I have written a C script to implement the inverse Vincenty's formula to calculate the distance between two sets of GPS coordinates based on the equations shown at https://en.wikipedia.org/wiki/Vincenty%27s_formulae
However, my results are different to the results given by this online calculator https://www.cqsrg.org/tools/GCDistance/ and Google maps. My results are consistently around 1.18 times the result of the online calculator.
My function is below, any tips on where I could be going wrong would be very much appreciated!
double get_distance(double lat1, double lon1, double lat2, double lon2)
{
double rad_eq = 6378137.0; //Radius at equator
double flattening = 1 / 298.257223563; //flattenig of earth
double rad_pol = (1 - flattening) * rad_eq; //Radius at poles
double U1,U2,L,lambda,old_lambda,sigma,sin_sig,cos_sig,alpha,cos2sigmam,A,B,C,u_sq,delta_s,dis;
//Convert to radians
lat1=M_PI*lat1/180.0;
lat2=M_PI*lat2/180.0;
lon1=M_PI*lon1/180.0;
lon2=M_PI*lon2/180.0;
//Calculate U1 and U2
U1=atan((1-flattening)*tan(lat1));
U2=atan((1-flattening)*tan(lat2));
L=lon2-lon1;
lambda=L;
double tolerance=pow(10.,-12.);//iteration tollerance should give 0.6mm
double diff=1.;
while (abs(diff)>tolerance)
{
sin_sig=sqrt(pow(cos(U2)*sin(lambda),2.)+pow(cos(U1)*sin(U2)-(sin(U1)*cos(U2)*cos(lambda)),2.));
cos_sig=sin(U1)*cos(U2)+cos(U1)*cos(U2)*cos(lambda);
sigma=atan(sin_sig/cos_sig);
alpha=asin((cos(U1)*cos(U2)*sin(lambda))/(sin_sig));
cos2sigmam=cos(sigma)-(2*sin(U1)*sin(U2))/((pow(cos(alpha),2.)));
C=(flattening/16)*pow(cos(alpha),2.)*(4+(flattening*(4-(3*pow(cos(alpha),2.)))));
old_lambda=lambda;
lambda=L+(1-C)*flattening*sin(alpha)*(sigma+C*sin_sig*(cos2sigmam+C*cos_sig*(-1+2*pow(cos2sigmam,2.))));
diff=abs(old_lambda-lambda);
}
u_sq=pow(cos(alpha),2.)*((pow(rad_eq,2.)-pow(rad_pol,2.))/(pow(rad_pol,2.)));
A=1+(u_sq/16384)*(4096+(u_sq*(-768+(u_sq*(320-(175*u_sq))))));
B=(u_sq/1024)*(256+(u_sq*(-128+(u_sq*(74-(47*u_sq))))));
delta_s=B*sin_sig*(cos2sigmam+(B/4)*(cos_sig*(-1+(2*pow(cos2sigmam,2.)))-(B/6)*cos2sigmam*(-3+(4*pow(sin_sig,2.)))*(-3+(4*pow(cos2sigmam,2.)))));
dis=rad_pol*A*(sigma-delta_s);
//Returns distance in metres
return dis;
}
This formula is not symmetric:
cos_sig = sin(U1)*cos(U2)
+ cos(U1)*cos(U2) * cos(lambda);
And turns out to be wrong, a sin is missing.
Another style of formatting (one including some whitespace) could also help.
Besides the fabs for abs and one sin for that cos I also changed the loop; there were two abs()-calls and diff had to be preset with the while-loop.
I inserted a printf to see how the value progresses.
Some parentheses can be left out. These formulas are really difficult to realize. Some more helper variables could be useful in this jungle of nested math operations.
do {
sin_sig = sqrt(pow( cos(U2) * sin(lambda), 2)
+ pow(cos(U1)*sin(U2)
- (sin(U1)*cos(U2) * cos(lambda))
, 2)
);
cos_sig = sin(U1) * sin(U2)
+ cos(U1) * cos(U2) * cos(lambda);
sigma = atan2(sin_sig, cos_sig);
alpha = asin(cos(U1) * cos(U2) * sin(lambda)
/ sin_sig
);
double cos2alpha = cos(alpha)*cos(alpha); // helper var.
cos2sigmam = cos(sigma) - 2*sin(U1)*sin(U2) / cos2alpha;
C = (flat/16) * cos2alpha * (4 + flat * (4 - 3*cos2alpha));
old_lambda = lambda;
lambda = L + (1-C) * flat * sin(alpha)
*(sigma + C*sin_sig
*(cos2sigmam + C*cos_sig
*(2 * pow(cos2sigmam, 2) - 1)
)
);
diff = fabs(old_lambda - lambda);
printf("%.12f\n", diff);
} while (diff > tolerance);
For 80,80, 0,0 the output is (in km):
0.000885870048
0.000000221352
0.000000000055
0.000000000000
9809.479224
which corresponds to the millimeter with WGS-84.
I'm writing a C program for a PIC micro-controller which needs to do a very specific exponential function. I need to calculate the following:
A = k . (1 - (p/p0)^0.19029)
k and p0 are constant, so it's all pretty simple apart from finding x^0.19029
(p/p0) ratio would always be in the range 0-1.
