在OpenCL中快速实现二进制求幂

Question

I've been trying to design a fast binary exponentiation implementation in OpenCL. My current implementation is very similar to the one in this book about pi .

// Returns 16^n mod ak
inline double expm (long n, double ak)
{
    double r = 16.0;
    long nt;

    if (ak == 1) return 0.;
    if (n == 0) return 1;
    if (n == 1) return fmod(16.0, ak);

    for (nt=1; nt <= n; nt <<=1);

    nt >>= 2;

    do
    {
        r = fmod(r*r, ak);
        if ((n & nt) != 0)
            r = fmod(16.0*r, ak);
        nt >>= 1;
    } while (nt != 0);
    return r;
}

Is there room for improvement? Right now my program is spending the vast majority of it's time in this function.

Answer 1

My first thought is to vectorize it, for a potential speed up of ~1.6x. This uses 5 multiplies per loop compared to 2 multiplies in the original, but with approximately a quarter the number of loops for sufficiently large N. Converting all the double s to long s, and swapping out the fmod s for % s may provide some speed up depending on the exact GPU used and whatever.

inline double expm(long n, double ak) {

    double4 r = (1.0, 1.0, 1.0, 1.0);
    long4 ns = n & (0x1111111111111111, 0x2222222222222222, 0x4444444444444444,
            0x8888888888888888);
    long nt;

    if(ak == 1) return 0.;

    for(nt=15; nt<n; nt<<=4); //This can probably be vectorized somehow as well.

    do {
        double4 tmp = r*r;
        tmp = tmp*tmp;
        tmp = tmp*tmp;
        r = fmod(tmp*tmp, ak); //Raise it to the 16th power, 
                                       //same as multiplying the exponent 
                                       //(of the result) by 16, same as
                                       //bitshifting the exponent to the right 4 bits.

        r = select(fmod(r*(16.0,256.0,65536.0, 4294967296.0), ak), r, (ns & nt) - 1);
        nt >>= 4;
    } while(nt != 0); //Process n four bits at a time.

    return fmod(r.x*r.y*r.z*r.w, ak); //And then combine all of them.
}

Edit: I'm pretty sure it works now.

Answer 2

• The loop to extract nt = log2(n); can be replaced by
  if (n & 1) ...; n >>= 1;
  in the do-while loop.
• Given that initially r = 16; , fmod(r*r, ak) vs fmod(16*r,ak) can be easily delayed to calculate the modulo only every Nth iteration or so -- Loop unrolling?
• Also why fmod?