fly/flightgear
1
0
Fork 0
Code Activity
flightgear/Math/MAT3inv.c
curt 935d4f3bda The MAT3 routines from SRGP.
1997-05-30 19:25:54 +00:00

312 lines
8.5 KiB
C
Raw Blame History

/* Copyright 1988, Brown Computer Graphics Group. All Rights Reserved. */
/* --------------------------------------------------------------------------
* This file contains routines that operate solely on matrices.
* -------------------------------------------------------------------------*/
#include "mat3defs.h"
/* -------------------------- Static Routines ---------------------------- */
#define SMALL 1e-20 /* Small enough to be considered zero */
/*
* Shuffles rows in inverse of 3x3. See comment in MAT3_inv3_second_col().
*/
static void
MAT3_inv3_swap( register double inv[3][3], int row0, int row1, int row2)
{
register int i, tempi;
double temp;
#define SWAP_ROWS(a, b) \
for (i = 0; i < 3; i++) SWAP(inv[a][i], inv[b][i], temp); \
SWAP(a, b, tempi)
if (row0 != 0){
if (row1 == 0) {
SWAP_ROWS(row0, row1);
}
else {
SWAP_ROWS(row0, row2);
}
}
if (row1 != 1) {
SWAP_ROWS(row1, row2);
}
}
/*
* Does Gaussian elimination on second column.
*/
static int
MAT3_inv3_second_col (register double source[3][3], register double inv[3][3], int row0)
{
register int row1, row2, i1, i2, i;
double temp;
double a, b;
/* Find which row to use */
if (row0 == 0) i1 = 1, i2 = 2;
else if (row0 == 1) i1 = 0, i2 = 2;
else i1 = 0, i2 = 1;
/* Find which is larger in abs. val.:the entry in [i1][1] or [i2][1] */
/* and use that value for pivoting. */
a = source[i1][1]; if (a < 0) a = -a;
b = source[i2][1]; if (b < 0) b = -b;
if (a > b) row1 = i1;
else row1 = i2;
row2 = (row1 == i1 ? i2 : i1);
/* Scale row1 in source */
if ((source[row1][1] < SMALL) && (source[row1][1] > -SMALL)) return(FALSE);
temp = 1.0 / source[row1][1];
source[row1][1] = 1.0;
source[row1][2] *= temp; /* source[row1][0] is zero already */
/* Scale row1 in inv */
inv[row1][row1] = temp; /* it used to be a 1.0 */
inv[row1][row0] *= temp;
/* Clear column one, source, and make corresponding changes in inv */
for (i = 0; i < 3; i++) if (i != row1) { /* for i = all rows but row1 */
temp = -source[i][1];
source[i][1] = 0.0;
source[i][2] += temp * source[row1][2];
inv[i][row1] = temp * inv[row1][row1];
inv[i][row0] += temp * inv[row1][row0];
}
/* Scale row2 in source */
if ((source[row2][2] < SMALL) && (source[row2][2] > -SMALL)) return(FALSE);
temp = 1.0 / source[row2][2];
source[row2][2] = 1.0; /* source[row2][*] is zero already */
/* Scale row2 in inv */
inv[row2][row2] = temp; /* it used to be a 1.0 */
inv[row2][row0] *= temp;
inv[row2][row1] *= temp;
/* Clear column one, source, and make corresponding changes in inv */
for (i = 0; i < 3; i++) if (i != row2) { /* for i = all rows but row2 */
temp = -source[i][2];
source[i][2] = 0.0;
inv[i][row0] += temp * inv[row2][row0];
inv[i][row1] += temp * inv[row2][row1];
inv[i][row2] += temp * inv[row2][row2];
}
/*
* Now all is done except that the inverse needs to have its rows shuffled.
* row0 needs to be moved to inv[0][*], row1 to inv[1][*], etc.
*
* We *didn't* do the swapping before the elimination so that we could more
* easily keep track of what ops are needed to be done in the inverse.
*/
MAT3_inv3_swap(inv, row0, row1, row2);
return(TRUE);
}
/*
* Fast inversion routine for 3 x 3 matrices. - Written by jfh.
*
* This takes 30 multiplies/divides, as opposed to 39 for Cramer's Rule.
* The algorithm consists of performing fast gaussian elimination, by never
* doing any operations where the result is guaranteed to be zero, or where
* one operand is guaranteed to be zero. This is done at the cost of clarity,
* alas.
*
* Returns 1 if the inverse was successful, 0 if it failed.
