I want to estimate multiplicity of polynomial roots.
I have found some info about it, choosed the test example and made c program
Here should be 4 roots. One simple root and one with multiplicity 3.
#include <complex.h>
#include <math.h>
#include <stdio.h>
complex long double z0 = +1.5; // exact period = 1 stability = 3.000000000000000000 multiplicity = ?
complex long double z1 = -0.5; // exact period = 2 stability = 0.999999999999900080 multiplicity = ?
complex long double c = -0.75; // parameter of the f function
/*
https://en.wikibooks.org/wiki/Fractals/Mathematics/Newton_method
*/
int GiveMultiplicity(const complex long double c, const complex long double z0 , const int pMax){
complex long double z = z0;
complex long double d = 1.0; /* d = first derivative with respect to z */
complex long double e = 0.0; // second derivative with respect to z
complex long double m;
int multiplicity;
int p;
for (p=0; p < pMax; p++){
d = 2*z*d; // f' = first derivative with respect to z */
e = 2*(d*d +z*e); // f'' = second derivative with respect to z
z = z*z +c ; // f = complex quadratic polynomial
}
m = (d*d)/(d*d -z*e);
multiplicity = (int) round(cabs(m));
return multiplicity;
}
int main(){
int m;
m = GiveMultiplicity(c, z0, 1);
printf("m = %d \n", m);
m = GiveMultiplicity(c, z1, 1);
printf("m = %d \n", m);
m = GiveMultiplicity(c, z1, 2);
printf("m = %d \n", m);
return 0;
}
The result is :
m=1
m=1
m=1
Is it good ? Maybe I should simply add the results ?
Good results using symbolic computations are roots: [ 3/2, -1/2] and its multiplicities : [1,3]
Here is a graph of the function f(z)= (z^2-0.75)^2-z-0.75 = z^4-1.5*z^2-z-3/16
Is it possibly to compute the similar values numerically ?
You do this with contour integration, see here. Software is available.
Summary of changes:
evaluate e before evaluating d inside the loop;
when subtracting z0 from z after the loop, you also need to subtract 1 from d to match;
perturb input a small amount from true root location to avoid 0/0 = NaN result: h must be small enough, but not too small...
Complete program:
#include <complex.h>
#include <math.h>
#include <stdio.h>
complex long double h = 1.0e-6; // perturb a little; not too big, not too small
complex long double z0 = +1.5; // exact period = 1 stability = 3.000000000000000000 multiplicity = ?
complex long double z1 = -0.5; // exact period = 2 stability = 0.999999999999900080 multiplicity = ?
complex long double c = -0.75; // parameter of the f function
/*
https://en.wikibooks.org/wiki/Fractals/Mathematics/Newton_method
*/
int GiveMultiplicity(const complex long double c, const complex long double z0, const int pMax){
complex long double z = z0;
complex long double d = 1.0; /* d = first derivative with respect to z */
complex long double e = 0.0; // second derivative with respect to z
complex long double m;
int multiplicity;
int p;
for (p=0; p < pMax; p++){
e = 2*(d*d +z*e); // f'' = second derivative with respect to z
d = 2*z*d; // f' = first derivative with respect to z */
z = z*z +c ; // f = complex quadratic polynomial
}
d = d - 1;
z = z - z0;
m = (d*d)/(d*d -z*e);
multiplicity = (int) round(cabs(m));
return multiplicity;
}
int main(){
int m;
m = GiveMultiplicity(c, z0 + h, 1);
printf("m = %d\n", m);
m = GiveMultiplicity(c, z1 + h, 1);
printf("m = %d\n", m);
m = GiveMultiplicity(c, z1 + h, 2);
printf("m = %d\n", m);
return 0;
}
Output:
m = 1
m = 1
m = 3
I have found one error im my initial program. Function for finding periodic points should be
f^n(z) - z
so
for (p=0; p < pMax; p++){
d = 2*z*d; // f' = first derivative with respect to z */
e = 2*(d*d +z*e); // f'' = second derivative with respect to z
z = z*z +c ; // f = complex quadratic polynomial
}
z = z - z0; // new line
I have choosed the method based on the geometrical notation of the root
It is described in The Fundamental Theorem of Algebra: A Visual Approach by Daniel J. Velleman
I count how many times color chages along a circle around root.
I use carg function which returns the phase angle of z in the interval [−π; π]. So count the sign change of the argument and divide it by 2. This estimates the multiplicity of the root.
It is probly the same method as above, but easier to understand and implement for me.
Here is the image of dynamical plane
before transformation:
and after f(z):
and the code:
// gcc p.c -Wall -lm
// ./a.out
#include <complex.h>
#include <math.h>
#include <stdio.h>
// parameter c of the function fc(z) = z^2+c is c = -0.7500000000000000 ; 0.0000000000000000
const long double pi = 3.1415926535897932384626433832795029L;
long double EPS2 = 1e-18L*1e-18L; //
complex double c = -0.75;
complex double z = 1.5; //-0.5;
//https://stackoverflow.com/questions/1903954/is-there-a-standard-sign-function-signum-sgn-in-c-c
int sign(long double x){
if (x > 0.0) return 1;
if (x < 0.0) return -1;
return 0;
}
int DifferentSign(long double x, long double y){
if (sign(x)!=sign(y)) return 1;
return 0;
}
long double complex Give_z0(long double InternalAngleInTurns, long double radius )
{
//0 <= InternalAngleInTurns <=1
long double a = InternalAngleInTurns *2.0*pi; // from turns to radians
long double Cx, Cy; /* C = Cx+Cy*i */
Cx = radius*cosl(a);
Cy = radius*sinl(a);
return Cx + Cy*I;
}
int GiveMultiplicity(complex long double zr, int pMax){
int s; // number of starting point z0
int sMax = 5*pMax; // it should be greater then 2*pMax
long double t= 0.0; // angle of circle around zr, measured in turns
long double dt = 1.0 / sMax; // t step
long double radius = 0.001; // radius should be smaller then minimal distance between roots
int p;
long double arg_old = 0.0;
long double arg_new = 0.0;
int change = 0;
complex long double z;
complex long double z0;
//complex long double zp;
//
for (s=0; s<sMax; ++s){
z0 = zr + Give_z0(t, radius); // z = point on the circle around root zr
// compute zp = f^p(z)
z = z0;
for (p=0; p < pMax; ++p){z = z*z + c ;} /* complex quadratic polynomial */
// turn (zp-z0)
z = z - z0; // equation for periodic_points of f for period p
arg_new = carg(z);
if (DifferentSign(arg_new, arg_old)) {change+=1;}
arg_old = arg_new;
//printf("z0 = %.16f %.16f zp = %.16f %.16f\n", creal(z0), cimag(z0), creal(zp), cimag(zp));
t += dt; // next angle using globl variable dt
}
return change/2;
}
int main(){
printf("multiplicity = %d\n", GiveMultiplicity(z,2));
return 0;
}
And here is the image of argument of z around root ( it uses carg )
I have some data and I need to fit a second order "polynomial" in 1/x to it using C and GSL, but I don't really understand how to do it.