It works well if I add in math.h and use the power function, except that uses up all of the available 16 kB of program memory. Talk about bloatware! (Rest of program without power function = ~20% flash memory usage; add math.h and power function, =100%).
I'd like the program to do some other things as well. I was wondering if I can write a special case implementation for x^0.19029, maybe involving iteration and some kind of lookup table.
My idea is to generate a look-up table for the function x^0.19029, with perhaps 10-100 values of x in the range 0-1. The code would find a close match, then (somehow) iteratively refine it by re-scaling the lookup table values. However, this is where I get lost because my tiny brain can't visualise the maths involved.
Could this approach work?
Alternatively, I've looked at using Exp(x) and Ln(x), which can be implemented with a Taylor expansion. b^x can the be found with:
b^x = (e^(ln b))^x = e^(x.ln(b))
(See: Wikipedia - Powers via Logarithms)
This looks a bit tricky and complicated to me, though. Am I likely to get the implementation smaller then the compiler's math library, and can I simplify it for my special case (i.e. base = 0-1, exponent always 0.19029)?
Note that RAM usage is OK at the moment, but I've run low on Flash (used for code storage). Speed is not critical. Somebody has already suggested that I use a bigger micro with more flash memory, but that sounds like profligate wastefulness!
[EDIT] I was being lazy when I said "(p/p0) ratio would always be in the range 0-1". Actually it will never reach 0, and I did some calculations last night and decided that in fact a range of 0.3 - 1 would be quite adequate! This mean that some of the simpler solutions below should be suitable. Also, the "k" in the above is 44330, and I'd like the error in the final result to be less than 0.1. I guess that means an error in the (p/p0)^0.19029 needs to be less than 1/443300 or 2.256e-6
Use splines. The relevant part of the function is shown in the figure below. It varies approximately like the 5th root, so the problematic zone is close to p / p0 = 0. There is mathematical theory how to optimally place the knots of splines to minimize the error (see Carl de Boor: A Practical Guide to Splines). Usually one constructs the spline in B form ahead of time (using toolboxes such as Matlab's spline toolbox - also written by C. de Boor), then converts to Piecewise Polynomial representation for fast evaluation.
In C. de Boor, PGS, the function g(x) = sqrt(x + 1) is actually taken as an example (Chapter 12, Example II). This is exactly what you need here. The book comes back to this case a few times, since it is admittedly a hard problem for any interpolation scheme due to the infinite derivatives at x = -1. All software from PGS is available for free as PPPACK in netlib, and most of it is also part of SLATEC (also from netlib).
Edit (Removed)
(Multiplying by x once does not significantly help, since it only regularizes the first derivative, while all other derivatives at x = 0 are still infinite.)
Edit 2
My feeling is that optimally constructed splines (following de Boor) will be best (and fastest) for relatively low accuracy requirements. If the accuracy requirements are high (say 1e-8), one may be forced to get back to the algorithms that mathematicians have been researching for centuries. At this point, it may be best to simply download the sources of glibc and copy (provided GPL is acceptable) whatever is in
glibc-2.19/sysdeps/ieee754/dbl-64/e_pow.c
Since we don't have to include the whole math.h, there shouldn't be a problem with memory, but we will only marginally profit from having a fixed exponent.
Edit 3
Here is an adapted version of e_pow.c from netlib, as found by #Joni. This seems to be the grandfather of glibc's more modern implementation mentioned above. The old version has two advantages: (1) It is public domain, and (2) it uses a limited number of constants, which is beneficial if memory is a tight resource (glibc's version defines over 10000 lines of constants!). The following is completely standalone code, which calculates x^0.19029 for 0 <= x <= 1 to double precision (I tested it against Python's power function and found that at most 2 bits differed):
#define __LITTLE_ENDIAN
#ifdef __LITTLE_ENDIAN
#define __HI(x) *(1+(int*)&x)
#define __LO(x) *(int*)&x
#else
#define __HI(x) *(int*)&x
#define __LO(x) *(1+(int*)&x)
#endif
static const double
bp[] = {1.0, 1.5,},
