# Fibonacci in Every Language

Sample Programs Challenges (2 Part Series)

Once again, I thought it would be fun to kick off one of these challenges in as many programming languages as possible. Last time, we tried our hand at Fizz Buzz in Every Language. If you haven't gotten a chance to share a solution in your favorite language, go for it! Otherwise, we're going to tackle Fibonacci in Every Language today.

## The Challenge

As many of you are probably aware, Fibonacci is a term often associated with a particular sequence of numbers where the nth term is the sum of the previous two terms—or more formally: Fn = Fn-1 + Fn-2. For our purposes, we'll start the sequence with a pair of ones:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...

The goal in this challenge is to write a program in a language of your choice to output the sequence up to the nth term. In other words, the program should be able to be executed like:

./fib 5


and, output the sequence above up to the fifth term:

1, 1, 2, 3, 5


If you're feeling adventurous, you could try making your program conform to the requirements outlined in the Sample Programs repo. In addition to outputting the sequence, you should also output the index of each term:

1: 1
2: 1
3: 2
4: 3
5: 5


Either style is fine for the purposes of this challenge. That said, if you follow the second convention, it'll be much easier to transition into the repo!

## The Solution

As always, I'll kick off the fun with a solution in Haskell brought to you by a couple of our contributors (Parker Johansen and Panagiotis Georgiadis):

module Main where

import System.Environment

fibonacci :: Int -> [Integer]
fibonacci = flip (take) fibs

fibs :: [Integer]
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

headMaybe :: [a] -> Maybe a

-- Takes a list of values and returns a list of strings in the format "ONE_BASED_INDEX: VALUE"
printWithIndex :: (Show a) => [a] -> [[Char]]
printWithIndex = zipWith (\i x -> (show i) ++ ": " ++ (show x)) [1..]

-- Prints out the first N numbers from the fibonacci sequence
-- where N equals to the first command line argument.
main :: IO ()
main = do
args <- getArgs
Nothing -> putStrLn "Usage: please input the count of fibonacci numbers to output"
Just n  -> mapM_ (putStrLn) \$ (printWithIndex . fibonacci) n


For those of you that maybe haven't seen functional programming before, this solution probably looks pretty wild. That said, I think Haskell is far more readable than anything Lisp-based, so you can probably get the gist of how this works.

Unfortunately, we don't currently have this solution documented in our collection, so maybe one of you can help out. We'd appreciate the support!

Sample Programs Challenges (2 Part Series)

### Discussion Python: (constant time solution)

import math
from decimal import Decimal, getcontext

getcontext().prec = 1000
ONE = Decimal(1)
SQRT_FIVE = Decimal(5).sqrt()
HALF = Decimal(0.5)
PHI = (ONE + SQRT_FIVE) * HALF

def fibonacci(n):
if n == 0:
return 0
return math.floor((PHI ** n) / SQRT_FIVE + HALF)


>>> fib.fibonacci(1000)
43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875


Edit: This is accurate until F4767 with the given precision.

Clever! I didn't realize there was an equation that could so closely approximate the series.

It turns out also in DSP, it is a classical example of use of z-trasform for beginners. Actually, the exact form has two terms: phi above and 1/phi and the n-th term is something like C*(phin +1/phin ) (more or less, I am going by memory and I am too lazy now to do the computation... ;-)). Of course, since 1/phi is smaller than one, 1/phin goes rapidly to zero and you get a wonderful approximation with only phin already for small n.

I only know it because my Theory of Computability professor used the “one trillionth Fibonacci number” as a limit to our newfound power understanding recursive solutions. (To find the Fn, you need to find Fn-1 + Fn-2. To find those two numbers, you need to find Fn-2 + 2*Fn-3+Fn-4. This gets towards heat death of the universe level amount of instructions fairly quickly. The loop with saved state gets you down to hours IIRC.)

Then he dropped that this equation existed and my mind kinda exploded.

It wasn’t until trying to get that Project Euler question 2 to complete in under 30 seconds that I ever implemented this. It’s simpler than I expected going in.

Yeah, it’s not too hard to put together an inefficient solution using recursion (since that’s how the equation is written). If you’re clever, you might save computations in a list. Or, maybe you write an iterative solution which only needs two variables at any give time.

That said, an equation like this is next level. Haha

Solution in Go language:

package main

import (
"fmt"
"os"
"strconv"
)

func fib(n int) {
a, b := 0, 1
for i := 0; i < n; i++ {
a, b = b, a+b
fmt.Printf("%d: %d\n", i+1, a)
}
}

const msg = "Usage: please input the count of fibonacci numbers to output"

func main() {
if len(os.Args) == 1 {
fmt.Println(msg)
return
}
if os.Args == "" {
fmt.Println(msg)
return
}
n, err := strconv.Atoi(os.Args)
if err != nil {
fmt.Println(msg)
return
}
fib(n)
}


with Ada.Command_Line;

procedure Fib is
Old   : Natural := 1;
Older : Natural := 0;
Limit : Positive;
Tmp   : Positive;

begin
if Command_Line.Argument_Count /= 1 Then
end if;

Limit := Positive'Value (Command_Line.Argument (1));

for K in 1..Limit Loop
Put_Line (Standard_Error, Integer'Image (K) & ":" & Integer'Image(Old));
Tmp := Old + Older;
Older := Old;
Old := Tmp;
end loop;
exception
Put_Line(Standard_Error,
"Usage : Please Input The Count of Fibonacci Numbers To Output");

Command_Line.Set_Exit_Status(Command_Line.Failure);
end Fib;


This is my first post here. I am intrigued by such kinds of problems. this is a simple enough JS code, I hope, to achieve the goal.
It is a recursive function, hence I am not expecting very match in terms of performance.

function fib(i){
if (i<=0) return -1;
else if (i==1) return 1;
else if (i==2) return 1;
else return fib(i-1)+fib(i-2)
}

// some examples
//fib(3) returns 2
//fib(5) returns 5



You may try this on chrome dev tools for quick tests.

Racket:

(define (fib n)
(define (fib-iter i n current previous)
(if (>= i n)
current
(fib-iter (+ i 1) n (+ current previous) current)))
(fib-iter 0 n 1 0))


As much as I like Haskell in general, monadic IO really isn't my cup of tea and one of the reasons I always go back to less pure FP languages.

fib :: Integer -> Integer
fib 0 = 0
fib 1 = 1
fib n = fib (n-1) + fib (n-2)


For anyone interested, this challenge is also available here in 257 languages : rosettacode.org/wiki/Fibonacci_seq...

Also related, Project Euler problem #2

Love that the example solution for this challenge was implemented in Haskell. Even being a beginner, I appreciate all that functional goodness.  