {
            sum += a;
        }
        let c = a+b;
        a = b;
        b = c;
    }
    sum
}

fn main() {
    println!("{}", fibo(1,2,4000000))  // 4613732
}

[Priyansh Jain]

def fibonacci(num, prev_num, sum=0):
    print(num, prev_num, sum)
    if num % 2 == 0:
        sum += num
    if num < 4000000:
        return fibonacci(num + prev_num, num, sum)
    return sum


print(fibonacci(1, 1))

Uses recursion, can someone explain how to calculate the complexity?

[Seiei Miyagi]

Ruby✨💎✨

puts Enumerator.new { |y|
  f1, f2 = 0, 1
  f1, f2 = f1 + f2, f1 while y << f1
}.take_while {|x| x <= 4_000_000 }.select(&:even?).sum

[Stephanie Handsteiner]

Solved a few of the Euler challenges a few years ago (when I was still in high school, because what else would you do?) using PHP.

Solution: 4613732

$sum = 0;
$curr = 2;
$prev = 1;
$goal = 4000000;
while ($curr < $goal) {
if($curr % 2 == 0) {
$sum += $curr;
}
$tmp = $prev;
$prev = $curr;
$curr += $tmp;
}
echo $sum;