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authorDaniel Baumann <daniel.baumann@progress-linux.org>2024-04-09 12:48:01 +0000
committerDaniel Baumann <daniel.baumann@progress-linux.org>2024-04-09 12:48:01 +0000
commitb2d2d555a704148968cb7e566735a2a1b1a2f189 (patch)
tree18549ff498338f40ecf7aa327620abf4c1c3ee43 /regress.c
parentInitial commit. (diff)
downloadchrony-b2d2d555a704148968cb7e566735a2a1b1a2f189.tar.xz
chrony-b2d2d555a704148968cb7e566735a2a1b1a2f189.zip
Adding upstream version 4.5.upstream/4.5
Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>
Diffstat (limited to 'regress.c')
-rw-r--r--regress.c704
1 files changed, 704 insertions, 0 deletions
diff --git a/regress.c b/regress.c
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+++ b/regress.c
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+/*
+ chronyd/chronyc - Programs for keeping computer clocks accurate.
+
+ **********************************************************************
+ * Copyright (C) Richard P. Curnow 1997-2003
+ * Copyright (C) Miroslav Lichvar 2011, 2016-2017
+ *
+ * This program is free software; you can redistribute it and/or modify
+ * it under the terms of version 2 of the GNU General Public License as
+ * published by the Free Software Foundation.
+ *
+ * This program is distributed in the hope that it will be useful, but
+ * WITHOUT ANY WARRANTY; without even the implied warranty of
+ * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
+ * General Public License for more details.
+ *
+ * You should have received a copy of the GNU General Public License along
+ * with this program; if not, write to the Free Software Foundation, Inc.,
+ * 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
+ *
+ **********************************************************************
+
+ =======================================================================
+
+ Regression algorithms.
+
+ */
+
+#include "config.h"
+
+#include "sysincl.h"
+
+#include "regress.h"
+#include "logging.h"
+#include "util.h"
+
+#define MAX_POINTS 64
+
+void
+RGR_WeightedRegression
+(double *x, /* independent variable */
+ double *y, /* measured data */
+ double *w, /* weightings (large => data
+ less reliable) */
+
+ int n, /* number of data points */
+
+ /* And now the results */
+
+ double *b0, /* estimated y axis intercept */
+ double *b1, /* estimated slope */
+ double *s2, /* estimated variance of data points */
+
+ double *sb0, /* estimated standard deviation of
+ intercept */
+ double *sb1 /* estimated standard deviation of
+ slope */
+
+ /* Could add correlation stuff later if required */
+)
+{
+ double P, Q, U, V, W;
+ double diff;
+ double u, ui, aa;
+ int i;
+
+ assert(n >= 3);
+
+ W = U = 0;
+ for (i=0; i<n; i++) {
+ U += x[i] / w[i];
+ W += 1.0 / w[i];
+ }
+
+ u = U / W;
+
+ /* Calculate statistics from data */
+ P = Q = V = 0.0;
+ for (i=0; i<n; i++) {
+ ui = x[i] - u;
+ P += y[i] / w[i];
+ Q += y[i] * ui / w[i];
+ V += ui * ui / w[i];
+ }
+
+ *b1 = Q / V;
+ *b0 = (P / W) - (*b1) * u;
+
+ *s2 = 0.0;
+ for (i=0; i<n; i++) {
