1,000,000th Fibonacci Number One-Liner in C

September 9, 2009

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!

Posted, by Erik in Uncategorized Tags: awesome, bash, c, code, linux, one-liner, programming, shell

Comments

Robert Ottignon

December 10, 2009 at 9:01 PM

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

# Fibonacci in Python 3.0
i=0
c=1000000
# Sets number you want program to stop at.
x=0
y=1
while i< =c:
	print ("  ")
	print ("  ")
	print ("Fibonacci number ",i," = ",x)
	y=y+x
	i=i+1
	if i<=c:
		print ("  ")
		print ("  ")
		print ("Fibonacci number ",i," = ",y)
		x=x+y
		i=i+1
	else:
		break

Erik

December 10, 2009 at 9:37 PM
Robert:

Thanks for the comment. That isn’t a bad way to generate the number, and if I slightly modify it, it does:
```
# Fibonacci in Python 3.0
i=0
c=1000000
# Sets number you want program to stop at.
x=0
y=1
while i<=c:
  y=y+x
  i=i+1
  if i<=c:
    x=x+y
    i=i+1
  else:
    break

print ("  ")
print ("  ")
print ("Fibonacci number ",i," = ",y)
```
Yet, it takes over 3 times as long as my example

Your script: 0m59.591s
My program: 0m17.435s
Ekharion

May 16, 2011 at 11:40 AM

Prolog:

%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:
Erik

May 21, 2011 at 12:15 PM

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

October 4, 2011 at 8:13 PM

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

1,000,000th Fibonacci Number One-Liner in C

Update May 21, 2011

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