Write the fastest Fibonacci

Question

This is yet another challenge about the Fibonacci numbers.

The goal is to compute the 20'000'000^th Fibonacii number as fast as possible. The decimal output is about 4 MiB large; it starts with:

28543982899108793710435526490684533031144309848579

The MD5 sum of the output is

fa831ff5dd57a830792d8ded4c24c2cb

You have to submit a program that calculates the number while running and puts the result to stdout. The fastest program, as measured on my own machine, wins.

Here are some additional rules:

• You have to submit the source code and a binary runnable on an x64 Linux
• The source code must be shorter than 1 MiB, in case of assembly it is also acceptable if only the binary is that small.
• You must not include the number to be computed in your binary, even in a disguised fashion. The number has to be calculated at runtime.
• My computer has two cores; you are allowed to use parallelism

I took a small implementation from the Internet which runs in about 4.5 seconds. It should not be very difficult to beat this, assuming that you have a good algorithm.

Answer 1

Sage

Hmm, you seem to assume that the fastest is going to be a compiled program. No binary for you!

print fibonacci(2000000)

On my machine, it takes 0.10 cpu seconds, 0.15 wall seconds.

edit: timed on the console, instead of the notebook

Answer 2

Haskell

This is my own try, although I did not wrote the algorithm by myself. I rather copied it from haskell.org and adapted it to use Data.Vector with its famous stream fusion:

import Data.Vector as V
import Data.Bits

main :: IO ()
main = print $ fib 20000000

fib :: Int -> Integer
fib n = snd . V.foldl' fib' (1,0) . V.dropWhile not $ V.map (testBit n) $ V.enumFromStepN (s-1) (-1) s
    where
        s = bitSize n
        fib' (f,g) p
            | p         = (f*(f+2*g),ss)
            | otherwise = (ss,g*(2*f-g))
            where ss = f*f+g*g

This takes around 4.5 seconds when compiled with GHC 7.0.3 and the following flags:

ghc -O3 -fllvm fib.hs

Answer 3

C with GMP, 3.6s

Gods, but GMP makes code ugly. With a Karatsuba-style trick, I managed to cut down to 2 multiplies per doubling step. Now that I'm reading FUZxxl's solution, I'm not the first to have the idea. I've got a couple more tricks up my sleeve... maybe I'll try 'em out later on.

#include <gmp.h>
#include <stdio.h>

#define DBL mpz_mul_2exp(u,a,1);mpz_mul_2exp(v,b,1);mpz_add(u,u,b);mpz_sub(v,a,v);mpz_mul(b,u,b);mpz_mul(a,v,a);mpz_add(a,b,a);
#define ADD mpz_add(a,a,b);mpz_swap(a,b);

int main(){
    mpz_t a,b,u,v;
    mpz_init(a);mpz_set_ui(a,0);
    mpz_init(b);mpz_set_ui(b,1);
    mpz_init(u);
    mpz_init(v);

    DBL
    DBL
    DBL ADD
    DBL ADD
    DBL
    DBL
    DBL
    DBL ADD
    DBL
    DBL
    DBL ADD
    DBL
    DBL ADD
    DBL ADD
    DBL
    DBL ADD
    DBL
    DBL
    DBL
    DBL
    DBL
    DBL
    DBL
    DBL /*Comment this line out for F(10M)*/

    mpz_out_str(stdout,10,b);
    printf("\n");
}

Built with gcc -O3 m.c -o m -lgmp.

Answer 4

COW

 MoO moO MoO mOo MOO OOM MMM moO moO
 MMM mOo mOo moO MMM mOo MMM moO moO
 MOO MOo mOo MoO moO moo mOo mOo moo

Moo! (Takes a while. Drink some milk...)

Answer 5

Mathematica, interpreted:

First@Timing[Fibonacci[2 10^6]]

Timed:

0.032 secs on my poor man's laptop.

And of course, no binary.

Answer 6

Ocaml, 0.856s on my laptop

Requires the zarith library. I used Big_int but it's dog slow compared to zarith. It took 10 minutes with the same code! Most of the time was spent printing the damn number (9½ minutes or so)!

module M = Map.Make
  (struct
    type t = int
    let compare = compare
   end)

let double b = Z.shift_left b 1
let ( +. ) b1 b2 = Z.add b1 b2
let ( *. ) b1 b2 = Z.mul b1 b2

let cache = ref M.empty 
let rec fib_log n =
  if n = 0
  then Z.zero
  else if n = 1
  then Z.one
  else if n mod 2 = 0
  then
    let f_n_half = fib_log_cached (n/2)
    and f_n_half_minus_one = fib_log_cached (n/2-1)
    in f_n_half *. (f_n_half +. double f_n_half_minus_one)
  else
    let f_n_half = fib_log_cached (n/2)
    and f_n_half_plus_one = fib_log_cached (n/2+1)
    in (f_n_half *. f_n_half) +.
    (f_n_half_plus_one *. f_n_half_plus_one)
and fib_log_cached n =
    try M.find n !cache
    with Not_found ->
      let res = fib_log n
      in cache := M.add n res !cache;
      res

let () =
  let res = fib_log 20_000_000 in
  Z.print res; print_newline ()

I can't believe how much a difference the library made!

