If that observation is correct (and it *seems* it is…), then another observation – the full SUM of N terms = N+2 – 1, then…, all that was left to do is find the last term below 4 mil (N = 34) and apply the solution function to find (N+2 – 1)/2 => (F(35) – 1)/2, So if correct that means that the only issue to solve is (efficiently) finding N = 33, rather than having to see the full fib sequence to *know* that… which shouldn’t be too hard if I didn’t feel like putting my feet up already. Extended to solve all test cases for Project Euler Problem 2.

Better way of calculating Project Euler #2 (Fibonacci sequence) 10. 8 (second even fibonacci number) = 2 * 1(fib) + 3 * 2(fib)

Sign up for the Mathblog newsletter, and get updates every two weeks. Reference: The On-Line Encyclopedia of Integer Sequences (OEIS) A000045. I think I found a little faster method: Therefore, sum(F(3i)) from i = 1 to n/3 = (F(n + 2) – 1)/2. I am starting with the calculation of F6 which means I need to initialize Fn-3 = F3=2 and Fn-6= F0= 0. Source: This function calls itself until it reaches the first two numbers. ( 832040 • 4 ) + 196418 = 3524578. This can be solved in O(log2(n)), assuming the function you use to calculate exponents has that time complexity. 2. They way you constructed the solution, you’re limiting the sum of those numbers to be under 4 million. # the three numbers of the sequence in a row will be a1, a2 and a3. Interestingly, the GMP library uses tables for Fibonacci numbers up to twelve quintillion (1018). The result of running the code on my machine, On my computer it went so fast that it wasn’t really possible to time it, but then again, we need no more than F34 to exceed 4,000,000.

As indicated in the article, the sum of all even Fibonacci numbers is 4613732. long sec=1; We have a closed form for fibonacci numbers, using the golden ratio, known as Binet's Formula: \[F_n = \frac{\Phi^n - \Psi^n}{\sqrt{5}}, \Phi = \frac{1+\sqrt{5}}{2}, \Psi = \frac{1-\sqrt{5}}{2}\]. It is beacuse you should print out the “summed” variable. Fibonacci numbers have the following generating rule: \[F_n=F_{n-1}+F_{n-2} \text{ with } F_0=1, F_1=2\]. I can’t quite understand why these numbers would be different given the exact same code.

var total = 0; The problem description of Problem 2 of Project Euler reads. If the calculations were not stored in longs, but rather large arrays, saving one of them would have been beneficial, and a method I would consider for use in high performance computing. return 1; int i = 6; As the Fibonacci sequence begins with two odd numbers and each subsequent Fibonacci number is the sum of the previous two, then the following term will be even. , where reliability below 70 is problematic, A mathematical approach (+ some Scala): Examples : Therefore (sorry, I don’t know C++, it’s python): I think a simpler approach would be to use the golden ratio as the driver and noting that every 3rd fib number is even. The solution to problem 2 of Project Euler: Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million. First of all, the fibonnaci series actually starts at 1 and the second number is also 1. ( 2 • 4 ) + 0 = 8 And if you're still not convinced, here’s a proof: We solve this problem by defining a new generalized Fibonacci sequence for even-valued Fibonacci numbers starting from zero as: {0, 2, 8, 34, 144, 610, … }. For the curious, my spreadsheet solution (with cell ‘pseudo variables) was simply: n=ROUND(LN(threshold * SQRT(5))/LN((1 + SQRT(5)) / 2)), answer = (((POWER((1+SQRT(5))/2,n+2) – POWER((1-SQRT(5))/2,n+2)) / SQRT(5)) – 1)/2, NB: … If it was, you would have got the wrong total…You should put (fib1+fib2<4000000) instead of (result<4000000). 2) By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. No real problem, with such a few calculations. A much faster method is to create a vector of Fibonacci numbers and add the even ones. var max = 4000000; The code then sums the even numbers in the sequence. 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.

Those links are … Updated | 16 July 2020 }. } Correction: Sorry, one line is missing in my previous comment. Maybe that can help you to resolve the problem faster, or not, I’m not a programmer by the way. Fibonacci odd numbers below 4000000

Then, we have to add 2 to 4613730. 4Fn-3 + Fn-6 (since Fn-4 + Fn-5 = Fn-3). This is cause quite a few writes. Problem 2. could you please help me with it?

(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL) Using brute force, we can implement the solution shown on this flowchart. Recursive functions are, however, very memory hungry and not very fast. const double golden_ratio = 1.618034; You can also save one long variable in this way. Project Euler #2 in Python. The first program, shown above, takes about 14 times longer to execute. The intuitive way to calculate Fibonacci numbers is to use a recursive function. Usually the first two numbers in the Fibonacci sequence is defined as F1 = F2 = 1. No memory requirements nor cpu intensive. This page lists all of my Project Euler solution code, along with other helpful information like bench­mark timings and my overall thoughts on the nature of math and programming in Project Euler. The Fibonacci sequence is formally defined by the recurrence relation: and generates the sequence: {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, …}. … temp = fib + fib2 Then I loop, until the calculated number exceeds 4,000,000. 430 words | This site aims to provide complete and accurate solution listings for Project Euler. And the result is then already 2 since Fn-3 is already calculated. The spiral staircase uses Fibonacci numbers as part of its geometry. The purpose of the problem is to sum the even-valued terms of the Fibonacci Sequence, as they are highlighted above, to an upper bound. I promise I will include cool tidbits for you.

