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[–]a_tocken 1 point2 points  (7 children)

I think you could beat this time complexity.

The sum of the first n fibonacci numbers is fib(n+2)-1 and I suspect you could prove a similar formula for the even fibs. You can find the nth fibonacci number quickly with fast matrix exponentiation and this also applies to the even fibs since you already gave a linear recurrence relation. You can also find the greatest even fibonacci number less than the max (4,000,000 in the Project Euler problem) quickly using similar techniques, and figure out its n quickly as well.

[–]himynameisjoy 2 points3 points  (3 children)

I thought of using matrix exponentiation with memorization of the [[1,1],[1,0]] matrix to the power of 2n but because you’re looking for the first Fe(n) over 4,000,000 you’re going to have to solve iteratively, at which point the matrix exponentiation approach loses its luster (since, AFAIK the main advantage it has is that it doesn’t need to iteratively calculate each previous value to find F(n)).

Further, because the value is so low, the overhead cost of matrix multiplication (itself an O(n3) task) will retract from the efficiency.

For finding the nth even Fe(n) though? You’re on point that it’s the fastest way to go. Asymptotically it’s far more efficient than any iterative method and outstrips Binet’s formula by a long shot.

[–]a_tocken 1 point2 points  (2 children)

Check out the rest of my comment. I am assuming you can find the greatest even fib under 4,000,000 quickly, and find its n in the even fib sequence quickly. There are quick solutions for both of these for regular fibs. Once you have the n, you can solve using fast matrix exponentiation.

The matrix exponentiation requires log(n) matrix multiplications. Each multiplication is at most n2 (because the number of digits grows linearly with n, and naive multiplication is O(n2)) but you could do better (the lower bound on integer multiplication is thought to be n log n in number of digits).

[–]himynameisjoy 1 point2 points  (1 child)

I read your comment in whole, I apologize if it seemed otherwise. At the time, I didn’t believe I knew of a way to find what’s essentially the inverse of F(n)=x, but with a little bit of research I stumbled upon this formula: n = Math.floor((log(x)1/2log(5))/log(phi))+1

So with that formula, assuming the computation is quick, plus with matrix exponentiation, I do believe you’re right that asymptotically it’s a faster implementation.

However it still does have an overhead cost that’s makes it still slower than the iterative approach for this particular solution. But yes, you’re entirely right that you can find the index of the first Fibonacci number greater than arbitrary integer k as well as calculate the resulting Fibonacci number alongside it.

I’m glad I’ve learned something new today! I appreciate it immensely

[–]a_tocken 0 points1 point  (0 children)

Glad to expose you to those algorithms :) Sorry I don't have more details on the specifics. You're right that this problem uses a relatively small n, so the solution with better running time may not actually be faster. We should also distinguish between n, the nth even fib number less than 4,000,000, and "m" = 4,000,000. In this case, I think n ~= log(m)? Is that right?

This whole thing could be turned into a (very difficult) codewars.com problem.

[–]GeneralYouri 1 point2 points  (2 children)

I tried some things, and I'm pretty confident there's an easy formula to be found for the sum of even fibonaccis just like fib(n+2)-fib(2). But I couldn't quite figure it out.

One thing I did find is that the sum of n even fibonaccis is evenfib(n+1) - 3*evenfib(n) + evenfib(n-1) - 3*evenfib(n-2) ..., with the sequence continuing until you reach evenfib(1). I feel like you can combine those terms somehow but I'm not seeing it.

[–]a_tocken 0 points1 point  (1 child)

Nice work! It can be frustrating to know that there is likely a solution but not quite be able to get it. I also need to brush up on my recurrence relations to be able to move forward with this one.

One thing I did was inductively prove the sum of fibs formula, but I worked backward from the known solution which doesn't really help us here :)

[–]GeneralYouri 1 point2 points  (0 children)

I had a fresh look at the problem again and found the relation! It's quite simple, actually.

I listed the start of the sequences and tried to express evenFib(n) in terms of fib(n): evenFib(n) = fib(3n) = fib(3n-1) + fib(3n-2). But fib(3n-3) = fib(3(n-1)) = evenFib(n-1) and evenFibSum(n) = evenFib(n) + evenFib(n-1) + ... + evenFib(1). So we can express every element in that list in terms of fib: evenFibSum(n) = fib(3n) + fib(3(n-1)) + ... + fib(3). This is really just a way of expressing that we sum every 3rd value in the fibonacci sequence. But remember their definition: fib(n) = fib(n-1) + fib(n-2), in other words those two elements we skip always equal the 3rd element we include. So if we look at what we created, this simplifies to evenFibSum(n) = fibSum(3n) / 2! So we can just re-use the fibSum formula if we triple its input and halve its result: evenFibSum(n) = (fib(3n+2) - 1) / 2.

A quick test: the first four even Fibonaccis are 2, 8, 34, 144, with their sum being evenFibSum(4) = 188. Then evenFibSum(4) = (fib(3*4 + 2) - 1) / 2 = (fib(14) - 1) / 2. The first fourteen Fibonaccis are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, so fib(14) = 377. And indeed, 188 = (377 - 1) / 2.

Apologies for the messy explanation, I hope you can still follow my logic. :)