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+/**
+ This file is part of Kig, a KDE program for Interactive Geometry...
+ Copyright (C) 2002 Maurizio Paolini <paolini@dmf.unicatt.it>
+
+ This program is free software; you can redistribute it and/or modify
+ it under the terms of the GNU General Public License as published by
+ the Free Software Foundation; either version 2 of the License, or
+ (at your option) any later version.
+
+ 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
+**/
+
+#include "kignumerics.h"
+#include "common.h"
+
+/*
+ * compute one of the roots of a cubic polynomial
+ * if xmin << 0 or xmax >> 0 then autocompute a bound for all the
+ * roots
+ */
+
+double calcCubicRoot ( double xmin, double xmax, double a,
+ double b, double c, double d, int root, bool& valid, int& numroots )
+{
+ // renormalize: positive a and infinity norm = 1
+
+ double infnorm = fabs(a);
+ if ( infnorm < fabs(b) ) infnorm = fabs(b);
+ if ( infnorm < fabs(c) ) infnorm = fabs(c);
+ if ( infnorm < fabs(d) ) infnorm = fabs(d);
+ if ( a < 0 ) infnorm = -infnorm;
+ a /= infnorm;
+ b /= infnorm;
+ c /= infnorm;
+ d /= infnorm;
+
+ const double small = 1e-7;
+ valid = false;
+ if ( fabs(a) < small )
+ {
+ if ( fabs(b) < small )
+ {
+ if ( fabs(c) < small )
+ { // degree = 0;
+ numroots = 0;
+ return 0.0;
+ }
+ // degree = 1
+ double rootval = -d/c;
+ numroots = 1;
+ if ( rootval < xmin || xmax < rootval ) numroots--;
+ if ( root > numroots ) return 0.0;
+ valid = true;
+ return rootval;
+ }
+ // degree = 2
+ if ( b < 0 ) { b = -b; c = -c; d = -d; }
+ double discrim = c*c - 4*b*d;
+ numroots = 2;
+ if ( discrim < 0 )
+ {
+ numroots = 0;
+ return 0.0;
+ }
+ discrim = sqrt(discrim)/(2*fabs(b));
+ double rootmiddle = -c/(2*b);
+ if ( rootmiddle - discrim < xmin ) numroots--;
+ if ( rootmiddle + discrim > xmax ) numroots--;
+ if ( rootmiddle + discrim < xmin ) numroots--;
+ if ( rootmiddle - discrim > xmax ) numroots--;
+ if ( root > numroots ) return 0.0;
+ valid = true;
+ if ( root == 2 || rootmiddle - discrim < xmin ) return rootmiddle + discrim;
+ return rootmiddle - discrim;
+ }
+
+ if ( xmin < -1e8 || xmax > 1e8 )
+ {
+
+ // compute a bound for all the real roots:
+
+ xmax = fabs(d/a);
+ if ( fabs(c/a) + 1 > xmax ) xmax = fabs(c/a) + 1;
+ if ( fabs(b/a) + 1 > xmax ) xmax = fabs(b/a) + 1;
+ xmin = -xmax;
+ }
+
+ // computing the coefficients of the Sturm sequence
+ double p1a = 2*b*b - 6*a*c;
+ double p1b = b*c - 9*a*d;
+ double p0a = c*p1a*p1a + p1b*(3*a*p1b - 2*b*p1a);
+
+ int varbottom = calcCubicVariations (xmin, a, b, c, d, p1a, p1b, p0a);
+ int vartop = calcCubicVariations (xmax, a, b, c, d, p1a, p1b, p0a);
+ numroots = vartop - varbottom;
+ valid = false;
+ if (root <= varbottom || root > vartop ) return 0.0;
+
+ valid = true;
+
+ // now use bisection to separate the required root
+ double dx = (xmax - xmin)/2;
+ while ( vartop - varbottom > 1 )
+ {
+ if ( fabs( dx ) < 1e-8 ) return (xmin + xmax)/2;
+ double xmiddle = xmin + dx;
+ int varmiddle = calcCubicVariations (xmiddle, a, b, c, d, p1a, p1b, p0a);
+ if ( varmiddle < root ) // I am below
+ {
+ xmin = xmiddle;
+ varbottom = varmiddle;
+ } else {
+ xmax = xmiddle;
+ vartop = varmiddle;
+ }
+ dx /= 2;
+ }
+
+ /*
+ * now [xmin, xmax] enclose a single root, try using Newton
+ */
+ if ( vartop - varbottom == 1 )
+ {
+ double fval1 = a; // double check...