dp_h[] = { 0.0, 5.84962487220764160156e-01,}, /* 0x3FE2B803, 0x40000000 */
dp_l[] = { 0.0, 1.35003920212974897128e-08,}, /* 0x3E4CFDEB, 0x43CFD006 */
zero = 0.0,
one = 1.0,
two = 2.0,
two53 = 9007199254740992.0, /* 0x43400000, 0x00000000 */
/* poly coefs for (3/2)*(log(x)-2s-2/3*s**3 */
L1 = 5.99999999999994648725e-01, /* 0x3FE33333, 0x33333303 */
L2 = 4.28571428578550184252e-01, /* 0x3FDB6DB6, 0xDB6FABFF */
L3 = 3.33333329818377432918e-01, /* 0x3FD55555, 0x518F264D */
L4 = 2.72728123808534006489e-01, /* 0x3FD17460, 0xA91D4101 */
L5 = 2.30660745775561754067e-01, /* 0x3FCD864A, 0x93C9DB65 */
L6 = 2.06975017800338417784e-01, /* 0x3FCA7E28, 0x4A454EEF */
P1 = 1.66666666666666019037e-01, /* 0x3FC55555, 0x5555553E */
P2 = -2.77777777770155933842e-03, /* 0xBF66C16C, 0x16BEBD93 */
P3 = 6.61375632143793436117e-05, /* 0x3F11566A, 0xAF25DE2C */
P4 = -1.65339022054652515390e-06, /* 0xBEBBBD41, 0xC5D26BF1 */
P5 = 4.13813679705723846039e-08, /* 0x3E663769, 0x72BEA4D0 */
lg2 = 6.93147180559945286227e-01, /* 0x3FE62E42, 0xFEFA39EF */
lg2_h = 6.93147182464599609375e-01, /* 0x3FE62E43, 0x00000000 */
lg2_l = -1.90465429995776804525e-09, /* 0xBE205C61, 0x0CA86C39 */
ovt = 8.0085662595372944372e-0017, /* -(1024-log2(ovfl+.5ulp)) */
cp = 9.61796693925975554329e-01, /* 0x3FEEC709, 0xDC3A03FD =2/(3ln2) */
cp_h = 9.61796700954437255859e-01, /* 0x3FEEC709, 0xE0000000 =(float)cp */
cp_l = -7.02846165095275826516e-09, /* 0xBE3E2FE0, 0x145B01F5 =tail of cp_h*/
ivln2 = 1.44269504088896338700e+00, /* 0x3FF71547, 0x652B82FE =1/ln2 */
ivln2_h = 1.44269502162933349609e+00, /* 0x3FF71547, 0x60000000 =24b 1/ln2*/
ivln2_l = 1.92596299112661746887e-08; /* 0x3E54AE0B, 0xF85DDF44 =1/ln2 tail*/
double pow0p19029(double x)
{
double y = 0.19029e+00;
double z,ax,z_h,z_l,p_h,p_l;
double y1,t1,t2,r,s,t,u,v,w;
int i,j,k,n;
int hx,hy,ix,iy;
unsigned lx,ly;
hx = __HI(x); lx = __LO(x);
hy = __HI(y); ly = __LO(y);
ix = hx&0x7fffffff; iy = hy&0x7fffffff;
ax = x;
/* special value of x */
if(lx==0) {
if(ix==0x7ff00000||ix==0||ix==0x3ff00000){
z = ax; /*x is +-0,+-inf,+-1*/
return z;
}
}
s = one; /* s (sign of result -ve**odd) = -1 else = 1 */
double ss,s2,s_h,s_l,t_h,t_l;
n = ((ix)>>20)-0x3ff;
j = ix&0x000fffff;
/* determine interval */
ix = j|0x3ff00000; /* normalize ix */
if(j<=0x3988E) k=0; /* |x|<sqrt(3/2) */
else if(j<0xBB67A) k=1; /* |x|<sqrt(3) */
else {k=0;n+=1;ix -= 0x00100000;}
__HI(ax) = ix;
/* compute ss = s_h+s_l = (x-1)/(x+1) or (x-1.5)/(x+1.5) */
u = ax-bp[k]; /* bp[0]=1.0, bp[1]=1.5 */
v = one/(ax+bp[k]);
ss = u*v;
s_h = ss;
__LO(s_h) = 0;
/* t_h=ax+bp[k] High */
t_h = zero;
__HI(t_h)=((ix>>1)|0x20000000)+0x00080000+(k<<18);
t_l = ax - (t_h-bp[k]);
s_l = v*((u-s_h*t_h)-s_h*t_l);
/* compute log(ax) */
s2 = ss*ss;
r = s2*s2*(L1+s2*(L2+s2*(L3+s2*(L4+s2*(L5+s2*L6)))));
r += s_l*(s_h+ss);
s2 = s_h*s_h;
t_h = 3.0+s2+r;
__LO(t_h) = 0;
t_l = r-((t_h-3.0)-s2);
/* u+v = ss*(1+...) */
u = s_h*t_h;
v = s_l*t_h+t_l*ss;
/* 2/(3log2)*(ss+...) */
p_h = u+v;
__LO(p_h) = 0;
p_l = v-(p_h-u);
z_h = cp_h*p_h; /* cp_h+cp_l = 2/(3*log2) */
z_l = cp_l*p_h+p_l*cp+dp_l[k];
/* log2(ax) = (ss+..)*2/(3*log2) = n + dp_h + z_h + z_l */
t = (double)n;
t1 = (((z_h+z_l)+dp_h[k])+t);
__LO(t1) = 0;
t2 = z_l-(((t1-t)-dp_h[k])-z_h);
/* split up y into y1+y2 and compute (y1+y2)*(t1+t2) */
y1 = y;
__LO(y1) = 0;
p_l = (y-y1)*t1+y*t2;
p_h = y1*t1;
z = p_l+p_h;
j = __HI(z);
i = __LO(z);
/*
* compute 2**(p_h+p_l)
*/
i = j&0x7fffffff;
k = (i>>20)-0x3ff;
n = 0;
if(i>0x3fe00000) { /* if |z| > 0.5, set n = [z+0.5] */
n = j+(0x00100000>>(k+1));
k = ((n&0x7fffffff)>>20)-0x3ff; /* new k for n */
t = zero;
__HI(t) = (n&~(0x000fffff>>k));
n = ((n&0x000fffff)|0x00100000)>>(20-k);
if(j<0) n = -n;
p_h -= t;
}
t = p_l+p_h;
__LO(t) = 0;
u = t*lg2_h;
v = (p_l-(t-p_h))*lg2+t*lg2_l;
z = u+v;
w = v-(z-u);
t = z*z;
t1 = z - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))));
r = (z*t1)/(t1-two)-(w+z*w);
z = one-(r-z);
__HI(z) += (n<<20);
return s*z;
}
Clearly, 50+ years of research have gone into this, so it's probably very hard to do any better. (One has to appreciate that there are 0 loops, only 2 divisions, and only 6 if statements in the whole algorithm!) The reason for this is, again, the behavior at x = 0, where all derivatives diverge, which makes it extremely hard to keep the error under control: I once had a spline representation with 18 knots that was good up to x = 1e-4, with absolute and relative errors < 5e-4 everywhere, but going to x = 1e-5 ruined everything again.