+ diff = y[i] - *b0 - *b1*x[i];
+ *s2 += diff*diff / w[i];
+ }
+
+ *s2 /= (double)(n-2);
+
+ *sb1 = sqrt(*s2 / V);
+ aa = u * (*sb1);
+ *sb0 = sqrt(*s2 / W + aa * aa);
+
+ *s2 *= (n / W); /* Giving weighted average of variances */
+}
+
+/* ================================================== */
+/* Get the coefficient to multiply the standard deviation by, to get a
+ particular size of confidence interval (assuming a t-distribution) */
+
+double
+RGR_GetTCoef(int dof)
+{
+ /* Assuming now the 99.95% quantile */
+ static const float coefs[] =
+ { 636.6, 31.6, 12.92, 8.61, 6.869,
+ 5.959, 5.408, 5.041, 4.781, 4.587,
+ 4.437, 4.318, 4.221, 4.140, 4.073,
+ 4.015, 3.965, 3.922, 3.883, 3.850,
+ 3.819, 3.792, 3.768, 3.745, 3.725,
+ 3.707, 3.690, 3.674, 3.659, 3.646,
+ 3.633, 3.622, 3.611, 3.601, 3.591,
+ 3.582, 3.574, 3.566, 3.558, 3.551};
+
+ if (dof <= 40) {
+ return coefs[dof-1];
+ } else {
+ return 3.5; /* Until I can be bothered to do something better */
+ }
+}
+
+/* ================================================== */
+/* Get 90% quantile of chi-square distribution */
+
+double
+RGR_GetChi2Coef(int dof)
+{
+ static const float coefs[] = {
+ 2.706, 4.605, 6.251, 7.779, 9.236, 10.645, 12.017, 13.362,
+ 14.684, 15.987, 17.275, 18.549, 19.812, 21.064, 22.307, 23.542,
+ 24.769, 25.989, 27.204, 28.412, 29.615, 30.813, 32.007, 33.196,
+ 34.382, 35.563, 36.741, 37.916, 39.087, 40.256, 41.422, 42.585,
+ 43.745, 44.903, 46.059, 47.212, 48.363, 49.513, 50.660, 51.805,
+ 52.949, 54.090, 55.230, 56.369, 57.505, 58.641, 59.774, 60.907,
+ 62.038, 63.167, 64.295, 65.422, 66.548, 67.673, 68.796, 69.919,
+ 71.040, 72.160, 73.279, 74.397, 75.514, 76.630, 77.745, 78.860
+ };
+
+ if (dof <= 64) {
+ return coefs[dof-1];
+ } else {
+ return 1.2 * dof; /* Until I can be bothered to do something better */
+ }
+}
+
+/* ================================================== */
+/* Critical value for number of runs of residuals with same sign.
+ 5% critical region for now. */
+
+static char critical_runs[] = {
+ 0, 0, 0, 0, 0, 0, 0, 0, 2, 3,
+ 3, 3, 4, 4, 5, 5, 5, 6, 6, 7,
+ 7, 7, 8, 8, 9, 9, 9, 10, 10, 11,
+ 11, 11, 12, 12, 13, 13, 14, 14, 14, 15,
+ 15, 16, 16, 17, 17, 18, 18, 18, 19, 19,
+ 20, 20, 21, 21, 21, 22, 22, 23, 23, 24,
+ 24, 25, 25, 26, 26, 26, 27, 27, 28, 28,
+ 29, 29, 30, 30, 30, 31, 31, 32, 32, 33,
+ 33, 34, 34, 35, 35, 35, 36, 36, 37, 37,
+ 38, 38, 39, 39, 40, 40, 40, 41, 41, 42,
+ 42, 43, 43, 44, 44, 45, 45, 46, 46, 46,
+ 47, 47, 48, 48, 49, 49, 50, 50, 51, 51,
+ 52, 52, 52, 53, 53, 54, 54, 55, 55, 56
+};
+
+/* ================================================== */
+
+static int
+n_runs_from_residuals(double *resid, int n)
+{
+ int nruns;
+ int i;
+
+ nruns = 1;
+ for (i=1; i<n; i++) {
+ if (((resid[i-1] < 0.0) && (resid[i] < 0.0)) ||
+ ((resid[i-1] > 0.0) && (resid[i] > 0.0))) {
+ /* Nothing to do */
+ } else {
+ nruns++;
+ }
+ }
+
+ return nruns;
+}
+
+/* ================================================== */
+/* Return a boolean indicating whether we had enough points for
+ regression */
+
+int
