## DEV Community is a community of 550,319 amazing developers

We're a place where coders share, stay up-to-date and grow their careers.

# Challenge: Write the recursive Fibonacci algorithm in a different language.

Hellow DEV community, how are you?

Can we challenge ourselves in this fun "game"?

Let's see how many languages we can enumerate by writing the Fibonacci recursive algorithm on the comments. Are you in?

EDIT: this is not about having the fast/better code, just to have some fun

## Discussion  Avalander

I'll continue with javascript.

``````const fibonacci = (n, i=0, curr=1, prev=0) =>
(i >= n ? curr : fibonacci(n, i + 1, curr + prev, curr))
``````

(Look mum, with a single expression function!) David Wickes

To be absolutely pedantic, this code is valid and will work with large numbers because it is written in EcmaScript 6. Part of the ES6 standard is for the language interpreter to implement tail call optimization (TCO). EcmaScript is a language defined by a standard, not by an implementation.

Now, at present, barely any interpreters implement TCO - here's a fun comparison table to see which do or don't. This makes me sad, but there we are.

But, as I say, the function above will work just fine according to the language standard. Whether this code blows the call stack of your interpreter is dependent on whether your interpreter implements the full ES6 standard. Which it probably doesn't.

To put it strongly, complaining that this code doesn't run on an JS interpreter without TCO is like complaining that it can't run by a Ruby interpreter. The fault is with the interpreter, not the code. Rob Hoelz

I haven't seen a SQL implementation yet, so here's one:

``````with recursive fib(a, b) as
(select 1 as a, 0 as b

union

select a + b as a, a as b from fib)

select a from fib limit 10;
`````` Slavius

Here's interesting one for SQL Server with maximum value before arithmetic overflow due to data type limit:

``````SET STATISTICS TIME,PROFILE ON

DECLARE @MAX SMALLINT
SET @MAX = 1474

SELECT TOP 1 s.Fib FROM (
SELECT
FLOOR(POWER(( 1 + SQRT(5) ) / 2.0, number) / SQRT(5) + 0.5) AS Fib
FROM master..spt_values
WHERE TYPE = 'p' AND number BETWEEN 1 AND @MAX
) AS s
ORDER BY s.Fib DESC
``````  Ross Henderson

Here it is in Oracle SQL:

``````with FIBONACCI (i, SPIRAL, PREV) as
(
select
1 i, 0 SPIRAL, cast(null as number) PREV
from
DUAL
union all
select
f1.i + 1 i, f1.SPIRAL + nvl(f1.PREV,1) SPIRAL, f1.SPIRAL PREV
from
FIBONACCI f1
where
f1.i < 605
)

select SPIRAL from FIBONACCI order by i;
``````

The limit of 605 is the limit before numeric overflow. M. Shemayev

Fuck yes SQL!

I feel like every time someone responds to one of these with a SQL implementation that a) my heart grows a few sizes and b) they earn extra special bonus points in my head because hell yeah.

⭐️

(nonironic gold star, fwiw) Slavius

I'll follow up in C# (and LINQ):

``````Console.WriteLine(
Enumerable.Range(1, 10)
.Skip(2)
.Aggregate(
new { Current = 1, Prev = 1 }, (x, index) => new {
Current = x.Prev + x.Current, Prev = x.Current }
).Current
);
``````

(Look ma', single expression lambda without recursion!) Slavius

I lied a bit. The recursion is there, hidden in the Aggregate function. It's the framework abstraction that masks it.

Otherwise it wouldn't qualify as an answer to OP challenge, or would it? ;) Avalander

I know nothing about C#, but I thought that the `Aggregate` function was somehow invoking an anonymous function here that given current and previous values returns the next element. If that's the case, and it's a function calling another function multiple times instead of the function invoking itself, can we still call it recursion? Slavius

The `Aggregate` function is member of the `Enumerable` type and applies an anonymous accumulator function looping all elements in that "array/set". Hard to tell if we can treat it as a recursion. I would have to look at the stack if there are pointers left to the originating caller.

Fibonacci's sequence is strictly sequential so it works well but for more parallel calculations, like SUM, AVG, MAX, MIN accumulator functions you can do:

``````var result = Enumerable.Range(1,100000)
.AsParallel()
.Aggregate(0, (sum, i) => { sum += i } );
``````

And it will multithread across many OS threads to achieve the best efficiency.

Edit: This is however not plain C# .Net anymore but LINQ ;) Avalander

Since I doubt anybody else is going to do it, here is an implementation in Scheme

``````(define (fib n)
(define (fib-iter a b count)
(if (= count n)
b
(fib-iter (+ a b) a (+ count 1))))
(fib-iter 1 0 0))
``````

Tested for values of `n` up to `10001`. Michael Kohl

Tail-recursive or not, no function accepts any input. Infinity is rather big, I can always find a number that not longer fits in the working memory of your computer. Does that make your solution incorrect?