Answer 7

Haskell

On my system, this runs almost as fast as FUZxxl's answer (~18 seconds instead of ~17 seconds).

main = print $ fst $ fib2 20000000

-- | fib2: Compute (fib n, fib (n+1)).
--
-- Having two adjacent Fibonacci numbers lets us
-- traverse up or down the series efficiently.
fib2 :: Int -> (Integer, Integer)

-- Guard against negative n.
fib2 n | n < 0 = error "fib2: negative index"

-- Start with a few base cases.
fib2 0 = (0, 1)
fib2 1 = (1, 1)
fib2 2 = (1, 2)
fib2 3 = (2, 3)

-- For larger numbers, derive fib2 n from fib2 (n `div` 2)
-- This takes advantage of the following identity:
--
--    fib(n) = fib(k)*fib(n-k-1) + fib(k+1)*fib(n-k)
--             where n > k
--               and k ≥ 0.
--
fib2 n =
    let (a, b) = fib2 (n `div` 2)
     in if even n
        then ((b-a)*a + a*b, a*a + b*b)
        else (a*a + b*b, a*b + b*(a+b))

Answer 8

C, naive algorithm

Was curious, and I hadn't used gmp before... so:

#include <stdio.h>
#include <stdlib.h>
#include <gmp.h>

int main(int argc, char *argv[]){
    int n = (argc>1)?atoi(argv[1]):0;

    mpz_t temp,prev,result;
    mpz_init(temp);
    mpz_init_set_ui(prev, 0);
    mpz_init_set_ui(result, 1);

    for(int i = 2; i <= n; i++) {
        mpz_add(temp, result, prev);
        mpz_swap(temp, result);
        mpz_swap(temp, prev);
    }

    printf("fib(%d) = %s\n", n, mpz_get_str (NULL, 10, result));

    return 0;
}

fib(1 million) takes about 7secs... so this algorithm won't win the race.

Answer 9

I implemented the matrix multiplication method (from sicp, http://sicp.org.ua/sicp/Exercise1-19) in SBCL but it takes about 30 seconds to finish. I ported it to C using GMP, and it returns the correct result in about 1.36 seconds on my machine. It's about as fast as boothby's answer.

#include <gmp.h>
#include <stdio.h>

int main()
{
  int n = 20000000;

  mpz_t a, b, p, q, psq, qsq, twopq, bq, aq, ap, bp;
  int count = n;

  mpz_init_set_si(a, 1);
  mpz_init_set_si(b, 0);
  mpz_init_set_si(p, 0);
  mpz_init_set_si(q, 1);
  mpz_init(psq);
  mpz_init(qsq);
  mpz_init(twopq);
  mpz_init(bq);
  mpz_init(aq);
  mpz_init(ap);
  mpz_init(bp);

  while(count > 0)
    {
      if ((count % 2) == 0)
        {
          mpz_mul(psq, p, p);
          mpz_mul(qsq, q, q);
          mpz_mul(twopq, p, q);
          mpz_mul_si(twopq, twopq, 2);

          mpz_add(p, psq, qsq);    // p -> (p * p) + (q * q)
          mpz_add(q, twopq, qsq);  // q -> (2 * p * q) + (q * q) 
          count/=2;
        }

      else
       {
          mpz_mul(bq, b, q);
          mpz_mul(aq, a, q);
          mpz_mul(ap, a, p);
          mpz_mul(bp, b, p);

          mpz_add(a, bq, aq);      // a -> (b * q) + (a * q)
          mpz_add(a, a, ap);       //              + (a * p)

          mpz_add(b, bp, aq);      // b -> (b * p) + (a * q)

          count--;
       }

    }

  gmp_printf("%Zd\n", b);
  return 0;
}

Answer 10

Java: 8 seconds to compute, 18 seconds to write

public static BigInteger fibonacci1(int n) {
    if (n < 0) explode("non-negative please");
    short charPos = 32;
    boolean[] buf = new boolean[32];
    do {
        buf[--charPos] = (n & 1) == 1;
        n >>>= 1;
    } while (n != 0);
    BigInteger a = BigInteger.ZERO;
    BigInteger b = BigInteger.ONE;
    BigInteger temp;
    do {
        if (buf[charPos++]) {
            temp = b.multiply(b).add(a.multiply(a));
            b = b.multiply(a.shiftLeft(1).add(b));
            a = temp;
        } else {
            temp = b.multiply(b).add(a.multiply(a));
            a = a.multiply(b.shiftLeft(1).subtract(a));
            b = temp;
        }
    } while (charPos < 32);
    return a;
}

public static void main(String[] args) {
    BigInteger f;
    f = fibonacci1(20000000);
    // about 8 seconds
    System.out.println(f.toString());
    // about 18 seconds
}

Answer 11

Go

It's embarrasingly slow. On my computer it takes a little less than 3 minutes. It's only 120 recursive calls, though (after adding the cache). Note that this may use a lot of memory (like 1.4 GiB)!

package main

import (
    "math/big"
    "fmt"
)

var cache = make(map[int64] *big.Int)

func fib_log_cache(n int64) *big.Int {
    if res, ok := cache[n]; ok {
        return res
    }
    res := fib_log(n)
    cache[n] = res
    return res
}

func fib_log(n int64) *big.Int {
    if n <= 1 {
        return big.NewInt(n)
    }

    if n % 2 == 0 {
        f_n_half := fib_log_cache(n/2)
        f_n_half_minus_one := fib_log_cache(n/2-1)
        res := new(big.Int).Lsh(f_n_half_minus_one, 1)
        res.Add(f_n_half, res)
        res.Mul(f_n_half, res)
        return res
    }
    f_n_half := fib_log_cache(n/2)
    f_n_half_plus_one := fib_log_cache(n/2+1)
    res := new(big.Int).Mul(f_n_half_plus_one, f_n_half_plus_one)
    tmp := new(big.Int).Mul(f_n_half, f_n_half)
    res.Add(res, tmp)
    return res
}

func main() {
    fmt.Println(fib_log(20000000))
}