The documentation for GSL is, unfortunately, not very helpful, I have read it for a few hours now, but I don't seem to be getting closer to the solution.
Google doesn't turn up anything useful either, and I really don't know what to do anymore.
Could you maybe give me some hints on how to accomplish this, or where even to look?
Thanks
Edit 1: The main problem basically is that
Sum n : a_n*x^(-1)
is not a polynomial, so basic fitting or solving algorithms won't work correctly. That's what I tried, using the code for quadratic fitting from this link, also substituting x->1/x, but it didn't work.
May be it's a bit too late for you to read this. However, I post my answer anyway for other people looking for enlightenment.
I suppose, that this basic example can help you. First of all, you have to read about this method of non-linear fitting since you have to adapt the code for any of your own problem.
Second, it's a bit not really clear for me from your post what function you use.
For the sake of clarity let's consider
a1/x + a2/x**2
where a1 and a2 - your parameters.
Using that slightly modified code from the link above ( I replaced 1/x with 1/(x + 0.1) to avoid singularities but it doesn't really change the picture):
#include <stdlib.h>
#include <stdio.h>
#include <gsl/gsl_rng.h>
#include <gsl/gsl_randist.h>
#include <gsl/gsl_matrix.h>
#include <gsl/gsl_vector.h>
#include <gsl/gsl_blas.h>
#include <gsl/gsl_multifit_nlinear.h>
/* number of data points to fit */
#define N 40
#define FIT(i) gsl_vector_get(w->x, i)
#define ERR(i) sqrt(gsl_matrix_get(covar,i,i))
struct data
{
size_t n;
double * y;
};
int expb_f (const gsl_vector * x, void *data, gsl_vector * f)
{
size_t n = ((struct data *)data)->n;
double *y = ((struct data *)data)->y;
double A_1 = gsl_vector_get (x, 0);
double A_2 = gsl_vector_get (x, 1);
size_t i;
for (i = 0; i < n; i++)
{
/* Model Yi = A_1 / x + A_2 / x**2 */
double t = i;
double Yi = A_1 / (t + 0.1) +A_2 / (t*t + 0.2*t + 0.01) ;
gsl_vector_set (f, i, Yi - y[i]);
}
return GSL_SUCCESS;
}
int expb_df (const gsl_vector * x, void *data, gsl_matrix * J)
{
size_t n = ((struct data *)data)->n;
double A_1 = gsl_vector_get (x, 0);
double A_2 = gsl_vector_get (x, 1);
size_t i;
for (i = 0; i < n; i++)
{
/* Jacobian matrix J(i,j) = dfi / dxj, */
/* where fi = (Yi - yi)/sigma[i], */
/* Yi = A_1 / (t + 0.1) +A_2 / (t*t + 0.2*t + 0.01) */
/* and the xj are the parameters (A_1,A_2) */
double t = i;
double e = 1 / (t + 0.1);
double e1 = 1 / (t*t + 0.2*t + 0.01);
gsl_matrix_set (J, i, 0, e);
gsl_matrix_set (J, i, 1, e1);
}
return GSL_SUCCESS;
}
void callback(const size_t iter, void *params, const gsl_multifit_nlinear_workspace *w)
{
gsl_vector *f = gsl_multifit_nlinear_residual(w);
gsl_vector *x = gsl_multifit_nlinear_position(w);
double rcond;
/* compute reciprocal condition number of J(x) */
gsl_multifit_nlinear_rcond(&rcond, w);
fprintf(stderr, "iter %2zu: A_1 = % e A_2 = % e cond(J) = % e, |f(x)| = % e \n", iter, gsl_vector_get(x, 0), gsl_vector_get(x, 1), 1.0 / rcond, gsl_blas_dnrm2(f));
}
int main (void)
{
const gsl_multifit_nlinear_type *T = gsl_multifit_nlinear_trust;
gsl_multifit_nlinear_workspace *w;
gsl_multifit_nlinear_fdf fdf;
gsl_multifit_nlinear_parameters fdf_params = gsl_multifit_nlinear_default_parameters();
const size_t n = N;
const size_t p = 2;
gsl_vector *f;
gsl_matrix *J;
gsl_matrix *covar = gsl_matrix_alloc (p, p);
double y[N], weights[N];
struct data d = { n, y };
double x_init[2] = { 1.0, 1.0 }; /* starting values */
gsl_vector_view x = gsl_vector_view_array (x_init, p);
gsl_vector_view wts = gsl_vector_view_array(weights, n);
gsl_rng * r;
double chisq, chisq0;
int status, info;
size_t i;
const double xtol = 1e-8;
const double gtol = 1e-8;
const double ftol = 0.0;
gsl_rng_env_setup();
r = gsl_rng_alloc(gsl_rng_default);
/* define the function to be minimized */
fdf.f = expb_f;
fdf.df = expb_df; /* set to NULL for finite-difference Jacobian */
fdf.fvv = NULL; /* not using geodesic acceleration */
fdf.n = n;
fdf.p = p;
fdf.params = &d;
/* this is the data to be fitted */
for (i = 0; i < n; i++)
{
double t = i;
double yi = (0.1 + 3.2/(t + 0.1))/(t + 0.1);
double si = 0.1 * yi;
double dy = gsl_ran_gaussian(r, si);
weights[i] = 1.0 / (si * si);
y[i] = yi + dy;
printf ("% e % e \n",t + 0.1, y[i]);
