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 << '\t' << std_root << '\t' << fixed_root << '\t' << 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!
The original implementation obviously has some problems. I became frustrated with trying to fix them all with the way the code is currently done and ended up going at it with a different approach. I could probably fix the original now, but I like my way better anyway.
I treat the input number as being in Q64 to start which is the same as shifting by 28 and then shifting back by 14 afterwards (the sqrt halves it). However, if you just do that, then the accuracy is limited to 1/2^14 = 6.1035e-5 because the last 14 bits will be 0. To remedy this, I then shift a and remainder correctly and to keep filling in digits I do the loop again. The code can be made more efficient and cleaner, but I'll leave that to someone else. The accuracy shown below is pretty much as good as you can get with Q36.28. If you compare the fixed point sqrt with the floating point sqrt of the input number after it has been truncated by fixed point(convert it to fixed point and back), then the errors are around 2e-9(I didn't do this in the code below, but it requires one line of change). This is right in line with the best accuracy for Q36.28 which is 1/2^28 = 3.7529e-9.
By the way, one big mistake in the original code is that the term where m = 0 is never considered so that bit can never be set. Anyway, here is the code. Enjoy!
#include <iostream>
#include <cmath>
typedef unsigned long uint64_t;
uint64_t sqrt(uint64_t in_val)
{
const uint64_t fixed_resolution_shift = 28;
const unsigned max_shift=62;
uint64_t a_squared=1ULL<<max_shift;
unsigned b_shift=(max_shift>>1) + 1;
uint64_t a=1ULL<<(b_shift - 1);
uint64_t x=in_val;
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=1ULL<<(2*(b_shift - 1));
uint64_t two_a_b=(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((remainder)>=delta && b_shift)
{
a+=(1ULL<<(b_shift - 1));
remainder-=delta;
--b_shift;
}
}
a <<= (fixed_resolution_shift/2);
b_shift = (fixed_resolution_shift/2) + 1;
remainder <<= (fixed_resolution_shift);
while(remainder && b_shift)
{
uint64_t b_squared=1ULL<<(2*(b_shift - 1));
uint64_t two_a_b=(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((remainder)>=delta && b_shift)
{
a+=(1ULL<<(b_shift - 1));
remainder-=delta;
--b_shift;
}
}
return a;
}
double fixed2float(uint64_t x)
{
return static_cast<double>(x) * pow(2.0, -28.0);
}
uint64_t float2fixed(double f)
{
return static_cast<uint64_t>(f * pow(2, 28.0));
}
void finderror(double num)
{
double root1 = fixed2float(sqrt(float2fixed(num)));
double root2 = pow(num, 0.5);
std::cout << "input: " << num << ", fixed sqrt: " << root1 << " " << ", float sqrt: " << root2 << ", finderror: " << root2 - root1 << std::endl;
}
main()
{
finderror(0);
finderror(1e-5);
finderror(2e-5);
finderror(3e-5);
finderror(4e-5);
finderror(5e-5);
finderror(pow(2.0,1));
finderror(1ULL<<35);
}
with the output of the program being
input: 0, fixed sqrt: 0 , float sqrt: 0, finderror: 0
input: 1e-05, fixed sqrt: 0.00316207 , float sqrt: 0.00316228, finderror: 2.10277e-07
input: 2e-05, fixed sqrt: 0.00447184 , float sqrt: 0.00447214, finderror: 2.97481e-07
input: 3e-05, fixed sqrt: 0.0054772 , float sqrt: 0.00547723, finderror: 2.43815e-08
input: 4e-05, fixed sqrt: 0.00632443 , float sqrt: 0.00632456, finderror: 1.26255e-07
input: 5e-05, fixed sqrt: 0.00707086 , float sqrt: 0.00707107, finderror: 2.06055e-07
input: 2, fixed sqrt: 1.41421 , float sqrt: 1.41421, finderror: 1.85149e-09
input: 3.43597e+10, fixed sqrt: 185364 , float sqrt: 185364, finderror: 2.24099e-09
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.
I'm not sure how you're getting the numbers from fixed::sqrt() shown in the table.