+ double fval2 = a;
+ fval1 = b + xmin*fval1;
+ fval2 = b + xmax*fval2;
+ fval1 = c + xmin*fval1;
+ fval2 = c + xmax*fval2;
+ fval1 = d + xmin*fval1;
+ fval2 = d + xmax*fval2;
+ assert ( fval1 * fval2 <= 0 );
+ return calcCubicRootwithNewton ( xmin, xmax, a, b, c, d, 1e-8 );
+ }
+ else // probably a double root here!
+ return ( xmin + xmax )/2;
+}
+
+/*
+ * computation of the number of sign changes in the sturm sequence for
+ * a third degree polynomial at x. This number counts the number of
+ * roots of the polynomial on the left of point x.
+ *
+ * a, b, c, d: coefficients of the third degree polynomial (a*x^3 + ...)
+ *
+ * the second degree polynomial in the sturm sequence is just minus the
+ * derivative, so we don't need to compute it.
+ *
+ * p1a*x + p1b: is the third (first degree) polynomial in the sturm sequence.
+ *
+ * p0a: is the (constant) fourth polynomial of the sturm sequence.
+ */
+
+int calcCubicVariations (double x, double a, double b, double c,
+ double d, double p1a, double p1b, double p0a)
+{
+ double fval, fpval;
+ fval = fpval = a;
+ fval = b + x*fval;
+ fpval = fval + x*fpval;
+ fval = c + x*fval;
+ fpval = fval + x*fpval;
+ fval = d + x*fval;
+
+ double f1val = p1a*x + p1b;
+
+ bool f3pos = fval >= 0;
+ bool f2pos = fpval <= 0;
+ bool f1pos = f1val >= 0;
+ bool f0pos = p0a >= 0;
+
+ int variations = 0;
+ if ( f3pos != f2pos ) variations++;
+ if ( f2pos != f1pos ) variations++;
+ if ( f1pos != f0pos ) variations++;
+ return variations;
+}
+
+/*
+ * use newton to solve a third degree equation with already isolated
+ * root
+ */
+
+inline void calcCubicDerivatives ( double x, double a, double b, double c,
+ double d, double& fval, double& fpval, double& fppval )
+{
+ fval = fpval = fppval = a;
+ fval = b + x*fval;
+ fpval = fval + x*fpval;
+ fppval = fpval + x*fppval; // this is really half the second derivative
+ fval = c + x*fval;
+ fpval = fval + x*fpval;
+ fval = d + x*fval;
+}
+
+double calcCubicRootwithNewton ( double xmin, double xmax, double a,
+ double b, double c, double d, double tol )
+{
+ double fval, fpval, fppval;
+
+ double fval1, fval2, fpval1, fpval2, fppval1, fppval2;
+ calcCubicDerivatives ( xmin, a, b, c, d, fval1, fpval1, fppval1 );
+ calcCubicDerivatives ( xmax, a, b, c, d, fval2, fpval2, fppval2 );
+ assert ( fval1 * fval2 <= 0 );
+
+ assert ( xmax > xmin );
+ while ( xmax - xmin > tol )
+ {
+ // compute the values of function, derivative and second derivative:
+ assert ( fval1 * fval2 <= 0 );
+ if ( fppval1 * fppval2 < 0 || fpval1 * fpval2 < 0 )
+ {
+ double xmiddle = (xmin + xmax)/2;
+ calcCubicDerivatives ( xmiddle, a, b, c, d, fval, fpval, fppval );
+ if ( fval1*fval <= 0 )
+ {
+ xmax = xmiddle;
+ fval2 = fval;
+ fpval2 = fpval;
+ fppval2 = fppval;
+ } else {
+ xmin = xmiddle;
+ fval1 = fval;
+ fpval1 = fpval;
+ fppval1 = fppval;
+ }
+ } else
+ {
+ // now we have first and second derivative of constant sign, we
+ // can start with Newton from the Fourier point.
+ double x = xmin;
+ if ( fval2*fppval2 > 0 ) x = xmax;
+ double p = 1.0;
+ int iterations = 0;
+ while ( fabs(p) > tol && iterations++ < 100 )
+ {
+ calcCubicDerivatives ( x, a, b, c, d, fval, fpval, fppval );
+ p = fval/fpval;
+ x -= p;
+ }
+ if( iterations >= 100 )
+ {
+ // Newton scheme did not converge..