};
/* allocate workspace with default parameters */
w = gsl_multifit_nlinear_alloc (T, &fdf_params, n, p);
/* initialize solver with starting point and weights */
gsl_multifit_nlinear_winit (&x.vector, &wts.vector, &fdf, w);
/* compute initial cost function */
f = gsl_multifit_nlinear_residual(w);
gsl_blas_ddot(f, f, &chisq0);
/* solve the system with a maximum of 20 iterations */
status = gsl_multifit_nlinear_driver(20, xtol, gtol, ftol, callback, NULL, &info, w);
/* compute covariance of best fit parameters */
J = gsl_multifit_nlinear_jac(w);
gsl_multifit_nlinear_covar (J, 0.0, covar);
/* compute final cost */
gsl_blas_ddot(f, f, &chisq);
fprintf(stderr, "summary from method '%s/%s'\n", gsl_multifit_nlinear_name(w), gsl_multifit_nlinear_trs_name(w));
fprintf(stderr, "number of iterations: %zu \n", gsl_multifit_nlinear_niter(w));
fprintf(stderr, "function evaluations: %zu \n", fdf.nevalf);
fprintf(stderr, "Jacobian evaluations: %zu \n", fdf.nevaldf);
fprintf(stderr, "reason for stopping: %s \n", (info == 1) ? "small step size" : "small gradient");
fprintf(stderr, "initial |f(x)| = % e \n", sqrt(chisq0));
fprintf(stderr, "final |f(x)| = % e \n", sqrt(chisq));
{
double dof = n - p;
double c = GSL_MAX_DBL(1, sqrt(chisq / dof));
fprintf(stderr, "chisq/dof = % e \n", chisq / dof);
fprintf (stderr, "A_1 = % f +/- % f \n", FIT(0), c*ERR(0));
fprintf (stderr, "A_2 = % f +/- % f \n", FIT(1), c*ERR(1));
}
fprintf (stderr, "status = %s \n", gsl_strerror (status));
gsl_multifit_nlinear_free (w);
gsl_matrix_free (covar);
gsl_rng_free (r);
return 0;
}
Results of simulations
Unfortunately, Gnuplot doesn't want to fit this data for some reason. Usually it gives the same function up to certain decimal numbers and helps to verify your code.
I need to write my own asin() function without math.h library with the use of Taylor series. It works fine for numbers between <-0.98;0.98> but when I am close to limits it stops with 1604 iterations and therefore is inaccurate.
I don't know how to make it more accurete. Any suggestions are very appreciated!
The code is following:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define EPS 0.000000000001
double my_arcsin(double x)
{
long double a, an, b, bn;
a = an = 1.0;
b = bn = 2.0;
long double n = 3.0;
double xn;
double xs = x;
double xp = x;
int iterace = 0;
xn = xs + (a/b) * (my_pow(xp,n) / n);
while (my_abs(xn - xs) >= EPS)
{
n += 2.0;
an += 2.0;
bn += 2.0;
a = a * an;
b = b * bn;
xs = xn;
xn = xs + (a/b) * (my_pow(xp,n) / n);
iterace++;
}
//printf("%d\n", iterace);
return xn;
}
int main(int argc, char* argv[])
{
double x = 0.0;
if (argc > 2)
x = strtod(argv[2], NULL);
if (strcmp(argv[1], "--asin") == 0)
{
if (x < -1 || x > 1)
printf("nan\n");
else
{
printf("%.10e\n", my_arcsin(x));
//printf("%.10e\n", asin(x));
}
return 0;
}
}
And also a short list of my values and expected ones:
My values Expected values my_asin(x)
5.2359877560e-01 5.2359877560e-01 0.5
1.5567132089e+00 1.5707963268e+00 1 //problem
1.4292568534e+00 1.4292568535e+00 0.99 //problem
1.1197695150e+00 1.1197695150e+00 0.9
1.2532358975e+00 1.2532358975e+00 0.95
Even though the convergence radius of the series expansion you are using is 1, therefore the series will eventually converge for -1 < x < 1, convergence is indeed painfully slow close to the limits of this interval. The solution is to somehow avoid these parts of the interval.
I suggest that you
use your original algorithm for |x| <= 1/sqrt(2),
use the identity arcsin(x) = pi/2 - arcsin(sqrt(1-x^2)) for 1/sqrt(2) < x <= 1.0,
use the identity arcsin(x) = -pi/2 + arcsin(sqrt(1-x^2)) for -1.0 <= x < -1/sqrt(2).
This way you can transform your input x into [-1/sqrt(2),1/sqrt(2)], where convergence is relatively fast.
PLEASE NOTICE: In this case I strongly recommend #Bence's method, since you can't expect a slowly convergent method with low data accuracy to obtain arbitrary precision.
However I'm willing to show you how to improve the result using your current algorithm.
The main problem is that a and b grows too fast and soon become inf (after merely about 150 iterations). Another similar problem is my_pow(xp,n) grows fast when n grows, however this doesn't matter much in this very case since we could assume the input data goes inside the range of [-1, 1].