So, unless the requirement to go arbitrarily close to zero is relaxed, I recommend using the adapted version of e_pow.c given above.
Edit 4
Now that we know that the domain 0.3 <= x <= 1 is sufficient, and that we have very low accuracy requirements, Edit 3 is clearly overkill. As #MvG has demonstrated, the function is so well behaved that a polynomial of degree 7 is sufficient to satisfy the accuracy requirements, which can be considered a single spline segment. #MvG's solution minimizes the integral error, which already looks very good.
The question arises as to how much better we can still do? It would be interesting to find the polynomial of a given degree that minimizes the maximum error in the interval of interest. The answer is the minimax
polynomial, which can be found using Remez' algorithm, which is implemented in the Boost library. I like #MvG's idea to clamp the value at x = 1 to 1, which I will do as well. Here is minimax.cpp:
#include <ostream>
#define TARG_PREC 64
#define WORK_PREC (TARG_PREC*2)
#include <boost/multiprecision/cpp_dec_float.hpp>
typedef boost::multiprecision::number<boost::multiprecision::cpp_dec_float<WORK_PREC> > dtype;
using boost::math::pow;
#include <boost/math/tools/remez.hpp>
boost::shared_ptr<boost::math::tools::remez_minimax<dtype> > p_remez;
dtype f(const dtype& x) {
static const dtype one(1), y(0.19029);
return one - pow(one - x, y);
}
void out(const char *descr, const dtype& x, const char *sep="") {
std::cout << descr << boost::math::tools::real_cast<double>(x) << sep << std::endl;
}
int main() {
dtype a(0), b(0.7); // range to optimise over
bool rel_error(false), pin(true);
int orderN(7), orderD(0), skew(0), brake(50);
int prec = 2 + (TARG_PREC * 3010LL)/10000;
std::cout << std::scientific << std::setprecision(prec);
p_remez.reset(new boost::math::tools::remez_minimax<dtype>(
&f, orderN, orderD, a, b, pin, rel_error, skew, WORK_PREC));
out("Max error in interpolated form: ", p_remez->max_error());
p_remez->set_brake(brake);
unsigned i, count(50);
for (i = 0; i < count; ++i) {
std::cout << "Stepping..." << std::endl;
dtype r = p_remez->iterate();
out("Maximum Deviation Found: ", p_remez->max_error());
out("Expected Error Term: ", p_remez->error_term());
out("Maximum Relative Change in Control Points: ", r);
}
boost::math::tools::polynomial<dtype> n = p_remez->numerator();
for(i = n.size(); i--; ) {
out("", n[i], ",");
}
}
Since all parts of boost that we use are header-only, simply build with:
c++ -O3 -I<path/to/boost/headers> minimax.cpp -o minimax
We finally get the coefficients, which are after multiplication by 44330:
24538.3409, -42811.1497, 34300.7501, -11284.1276, 4564.5847, 3186.7541, 8442.5236, 0.
The following error plot demonstrates that this is really the best possible degree-7 polynomial approximation, since all extrema are of equal magnitude (0.06659):
Should the requirements ever change (while still keeping well away from 0!), the C++ program above can be simply adapted to spit out the new optimal polynomial approximation.