Excuse the silly hyperbole, but on a site that also attracts a lot of new(ish) developers your self-indulgent overcorrectness on what OP explicitly called a fun "game" is a bit off-putting to be honest. Slavius

Please do not feed this troll.

He picked fib(10000) because it blows Int64 and now he's trying to victoriously convince everyone their programming language of choice is crap.

The same as if I would say I can compute and save a file containing Pi to 1e+13 numbers and don't tell anyone I have 12TB NAS storage attached... Avalander

Aleksei, if you try to work with very large numbers in Javascript, at some point (before the 10000th element in the fibonacci series) it just returns `Infinity` because it can't deal with them. Avalander

If you think it's important enough for a code challenge to have a Javascript implementation that returns `Infinity` instead of blowing the stack for large values, be my guest and suggest an alternative implementation.

Just posting the console output of calling my function with an arbitrarily long value seems rather pointless. I agree my solution is not perfect but I'd appreciate a better solution more than an error message. Felix Edelmann

I did this some time ago in Rust.

My first approach (the naive one) is very similar to @rapidnerd 's (George Marr's) solution in R and @andreanidouglas ' (Douglas R Andreani's) solution in Python:

``````fn fib(n: u64) -> u64 {
match n {
0 => 0,
1 => 1,
_ => fib(n-1) + fib(n-2),
}
}
``````

I realized that this solution quickly exceeded in execution time for n > 40, so I optimized it using a mathematical theorem:

Given that n and p are integers, and that n = 2p + 1, i. e. n is odd, the n-th fibonacci number is then

fib(n) = fib(p + (p+1)) = fib(p)² + fib(p+1)²

This means, we can break down numbers with odd indices into to numbers with about the half of the index – and do this again and again and again. This dramatically reduces execution time:

``````fn fib_fast(n: u64) -> u64 {
match n {
0 => 0,
1 => 1,

// even
_ if n % 2 == 0 => fib_fast(n-1) + fib_fast(n-2),

// odd
_ => {
let a = (n-1)/2;
let b = n-a;
let fib_a = fib_fast(a);
let fib_b = fib_fast(b);
return fib_a*fib_a + fib_b*fib_b;
},
}
}
``````

Inspired by @avalander 's answer, I also added this even faster algorithm:

``````fn fib_super_fast(n: u64, i: u64, curr: u64, prev: u64) -> u64 {
if i >= n {
return curr;
}

return fib_super_fast(n, i + 1, curr + prev, curr);
}
``````

`fib_fast` and `fib_super_fast` execute within microseconds even for higher indices. However, indices of 94 and higher cause a crash:

Some performance measurements:

``````Enter n:
5
fib_super_fast(n) = 5 (0µs)
fib_fast(n)       = 5 (0µs)
fib(n)            = 5 (1µs)

Enter n:
20
fib_super_fast(n) = 6765 (1µs)
fib_fast(n)       = 6765 (10µs)
fib(n)            = 6765 (536µs)

Enter n:
40
fib_super_fast(n) = 102334155 (1µs)
fib_fast(n)       = 102334155 (54µs)
fib(n)            = 102334155 (1908011µs)

Enter n:
45
fib_super_fast(n) = 1134903170 (2µs)
fib_fast(n)       = 1134903170 (15µs)
fib(n)            = 1134903170 (21216762µs)

Enter n:
90
fib_super_fast(n) = 2880067194370816120 (2µs)
fib_fast(n)       = 2880067194370816120 (958µs)

Enter n:
93
fib_super_fast(n) = 12200160415121876738 (3µs)
fib_fast(n)       = 12200160415121876738 (152µs)

Enter n:
94
`````` David Wickes

Just for a nice style, you could factor out the `return`s as `if` is an expression...

``````fn fib_super_fast(n: u64, curr: u64, prev: u64) -> u64 {
if n == 0 {
curr
} else {
fib_super_fast(n - 1, curr + prev, curr)
}
}
`````` Avalander

Nice!