So I've just changed the method you deal with a/b by introducing ab_ratio, see my edited code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define EPS 0.000000000001
#include <math.h>
#define my_pow powl
#define my_abs fabsl
double my_arcsin(double x)
{
#if 0
long double a, an, b, bn;
a = an = 1.0;
b = bn = 2.0;
#endif
unsigned long _n = 0;
long double ab_ratio = 0.5;
long double n = 3.0;
long double xn;
long double xs = x;
long double xp = x;
int iterace = 0;
xn = xs + ab_ratio * (my_pow(xp,n) / n);
long double step = EPS;
#if 0
while (my_abs(step) >= EPS)
#else
while (1) /* manually stop it */
#endif
{
n += 2.0;
#if 0
an += 2.0;
bn += 2.0;
a = a * an;
b = b * bn;
#endif
_n += 1;
ab_ratio *= (1.0 + 2.0 * _n) / (2.0 + 2.0 * _n);
xs = xn;
step = ab_ratio * (my_pow(xp,n) / n);
xn = xs + step;
iterace++;
if (_n % 10000000 == 0)
printf("%lu %.10g %g %g %g %g\n", _n, (double)xn, (double)ab_ratio, (double)step, (double)xn, (double)my_pow(xp, n));
}
//printf("%d\n", iterace);
return xn;
}
int main(int argc, char* argv[])
{
double x = 0.0;
if (argc > 2)
x = strtod(argv[2], NULL);
if (strcmp(argv[1], "--asin") == 0)
{
if (x < -1 || x > 1)
printf("nan\n");
else
{
printf("%.10e\n", my_arcsin(x));
//printf("%.10e\n", asin(x));
}
return 0;
}
}
For 0.99 (and even 0.9999999) it soon gives correct results with more than 10 significant digits. However it gets slow when getting near to 1.
Actually the process has been running for nearly 12 minutes on my laptop calculating --asin 1, and the current result is 1.570786871 after 3560000000 iterations.
UPDATED: It's been 1h51min now and the result 1.570792915 and iteration count is 27340000000.
I am trying to implement a linear least squares fit onto 2 arrays of data: time vs amplitude. The only technique I know so far is to test all of the possible m and b points in (y = m*x+b) and then find out which combination fits my data best so that it has the least error. However, I think iterating so many combinations is sometimes useless because it tests out everything. Are there any techniques to speed up the process that I don't know about? Thanks.
Try this code. It fits y = mx + b to your (x,y) data.
The arguments to linreg are
linreg(int n, REAL x[], REAL y[], REAL* b, REAL* m, REAL* r)
n = number of data points
x,y = arrays of data
*b = output intercept
*m = output slope
*r = output correlation coefficient (can be NULL if you don't want it)
The return value is 0 on success, !=0 on failure.
Here's the code
#include "linreg.h"
#include <stdlib.h>
#include <math.h> /* math functions */
//#define REAL float
#define REAL double
inline static REAL sqr(REAL x) {
return x*x;
}
int linreg(int n, const REAL x[], const REAL y[], REAL* m, REAL* b, REAL* r){
REAL sumx = 0.0; /* sum of x */
REAL sumx2 = 0.0; /* sum of x**2 */
REAL sumxy = 0.0; /* sum of x * y */
REAL sumy = 0.0; /* sum of y */
REAL sumy2 = 0.0; /* sum of y**2 */
for (int i=0;i<n;i++){
sumx += x[i];
sumx2 += sqr(x[i]);
sumxy += x[i] * y[i];
sumy += y[i];
sumy2 += sqr(y[i]);
}
REAL denom = (n * sumx2 - sqr(sumx));
if (denom == 0) {
// singular matrix. can't solve the problem.
*m = 0;
*b = 0;
if (r) *r = 0;
return 1;
}
*m = (n * sumxy - sumx * sumy) / denom;
*b = (sumy * sumx2 - sumx * sumxy) / denom;
if (r!=NULL) {
*r = (sumxy - sumx * sumy / n) / /* compute correlation coeff */
sqrt((sumx2 - sqr(sumx)/n) *
(sumy2 - sqr(sumy)/n));
}
return 0;
}
Example
You can run this example online.
int main()
{
int n = 6;
REAL x[6]= {1, 2, 4, 5, 10, 20};
REAL y[6]= {4, 6, 12, 15, 34, 68};
REAL m,b,r;
linreg(n,x,y,&m,&b,&r);
printf("m=%g b=%g r=%g\n",m,b,r);
return 0;
}
Here is the output
m=3.43651 b=-0.888889 r=0.999192
Here is the Excel plot and linear fit (for verification).
All values agree exactly with the C code above (note C code returns r while Excel returns R**2).
There are efficient algorithms for least-squares fitting; see Wikipedia for details. There are also libraries that implement the algorithms for you, likely more efficiently than a naive implementation would do; the GNU Scientific Library is one example, but there are others under more lenient licenses as well.
From Numerical Recipes: The Art of Scientific Computing in (15.2) Fitting Data to a Straight Line:
Linear Regression:
Consider the problem of fitting a set of N data points (xi, yi) to a straight-line model:
Assume that the uncertainty: sigmai associated with each yi and that the xi’s (values of the dependent variable) are known exactly. To measure how well the model agrees with the data, we use the chi-square function, which in this case is:
The above equation is minimized to determine a and b. This is done by finding the derivative of the above equation with respect to a and b, equate them to zero and solve for a and b. Then we estimate the probable uncertainties in the estimates of a and b, since obviously the measurement errors in the data must introduce some uncertainty in the determination of those parameters. Additionally, we must estimate the goodness-of-fit of the data to the
model. Absent this estimate, we have not the slightest indication that the parameters a and b in the model have any meaning at all.
The below struct performs the mentioned calculations:
struct Fitab {
// Object for fitting a straight line y = a + b*x to a set of
// points (xi, yi), with or without available
// errors sigma i . Call one of the two constructors to calculate the fit.
// The answers are then available as the variables:
// a, b, siga, sigb, chi2, and either q or sigdat.
int ndata;
double a, b, siga, sigb, chi2, q, sigdat; // Answers.
vector<double> &x, &y, &sig;
// Constructor.
Fitab(vector<double> &xx, vector<double> &yy, vector<double> &ssig)
: ndata(xx.size()), x(xx), y(yy), sig(ssig), chi2(0.), q(1.), sigdat(0.)
