Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
158 views
in Technique[技术] by (71.8m points)

c++ - How to improve fixed point square-root for small values

I am using Anthony Williams' fixed point library described in the Dr Dobb's article "Optimizing Math-Intensive Applications with Fixed-Point Arithmetic" to calculate the distance between two geographical points using the Rhumb Line method.

This works well enough when the distance between the points is significant (greater than a few kilometers), but is very poor at smaller distances. The worst case being when the two points are equal or near equal, the result is a distance of 194 meters, while I need precision of at least 1 metre at distances >= 1 metre.

By comparison with a double precision floating-point implementation, I have located the problem to the fixed::sqrt() function, which performs poorly at small values:

x       std::sqrt(x)    fixed::sqrt(x)  error
----------------------------------------------------
0       0               3.05176e-005    3.05176e-005
1e-005  0.00316228      0.00316334      1.06005e-006
2e-005  0.00447214      0.00447226      1.19752e-007
3e-005  0.00547723      0.0054779       6.72248e-007
4e-005  0.00632456      0.00632477      2.12746e-007
5e-005  0.00707107      0.0070715       4.27244e-007
6e-005  0.00774597      0.0077467       7.2978e-007
7e-005  0.0083666       0.00836658      1.54875e-008
8e-005  0.00894427      0.00894427      1.085e-009

Correcting the result for fixed::sqrt(0) is trivial by treating it as a special case, but that will not solve the problem for small non-zero distances, where the error starts at 194 metres and converges toward zero with increasing distance. I probably need at least an order of maginitude improvement in precision toward zero.

The fixed::sqrt() algorithim is briefly explained on page 4 of the article linked above, but I am struggling to follow it let alone determine whether it is possible to improve it. The code for the function is reproduced below:

fixed fixed::sqrt() const
{
    unsigned const max_shift=62;
    uint64_t a_squared=1LL<<max_shift;
    unsigned b_shift=(max_shift+fixed_resolution_shift)/2;
    uint64_t a=1LL<<b_shift;

    uint64_t x=m_nVal;

    while(b_shift && a_squared>x)
    {
        a>>=1;
        a_squared>>=2;
        --b_shift;
    }

    uint64_t remainder=x-a_squared;
    --b_shift;

    while(remainder && b_shift)
    {
        uint64_t b_squared=1LL<<(2*b_shift-fixed_resolution_shift);
        int const two_a_b_shift=b_shift+1-fixed_resolution_shift;
        uint64_t two_a_b=(two_a_b_shift>0)?(a<<two_a_b_shift):(a>>-two_a_b_shift);

        while(b_shift && remainder<(b_squared+two_a_b))
        {
            b_squared>>=2;
            two_a_b>>=1;
            --b_shift;
        }
        uint64_t const delta=b_squared+two_a_b;
        if((2*remainder)>delta)
        {
            a+=(1LL<<b_shift);
            remainder-=delta;
            if(b_shift)
            {
                --b_shift;
            }
        }
    }
    return fixed(internal(),a);
}

Note that m_nVal is the internal fixed point representation value, it is an int64_t and the representation uses Q36.28 format (fixed_resolution_shift = 28). The representation itself has enough precision for at least 8 decimal places, and as a fraction of equatorial arc is good for distances of around 0.14 metres, so the limitation is not the fixed-point representation.

Use of the rhumb line method is a standards body recommendation for this application so cannot be changed, and in any case a more accurate square-root function is likely to be required elsewhere in the application or in future applications.

Question: Is it possible to improve the accuracy of the fixed::sqrt() algorithm for small non-zero values while still maintaining its bounded and deterministic convergence?

Additional Information The test code used to generate the table above:

#include <cmath>
#include <iostream>
#include "fixed.hpp"

int main()
{
    double error = 1.0 ;
    for( double x = 0.0; error > 1e-8; x += 1e-5 )
    {
        double fixed_root = sqrt(fixed(x)).as_double() ;
        double std_root = std::sqrt(x) ;
        error = std::fabs(fixed_root - std_root) ;
        std::cout << x << '' << std_root << '' << fixed_root << '' << error << std::endl ;
    }
}

Conclusion In the light of Justin Peel's solution and analysis, and comparison with the algorithm in "The Neglected Art of Fixed Point Arithmetic", I have adapted the latter as follows:

fixed fixed::sqrt() const
{
    uint64_t a = 0 ;            // root accumulator
    uint64_t remHi = 0 ;        // high part of partial remainder
    uint64_t remLo = m_nVal ;   // low part of partial remainder
    uint64_t testDiv ;
    int count = 31 + (fixed_resolution_shift >> 1); // Loop counter
    do 
    {
        // get 2 bits of arg
        remHi = (remHi << 2) | (remLo >> 62); remLo <<= 2 ;

        // Get ready for the next bit in the root
        a <<= 1;   

        // Test radical
        testDiv = (a << 1) + 1;    
        if (remHi >= testDiv) 
        {
            remHi -= testDiv;
            a += 1;
        }

    } while (count-- != 0);

    return fixed(internal(),a);
}

While this gives far greater precision, the improvement I needed is not to be achieved. The Q36.28 format alone just about provides the precision I need, but it is not possible to perform a sqrt() without loss of a few bits of precision. However some lateral thinking provides a better solution. My application tests the calculated distance against some distance limit. The rather obvious solution in hindsight is to test the square of the distance against the square of the limit!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Given that sqrt(ab) = sqrt(a)sqrt(b), then can't you just trap the case where your number is small and shift it up by a given number of bits, compute the root and shift that back down by half the number of bits to get the result?

I.e.

 sqrt(n) = sqrt(n.2^k)/sqrt(2^k)
         = sqrt(n.2^k).2^(-k/2)

E.g. Choose k = 28 for any n less than 2^8.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...