Since writing my implementation I've remembered that the parameter `i` is unnecessary and you can decrease `n` instead until it's 0.

``````fn fib_super_fast(n: u64, curr: u64, prev: u64) -> u64 {
if n <= 0 {
return curr;
}

return fib_super_fast(n - 1, curr + prev, curr);
}
``````

By the way, you didn't set default values for `curr` and `prev`, how does that work, do you need to call it with `fib_super_fast(n, 1, 0)`? Felix Edelmann

Yes, I do. But I have to use fib_super_fast(n, 1, 1, 0) as my other implementations also begin with 1. The code for calling the fibonacci methods is:

``````fn main() {
loop {
println!("Enter n:");

let mut n_str = String::new();
let n = user_input_to_int(n_str);

let t0 = SystemTime::now();
let fib_super_fast_n = fib_super_fast(n, 1, 1, 0);
let t1 = SystemTime::now();
println!("fib_super_fast(n) = {} ({}µs)", fib_super_fast_n, time_difference(t0, t1));

let t0 = SystemTime::now();
let fib_fast_n = fib_fast(n);
let t1 = SystemTime::now();
println!("fib_fast(n)       = {} ({}µs)", fib_fast_n, time_difference(t0, t1));

let t0 = SystemTime::now();
let fib_n = fib(n);
let t1 = SystemTime::now();
println!("fib(n)            = {} ({}µs)", fib_n, time_difference(t0, t1));
}
}
`````` Marvodor

Here's a Python variant for quite large numbers. Alas, due to recursion depth restrictions, any new calculated Fibonacci number should not have a larger sequence position than about 500 more than that of the highest already calculated.

``````def fib(a, F={}):
if a in F:
return F[a]
if a < 2:
return 1
f = fib(a-1)
F[a-1] = f
return f + fib(a-2)

for i in range(0,50001,500):
print("fib(%i) = %i"%(i,fib(i)))
`````` Fabian

Technically lazy evaluation of a endless stream is iterative, not recursive, but anyway, here is a solution in Haskell:

``````fibonnaciNums = 0 : 1 : zipWith (+) fibonnaciNums (tail fibonnaciNums)

fibonnaci n        = last \$ take n fibonnaciNums
-- or even shorter:  fibonnaciNums !! n
`````` Dave Cridland
``````#include <iostream>

template<long long i> long long fibonacci() {
return fibonacci<i-1>() + fibonacci<i-2>();
}

template<> long long fibonacci<1>() {
return 1;
}

template<> long long fibonacci<2>() {
return 1;
}

int main() {
std::cout << "Fibonacci of 27 is " << fibonacci<27>() << std::endl;
return 0;
}
``````

C++ can, of course, do it at build time with a bit of templates.

Here, we're definitely recursing - twice, because it's simpler. At runtime, it's just printing the value out that's been calculated by the compiler during the build. Slavius

I see what you did there. Trying to troll languages that you think may not have Int128 support :D. Nice try!

``````using System;
using System.Linq;
using System.Numerics;

namespace ConsoleApp1
{
class Program
{
static void Main(string[] args)
{
var start = DateTime.Now;
Console.WriteLine(
Enumerable.Range(1, 10000)
.Skip(2)
.Aggregate(
new { Current = (BigInteger)1, Prev = (BigInteger)1 }, (x, index) => new {
Current = x.Prev + x.Current,
Prev = x.Current
}
).Current
);
Console.WriteLine(DateTime.Now - start);
}
}
}
``````
``````33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875
00:00:00.0245407
`````` David Wickes

Common Lisp!

``````(defun fib (n &optional (curr 0) (next 1))
(if (zerop n)
curr
(fib (1- n) next (+ curr next))))
``````

(and for the TCO crowd... it's implementation dependent, so YMMV) Michael Kohl

Thanks for taking the time to type out this response!

Like you I'm not a native English speaker, and my native language (and culture) generally is also much more direct and — for lack of a better word — "confrontational" than English. That said I moved a lot over the last 16 years and spent almost half of that time in East and Southeast Asia, where communication tends to be very indirect and high context, and overall culture is extremely non-confrontational. This definitely had some influence on how I communicate nowadays.

I don't doubt that you have only the best intentions, and will keep that in mind for future occasions where I may feel tempted to butt in. stereobooster

No worries @avalander . You all did right, it is just JS doesn't support TCO there were some attempts (as far as I remember it is hold back by Microsoft member or committee or something like that). There is still way around - you can use trampoline

``````const trampoline = (fn) => {
let newFn = eval(fn.toString().replace(new RegExp(fn.name + '\\(([^\\)]*)\\)'), '[\$1]'))
return function() {
let result = newFn.apply(null, arguments)
while (result instanceof Array) result = newFn.apply(null, result)
return result
}
};

const _fibonacciIter = (n, i, curr, prev) =>
(i >= n ? curr : _fibonacciIter(n, i + 1, curr + prev, curr));

const fibonacciIter = trampoline(_fibonacciIter);

const fibonacci = n => fibonacciIter(n, 0, 1, 0);
``````

I doubt you would want to use this in production, but just to make a point.

TCO - stands for tail call optimisation, to prevent Maximum call stack error
Trampoline is the way how some compilers implement TCO, they convert recursion to a loop Slavius

I'm just letting everyone (else) know I'm not reacting to his childish unsuccessfull trolling attempts anymore... I'm not his mother to raise him to his adulthood.