{
// Given a set of data points x[0..ndata-1], y[0..ndata-1]
// with individual standard deviations sig[0..ndata-1],
// sets a,b and their respective probable uncertainties
// siga and sigb, the chi-square: chi2, and the goodness-of-fit
// probability: q
Gamma gam;
int i;
double ss=0., sx=0., sy=0., st2=0., t, wt, sxoss; b=0.0;
for (i=0;i < ndata; i++) { // Accumulate sums ...
wt = 1.0 / SQR(sig[i]); //...with weights
ss += wt;
sx += x[i]*wt;
sy += y[i]*wt;
}
sxoss = sx/ss;
for (i=0; i < ndata; i++) {
t = (x[i]-sxoss) / sig[i];
st2 += t*t;
b += t*y[i]/sig[i];
}
b /= st2; // Solve for a, b, sigma-a, and simga-b.
a = (sy-sx*b) / ss;
siga = sqrt((1.0+sx*sx/(ss*st2))/ss);
sigb = sqrt(1.0/st2); // Calculate chi2.
for (i=0;i<ndata;i++) chi2 += SQR((y[i]-a-b*x[i])/sig[i]);
if (ndata>2) q=gam.gammq(0.5*(ndata-2),0.5*chi2); // goodness of fit
}
// Constructor.
Fitab(vector<double> &xx, vector<double> &yy)
: ndata(xx.size()), x(xx), y(yy), sig(xx), chi2(0.), q(1.), sigdat(0.)
{
// As above, but without known errors (sig is not used).
// The uncertainties siga and sigb are estimated by assuming
// equal errors for all points, and that a straight line is
// a good fit. q is returned as 1.0, the normalization of chi2
// is to unit standard deviation on all points, and sigdat
// is set to the estimated error of each point.
int i;
double ss,sx=0.,sy=0.,st2=0.,t,sxoss;
b=0.0; // Accumulate sums ...
for (i=0; i < ndata; i++) {
sx += x[i]; // ...without weights.
sy += y[i];
}
ss = ndata;
sxoss = sx/ss;
for (i=0;i < ndata; i++) {
t = x[i]-sxoss;
st2 += t*t;
b += t*y[i];
}
b /= st2; // Solve for a, b, sigma-a, and sigma-b.
a = (sy-sx*b)/ss;
siga=sqrt((1.0+sx*sx/(ss*st2))/ss);
sigb=sqrt(1.0/st2); // Calculate chi2.
for (i=0;i<ndata;i++) chi2 += SQR(y[i]-a-b*x[i]);
if (ndata > 2) sigdat=sqrt(chi2/(ndata-2));
// For unweighted data evaluate typical
// sig using chi2, and adjust
// the standard deviations.
siga *= sigdat;
sigb *= sigdat;
}
};
where struct Gamma:
struct Gamma : Gauleg18 {
// Object for incomplete gamma function.
// Gauleg18 provides coefficients for Gauss-Legendre quadrature.
static const Int ASWITCH=100; When to switch to quadrature method.
static const double EPS; // See end of struct for initializations.
static const double FPMIN;
double gln;
double gammp(const double a, const double x) {
// Returns the incomplete gamma function P(a,x)
if (x < 0.0 || a <= 0.0) throw("bad args in gammp");
if (x == 0.0) return 0.0;
else if ((Int)a >= ASWITCH) return gammpapprox(a,x,1); // Quadrature.
else if (x < a+1.0) return gser(a,x); // Use the series representation.
else return 1.0-gcf(a,x); // Use the continued fraction representation.
}
double gammq(const double a, const double x) {
// Returns the incomplete gamma function Q(a,x) = 1 - P(a,x)
if (x < 0.0 || a <= 0.0) throw("bad args in gammq");
if (x == 0.0) return 1.0;
else if ((Int)a >= ASWITCH) return gammpapprox(a,x,0); // Quadrature.
else if (x < a+1.0) return 1.0-gser(a,x); // Use the series representation.
else return gcf(a,x); // Use the continued fraction representation.
}
double gser(const Doub a, const Doub x) {
// Returns the incomplete gamma function P(a,x) evaluated by its series representation.
// Also sets ln (gamma) as gln. User should not call directly.
double sum,del,ap;
gln=gammln(a);
ap=a;
del=sum=1.0/a;
for (;;) {
++ap;
del *= x/ap;
sum += del;
if (fabs(del) < fabs(sum)*EPS) {
return sum*exp(-x+a*log(x)-gln);
}
}
}
double gcf(const Doub a, const Doub x) {
// Returns the incomplete gamma function Q(a, x) evaluated
// by its continued fraction representation.
// Also sets ln (gamma) as gln. User should not call directly.
int i;
double an,b,c,d,del,h;
gln=gammln(a);
b=x+1.0-a; // Set up for evaluating continued fraction
// by modified Lentz’s method with with b0 = 0.
c=1.0/FPMIN;
d=1.0/b;
h=d;
for (i=1;;i++) {
// Iterate to convergence.
an = -i*(i-a);
b += 2.0;
d=an*d+b;
if (fabs(d) < FPMIN) d=FPMIN;
c=b+an/c;
if (fabs(c) < FPMIN) c=FPMIN;
d=1.0/d;
del=d*c;
h *= del;
if (fabs(del-1.0) <= EPS) break;
}
return exp(-x+a*log(x)-gln)*h; Put factors in front.
}
double gammpapprox(double a, double x, int psig) {
// Incomplete gamma by quadrature. Returns P(a,x) or Q(a, x),
// when psig is 1 or 0, respectively. User should not call directly.
int j;
double xu,t,sum,ans;
double a1 = a-1.0, lna1 = log(a1), sqrta1 = sqrt(a1);
gln = gammln(a);
// Set how far to integrate into the tail:
if (x > a1) xu = MAX(a1 + 11.5*sqrta1, x + 6.0*sqrta1);
else xu = MAX(0.,MIN(a1 - 7.5*sqrta1, x - 5.0*sqrta1));
sum = 0;
for (j=0;j<ngau;j++) { // Gauss-Legendre.
t = x + (xu-x)*y[j];
sum += w[j]*exp(-(t-a1)+a1*(log(t)-lna1));
}
ans = sum*(xu-x)*exp(a1*(lna1-1.)-gln);
return (psig?(ans>0.0? 1.0-ans:-ans):(ans>=0.0? ans:1.0+ans));
}
double invgammp(Doub p, Doub a);
// Inverse function on x of P(a,x) .