Instead of a lookup table, I'd use a polynomial approximation:
1 - x0.19029 ≈ - 1073365.91783x15 + 8354695.40833x14 - 29422576.6529x13 + 61993794.537x12 - 87079891.4988x11 + 86005723.842x10 - 61389954.7459x9 + 32053170.1149x8 - 12253383.4372x7 + 3399819.97536x6 - 672003.142815x5 + 91817.6782072x4 - 8299.75873768x3 + 469.530204564x2 - 16.6572179869x + 0.722044145701
Or in code:
double f(double x) {
double fx;
fx = - 1073365.91783;
fx = fx*x + 8354695.40833;
fx = fx*x - 29422576.6529;
fx = fx*x + 61993794.537;
fx = fx*x - 87079891.4988;
fx = fx*x + 86005723.842;
fx = fx*x - 61389954.7459;
fx = fx*x + 32053170.1149;
fx = fx*x - 12253383.4372;
fx = fx*x + 3399819.97536;
fx = fx*x - 672003.142815;
fx = fx*x + 91817.6782072;
fx = fx*x - 8299.75873768;
fx = fx*x + 469.530204564;
fx = fx*x - 16.6572179869;
fx = fx*x + 0.722044145701;
return fx;
}
I computed this in sage using the least squares approach:
f(x) = 1-x^(19029/100000) # your function
d = 16 # number of terms, i.e. degree + 1
A = matrix(d, d, lambda r, c: integrate(x^r*x^c, (x, 0, 1)))
b = vector([integrate(x^r*f(x), (x, 0, 1)) for r in range(d)])
A.solve_right(b).change_ring(RDF)
Here is a plot of the error this will entail:
Blue is the error from my 16 term polynomial, while red is the error you'd get from piecewise linear interpolation with 16 equidistant values. As you can see, both errors are quite small for most parts of the range, but will become really huge close to x=0. I actually clipped the plot there. If you can somehow narrow the range of possible values, you could use that as the domain for the integration, and obtain an even better fit for the relevant range. At the cost of worse fit outside, of course. You could also increase the number of terms to obtain a closer fit, although that might also lead to higher oscillations.
I guess you can also combine this approach with the one Stefan posted: use his to split the domain into several parts, then use mine to find a close low degree polynomial for each part.
Update
Since you updated the specification of your question, with regard to both the domain and the error, here is a minimal solution to fit those requirements:
44330(1 - x0.19029) ≈ + 23024.9160933(1-x)7 - 39408.6473636(1-x)6 + 31379.9086193(1-x)5 - 10098.7031260(1-x)4 + 4339.44098317(1-x)3 + 3202.85705860(1-x)2 + 8442.42528906(1-x)
double f(double x) {
double fx, x1 = 1. - x;
fx = + 23024.9160933;
fx = fx*x1 - 39408.6473636;
fx = fx*x1 + 31379.9086193;
fx = fx*x1 - 10098.7031260;
fx = fx*x1 + 4339.44098317;
fx = fx*x1 + 3202.85705860;
fx = fx*x1 + 8442.42528906;
fx = fx*x1;
return fx;
}
I integrated x from 0.293 to 1 or equivalently 1 - x from 0 to 0.707 to keep the worst oscillations outside the relevant domain. I also omitted the constant term, to ensure an exact result at x=1. The maximal error for the range [0.3, 1] now occurs at x=0.3260 and amounts to 0.0972 < 0.1. Here is an error plot, which of course has bigger absolute errors than the one above due to the scale factor k=44330 which has been included here.
I can also state that the first three derivatives of the function will have constant sign over the range in question, so the function is monotonic, convex, and in general pretty well-behaved.
Not meant to answer the question, but it illustrates the Road Not To Go, and thus may be helpful:
This quick-and-dirty C code calculates pow(i, 0.19029) for 0.000 to 1.000 in steps of 0.01. The first half displays the error, in percents, when stored as 1/65536ths (as that theoretically provides slightly over 4 decimals of precision). The second half shows both interpolated and calculated values in steps of 0.001, and the difference between these two.
It kind of looks okay if you read from the bottom up, all 100s and 99.99s there, but about the first 20 values from 0.001 to 0.020 are worthless.
#include <stdio.h>
#include <math.h>
float powers[102];
int main (void)
{
int i, as_int;
double as_real, low, high, delta, approx, calcd, diff;
printf ("calculating and storing:\n");
for (i=0; i<=101; i++)
{
as_real = pow(i/100.0, 0.19029);
as_int = (int)round(65536*as_real);
powers[i] = as_real;
diff = 100*as_real/(as_int/65536.0);
printf ("%.5f %.5f %.5f ~ %.3f\n", i/100.0, as_real, as_int/65536.0, diff);
}
printf ("\n");
printf ("-- interpolating in 1/10ths:\n");
for (i=0; i<1000; i++)
{
as_real = i/1000.0;
low = powers[i/10];
high = powers[1+i/10];
delta = (high-low)/10.0;
approx = low + (i%10)*delta;
calcd = pow(as_real, 0.19029);
diff = 100.0*approx/calcd;
printf ("%.5f ~ %.5f = %.5f +/- %.5f%%\n", as_real, approx, calcd, diff);
}
return 0;
}
You can find a complete, correct standalone implementation of pow in fdlibm. It's about 200 lines of code, about half of which deal with special cases. If you remove the code that deals with special cases you're not interested in I doubt you'll have problems including it in your program.
LutzL's answer is a really good one: Calculate your power as (x^1.52232)^(1/8), computing the inner power by spline interpolation or another method. The eighth root deals with the pathological non-differentiable behavior near zero. I took the liberty of mocking up an implementation this way. The below, however, only does a linear interpolation to do x^1.52232, and you'd need to get the full coefficients using your favorite numerical mathematics tools. You'll adding scarcely 40 lines of code to get your needed power, plus however many knots you choose to use for your spline, as dicated by your required accuracy.