Here's what I do:
#include <stdio.h>
#include <math.h>
#define __int64 long long // gcc doesn't know __int64
typedef __int64 fixed;
#define FRACT 28
#define DBL2FIX(x) ((fixed)((double)(x) * (1LL << FRACT)))
#define FIX2DBL(x) ((double)(x) / (1LL << FRACT))
// De-++-ified code from
// http://www.justsoftwaresolutions.co.uk/news/optimizing-applications-with-fixed-point-arithmetic.html
fixed sqrtfix0(fixed num)
{
static unsigned const fixed_resolution_shift=FRACT;
unsigned const max_shift=62;
unsigned __int64 a_squared=1LL<<max_shift;
unsigned b_shift=(max_shift+fixed_resolution_shift)/2;
unsigned __int64 a=1LL<<b_shift;
unsigned __int64 x=num;
unsigned __int64 remainder;
while(b_shift && a_squared>x)
{
a>>=1;
a_squared>>=2;
--b_shift;
}
remainder=x-a_squared;
--b_shift;
while(remainder && b_shift)
{
unsigned __int64 b_squared=1LL<<(2*b_shift-fixed_resolution_shift);
int const two_a_b_shift=b_shift+1-fixed_resolution_shift;
unsigned __int64 two_a_b=(two_a_b_shift>0)?(a<<two_a_b_shift):(a>>-two_a_b_shift);
unsigned __int64 delta;
while(b_shift && remainder<(b_squared+two_a_b))
{
b_squared>>=2;
two_a_b>>=1;
--b_shift;
}
delta=b_squared+two_a_b;
if((2*remainder)>delta)
{
a+=(1LL<<b_shift);
remainder-=delta;
if(b_shift)
{
--b_shift;
}
}
}
return (fixed)a;
}
// Adapted code from
// http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit-by-digit_calculation
fixed sqrtfix1(fixed num)
{
fixed res = 0;
fixed bit = (fixed)1 << 62; // The second-to-top bit is set
int s = 0;
// Scale num up to get more significant digits
while (num && num < bit)
{
num <<= 1;
s++;
}
if (s & 1)
{
num >>= 1;
s--;
}
s = 14 - (s >> 1);
while (bit != 0)
{
if (num >= res + bit)
{
num -= res + bit;
res = (res >> 1) + bit;
}
else
{
res >>= 1;
}
bit >>= 2;
}
if (s >= 0) res <<= s;
else res >>= -s;
return res;
}
int main(void)
{
double testData[] =
{
0,
1e-005,
2e-005,
3e-005,
4e-005,
5e-005,
6e-005,
7e-005,
8e-005,
};
int i;
for (i = 0; i < sizeof(testData) / sizeof(testData[0]); i++)
{
double x = testData[i];
fixed xf = DBL2FIX(x);
fixed sqf0 = sqrtfix0(xf);
fixed sqf1 = sqrtfix1(xf);
double sq0 = FIX2DBL(sqf0);
double sq1 = FIX2DBL(sqf1);
printf("%10.8f: "
"sqrtfix0()=%10.8f / err=%e "
"sqrt()=%10.8f "
"sqrtfix1()=%10.8f / err=%e\n",
x,
sq0, fabs(sq0 - sqrt(x)),
sqrt(x),
sq1, fabs(sq1 - sqrt(x)));
}
printf("sizeof(double)=%d\n", (int)sizeof(double));
return 0;
}
And here's what I get (with gcc and Open Watcom):
0.00000000: sqrtfix0()=0.00003052 / err=3.051758e-05 sqrt()=0.00000000 sqrtfix1()=0.00000000 / err=0.000000e+00
0.00001000: sqrtfix0()=0.00311279 / err=4.948469e-05 sqrt()=0.00316228 sqrtfix1()=0.00316207 / err=2.102766e-07
0.00002000: sqrtfix0()=0.00445557 / err=1.656955e-05 sqrt()=0.00447214 sqrtfix1()=0.00447184 / err=2.974807e-07
0.00003000: sqrtfix0()=0.00543213 / err=4.509667e-05 sqrt()=0.00547723 sqrtfix1()=0.00547720 / err=2.438148e-08
0.00004000: sqrtfix0()=0.00628662 / err=3.793423e-05 sqrt()=0.00632456 sqrtfix1()=0.00632443 / err=1.262553e-07
0.00005000: sqrtfix0()=0.00701904 / err=5.202484e-05 sqrt()=0.00707107 sqrtfix1()=0.00707086 / err=2.060551e-07
0.00006000: sqrtfix0()=0.00772095 / err=2.501943e-05 sqrt()=0.00774597 sqrtfix1()=0.00774593 / err=3.390476e-08
0.00007000: sqrtfix0()=0.00836182 / err=4.783859e-06 sqrt()=0.00836660 sqrtfix1()=0.00836649 / err=1.086198e-07