+RGR_FindBestRegression
+(double *x, /* independent variable */
+ double *y, /* measured data */
+ double *w, /* weightings (large => data
+ less reliable) */
+
+ int n, /* number of data points */
+ int m, /* number of extra samples in x and y arrays
+ (negative index) which can be used to
+ extend runs test */
+ int min_samples, /* minimum number of samples to be kept after
+ changing the starting index to pass the runs
+ test */
+
+ /* And now the results */
+
+ double *b0, /* estimated y axis intercept */
+ double *b1, /* estimated slope */
+ double *s2, /* estimated variance of data points */
+
+ double *sb0, /* estimated standard deviation of
+ intercept */
+ double *sb1, /* estimated standard deviation of
+ slope */
+
+ int *new_start, /* the new starting index to make the
+ residuals pass the two tests */
+
+ int *n_runs, /* number of runs amongst the residuals */
+
+ int *dof /* degrees of freedom in statistics (needed
+ to get confidence intervals later) */
+
+)
+{
+ double P, Q, U, V, W; /* total */
+ double resid[MAX_POINTS * REGRESS_RUNS_RATIO];
+ double ss;
+ double a, b, u, ui, aa;
+
+ int start, resid_start, nruns, npoints;
+ int i;
+
+ assert(n <= MAX_POINTS && m >= 0);
+ assert(n * REGRESS_RUNS_RATIO < sizeof (critical_runs) / sizeof (critical_runs[0]));
+
+ if (n < MIN_SAMPLES_FOR_REGRESS) {
+ return 0;
+ }
+
+ start = 0;
+ do {
+
+ W = U = 0;
+ for (i=start; i<n; i++) {
+ U += x[i] / w[i];
+ W += 1.0 / w[i];
+ }
+
+ u = U / W;
+
+ P = Q = V = 0.0;
+ for (i=start; i<n; i++) {
+ ui = x[i] - u;
+ P += y[i] / w[i];
+ Q += y[i] * ui / w[i];
+ V += ui * ui / w[i];
+ }
+
+ b = Q / V;
+ a = (P / W) - (b * u);
+
+ /* Get residuals also for the extra samples before start */
+ resid_start = n - (n - start) * REGRESS_RUNS_RATIO;
+ if (resid_start < -m)
+ resid_start = -m;
+
+ for (i=resid_start; i<n; i++) {
+ resid[i - resid_start] = y[i] - a - b*x[i];
+ }
+
+ /* Count number of runs */
+ nruns = n_runs_from_residuals(resid, n - resid_start);
+
+ if (nruns > critical_runs[n - resid_start] ||
+ n - start <= MIN_SAMPLES_FOR_REGRESS ||
+ n - start <= min_samples) {
+ if (start != resid_start) {
+ /* Ignore extra samples in returned nruns */
+ nruns = n_runs_from_residuals(resid + (start - resid_start), n - start);
+ }
+ break;
+ } else {
+ /* Try dropping one sample at a time until the runs test passes. */
+ ++start;
+ }
+
+ } while (1);
+
+ /* Work out statistics from full dataset */
+ *b1 = b;
+ *b0 = a;
+
+ ss = 0.0;
+ for (i=start; i<n; i++) {
+ ss += resid[i - resid_start]*resid[i - resid_start] / w[i];
+ }
+
+ npoints = n - start;
+ ss /= (double)(npoints - 2);
+ *sb1 = sqrt(ss / V);
+ aa = u * (*sb1);
+ *sb0 = sqrt((ss / W) + (aa * aa));
+ *s2 = ss * (double) npoints / W;
+
+ *new_start = start;
+ *dof = npoints - 2;
+ *n_runs = nruns;
+
+ return 1;
+
+}
+
+/* ================================================== */
+
+#define EXCH(a,b) temp=(a); (a)=(b); (b)=temp
+
+/* ================================================== */
+/* Find the index'th biggest element in the array x of n elements.