Don't be scared by the #include <math.h>; it's just for benchmarking the code.
#include <stdio.h>
#include <math.h>
double my_sqrt(double x) {
/* Newton's method for a square root. */
int i = 0;
double res = 1.0;
if (x > 0) {
for (i = 0; i < 10; i++) {
res = 0.5 * (res + x / res);
}
} else {
res = 0.0;
}
return res;
}
double my_152232(double x) {
/* Cubic spline interpolation for x ** 1.52232. */
int i = 0;
double res = 0.0;
/* coefs[i] will give the cubic polynomial coefficients between x =
i and x = i+1. Out of laziness, the below numbers give only a
linear interpolation. You'll need to do some work and research
to get the spline coefficients. */
double coefs[3][4] = {{0.0, 1.0, 0.0, 0.0},
{-0.872526, 1.872526, 0.0, 0.0},
{-2.032706, 2.452616, 0.0, 0.0}};
if ((x >= 0) && (x < 3.0)) {
i = (int) x;
/* Horner's method cubic. */
res = (((coefs[i][3] * x + coefs[i][2]) * x) + coefs[i][1] * x)
+ coefs[i][0];
} else if (x >= 3.0) {
/* Scaled x ** 1.5 once you go off the spline. */
res = 1.024824 * my_sqrt(x * x * x);
}
return res;
}
double my_019029(double x) {
return my_sqrt(my_sqrt(my_sqrt(my_152232(x))));
}
int main() {
int i;
double x = 0.0;
for (i = 0; i < 1000; i++) {
x = 1e-2 * i;
printf("%f %f %f \n", x, my_019029(x), pow(x, 0.19029));
}
return 0;
}
EDIT: If you're just interested in a small region like [0,1], even simpler is to peel off one sqrt(x) and compute x^1.02232, which is quite well behaved, using a Taylor series:
double my_152232(double x) {
double part_050000 = my_sqrt(x);
double part_102232 = 1.02232 * x + 0.0114091 * x * x - 3.718147e-3 * x * x * x;
return part_102232 * part_050000;
}
This gets you within 1% of the exact power for approximately [0.1,6], though getting the singularity exactly right is always a challenge. Even so, this three-term Taylor series gets you within 2.3% for x = 0.001.
I have a O(N^4) image processing loop and after profiling it (Using Intel Vtune 2013), I see that the number of Instructions retired is reduced drastically. I need help understanding this behavior on a multicore architecture. (I'm using Intel Xeon x5365- has 8 cores with shared L2 cache for every 2 cores). And also the no of branch mis-predictions have increased drastically!!
///////////////EDITS/////////// A sample of my non-Unrolled code is shown below:
for(imageNo =0; imageNo<496;imageNo++){
for (unsigned int k=0; k<256; k++)
{
double z = O_L + (double)k * R_L;
for (unsigned int j=0; j<256; j++)
{
double y = O_L + (double)j * R_L;
for (unsigned int i=0; i<256; i++)
{
double x[1] = {O_L + (double)i * R_L} ;
double w_n = (A_n[2] * x[0] + A_n[5] * y + A_n[8] * z + A_n[11]) ;
double u_n = ((A_n[0] * x[0] + A_n[3] * y + A_n[6] * z + A_n[9] ) / w_n);
double v_n = ((A_n[1] * x[0] + A_n[4] * y + A_n[7] * z + A_n[10]) / w_n);
for(int loop=0; loop<1;loop++)
{
px_x[loop] = (int) floor(u_n);
px_y[loop] = (int) floor(v_n);
alpha[loop] = u_n - px_x[loop] ;
beta[loop] = v_n - px_y[loop] ;
}
///////////////////(i,j) pixels ///////////////////////////////
if (px_x[0]>=0 && px_x[0]<(int)threadCopy[0].S_x && px_y[0]>=0 && px_y[0]<(int)threadCopy[0].S_y)
pixel_1[0] = threadCopy[0].I_n[px_y[0] * threadCopy[0].S_x + px_x[0]];
else
pixel_1[0] = 0.0;
if (px_x[0]+1>=0 && px_x[0]+1<(int)threadCopy[0].S_x && px_y[0]>=0 && px_y[0]<(int)threadCopy[0].S_y)
pixel_1[2] = threadCopy[0].I_n[px_y[0] * threadCopy[0].S_x + (px_x[0]+1)];
else
pixel_1[2] = 0.0;
/////////////////// (i+1, j) pixels/////////////////////////
if (px_x[0]>=0 && px_x[0]<(int)threadCopy[0].S_x && px_y[0]+1>=0 && px_y[0]+1<(int)threadCopy[0].S_y)
pixel_1[1] = threadCopy[0].I_n[(px_y[0]+1) * threadCopy[0].S_x + px_x[0]];
else
pixel_1[1] = 0.0;
if (px_x[0]+1>=0 && px_x[0]+1<(int)threadCopy[0].S_x && px_y[0]+1>=0 && px_y[0]+1<(int)threadCopy[0].S_y)
pixel_1[3] = threadCopy[0].I_n[(px_y[0]+1) * threadCopy[0].S_x + (px_x[0]+1)];
else
pixel_1[3] = 0.0;
pix_1 = (1.0 - alpha[0]) * (1.0 - beta[0]) * pixel_1[0] + (1.0 - alpha[0]) * beta[0] * pixel_1[1]
+ alpha[0] * (1.0 - beta[0]) * pixel_1[2] + alpha[0] * beta[0] * pixel_1[3];
f_L[k * L * L + j * L + i] += (float)(1.0 / (w_n * w_n) * pix_1);
}
}
}
}
I'm unrolling the inner most loop by 4 iterations.(You will have a general ideal how I stripped the loop. Basically i created an array of Array[4] and filled respective vales in it.) Doing the math, I'm reducing the total no of iterations by 75%. Say there are 4 loop handling instructions for every loop (load i, inc i, cmp i, jle loop), the total no of instructions after unrolling should reduce by (256-64)*4*256*256*496=24.96G.
