Solution. The spiral staircase uses Fibonacci numbers as part of its geometry. 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. … if temp % 2 > 0: Project Euler - Problem 2 Problem #2. Remember that adding two odd numbers together always sums to an even number just as adding an even and odd number will sum to an odd number. Correction: Sorry, one line is missing in my previous comment., Sum of all odd Fibonacci numbers as obtained with Python (v3.6.1), >>> fib = 1 In other words, for each three numbers in the sequence, one of them is even. This will yield two odd numbers after every even because an odd plus an even will be odd. Project Euler Problem 3: Largest prime factor, Project Euler Problem 1: Multiples of 3 and 5. 317811, 514229 (1089154); 1346269, 2178309 (4613732). First of all, the fibonnaci series actually starts at 1 and the second number is also 1. Maintaining the Single Responsibility Principle with Project Euler #2. I promise I will include cool tidbits for you. Fn-3 + Fn-4 + Fn-3 +Fn-3 +Fn-4 = 3Fn-3 + 2Fn-4 = 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)). >>> total = 0 const double golden_ratio = 1.618034; 2) By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. Getting a little extra once you figured it out yourself. I doubt it is measurable, but now the method is covered, since I might want to use it again. The correct Fibonacci series is 1, 1, 2, 3, 5, 8, 13, 21 and so on. Each new term in the Fibonacci sequence is generated by adding the previous two terms. 1. ## Project Euler 2: Sum of Even Fibonacci Numbers, Creative Commons Attribution-ShareAlike 4.0 International. There are likely a multitude of other ways, to speed up the calculation, so feel free to ask questions or comment on the post. # 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. We could generate only the even numbers as follows: \[\begin{array}{rl} F_{n}&= F_{n-1} + F_{n-2}\\&= F_{n-2} + F_{n-3} + F_{n-3} +F_{n-4}\\&= F_{n-3} + F_{n-4} + F_{n-3} +F_{n-3} +F_{n-4}\\&= 3F_{n-3} + 2F_{n-4 }\\&= 3F_{n-3} + F_{n-4} + F_{n-5} + F_{n-6}\\&= 4F_{n-3} + F_{n-6}\\\end{array}\]. 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,…. What I have done here is removed one of the longs, and replaced it with an integer counter. >>> fib2 = 2 Recursive functions are, however, very memory hungry and not very fast. Note:Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.In my opinion, Hackerrank's modified problems are usually a lot harder to solve. total += stack[i] No memory requirements nor cpu intensive. Fn-2 + Fn-3 + Fn-3 +Fn-4 = (since Fn-1 = Fn-2 + Fn-3 and so on) The GMP package for Multiple Precision Arithmetic provides a function to calculate the nth Fibonacci number. The correct answer (if I read the problem correctly) should be greater than 4 million (e.g., 4,613,732). return 1; Examples : 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.

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