+ // we should end up with an invalid Coordinate
+ return double_inf;
+ };
+ return x;
+ }
+ }
+
+ // we cannot apply Newton, (perhaps we are at an inflection point)
+
+ return (xmin + xmax)/2;
+}
+
+/*
+ * This function computes the LU factorization of a mxn matrix, with
+ * m typically less then n. This is done with complete pivoting; the
+ * exchanges in columns are recorded in the integer vector "exchange"
+ */
+bool GaussianElimination( double *matrix[], int numrows,
+ int numcols, int exchange[] )
+{
+ // start gaussian elimination
+ for ( int k = 0; k < numrows; ++k )
+ {
+ // ricerca elemento di modulo massimo
+ double maxval = -double_inf;
+ int imax = k;
+ int jmax = k;
+ for( int i = k; i < numrows; ++i )
+ {
+ for( int j = k; j < numcols; ++j )
+ {
+ if (fabs(matrix[i][j]) > maxval)
+ {
+ maxval = fabs(matrix[i][j]);
+ imax = i;
+ jmax = j;
+ }
+ }
+ }
+
+ // row exchange
+ if ( imax != k )
+ for( int j = k; j < numcols; ++j )
+ {
+ double t = matrix[k][j];
+ matrix[k][j] = matrix[imax][j];
+ matrix[imax][j] = t;
+ }
+
+ // column exchange
+ if ( jmax != k )
+ for( int i = 0; i < numrows; ++i )
+ {
+ double t = matrix[i][k];
+ matrix[i][k] = matrix[i][jmax];
+ matrix[i][jmax] = t;
+ }
+
+ // remember this column exchange at step k
+ exchange[k] = jmax;
+
+ // we can't usefully eliminate a singular matrix..
+ if ( maxval == 0. ) return false;
+
+ // ciclo sulle righe
+ for( int i = k+1; i < numrows; ++i)
+ {
+ double mik = matrix[i][k]/matrix[k][k];
+ matrix[i][k] = mik; //ricorda il moltiplicatore... (not necessary)
+ // ciclo sulle colonne
+ for( int j = k+1; j < numcols; ++j )
+ {
+ matrix[i][j] -= mik*matrix[k][j];
+ }
+ }
+ }
+ return true;
+}
+
+/*
+ * solve an undetermined homogeneous triangular system. the matrix is nonzero
+ * on its diagonal. The last unknown(s) are chosen to be 1. The
+ * vector "exchange" contains exchanges to be performed on the
+ * final solution components.
+ */
+
+void BackwardSubstitution( double *matrix[], int numrows, int numcols,
+ int exchange[], double solution[] )
+{
+ // the system is homogeneous and underdetermined, the last unknown(s)
+ // are chosen = 1
+ for ( int j = numrows; j < numcols; ++j )
+ {
+ solution[j] = 1.0; // other choices are possible here
+ };
+
+ for( int k = numrows - 1; k >= 0; --k )
+ {
+ // backward substitution
+ solution[k] = 0.0;
+ for ( int j = k+1; j < numcols; ++j)
+ {
+ solution[k] -= matrix[k][j]*solution[j];
+ }
+ solution[k] /= matrix[k][k];
+ }
+
+ // ultima fase: riordinamento incognite
+
+ for( int k = numrows - 1; k >= 0; --k )
+ {
+ int jmax = exchange[k];
+ double t = solution[k];
+ solution[k] = solution[jmax];
+ solution[jmax] = t;
+ }
+}
+
+bool Invert3by3matrix ( const double m[3][3], double inv[3][3] )
+{
+ double det = m[0][0]*(m[1][1]*m[2][2] - m[1][2]*m[2][1]) -
+ m[0][1]*(m[1][0]*m[2][2] - m[1][2]*m[2][0]) +
+ m[0][2]*(m[1][0]*m[2][1] - m[1][1]*m[2][0]);
+ if (det == 0) return false;
+
+ for (int i=0; i < 3; i++)
+ {
+ for (int j=0; j < 3; j++)
+ {
+ int i1 = (i+1)%3;
+ int i2 = (i+2)%3;
+ int j1 = (j+1)%3;
+ int j2 = (j+2)%3;
+ inv[j][i] = (m[i1][j1]*m[i2][j2] - m[i1][j2]*m[i2][j1])/det;
+ }
+ }
+ return true;
+}
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