The profiled results are:
Before UnRolling: Instr retired: 3.1603T no of branch mis-predictions: 96 million
After UnRolling: Instr retired: 2.642240T no of branch mis-predictions: 144 million
The no instr retired decreased by 518.06G . I have no clue how this is happening. I would appreciate any help regarding this (Even if it is remote possibility for its occurrence) . Also, I would like to know why are branch mis-predictions increasing. Thanks in advance!
It is not clear where gcc would be reducing the number of instructions. It is possible that increased register pressure might encourage gcc to use load+operate instructions (so the same number of primitive operations but fewer instructions). The index for f_L would only be incremented once per innermost loop, but this would only save 6.2G (3*64*256*256*496) instructions. (By the way, the loop overhead should only be three instructions since i should remain in a register.)
The following pseudo-assembly (for a RISC-like ISA) using a two-way unrolling shows how an increment can be saved:
// the address of f_L[k * L * L + j * L + i] is in r1
// (float)(1.0 / (w_n * w_n) * pix_1) results are in f1 and f2
load-single f9 [r1]; // load float at address in r1 to register f9
add-single f9 f9 f1; // f9 = f9 + f1
store-single [r1] f9; // store float in f9 to address in r1
load-single f10 4[r1]; // load float at address of r1+4 to f10
add-single f10 f10 f2; // f10 = f10 + f2
store-single 4[r1] f10; // store float in f10 to address of r1+4
add r1 r1 #8; // increase the address by 8 bytes
The trace of two iterations of the non-unrolled version would look more like:
load-single f9 [r1]; // load float at address of r1 to f9
add-single f9 f9 f2; // f9 = f9 + f2
store-single [r1] f9; // store float in f9 to address of r1
add r1 r1 #4; // increase the address by 4 bytes
...
load-single f9 [r1]; // load float at address of r1 to f9
add-single f9 f9 f2; // f9 = f9 + f2
store-single [r1] f9; // store float in f9 to address of r1
add r1 r1 #4; // increase the address by 4 bytes
Because memory addressing instructions commonly include adding an immediate offset (Itanium is an unusual exception) and the pipelines are not generally implemented to optimize the case when the immediate is zero, using a non-zero immediate offset is generally "free". (It certainly reduces the number of instructions—7 vs. 8 in this case—, but generally it also improves performance.)
With respect to branch prediction, the according to Agner Fog's The microarchitecture of Intel, AMD and VIA CPUs: An optimization guide for assembly programmers and compiler makers(PDF) the Core2 microarchitecture's branch predictor uses an 8 bit global history. This means that it tracks the results for the last 8 branches and uses these 8 bits (along with bits from the instruction address) to index a table. This allows correlations between nearby branches to be recognized.
For your code, the branch corresponding to, e.g., the 8th previous branch is not the same branch in each iteration (since short-circuiting is used), so it is not easy to conceptualize how well correlations would be recognized.
Some correlations in branches are obvious. If px_x[0]>=0 is true, px_x[0]+1>=0 will also be true. If px_x[0] <(int)threadCopy[0].S_x is true, then px_x[0]+1 <(int)threadCopy[0].S_x is likely to be true.
If the unrolling is done such that px_x[n] is tested for all four values of n then these correlations would be pushed farther away so that the results are not used by the branch predictor.
Some optimization possibilities
Although you did not ask about any optimization possibilities, I am going to offer some avenues for exploration.
First, for the branches, if not being strictly portable is OK, the test x>=0 && x<y can be simplified to (unsigned)x<(unsigned)y. This is not strictly portable because, e.g., a machine could theoretically represent negative numbers in a sign-magnitude format with the most significant bit as the sign and negative indicated by a zero bit. For the common representations of signed integers, such a reinterpreting cast will work as long as y is a positive signed integer since a negative x value will have the most significant bit set and so be larger than y interpreted as an unsigned integer.