+ flags is an array where a 1 indicates that the corresponding entry
+ in x is known to be sorted into its correct position and a 0
+ indicates that the corresponding entry is not sorted. However, if
+ flags[m] = flags[n] = 1 with m<n, then x[m] must be <= x[n] and for
+ all i with m<i<n, x[m] <= x[i] <= x[n]. In practice, this means
+ flags[] has to be the result of a previous call to this routine
+ with the same array x, and is used to remember which parts of the
+ x[] array we have already sorted.
+
+ The approach used is a cut-down quicksort, where we only bother to
+ keep sorting the partition that contains the index we are after.
+ The approach comes from Numerical Recipes in C (ISBN
+ 0-521-43108-5). */
+
+static double
+find_ordered_entry_with_flags(double *x, int n, int index, char *flags)
+{
+ int u, v, l, r;
+ double temp;
+ double piv;
+ int pivind;
+
+ assert(index >= 0);
+
+ /* If this bit of the array is already sorted, simple! */
+ if (flags[index]) {
+ return x[index];
+ }
+
+ /* Find subrange to look at */
+ u = v = index;
+ while (u > 0 && !flags[u]) u--;
+ if (flags[u]) u++;
+
+ while (v < (n-1) && !flags[v]) v++;
+ if (flags[v]) v--;
+
+ do {
+ if (v - u < 2) {
+ if (x[v] < x[u]) {
+ EXCH(x[v], x[u]);
+ }
+ flags[v] = flags[u] = 1;
+ return x[index];
+ } else {
+ pivind = (u + v) >> 1;
+ EXCH(x[u], x[pivind]);
+ piv = x[u]; /* New value */
+ l = u + 1;
+ r = v;
+ do {
+ while (l < v && x[l] < piv) l++;
+ while (x[r] > piv) r--;
+ if (r <= l) break;
+ EXCH(x[l], x[r]);
+ l++;
+ r--;
+ } while (1);
+ EXCH(x[u], x[r]);
+ flags[r] = 1; /* Pivot now in correct place */
+ if (index == r) {
+ return x[r];
+ } else if (index < r) {
+ v = r - 1;
+ } else if (index > r) {
+ u = l;
+ }
+ }
+ } while (1);
+}
+
+/* ================================================== */
+
+#if 0
+/* Not used, but this is how it can be done */
+static double
+find_ordered_entry(double *x, int n, int index)
+{
+ char flags[MAX_POINTS];
+
+ memset(flags, 0, n * sizeof (flags[0]));
+ return find_ordered_entry_with_flags(x, n, index, flags);
+}
+#endif
+
+/* ================================================== */
+/* Find the median entry of an array x[] with n elements. */
+
+static double
+find_median(double *x, int n)
+{
+ int k;
+ char flags[MAX_POINTS];
+
+ memset(flags, 0, n * sizeof (flags[0]));
+ k = n>>1;
+ if (n&1) {
+ return find_ordered_entry_with_flags(x, n, k, flags);
+ } else {
+ return 0.5 * (find_ordered_entry_with_flags(x, n, k, flags) +
+ find_ordered_entry_with_flags(x, n, k-1, flags));
+ }
+}
+
+/* ================================================== */
+
+double
+RGR_FindMedian(double *x, int n)
+{
+ double tmp[MAX_POINTS];
+
+ assert(n > 0 && n <= MAX_POINTS);
+ memcpy(tmp, x, n * sizeof (tmp[0]));
+
+ return find_median(tmp, n);
+}
+
+/* ================================================== */
+/* This function evaluates the equation
+
+ \sum_{i=0}^{n-1} x_i sign(y_i - a - b x_i)
+
+ and chooses the value of a that minimises the absolute value of the