};
const Doub Gamma::EPS = numeric_limits<Doub>::epsilon();
const Doub Gamma::FPMIN = numeric_limits<Doub>::min()/EPS
and stuct Gauleg18:
struct Gauleg18 {
// Abscissas and weights for Gauss-Legendre quadrature.
static const Int ngau = 18;
static const Doub y[18];
static const Doub w[18];
};
const Doub Gauleg18::y[18] = {0.0021695375159141994,
0.011413521097787704,0.027972308950302116,0.051727015600492421,
0.082502225484340941, 0.12007019910960293,0.16415283300752470,
0.21442376986779355, 0.27051082840644336, 0.33199876341447887,
0.39843234186401943, 0.46931971407375483, 0.54413605556657973,
0.62232745288031077, 0.70331500465597174, 0.78649910768313447,
0.87126389619061517, 0.95698180152629142};
const Doub Gauleg18::w[18] = {0.0055657196642445571,
0.012915947284065419,0.020181515297735382,0.027298621498568734,
0.034213810770299537,0.040875750923643261,0.047235083490265582,
0.053244713977759692,0.058860144245324798,0.064039797355015485
0.068745323835736408,0.072941885005653087,0.076598410645870640,
0.079687828912071670,0.082187266704339706,0.084078218979661945,
0.085346685739338721,0.085983275670394821};
and, finally fuinction Gamma::invgamp():
double Gamma::invgammp(double p, double a) {
// Returns x such that P(a,x) = p for an argument p between 0 and 1.
int j;
double x,err,t,u,pp,lna1,afac,a1=a-1;
const double EPS=1.e-8; // Accuracy is the square of EPS.
gln=gammln(a);
if (a <= 0.) throw("a must be pos in invgammap");
if (p >= 1.) return MAX(100.,a + 100.*sqrt(a));
if (p <= 0.) return 0.0;
if (a > 1.) {
lna1=log(a1);
afac = exp(a1*(lna1-1.)-gln);
pp = (p < 0.5)? p : 1. - p;
t = sqrt(-2.*log(pp));
x = (2.30753+t*0.27061)/(1.+t*(0.99229+t*0.04481)) - t;
if (p < 0.5) x = -x;
x = MAX(1.e-3,a*pow(1.-1./(9.*a)-x/(3.*sqrt(a)),3));
} else {
t = 1.0 - a*(0.253+a*0.12); and (6.2.9).
if (p < t) x = pow(p/t,1./a);
else x = 1.-log(1.-(p-t)/(1.-t));
}
for (j=0;j<12;j++) {
if (x <= 0.0) return 0.0; // x too small to compute accurately.
err = gammp(a,x) - p;
if (a > 1.) t = afac*exp(-(x-a1)+a1*(log(x)-lna1));
else t = exp(-x+a1*log(x)-gln);
u = err/t;
// Halley’s method.
x -= (t = u/(1.-0.5*MIN(1.,u*((a-1.)/x - 1))));
// Halve old value if x tries to go negative.
if (x <= 0.) x = 0.5*(x + t);
if (fabs(t) < EPS*x ) break;
}
return x;
}
Here is my version of a C/C++ function that does simple linear regression. The calculations follow the wikipedia article on simple linear regression. This is published as a single-header public-domain (MIT) library on github: simple_linear_regression. The library (.h file) is tested to work on Linux and Windows, and from C and C++ using -Wall -Werror and all -std versions supported by clang/gcc.
#define SIMPLE_LINEAR_REGRESSION_ERROR_INPUT_VALUE -2
#define SIMPLE_LINEAR_REGRESSION_ERROR_NUMERIC -3
int simple_linear_regression(const double * x, const double * y, const int n, double * slope_out, double * intercept_out, double * r2_out) {
double sum_x = 0.0;
double sum_xx = 0.0;
double sum_xy = 0.0;
double sum_y = 0.0;
double sum_yy = 0.0;
double n_real = (double)(n);
int i = 0;
double slope = 0.0;
double denominator = 0.0;
if (x == NULL || y == NULL || n < 2) {
return SIMPLE_LINEAR_REGRESSION_ERROR_INPUT_VALUE;
}
for (i = 0; i < n; ++i) {
sum_x += x[i];
sum_xx += x[i] * x[i];
sum_xy += x[i] * y[i];
sum_y += y[i];
sum_yy += y[i] * y[i];
}
denominator = n_real * sum_xx - sum_x * sum_x;
if (denominator == 0.0) {
return SIMPLE_LINEAR_REGRESSION_ERROR_NUMERIC;
}
slope = (n_real * sum_xy - sum_x * sum_y) / denominator;
if (slope_out != NULL) {
*slope_out = slope;
}
if (intercept_out != NULL) {
*intercept_out = (sum_y - slope * sum_x) / n_real;
}
if (r2_out != NULL) {
denominator = ((n_real * sum_xx) - (sum_x * sum_x)) * ((n_real * sum_yy) - (sum_y * sum_y));
if (denominator == 0.0) {
return SIMPLE_LINEAR_REGRESSION_ERROR_NUMERIC;
}
*r2_out = ((n_real * sum_xy) - (sum_x * sum_y)) * ((n_real * sum_xy) - (sum_x * sum_y)) / denominator;
}
return 0;
}
Usage example:
#define SIMPLE_LINEAR_REGRESSION_IMPLEMENTATION
#include "simple_linear_regression.h"
#include <stdio.h>
/* Some data that we want to find the slope, intercept and r2 for */
static const double x[] = { 1.47, 1.50, 1.52, 1.55, 1.57, 1.60, 1.63, 1.65, 1.68, 1.70, 1.73, 1.75, 1.78, 1.80, 1.83 };
static const double y[] = { 52.21, 53.12, 54.48, 55.84, 57.20, 58.57, 59.93, 61.29, 63.11, 64.47, 66.28, 68.10, 69.92, 72.19, 74.46 };
int main() {
double slope = 0.0;
double intercept = 0.0;
double r2 = 0.0;
int res = 0;
res = simple_linear_regression(x, y, sizeof(x) / sizeof(x[0]), &slope, &intercept, &r2);
if (res < 0) {
printf("Error: %s\n", simple_linear_regression_error_string(res));
return res;
}
printf("slope: %f\n", slope);
printf("intercept: %f\n", intercept);
printf("r2: %f\n", r2);
return 0;
}
The original example above worked well for me with slope and offset but I had a hard time with the corr coef. Maybe I don't have my parenthesis working the same as the assumed precedence? Anyway, with some help of other web pages I finally got values that match the linear trend-line in Excel. Thought I would share my code using Mark Lakata's variable names. Hope this helps.