I know my shit, anyone who writes C# can see it as well. I will now just silently laugh... Dave Cridland
``````def fib(arg):
curr, prev = 1, 1
n = int(arg)
if n != arg or n <= 0:
raise ValueError("Argument must be positive integer")
if n <= 2:
return 1
for count in range(n - 2):
curr, prev = curr + prev, curr
return curr
``````

Quick bit of Python. The error handling probably isn't complete, but Python has the useful benefit here that it has native, transparent, Big Number support - you can do fib(10000) if you want (or much higher if you have the memory).

But yeah, I just noticed Douglas wanted a recusrive algorithm, whereas I've done it iteratively without thinking. In general, iterative solutions will execute faster - though this isn't always the case. Michael Kohl

For people to grow from mistakes they need to listen to the person who points them out. To be that person, I'd reconsider my communication style.

There's nothing wrong with posting your own answer and discussing the merits of tail recursion and the problems with other solutions there. Posting a comment under every single answer seems a bit...too much.

Anyway, it's not my site, you're of course free to post whatever you want. But that may also invite people to comment on the limits of your communication style, so as long as you're cool with that ¯\(ツ) Jan van Brügge

I have haskell here with the best form of recursion: Self-recursion!

``````fibs = 1:1:zipWith (+) fibs (tail fibs)
fib n = fibs !! n
``````

Here `fibs` is an infinite List of all fibonacci numbers, the `!!` function returns the nth element of the list. Slavius
``````> #reset
Resetting execution engine.
> var start = DateTime.Now;
. Console.WriteLine(
.   Enumerable.Range(1, 10000)
.   .Skip(2)
.   .Aggregate(
.     new { Current = 1, Prev = 1 }, (x, index) => new {
.         Current = x.Prev + x.Current,
.         Prev = x.Current
.     }
.   ).Current
. );
. Console.WriteLine(DateTime.Now - start);
1242044891
00:00:00.0025070
>
`````` Avalander

Language limitations are the worst excuse for the developer, aren’t they?

It's also rather convenient to claim that other people's proposal is incorrect because they have to implement things that your language of choice gives you for free (or, in the case of Ruby, you need to enable a runtime option; still beats Javascript's strategy of not having it at all).

According to you I should not only hack tail-call optimisation into Javascript somehow (most current runtimes don't do any kind of tail-call optimisation, ever), but roll my own custom system to handle arbitrarily large numbers. I don't really think this was the point of the challenge.

Also, let's keep in mind that my solution blows when the stack isn't big enough to keep track of all the function calls, and yours blow when the number is too large to be kept in memory. So, while in practice my solution blows much, much earlier than yours, both are, technically, incorrect if the definition of correct is that it works for all natural numbers, and they are both incorrect for the same reason: computer limitations apply. stereobooster

So, while in practice my solution blows much, much earlier than yours, both are, technically, incorrect if the definition of correct is that it works for all natural numbers, and they are both incorrect for the same reason: computer limitations apply.

The word you are looking for is "totality". Function is total if it returns a value for each input in the domain Yoandy Rodriguez Martinez

An Emacs Lisp version. Warning!! This will likely freeze your Emacs!!

``````(defun fib(n)
(cond
((= n 0) 0)
((= n 1) 1)
(t (+ (fib (- n 1)) (fib (- n 2))))))
``````

UPDATE
Just realized it counts as a Common Lisp version too Nested Software

Respectfully, if you want to help newbies on a general issue such as recursion, just make a top-level comment addressing this problem. Show an example of the problem and suggest ways it can be fixed. That way, if the community deems the comment to be helpful, it will bubble up near the top of the discussion and it will be one of the first comments that newbies encounter.

In this manner, you won't be repeating the same comment all over the place, and you will have something clear that novices can wrap their minds around. In my opinion, making grumpy statements to the effect of "this crashes when the input reaches X" doesn't really help the very novices you say you most want to reach.

Edit: Added "in my opinion" :) Michael Kohl

Because different language runtimes have different execution models, so doing it automatically might not always be possible/feasible. Given that Scala — another JVM language — also has a `@tailrec` annotation I'd guess it's hard to efficiently detect tail recursion in the generated byte code, probably because the JVM was originally built without support for proper tail calls (unlike for example Scheme where tail call elimination is part of the standard since R5RS). Dave Cridland
``````#include <iostream>

constexpr long long fibonacci(long long i) {
if (i <= 2) {
return 1LL;
}
return fibonacci(i - 1) + fibonacci(i - 2);
}

int main() {
std::cout << "Fibonacci of 27 is " << fibonacci(27) << std::endl;
return 0;
}
``````

Modern C++ can do it more simply, though - this uses constexpr to tell the compiler that it can calculate these at build time if it wants. Using a constexpr arg (here a literal), it probably will do, but we could also pass in a variable to make it calculate it at runtime.