Unconventional tiny solution to fib(n) Fibonacci sequence programming problem c. 2005-04-19

fibalt.c

#include <math.h>
#include <stdio.h>
unsigned int fibalt(unsigned int n) { return pow(1.618033988749, n) / 2.2360679774 + 0.5; }
/*unsigned int fibalt(unsigned int n) { return pow(1.618033988749, n) / sqrt(5) + 0.5; }*/

unsigned int fib(unsigned int n) {
  unsigned int a[2] = { 0, 1 };
  unsigned int i = 0;

  for (; n > 0; --n) {
    i = !i;
    a[i] = a[0] + a[1];
  }

  return a[i];
}

void test(unsigned int n, unsigned int f) {
  unsigned int g = fib(n);

  if (g == f) {
    printf("[OK]     %2u == fib(%u)\n", f, n);
  }
  else {
    printf("[NOT OK] %2u != fib(%u) (== %u)\n", f, n, g);
  }
}

int main(int argc, char * argv[])
{
  printf("sizeof(unsigned int) == %lu\n\n", sizeof(unsigned int));

  unsigned int i;
  for (i = 0; i < 48; i++) {
    unsigned int f = fib(i);
    unsigned int g = fibalt(i);
    
    if (f == g) {
      printf("fib(%3u) = %12u == %12u)\n", i, f, g);
    }
    else {
      printf("fib(%3u) = %12u != %12u)\n", i, f, g);
    }
  }

  return 0;
}