fibonacci

I am really confused. I am trying to calculate Fibonacci numbers, but as they get larger and large the numbers start to become wrong. and I do not know why. How do you calculate accurate Fibonacci numbers using Binet's Formula, it is my understanding that this should always return a integer? Here is what I have been trying to work with. http://ideone.com/e6t6h See as the number goes up. it gets all weird? here I print it out with a cout.precision(15); http://ideone.com/XREh2 here I print it out with cout << fixed << blah blah; Here I have used a procedural loop to calculate it by going though

It is possible to improve "raw" Fibonacci recursive procedure Fib[n_] := If[n < 2, n, Fib[n - 1] + Fib[n - 2]] with Fib[n_] := Fib[n] = If[n < 2, n, Fib[n - 1] + Fib[n - 2]] in Wolfram Mathematica. First version will suffer from exponential explosion while second one will not since Mathematica will see repeating function calls in expression and memoize (reuse) them. Is it possible to do the same in OCaml? How to improve let rec fib n = if n<2 then n else fib (n-1) + fib (n-2);; in the same manner? You pretty much do what the mathematica version does but manually: let rec fib = let cache =

I have a problem about making a fibonacci sequence to a list, I'm just new to python someone help me please. This is my code. I know this is looking wrong or something because it says invalid syntax. I don't know what to do about this really :( This code works for a normal code without using a list! myArray1 = [0] myArray2 = [1] while myArray2 < 700: myArray1, myArray2 = b[i], myArray1+myArray2[i] print(myArray2) LeartS This code puts the first 700 fibonacci numbers in a list. Using meaningful variable names helps improve readability! fibonacci_numbers = [0, 1] for i in range(2,700): fibonacci

I'm doing a task in a subject were fib(0) is defined to = 1. But that can't be right? fib(0) is 0? Program with fib(0) = 1; spits out fib(4) = 5 Program with fib(0) = 0; spits out fib(3) = 3 What is the correct definition? Noldorin You're correct. The Fibonacci sequence is defined with seed values fib(0) = 0 and fib(1) = 1 . This is a requirement for the rest of the sequence to be correct. The only condition under which fib(0) = 1 could work is if you defined a "-1 based counting system" (as opposed to the usual conventions of 0-based and 1-based). This would be pretty wacky however, I'm sure

I'm having trouble dealing with stacks recursively in MIPS. I get the concept, but my program isn't reacting as I mean it to. My goal is to take user input as n and print the Fibonacci number at n. What I have so far is below. (I'm fairly certain the problem is in the actual calculation of the number in the fib function.) Thanks for any help! :) .text main: # Prompt user to input non-negative number la $a0,prompt li $v0,4 syscall li $v0,5 syscall move $t2,$v0 # Call function to get fibonnacci #n move $a0,$t2 move $v0,$t2 jal fib move $t3,$v0 # Output message and n la $a0,result li $v0,4

The term fib(N,F) is true when F is the N th Fibonacci number. The following Prolog code is generally working for me: :-use_module(library(clpfd)). fib(0,0). fib(1,1). fib(N,F) :- N #> 1, N #=< F + 1, F #>= N - 1, F #> 0, N1 #= N - 1, N2 #= N - 2, F1 #=< F, F2 #=< F, F #= F1 + F2, fib(N1,F1), fib(N2,F2). When executing this query (in SICStus Prolog), the first (and correct) match is found for N (rather instantly): | ?- fib(X,377). X = 14 ? When proceeding (by entering ";") to see if there are any further matches (which is impossible by definition), it takes an enormous amount of time (compared

I have a homework assignment that I'm stumped on. I'm trying to write a program that outputs the fibonacci sequence up the nth number. Here's what I have so far: def fib(): n = int(input("Please Enter a number: ")) if n == 1: return(1) elif n == 0: return(0) else: return (n-1) + (n-2) mylist = range[0:n] print(mylist) I'm thinking I could use separate functions but I can't figure out how to pass the argument that calculates the fibonacci sequence. Then the next step would be to to print out the sequence of numbers up to that number. J0HN Non-recursive solution def fib(n): cur = 1 old = 1 i = 1

I have a function that tells me the nth number in a Fibonacci sequence. The problem is it becomes very slow when trying to find larger numbers in the Fibonacci sequence does anyone know how I can fix this? function f = rtfib(n) if (n==1) f= 1; elseif (n == 2) f = 2; else f =rtfib(n-1) + rtfib(n-2); end The Results, tic; rtfib(20), toc ans = 10946 Elapsed time is 0.134947 seconds. tic; rtfib(30), toc ans = 1346269 Elapsed time is 16.6724 seconds. I can't even get a value after 5 mins doing rtfib(100) PS: I'm using octave 3.8.1 If time is important (not programming techniques): function f = fib