This is possibly the best one-liner I’ve ever written:

```
```gcc -x c -o /tmp/out - -lgmp <<< '#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdint.h>
#include <gmp.h>
void omg_i_love_leonardo_of_pisa(uint32_t num, mpz_t * result) { mpz_t retval, last, tmp; mpz_init(retval);
mpz_init(last); mpz_init(tmp); uint32_t i = 1; if(num == 0) return; mpz_set_ui(retval, 1U);
mpz_set_ui(last, 0U); for(; i < num; i++) { mpz_set(tmp, retval); mpz_add(retval, retval, last);
mpz_set(last, tmp); } mpz_set(*result, retval); } int main() { uint32_t num; mpz_t fibo; mpz_init(fibo);
omg_i_love_leonardo_of_pisa(1000001, &fibo); mpz_out_str(stdout, 10, fibo); printf("\n"); return 1; }
' && time /tmp/out

It compiles a C program given from `STDIN`

, puts it in `/tmp/out`

, and runs it with time to find the time it takes to run. It generates the 1,000,000th Fibonacci number. **Try it!**

## Update May 21, 2011

I changed the algorithm to do a matrix multiplication trick. The only problem is it goes over the number you ask for currently. I'm going to fix this with memoization soon.

gcc -x c -o /tmp/out - -lgmp <<< '#include <stdio.h> #include <string.h> #include <stdlib.h> #include <stdint.h> #include <gmp.h> void print_state(mpz_t* fm2, mpz_t* fm1, mpz_t* f, uint32_t n){gmp_printf("fib(%d) = %Zd\n", n, f);} #define NEXT_FIB() mpz_set(oldfm1, fm1);mpz_set(oldf, f);mpz_mul(f, f, f);mpz_mul(tmp, fm1, fm1);\ mpz_add(f, f, tmp);mpz_mul(fm1, oldf, fm1);mpz_mul(tmp, oldfm1, fm2);mpz_add(fm1, fm1, tmp); \ mpz_set(tmp, fm2);mpz_mul(fm2, oldfm1, oldfm1);mpz_mul(tmp, tmp, tmp);mpz_add(fm2, fm2, tmp);\ n += i;i *= 2; int main(){mpz_t fm2, fm1, f;uint32_t n = 2;uint32_t i = 1;mpz_inits(fm2, fm1, f, NULL);mpz_set_si(fm2, 0);mpz_set_si(fm1, 1);mpz_set_si(f, 1);mpz_t oldf, oldfm1, tmp;mpz_inits(oldf, oldfm1, tmp, NULL); uint32_t g = 1000000;while(n<g){NEXT_FIB();}print_state(&fm2, &fm1, &f, n);return 0;}' && time /tmp/out

This outputs almost immediately on my Intel Atom:

fib(1048577) = 19202837189514814................. real 0m0.840s user 0m0.280s sys 0m0.010s

The code is here. Feel free to fork and improve!

## Update August 30, 2013

I ended up looking at this again, and I improved it immensely. Apparently, gmp has built-in fibo functions (!!):

#include#include #include #include int main() { int n = 1000000; mpz_t fm2; mpz_inits(fm2, NULL); mpz_fib_ui(fm2, n); gmp_printf("fib(%d) = %Zd\n", n, fm2); return 1; }

This produces a number MUCH faster than the above implementation. It also makes a nicer oneliner:

gcc -x c -o /tmp/out - -lgmp <<< '#include <string.h> #include <stdlib.h> #include <stdint.h> #include <gmp.h> int main() { int n = 1000000; mpz_t fm2; mpz_inits(fm2, NULL); mpz_fib_ui(fm2, n); gmp_printf(\"fib(%d) = %Zd\n\", n, fm2); return 1; }" && time /tmp/out real 0m0.057s user 0m0.040s sys 0m0.008s

Robert OttignonPython code for generating 1000,000th number in the Fibonacci sequence (or any number, the attribute of c):

ErikPost authorRobert:

Thanks for the comment. That isn’t a bad way to generate the number, and if I slightly modify it, it does:

Yet, it takes over 3 times as long as my example

Your script: 0m59.591s

My program: 0m17.435s

EkharionProlog:

%fib(IN,OUT,P).

fib(0,0,0).

fib(1,1,0).

fib(IN,OUT,P) :- IN2 is IN -1, fib(IN2,P,P2), OUT is P + P2.

:mrgreen: :wink:

ErikPost authorYeah, you can do that. I recently found a huge optimization and I need to update the code here. Cheers.

micaleaQuisiera saber el numero de linner, del acido alfa naftalenacetico. Gracias.

Reuben Willsondef fib(n):

a,b=1,0

for x in range(n):

a,b=a+b,a

# the following comments make it easy to track where it’s at

#r,t = divmod(x+1,10)

#if t==0:

#print(x+1,a)

return(a)

ErikPost authorNice – what’s the runtime on that compared to the C version I’ve described in the article?