Second, the number of branches can be significantly reduced by using the 100% correlations for either px_x or px_y:
if ((unsigned) px_y[0]<(unsigned int)threadCopy[0].S_y)
{
if ((unsigned)px_x[0]<(unsigned int)threadCopy[0].S_x)
pixel_1[0] = threadCopy[0].I_n[px_y[0] * threadCopy[0].S_x + px_x[0]];
else
pixel_1[0] = 0.0;
if ((unsigned)px_x[0]+1<(unsigned int)threadCopy[0].S_x)
pixel_1[2] = threadCopy[0].I_n[px_y[0] * threadCopy[0].S_x + (px_x[0]+1)];
else
pixel_1[2] = 0.0;
}
if ((unsigned)px_y[0]+1<(unsigned int)threadCopy[0].S_y)
{
if ((unsigned)px_x[0]<(unsigned int)threadCopy[0].S_x)
pixel_1[1] = threadCopy[0].I_n[(px_y[0]+1) * threadCopy[0].S_x + px_x[0]];
else
pixel_1[1] = 0.0;
if ((unsigned)px_x[0]+1<(unsigned int)threadCopy[0].S_x)
pixel_1[3] = threadCopy[0].I_n[(px_y[0]+1) * threadCopy[0].S_x + (px_x[0]+1)];
else
pixel_1[3] = 0.0;
}
(If the above section of code is replicated for unrolling, it should probably be replicated as a block rather than interleaving tests for different px_x and px_y values to allow the px_y branch to be near the px_y+1 branch and the first px_x branch to be near the other px_x branch and the px_x+1 branches.)
Another possible optimization is changing the calculation of w_n into a calculation of its reciprocal. This would change a multiply and three divisions into four multiplies and one division. Division is much more expensive than multiplication. In addition, calculating an approximate reciprocal is much more SIMD-friendly since there are usually reciprocal estimate instructions that provide a starting point which can be refined by the Newton-Raphson method.
If even worse obfuscation of the code is acceptable, you might consider changing code like double y = O_L + (double)j * R_L; into double y = O_L; ... y += R_L;. (I ran a test, and gcc does not seem to recognize this optimization, probably because of the use of floating point and the cast to double.) Thus:
for(int imageNo =0; imageNo<496;imageNo++){
double z = O_L;
for (unsigned int k=0; k<256; k++)
{
double y = O_L;
for (unsigned int j=0; j<256; j++)
{
double x[1]; x[0] = O_L;
for (unsigned int i=0; i<256; i++)
{
...
x[0] += R_L ;
} // end of i loop
y += R_L;
} // end of j loop
z += R_L;
} // end of k loop
} // end of imageNo loop
I am guessing that this would only modest improve performance, so the obfuscation cost would be higher relative to the benefit.
Another change that might be worth trying is incorporating some of the pix_1 calculation into the sections conditionally setting pixel_1[] values. This would significantly obfuscate the code and might not have much benefit. In addition, it might make autovectorization by the compiler more difficult. (With conditionally setting the values to the appropriate I_n or to zero, an SIMD comparison could set each element to -1 or 0 and a simple and with the I_n value would provide the correct value. In this case, the overhead of forming the I_n vector would probably not be worthwhile given that Core2 only supports 2-wide double precision SIMD, but with gather support or even just a longer vector the tradeoffs might change.)
However, this change would increase the size of basic blocks and reduce the amount of computation when any of px_x and px_y are out of range (I am guessing this is uncommon, so the benefit would be very small at best).
double pix_1 = 0.0;
double alpha_diff = 1.0 - alpha;
if ((unsigned) px_y[0]<(unsigned int)threadCopy[0].S_y)
{
double beta_diff = 1.0 - beta;
if ((unsigned)px_x[0]<(unsigned int)threadCopy[0].S_x)
pix1 += alpha_diff * beta_diff
* threadCopy[0].I_n[px_y[0] * threadCopy[0].S_x + px_x[0]];
// no need for else statement since pix1 is already zeroed and not
// adding the pixel_1[0] factor is the same as zeroing pixel_1[0]
if ((unsigned)px_x[0]+1<(unsigned int)threadCopy[0].S_x)
pix1 += alpha * beta_diff
* threadCopy[0].I_n[px_y[0] * threadCopy[0].S_x + (px_x[0]+1)];
}
if ((unsigned)px_y[0]+1<(unsigned int)threadCopy[0].S_y)
{
if ((unsigned)px_x[0]<(unsigned int)threadCopy[0].S_x)
pix1 += alpha_diff * beta
* threadCopy[0].I_n[(px_y[0]+1) * threadCopy[0].S_x + px_x[0]];
if ((unsigned)px_x[0]+1<(unsigned int)threadCopy[0].S_x)
pix1 += alpha * beta
* threadCopy[0].I_n[(px_y[0]+1) * threadCopy[0].S_x + (px_x[0]+1)];
}
Ideally, code like yours would be vectorized, but I do not know how to get gcc to recognize the opportunities, how to express the opportunities using intrinsics, nor whether significant effort at manually vectorizing this code would be worthwhile with an SIMD width of only two.
I am not a programmer (just someone who likes learning and thinking about computer architecture) and I have a significant inclination toward micro-optimization (as clear from the above), so the above proposals should be considered in that light.