fibonacci

I wrote the program Fibonacci number calculation in compile time (constexpr) problem using the template metaprogramming techniques supported in C++11. The purpose of this is to calculate the difference in the run-time between the template metaprogramming approach and the old conventional approach. // Template Metaprograming Approach template<int N> constexpr int fibonacci() {return fibonacci<N-1>() + fibonacci<N-2>(); } template<> constexpr int fibonacci<1>() { return 1; } template<> constexpr int fibonacci<0>() { return 0; } // Conventional Approach int fibonacci(int N) { if ( N == 0 ) return

I'm trying to recall an algorithm on Fibonacci recursion. The following: public int fibonacci(int n) { if(n == 0) return 0; else if(n == 1) return 1; else return fibonacci(n - 1) + fibonacci(n - 2); } is not what I'm looking for because it's greedy. This will grow exponentially (just look at Java recursive Fibonacci sequence - the bigger the initial argument the more useless calls will be made). There is probably something like a "cyclic argument shift", where calling previous Fibonacci value will retrieve value instead of calculating it again. maybe like this: int fib(int term, int val = 1,

I'm having a hard time understanding why #include <iostream> using namespace std; int fib(int x) { if (x == 1) { return 1; } else { return fib(x-1)+fib(x-2); } } int main() { cout << fib(5) << endl; } results in a segmentation fault. Once x gets down to 1 shouldn't it eventually return? 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... The reason is because Fibonacci sequence starts with two known entities, 0 and 1. Your code only checks for one of them (being one). Change your code to int fib(int x) { if (x == 0) return 0;

I know there is nothing wrong with writing with proper function structure, but I would like to know how can I find nth fibonacci number with most Pythonic way with a one-line. I wrote that code, but It didn't seem to me best way: >>> fib=lambda n:reduce(lambda x,y:(x[0]+x[1],x[0]),[(1,1)]*(n-2))[0] >>> fib(8) 13 How could it be better and simplier? 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)

Any ideas why it works fine for values like 0, 1, 2, 3, 4... and seg faults for values like >15? #include #include #include void *fib(void *fibToFind); main(){ pthread_t mainthread; long fibToFind = 15; long finalFib; pthread_create(&mainthread,NULL,fib,(void*) fibToFind); pthread_join(mainthread,(void*)&finalFib); printf("The number is: %d\n",finalFib); } void *fib(void *fibToFind){ long retval; long newFibToFind = ((long)fibToFind); long returnMinusOne; long returnMinustwo; pthread_t minusone; pthread_t minustwo; if(newFibToFind == 0 || newFibToFind == 1) return newFibToFind; else{ long

So, we see a lot of fibonacci questions. I, personally, hate them. A lot. More than a lot. I thought it'd be neat if maybe we could make it impossible for anyone to ever use it as an interview question again. Let's see how close to O(1) we can get fibonacci. Here's my kick off, pretty much crib'd from Wikipedia, with of course plenty of headroom. Importantly, this solution will detonate for any particularly large fib, and it contains a relatively naive use of the power function, which places it at O(log(n)) at worst, if your libraries aren't good. I suspect we can get rid of the power function

The following takes about 30 seconds to run whereas I would expect it to be nearly instant. Is there a problem with my code? x <- fibonacci(35); fibonacci <- function(seq) { if (seq == 1) return(1); if (seq == 2) return(2); return (fibonacci(seq - 1) + fibonacci(seq - 2)); } Patrick Burns gives an example in R Inferno of one way to do memoization in R with local() and <<- . In fact, it's a fibonacci: fibonacci <- local({ memo <- c(1, 1, rep(NA, 100)) f <- function(x) { if(x == 0) return(0) if(x < 0) return(NA) if(x > length(memo)) stop("’x’ too big for implementation") if(!is.na(memo[x]))

This question already has an answer here: What is the fastest way to write Fibonacci function in Scala? 8 answers def fibSeq(n: Int): List[Int] = { var ret = scala.collection.mutable.ListBuffer[Int](1, 2) while (ret(ret.length - 1) < n) { val temp = ret(ret.length - 1) + ret(ret.length - 2) if (temp >= n) { return ret.toList } ret += temp } ret.toList } So the above is my code to generate a Fibonacci sequence using Scala to a value n . I am wondering if there is a more elegant way to do this in Scala? There are many ways to define the Fibonacci sequence, but my favorite is this one: val fibs