Efficient way to compute geometric mean of many numbers

Question

I need to compute the geometric mean of a large set of numbers, whose values are not a priori limited. The naive way would be

double geometric_mean(std::vect         


        

Answer 1

A different approach which would give better accuracy and performance than the logarithm method would be to compensate out-of-range exponents by a fixed amount, maintaining an exact logarithm of the cancelled excess. Like so:

const int EXP = 64; // maximal/minimal exponent
const double BIG = pow(2, EXP); // overflow threshold
const double SMALL = pow(2, -EXP); // underflow threshold

double product = 1;
int excess = 0; // number of times BIG has been divided out of product

for(int i=0; i<n; i++)
{
    product *= A[i];
    while(product > BIG)
    {
        product *= SMALL;
        excess++;
    }
    while(product < SMALL)
    {
        product *= BIG;
        excess--;
    }
}

double mean = pow(product, 1.0/n) * pow(BIG, double(excess)/n);

All multiplications by BIG and SMALL are exact, and there's no calls to log (a transcendental, and therefore particularly imprecise, function).