Douglas R Andreani

Posted on Aug 21, 2018

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

#challenge #beginners #fun

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

Top comments (51)

Avalander • Aug 21 '18 • Edited

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 • Aug 23 '18 • Edited

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.

David Wickes • Aug 21 '18

I like this! Don't think you need the outer parens though :D

Avalander • Aug 21 '18

Should work without, I'm just used to wrapping ternary expressions between them for some reason and I find it weird without :)

Rob Hoelz • Aug 21 '18

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 • Aug 22 '18

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 • Aug 23 '18

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.

Avalander • Aug 22 '18

👏👏 I never thought you could do that with SQL!

Max Cerrina • Aug 21 '18

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 • Aug 21 '18

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!)

Avalander • Aug 21 '18

Imaginative, I like it (even if it's not recursive :P)!

Slavius • Aug 21 '18

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 • Aug 21 '18

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 • Aug 21 '18 • Edited

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 • Aug 21 '18 • Edited

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.

Slavius • Aug 21 '18 • Edited

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 • Aug 21 '18

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.

Douglas R Andreani • Aug 21 '18

man, just sit down and relax. there is no point for all this harassement.

Avalander • Aug 21 '18

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 • Aug 22 '18

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:

thread 'main' panicked at 'attempt to add with overflow', src/main.rs:42:35

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
thread 'main' panicked at 'attempt to add with overflow', src/main.rs:42:35

David Wickes • Aug 22 '18

Just for a nice style, you could factor out the returns 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 • Aug 22 '18 • Edited

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 • Aug 22 '18

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();
    io::stdin().read_line(&mut n_str).expect("Failed to read line");
    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 • Aug 22 '18

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 Holzer • Aug 21 '18

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

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

Top comments (51)

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