0.00008000: sqrtfix0()=0.00894165 / err=2.621519e-06 sqrt()=0.00894427 sqrtfix1()=0.00894409 / err=1.777289e-07
sizeof(double)=8
EDIT:
I've missed the fact that the above sqrtfix1() won't work well with large arguments. It can be fixed by appending 28 zeroes to the argument and essentially calculating the exact integer square root of that. This comes at the expense of doing internal calculations in 128-bit arithmetic, but it's pretty straightforward:
fixed sqrtfix2(fixed num)
{
unsigned __int64 numl, numh;
unsigned __int64 resl = 0, resh = 0;
unsigned __int64 bitl = 0, bith = (unsigned __int64)1 << 26;
numl = num << 28;
numh = num >> (64 - 28);
while (bitl | bith)
{
unsigned __int64 tmpl = resl + bitl;
unsigned __int64 tmph = resh + bith + (tmpl < resl);
tmph = numh - tmph - (numl < tmpl);
tmpl = numl - tmpl;
if (tmph & 0x8000000000000000ULL)
{
resl >>= 1;
if (resh & 1) resl |= 0x8000000000000000ULL;
resh >>= 1;
}
else
{
numl = tmpl;
numh = tmph;
resl >>= 1;
if (resh & 1) resl |= 0x8000000000000000ULL;
resh >>= 1;
resh += bith + (resl + bitl < resl);
resl += bitl;
}
bitl >>= 2;
if (bith & 1) bitl |= 0x4000000000000000ULL;
if (bith & 2) bitl |= 0x8000000000000000ULL;
bith >>= 2;
}
return resl;
}
And it gives pretty much the same results (slightly better for 3.43597e+10) than this answer:
0.00000000: sqrtfix0()=0.00003052 / err=3.051758e-05 sqrt()=0.00000000 sqrtfix2()=0.00000000 / err=0.000000e+00
0.00001000: sqrtfix0()=0.00311279 / err=4.948469e-05 sqrt()=0.00316228 sqrtfix2()=0.00316207 / err=2.102766e-07
0.00002000: sqrtfix0()=0.00445557 / err=1.656955e-05 sqrt()=0.00447214 sqrtfix2()=0.00447184 / err=2.974807e-07
0.00003000: sqrtfix0()=0.00543213 / err=4.509667e-05 sqrt()=0.00547723 sqrtfix2()=0.00547720 / err=2.438148e-08
0.00004000: sqrtfix0()=0.00628662 / err=3.793423e-05 sqrt()=0.00632456 sqrtfix2()=0.00632443 / err=1.262553e-07
0.00005000: sqrtfix0()=0.00701904 / err=5.202484e-05 sqrt()=0.00707107 sqrtfix2()=0.00707086 / err=2.060551e-07
0.00006000: sqrtfix0()=0.00772095 / err=2.501943e-05 sqrt()=0.00774597 sqrtfix2()=0.00774593 / err=3.390476e-08
0.00007000: sqrtfix0()=0.00836182 / err=4.783859e-06 sqrt()=0.00836660 sqrtfix2()=0.00836649 / err=1.086198e-07
0.00008000: sqrtfix0()=0.00894165 / err=2.621519e-06 sqrt()=0.00894427 sqrtfix2()=0.00894409 / err=1.777289e-07
2.00000000: sqrtfix0()=1.41419983 / err=1.373327e-05 sqrt()=1.41421356 sqrtfix2()=1.41421356 / err=1.851493e-09
34359700000.00000000: sqrtfix0()=185363.69654846 / err=5.097361e-06 sqrt()=185363.69655356 sqrtfix2()=185363.69655356 / err=1
.164153e-09
Many many years ago I worked on a demo program for a small computer our outfit had built. The computer had a built-in square-root instruction, and we built a simple program to demonstrate the computer doing 16-bit add/subtract/multiply/divide/square-root on a TTY. Alas, it turned out that there was a serious bug in the square root instruction, but we had promised to demo the function. So we created an array of the squares of the values 1-255, then used a simple lookup to match the value typed in to one of the array values. The index was the square root.
来源：https://stackoverflow.com/questions/8721022/how-to-improve-fixed-point-square-root-for-small-values