+ result. (See pp703-704 of Numerical Recipes in C). */
+
+static void
+eval_robust_residual
+(double *x, /* The independent points */
+ double *y, /* The dependent points */
+ int n, /* Number of points */
+ double b, /* Slope */
+ double *aa, /* Intercept giving smallest absolute
+ value for the above equation */
+ double *rr /* Corresponding value of equation */
+)
+{
+ int i;
+ double a, res, del;
+ double d[MAX_POINTS];
+
+ for (i=0; i<n; i++) {
+ d[i] = y[i] - b * x[i];
+ }
+
+ a = find_median(d, n);
+
+ res = 0.0;
+ for (i=0; i<n; i++) {
+ del = y[i] - a - b * x[i];
+ if (del > 0.0) {
+ res += x[i];
+ } else if (del < 0.0) {
+ res -= x[i];
+ }
+ }
+
+ *aa = a;
+ *rr = res;
+}
+
+/* ================================================== */
+/* This routine performs a 'robust' regression, i.e. one which has low
+ susceptibility to outliers amongst the data. If one thinks of a
+ normal (least squares) linear regression in 2D being analogous to
+ the arithmetic mean in 1D, this algorithm in 2D is roughly
+ analogous to the median in 1D. This algorithm seems to work quite
+ well until the number of outliers is approximately half the number
+ of data points.
+
+ The return value is a status indicating whether there were enough
+ data points to run the routine or not. */
+
+int
+RGR_FindBestRobustRegression
+(double *x, /* The independent axis points */
+ double *y, /* The dependent axis points (which
+ may contain outliers). */
+ int n, /* The number of points */
+ double tol, /* The tolerance required in
+ determining the value of b1 */
+ double *b0, /* The estimated Y-axis intercept */
+ double *b1, /* The estimated slope */
+ int *n_runs, /* The number of runs of residuals */
+ int *best_start /* The best starting index */
+)
+{
+ int i;
+ int start;
+ int n_points;
+ double a, b;
+ double P, U, V, W, X;
+ double resid, resids[MAX_POINTS];
+ double blo, bhi, bmid, rlo, rhi, rmid;
+ double s2, sb, incr;
+ double mx, dx, my, dy;
+ int nruns = 0;
+
+ assert(n <= MAX_POINTS);
+
+ if (n < 2) {
+ return 0;
+ } else if (n == 2) {
+ /* Just a straight line fit (we need this for the manual mode) */
+ *b1 = (y[1] - y[0]) / (x[1] - x[0]);
+ *b0 = y[0] - (*b1) * x[0];
+ *n_runs = 0;
+ *best_start = 0;
+ return 1;
+ }
+
+ /* else at least 3 points, apply normal algorithm */
+
+ start = 0;
+
+ /* Loop to strip oldest points that cause the regression residuals
+ to fail the number of runs test */
+ do {
+
+ n_points = n - start;
+
+ /* Use standard least squares regression to get starting estimate */
+
+ P = U = 0.0;
+ for (i=start; i<n; i++) {
+ P += y[i];
+ U += x[i];
+ }
+
+ W = (double) n_points;
+
+ my = P/W;
+ mx = U/W;
+
+ X = V = 0.0;
+ for (i=start; i<n; i++) {
+ dy = y[i] - my;
+ dx = x[i] - mx;
+ X += dy * dx;
+ V += dx * dx;
+ }
+
+ b = X / V;
+ a = my - b*mx;
+
+ s2 = 0.0;
+ for (i=start; i<n; i++) {
+ resid = y[i] - a - b * x[i];
+ s2 += resid * resid;
+ }
+
+ /* Need to expand range of b to get a root in the interval.