double slope = ((n * sumxy) - (sumx * sumy )) / denom;
double intercept = ((sumy * sumx2) - (sumx * sumxy)) / denom;
double term1 = ((n * sumxy) - (sumx * sumy));
double term2 = ((n * sumx2) - (sumx * sumx));
double term3 = ((n * sumy2) - (sumy * sumy));
double term23 = (term2 * term3);
double r2 = 1.0;
if (fabs(term23) > MIN_DOUBLE) // Define MIN_DOUBLE somewhere as 1e-9 or similar
r2 = (term1 * term1) / term23;
as an assignment I had to code in C a simple linear regression using RMSE loss function. The program is dynamic and you can enter your own values and choose your own loss function which is for now limited to Root Mean Square Error. But first here are the algorithms I used:
now the code... you need gnuplot to display the chart, sudo apt install gnuplot
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <sys/types.h>
#define BUFFSIZE 64
#define MAXSIZE 100
static double vector_x[MAXSIZE] = {0};
static double vector_y[MAXSIZE] = {0};
static double vector_predict[MAXSIZE] = {0};
static double max_x;
static double max_y;
static double mean_x;
static double mean_y;
static double teta_0_intercept;
static double teta_1_grad;
static double RMSE;
static double r_square;
static double prediction;
static char intercept[BUFFSIZE];
static char grad[BUFFSIZE];
static char xrange[BUFFSIZE];
static char yrange[BUFFSIZE];
static char lossname_RMSE[BUFFSIZE] = "Simple Linear Regression using RMSE'";
static char cmd_gnu_0[BUFFSIZE] = "set title '";
static char cmd_gnu_1[BUFFSIZE] = "intercept = ";
static char cmd_gnu_2[BUFFSIZE] = "grad = ";
static char cmd_gnu_3[BUFFSIZE] = "set xrange [0:";
static char cmd_gnu_4[BUFFSIZE] = "set yrange [0:";
static char cmd_gnu_5[BUFFSIZE] = "f(x) = (grad * x) + intercept";
static char cmd_gnu_6[BUFFSIZE] = "plot f(x), 'data.temp' with points pointtype 7";
static char const *commands_gnuplot[] = {
cmd_gnu_0,
cmd_gnu_1,
cmd_gnu_2,
cmd_gnu_3,
cmd_gnu_4,
cmd_gnu_5,
cmd_gnu_6,
};
static size_t size;
static void user_input()
{
printf("Enter x,y vector size, MAX = 100\n");
scanf("%lu", &size);
if (size > MAXSIZE) {
printf("Wrong input size is too big\n");
user_input();
}
printf("vector's size is %lu\n", size);
size_t i;
for (i = 0; i < size; i++) {
printf("Enter vector_x[%ld] values\n", i);
scanf("%lf", &vector_x[i]);
}
for (i = 0; i < size; i++) {
printf("Enter vector_y[%ld] values\n", i);
scanf("%lf", &vector_y[i]);
}
}
static void display_vector()
{
size_t i;
for (i = 0; i < size; i++){
printf("vector_x[%lu] = %lf\t", i, vector_x[i]);
printf("vector_y[%lu] = %lf\n", i, vector_y[i]);
}
}
static void concatenate(char p[], char q[]) {
int c;
int d;
c = 0;
while (p[c] != '\0') {
c++;
}
d = 0;
while (q[d] != '\0') {
p[c] = q[d];
d++;
c++;
}
p[c] = '\0';
}
static void compute_mean_x_y()
{
size_t i;
double tmp_x = 0.0;
double tmp_y = 0.0;
for (i = 0; i < size; i++) {
tmp_x += vector_x[i];
tmp_y += vector_y[i];
}
mean_x = tmp_x / size;
mean_y = tmp_y / size;
printf("mean_x = %lf\n", mean_x);
printf("mean_y = %lf\n", mean_y);
}
static void compute_teta_1_grad()
{
double numerator = 0.0;
double denominator = 0.0;
double tmp1 = 0.0;
double tmp2 = 0.0;
size_t i;
for (i = 0; i < size; i++) {
numerator += (vector_x[i] - mean_x) * (vector_y[i] - mean_y);
}
for (i = 0; i < size; i++) {
tmp1 = vector_x[i] - mean_x;
tmp2 = tmp1 * tmp1;
denominator += tmp2;
}
teta_1_grad = numerator / denominator;
printf("teta_1_grad = %lf\n", teta_1_grad);
}
static void compute_teta_0_intercept()
{
teta_0_intercept = mean_y - (teta_1_grad * mean_x);
printf("teta_0_intercept = %lf\n", teta_0_intercept);
}
static void compute_prediction()
{
size_t i;
for (i = 0; i < size; i++) {
vector_predict[i] = teta_0_intercept + (teta_1_grad * vector_x[i]);
printf("y^[%ld] = %lf\n", i, vector_predict[i]);
}
printf("\n");
}
static void compute_RMSE()
{
compute_prediction();
double error = 0;
size_t i;
for (i = 0; i < size; i++) {
error = (vector_predict[i] - vector_y[i]) * (vector_predict[i] - vector_y[i]);
printf("error y^[%ld] = %lf\n", i, error);
RMSE += error;
}
/* mean */
RMSE = RMSE / size;
/* square root mean */
RMSE = sqrt(RMSE);
printf("\nRMSE = %lf\n", RMSE);
}
static void compute_loss_function()
{
int input = 0;
printf("Which loss function do you want to use?\n");
printf(" 1 - RMSE\n");
scanf("%d", &input);
switch(input) {
case 1:
concatenate(cmd_gnu_0, lossname_RMSE);
compute_RMSE();
printf("\n");
break;
default:
printf("Wrong input try again\n");
compute_loss_function(size);
}
}
