Re: constexpr is really very smart!

Michael S <already5chosen@yahoo.com> writes:

On Wed, 18 Dec 2024 01:33:42 +0200
Michael S <already5chosen@yahoo.com> wrote:
>

I am pretty sure that better variant exists, but for tonight it's
enough.

>
Morning fibs.
>
Recursive:
>
struct pair_t {
  long long a,b;
};
>
// return .a=fib(n), .b=fib(n+1)
static struct pair_t fib2(long n)
{
  if (n <= 0) {
    struct pair_t ret = { .a = 0, .b = n==0 ? 1 : 0 };
    return ret;
  }
  // for n > 0
  // fib(n*2)   = (fib(n-1)+fib(n+1))*fib(n)
  // fib(n*2+1) = fib(n)*fib(n)+fib(n+1)*fib(n+1)
  long m = (n-1) >> 1;
  struct pair_t ret = fib2(m);
  long long a = ret.a, b = ret.b;
  long long c = a + b;
  if ((n & 1)==0) { // (m,m+1) => ((m+1)*2,(m+1)*2*2+1)
    ret.a = (a + c)*b;
    ret.b = b*b + c*c;
  } else {          // (m,m+1) => (m*2+1,(m+1)*2)
    ret.a = a*a + b*b;
    ret.b = (a + c)*b;
  }
  return ret;
}
>
static long long fib(long n)
{
  struct pair_t x = fib2(n-1);
  return x.b;
}

I may have inadvertently misled you.  Here is a simple linear
formulation that uses recursion:

    typedef  unsigned long long  NN;

    static NN fibonacci_3( NN, NN, unsigned );

    NN
    fibonacci( unsigned n ){
        return  fibonacci_3( 1, 0, n );
    }

    NN
    fibonacci_3( NN a, NN b, unsigned n ){
        return  n == 0  ? b  :  fibonacci_3( b, a+b, n-1 );
    }

Can you apply this idea to your logarithmic scheme?

Iterative:
>
static long long fib(long n)
{
  if (n <= 0)
    return 0;
  // find MS bit
  unsigned long tmp = n;
  unsigned long bit = 1;
  while (tmp > 1) {
    bit += bit;
    tmp /= 2;
  }
  // for n > 0
  // fib(n*2)   = (fib(n-1)+fib(n+1))*fib(n)
  // fib(n*2+1) = fib(n)*fib(n)+fib(n+1)*fib(n+1)
  long long a = 0, b = 1;
  while (bit > 1) {
    long long c = a + b; // fib(n+1)
    bit /= 2;
    if ((n & bit)==0) { // (n-1,n) => (n*2-1,n*2)
      c += a;
      a = a*a + b*b;
      b = c*b;
    } else {            // (n-1,n) => (n*2,n*2+1)
      a = (a + c)*b;
      b = b*b + c*c;
    }
  }
  return b;
}
>
>
Both variants above are O(log(n)).
In practice for n in range 0 to 92 my measurement gear does not detect
differences in speed vs simple loop.  If anything, simple loop is a
little faster.

Yes, the log(n) can be beaten by simpler schemes for small n.

I wrote a (recursive) log(n) fibonacci in python.  It was able
to compute fibonacci( 10000000 ) in just under 3 seconds.