You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
tdeedu/kig/misc/kignumerics.cpp

390 lines
10 KiB

/**
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" tqcontains 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;
}