static void compute_r_square(size_t size)
{
double num_err = 0.0;
double den_err = 0.0;
size_t i;
for (i = 0; i < size; i++) {
num_err += (vector_y[i] - vector_predict[i]) * (vector_y[i] - vector_predict[i]);
den_err += (vector_y[i] - mean_y) * (vector_y[i] - mean_y);
}
r_square = 1 - (num_err/den_err);
printf("R_square = %lf\n", r_square);
}
static void compute_predict_for_x()
{
double x = 0.0;
printf("Please enter x value\n");
scanf("%lf", &x);
prediction = teta_0_intercept + (teta_1_grad * x);
printf("y^ if x = %lf -> %lf\n",x, prediction);
}
static void compute_max_x_y()
{
size_t i;
double tmp1= 0.0;
double tmp2= 0.0;
for (i = 0; i < size; i++) {
if (vector_x[i] > tmp1) {
tmp1 = vector_x[i];
max_x = vector_x[i];
}
if (vector_y[i] > tmp2) {
tmp2 = vector_y[i];
max_y = vector_y[i];
}
}
printf("vector_x max value %lf\n", max_x);
printf("vector_y max value %lf\n", max_y);
}
static void display_model_line()
{
sprintf(intercept, "%0.7lf", teta_0_intercept);
sprintf(grad, "%0.7lf", teta_1_grad);
sprintf(xrange, "%0.7lf", max_x + 1);
sprintf(yrange, "%0.7lf", max_y + 1);
concatenate(cmd_gnu_1, intercept);
concatenate(cmd_gnu_2, grad);
concatenate(cmd_gnu_3, xrange);
concatenate(cmd_gnu_3, "]");
concatenate(cmd_gnu_4, yrange);
concatenate(cmd_gnu_4, "]");
printf("grad = %s\n", grad);
printf("intercept = %s\n", intercept);
printf("xrange = %s\n", xrange);
printf("yrange = %s\n", yrange);
printf("cmd_gnu_0: %s\n", cmd_gnu_0);
printf("cmd_gnu_1: %s\n", cmd_gnu_1);
printf("cmd_gnu_2: %s\n", cmd_gnu_2);
printf("cmd_gnu_3: %s\n", cmd_gnu_3);
printf("cmd_gnu_4: %s\n", cmd_gnu_4);
printf("cmd_gnu_5: %s\n", cmd_gnu_5);
printf("cmd_gnu_6: %s\n", cmd_gnu_6);
/* print plot */
FILE *gnuplot_pipe = (FILE*)popen("gnuplot -persistent", "w");
FILE *temp = (FILE*)fopen("data.temp", "w");
/* create data.temp */
size_t i;
for (i = 0; i < size; i++)
{
fprintf(temp, "%f %f \n", vector_x[i], vector_y[i]);
}
/* display gnuplot */
for (i = 0; i < 7; i++)
{
fprintf(gnuplot_pipe, "%s \n", commands_gnuplot[i]);
}
}
int main(void)
{
printf("===========================================\n");
printf("INPUT DATA\n");
printf("===========================================\n");
user_input();
display_vector();
printf("\n");
printf("===========================================\n");
printf("COMPUTE MEAN X:Y, TETA_1 TETA_0\n");
printf("===========================================\n");
compute_mean_x_y();
compute_max_x_y();
compute_teta_1_grad();
compute_teta_0_intercept();
printf("\n");
printf("===========================================\n");
printf("COMPUTE LOSS FUNCTION\n");
printf("===========================================\n");
compute_loss_function();
printf("===========================================\n");
printf("COMPUTE R_square\n");
printf("===========================================\n");
compute_r_square(size);
printf("\n");
printf("===========================================\n");
printf("COMPUTE y^ according to x\n");
printf("===========================================\n");
compute_predict_for_x();
printf("\n");
printf("===========================================\n");
printf("DISPLAY LINEAR REGRESSION\n");
printf("===========================================\n");
display_model_line();
printf("\n");
return 0;
}
Look at Section 1 of this paper. This section expresses a 2D linear regression as a matrix multiplication exercise. As long as your data is well-behaved, this technique should permit you to develop a quick least squares fit.
Depending on the size of your data, it might be worthwhile to algebraically reduce the matrix multiplication to simple set of equations, thereby avoiding the need to write a matmult() function. (Be forewarned, this is completely impractical for more than 4 or 5 data points!)
The fastest, most efficient way to solve least squares, as far as I am aware, is to subtract (the gradient)/(the 2nd order gradient) from your parameter vector. (2nd order gradient = i.e. the diagonal of the Hessian.)
Here is the intuition:
Let's say you want to optimize least squares over a single parameter. This is equivalent to finding the vertex of a parabola. Then, for any random initial parameter, x0, the vertex of the loss function is located at x0 - f(1) / f(2). That's because adding - f(1) / f(2) to x will always zero out the derivative, f(1).
Side note: Implementing this in Tensorflow, the solution appeared at w0 - f(1) / f(2) / (number of weights), but I'm not sure if that's due to Tensorflow or if it's due to something else..