+ Estimate standard deviation of b and expand range about b based
+ on that. */
+ sb = sqrt(s2 * W/V);
+ incr = MAX(sb, tol);
+
+ do {
+ incr *= 2.0;
+
+ /* Give up if the interval is too large */
+ if (incr > 100.0)
+ return 0;
+
+ blo = b - incr;
+ bhi = b + incr;
+
+ /* We don't want 'a' yet */
+ eval_robust_residual(x + start, y + start, n_points, blo, &a, &rlo);
+ eval_robust_residual(x + start, y + start, n_points, bhi, &a, &rhi);
+
+ } while (rlo * rhi >= 0.0); /* fn vals have same sign or one is zero,
+ i.e. root not in interval (rlo, rhi). */
+
+ /* OK, so the root for b lies in (blo, bhi). Start bisecting */
+ do {
+ bmid = 0.5 * (blo + bhi);
+ if (!(blo < bmid && bmid < bhi))
+ break;
+ eval_robust_residual(x + start, y + start, n_points, bmid, &a, &rmid);
+ if (rmid == 0.0) {
+ break;
+ } else if (rmid * rlo > 0.0) {
+ blo = bmid;
+ rlo = rmid;
+ } else if (rmid * rhi > 0.0) {
+ bhi = bmid;
+ rhi = rmid;
+ } else {
+ assert(0);
+ }
+ } while (bhi - blo > tol);
+
+ *b0 = a;
+ *b1 = bmid;
+
+ /* Number of runs test, but not if we're already down to the
+ minimum number of points */
+ if (n_points == MIN_SAMPLES_FOR_REGRESS) {
+ break;
+ }
+
+ for (i=start; i<n; i++) {
+ resids[i] = y[i] - a - bmid * x[i];
+ }
+
+ nruns = n_runs_from_residuals(resids + start, n_points);
+
+ if (nruns > critical_runs[n_points]) {
+ break;
+ } else {
+ start++;
+ }
+
+ } while (1);
+
+ *n_runs = nruns;
+ *best_start = start;
+
+ return 1;
+
+}
+
+/* ================================================== */
+/* This routine performs linear regression with two independent variables.
+ It returns non-zero status if there were enough data points and there
+ was a solution. */
+
+int
+RGR_MultipleRegress
+(double *x1, /* first independent variable */
+ double *x2, /* second independent variable */
+ double *y, /* measured data */
+
+ int n, /* number of data points */
+
+ /* The results */
+ double *b2 /* estimated second slope */
+ /* other values are not needed yet */
+)
+{
+ double Sx1, Sx2, Sx1x1, Sx1x2, Sx2x2, Sx1y, Sx2y, Sy;
+ double U, V, V1, V2, V3;
+ int i;
+
+ if (n < 4)
+ return 0;
+
+ Sx1 = Sx2 = Sx1x1 = Sx1x2 = Sx2x2 = Sx1y = Sx2y = Sy = 0.0;
+
+ for (i = 0; i < n; i++) {
+ Sx1 += x1[i];
+ Sx2 += x2[i];
+ Sx1x1 += x1[i] * x1[i];
+ Sx1x2 += x1[i] * x2[i];
+ Sx2x2 += x2[i] * x2[i];
+ Sx1y += x1[i] * y[i];
+ Sx2y += x2[i] * y[i];
+ Sy += y[i];
+ }
+
+ U = n * (Sx1x2 * Sx1y - Sx1x1 * Sx2y) +
+ Sx1 * Sx1 * Sx2y - Sx1 * Sx2 * Sx1y +
+ Sy * (Sx2 * Sx1x1 - Sx1 * Sx1x2);
+
+ V1 = n * (Sx1x2 * Sx1x2 - Sx1x1 * Sx2x2);
+ V2 = Sx1 * Sx1 * Sx2x2 + Sx2 * Sx2 * Sx1x1;
+ V3 = -2.0 * Sx1 * Sx2 * Sx1x2;
+ V = V1 + V2 + V3;
+
+ /* Check if there is a (numerically stable) solution */
+ if (fabs(V) * 1.0e10 <= -V1 + V2 + fabs(V3))
+ return 0;
+
+ *b2 = U / V;
+
+ return 1;
+}