Next » solution Project Euler Problem 3: Largest prime factor There are likely a multitude of other ways, to speed up the calculation, so feel free to ask questions or comment on the post. Fibonacci odd numbers (cumulative values) Before heading on with a solution, I will make a small comment on the problem formulation. I solve Project Euler problemsto practice and extend my math and program­ming skills, all while having fun at the same time. Solution. Let’s not forget, nevertheless, that we do not have a true, legitimate Fibonacci sequence in this problem. What is the purpose of this site? This package is also able to work with huge numbers. O good god.. no i got it thank you ….

This function calculates the \(n^{th}\) Fibonacci number by calling itself with a reduced index until it reaches 1. Solution ( 34 • 4 ) + 8 = 144 4613730. I will be working through it and honing my C# skills along the way, Thanks again.

So through the counter variable which runs from 0 to 1, I can change the variable I am writing to. if(stack[i] % 2 == 0) { © 2020 mathblog.dk.

Hackerrank describes this problem as easy. The code then sums the even numbers in the sequence. Previous « solution Project Euler Problem 1: Multiples of 3 and 5, Project Euler Problem 2: Even Fibonacci numbers Python source, Run Project Euler Problem 2 using Python on repl.it. This number sequence seems to describe our sense of natural beauty and aesthetics. >>> fib2 = 2

This can be proven through induction. # let's consider this special type of Fibonacci sequence that starts with 1,2,3,5,... # here we start the sum with the first even term, 2. 4181, 6765 (14328); 17711, 28657 (60696); 75025, 121393 (257114); sum += fib; break; This page presents solutions to Project Euler Problem 2 in Clojure, Go, Haskell, JavaScript, Ruby and Rust. In C++, you can write shortly i = !i; instead of i = (i + 1) % 2; and fib[!i] instead of fib[(i + 1) % 2] .. C++ FTW , function fib(stack, n) { … fib2 = temp For a thorough exposition of solutions, I recommend Project Nayuki , which solves about 200 of the problems using Java, Python, Mathematica, and Haskell. We could generate only the even numbers as follows: \[\begin{array}{rl} F_{n}&= F_{n-1} + F_{n-2}\\&= F_{n-2} + F_{n-3} + F_{n-3} +F_{n-4}\\&= F_{n-3} + F_{n-4} + F_{n-3} +F_{n-3} +F_{n-4}\\&= 3F_{n-3} + 2F_{n-4 }\\&= 3F_{n-3} + F_{n-4} + F_{n-5} + F_{n-6}\\&= 4F_{n-3} + F_{n-6}\\\end{array}\]. That’s fair. Each problem that I solved always includes a Java program. Fn-3 + Fn-4 + Fn-3 +Fn-3 +Fn-4 = 3Fn-3 + 2Fn-4 = At the bottom of the loop we do a bit of moving around to keep the fib1 and fib2 variables updated. … temp = fib2 A different, more elegant approach would be to realize that, in a Fibonacci sequence, the even and odd numbers appear on the following fasion: odd, odd, even, odd, odd, even, odd, odd, even,…. The problem reads. Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.You will probably stumble upon better solutions when searching on your own.Maybe not all linked resources produce the correct result and/or exceed time/memory limits. 8 December 2016 46368 (60696); 196418 (257114); 832040 (1089154); 3524578 (4613732).

The naive way to implement the task is looping through all fibonacci numbers this way and adding the values, if they're even: As fibonacci numbers grow quite fast, it takes only 32 iterations to get to the solution. Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1. Really it saves so much of the memory. I doubled checked my code and couldn’t find any errors, so I commented it out and copied yours, still get 5702887. Made by Bruno Silva. Project Euler Problem 3 Statement. Let x = sum(F(3i)) from i = 1 to n/3 and let y = sum(Fi) from i = 1 to n. We want to solve x. x = y – (F1 + F2 + F4 + F5 + …) = y – (F3 + F6 + …) = y – x = y/2. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …. https://en.wikipedia.org/wiki/Fibonacci_number. long fib = 0; A website with the solutions to a few of the Project Euler's problems. By starting with 1 and 2, the first 10 terms will be: Find the sum of all the even-valued terms in the sequence which do not exceed four million. Project Euler 2 looks at Fibonacci numbers.

Note, that this n might not be for an even Fibonacci number. while (n<40000){ 12. I covered the brute force, a more clever brute force method, and a bit about lowering the amount of house keeping and lowering the footprint. … total += temp Euler Problem 2 is a bit less poetic as it only asks to generate and sum even numbers. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … Find the sum of all the even-valued terms in the sequence which do not exceed four million. 6.

… fib2 = temp

First note that any Fibonacci number Fn can be calculated using the formula: Fn = (a^n – B^n)/sqrt(5) where a = (1 + sqrt(5))/2 and B = (1 – sqrt(5))/2, Since 0 < |B| < 1, then 0 < |B^n| < 1, so, |Fn – a^n/sqrt(5)| = |B^n/sqrt(5)| < 1/sqrt(5) < 1/sqrt(4) = 1/2, Thus, F(n + 1) > a^(n + 1)/sqrt(5) – 1/2 > N >= Fn > a^n/sqrt(5) – 1/2, Solve for n and you get: n + 1 > ln((N + 1/2) * sqrt(5))/ln(a) > n. Thus we can find the n such that F(n + 1) > N >= Fn by solving floor(ln((N + 1/2) * sqrt(5))/ln(a)). Peter Prevos |

… As we floor the result, it works for arbitrary numbers and we get the closest fibonacci number below that limit, in our case four million. In this case, the solution makes almost no difference. Each new term in the Fibonacci sequence is generated by adding the previous two terms. fib = static_cast((::pow(golden_ratio, i) - ::pow(1.0 - golden_ratio, i)) / sqrt5); (And assuming I didn’t make any mistakes…). Let’s check which solution has the best performance: It seems like both solutions have the same performance. 1) Fibonacci number