Why is my Recursive Fibonacci implementation written in C++ segfaulting?
When x==2 you call fib(1) and fib(0): return fib(2-1)+fib(2-2); Consider what will happen when fib(0) is evaluated…
When x==2 you call fib(1) and fib(0): return fib(2-1)+fib(2-2); Consider what will happen when fib(0) is evaluated…
To answer you first question: (defn fib ([n] (fib [0 1] n)) ([x, n] (if (< (count x) n) (fib (conj x (+ (last x) (nth x (- (count x) 2)))) n) x))) This type of function definition is called multi-arity function definition. You can learn more about it here: http://clojure.org/functional_programming As for a better … Read more
Here is a near O(1) solution for a Fibonacci sequence term. Admittedly, O(log n) depending on the system Math.pow() implementation, but it is Fibonacci w/o a visible loop, if your interviewer is looking for that. The ceil() was due to rounding precision on larger values returning .9 repeating. Example in JS: function fib (n) { … Read more
Dynamic programming Idea:Instead of recomputing the same value multiple times you just store the value calculated and use them as you go along. f(n)=f(n-1)+f(n-2) with f(0)=0,f(1)=1. So at the point when you have calculated f(n-1) you can easily calculate f(n) if you store the values of f(n) and f(n-1). Let’s take an array of Bignums … Read more
I would use this method: Python 2 a = int(raw_input(‘Give amount: ‘)) def fib(n): a, b = 0, 1 for _ in xrange(n): yield a a, b = b, a + b print list(fib(a)) Python 3 a = int(input(‘Give amount: ‘)) def fib(n): a, b = 0, 1 for _ in range(n): yield a a, … Read more
When x==2 you call fib(1) and fib(0): return fib(2-1)+fib(2-2); Consider what will happen when fib(0) is evaluated…
fib = lambda n:reduce(lambda x,n:[x[1],x[0]+x[1]], range(n),[0,1])[0] (this maintains a tuple mapped from [a,b] to [b,a+b], initialized to [0,1], iterated N times, then takes the first tuple element) >>> fib(1000) 43466557686937456435688527675040625802564660517371780402481729089536555417949051 89040387984007925516929592259308032263477520968962323987332247116164299644090653 3187938298969649928516003704476137795166849228875L (note that in this numbering, fib(0) = 0, fib(1) = 1, fib(2) = 1, fib(3) = 2, etc.) (also note: reduce is a … Read more
You have never declared fib to be an array. Use var fib = []; to solve this. Also, you’re never modifying the y variable, neither using it. The code below makes more sense, plus, it doesn’t create unused variables: var i; var fib = [0, 1]; // Initialize array! for (i = 2; i <= … Read more
The definition with Fib(0) = 1 is known as the combinatorial definition, and Fib(0) = 0 is the classical definition. Both are used in the Fibonacci Quarterly, though authors that use the combinatorial definition need to add a sentence of explanation. Benjamin and Quinn in Proofs that Really Count use f_n for the nth combinatorial … Read more
Here’s a different and simpler function that calculates the n’th Fibonacci number: fib :: Integer -> Integer fib 0 = 0 fib 1 = 1 fib n = fib (n-1) + fib (n-2) The implementation you are referring to relays on some observations about how values in Fibonacci relate to each other (and how Haskell … Read more