*/
static int
MAT3_invert3 (register double source[3][3], register double inv[3][3])
{
register int i, row0;
double temp;
double a, b, c;
inv[0][0] = inv[1][1] = inv[2][2] = 1.0;
inv[0][1] = inv[0][2] = inv[1][0] = inv[1][2] = inv[2][0] = inv[2][1] = 0.0;
/* attempt to find the largest entry in first column to use as pivot */
a = source[0][0]; if (a < 0) a = -a;
b = source[1][0]; if (b < 0) b = -b;
c = source[2][0]; if (c < 0) c = -c;
if (a > b) {
if (a > c) row0 = 0;
else row0 = 2;
}
else {
if (b > c) row0 = 1;
else row0 = 2;
}
/* Scale row0 of source */
if ((source[row0][0] < SMALL) && (source[row0][0] > -SMALL)) return(FALSE);
temp = 1.0 / source[row0][0];
source[row0][0] = 1.0;
source[row0][1] *= temp;
source[row0][2] *= temp;
/* Scale row0 of inverse */
inv[row0][row0] = temp; /* other entries are zero -- no effort */
/* Clear column zero of source, and make corresponding changes in inverse */
for (i = 0; i < 3; i++) if (i != row0) { /* for i = all rows but row0 */
temp = -source[i][0];
source[i][0] = 0.0;
source[i][1] += temp * source[row0][1];
source[i][2] += temp * source[row0][2];
inv[i][row0] = temp * inv[row0][row0];
}
/*
* We've now done gaussian elimination so that the source and
* inverse look like this:
*
* 1 * * * 0 0
* 0 * * * 1 0
* 0 * * * 0 1
*
* We now proceed to do elimination on the second column.
*/
if (! MAT3_inv3_second_col(source, inv, row0)) return(FALSE);
return(TRUE);
}
/*
* Finds a new pivot for a non-simple 4x4. See comments in MAT3invert().
*/
static int
MAT3_inv4_pivot (register MAT3mat src, MAT3vec r, double *s, int *swap)
{
register int i, j;
double temp, max;
*swap = -1;
if (MAT3_IS_ZERO(src[3][3])) {
/* Look for a different pivot element: one with largest abs value */
max = 0.0;
for (i = 0; i < 4; i++) {
if (src[i][3] > max) max = src[*swap = i][3];
else if (src[i][3] < -max) max = -src[*swap = i][3];
}
/* No pivot element available ! */
if (*swap < 0) return(FALSE);
else for (j = 0; j < 4; j++) SWAP(src[*swap][j], src[3][j], temp);
}
MAT3_SET_VEC (r, -src[0][3], -src[1][3], -src[2][3]);
*s = 1.0 / src[3][3];
src[0][3] = src[1][3] = src[2][3] = 0.0;
src[3][3] = 1.0;
MAT3_SCALE_VEC(src[3], src[3], *s);
for (i = 0; i < 3; i++) {
src[0][i] += r[0] * src[3][i];
src[1][i] += r[1] * src[3][i];
src[2][i] += r[2] * src[3][i];
}
return(TRUE);
}
/* ------------------------- Internal Routines --------------------------- */
/* -------------------------- Public Routines ---------------------------- */
/*
* This returns the inverse of the given matrix. The result matrix
* may be the same as the one to invert.
*
* Fast inversion routine for 4 x 4 matrices, written by jfh.
*
* Returns 1 if the inverse was successful, 0 if it failed.
*
* This routine has been specially tweaked to notice the following:
* If the matrix has the form
* * * * 0
* * * * 0
* * * * 0
* * * * 1
*
* (as do many matrices in graphics), then we compute the inverse of
* the upper left 3x3 matrix and use this to find the general inverse.
*
* In the event that the right column is not 0-0-0-1, we do gaussian
* elimination to make it so, then use the 3x3 inverse, and then do
* our gaussian elimination.
*/
int
MAT3invert(result_mat, mat)
MAT3mat result_mat, mat;
{
MAT3mat src, inv;
register int i, j, simple;
double m[3][3], inv3[3][3], s, temp;
MAT3vec r, t;
int swap;
MAT3copy(src, mat);
MAT3identity(inv);
/* If last column is not (0,0,0,1), use special code */
simple = (mat[0][3] == 0.0 && mat[1][3] == 0.0 &&
mat[2][3] == 0.0 && mat[3][3] == 1.0);
if (! simple && ! MAT3_inv4_pivot(src, r, &s, &swap)) return(FALSE);
MAT3_COPY_VEC(t, src[3]); /* Translation vector */
/* Copy upper-left 3x3 matrix */
for (i = 0; i < 3; i++) for (j = 0; j < 3; j++) m[i][j] = src[i][j];
if (! MAT3_invert3(m, inv3)) return(FALSE);
for (i = 0; i < 3; i++) for (j = 0; j < 3; j++) inv[i][j] = inv3[i][j];
for (i = 0; i < 3; i++) for (j = 0; j < 3; j++)
inv[3][i] -= t[j] * inv3[j][i];
if (! simple) {
/* We still have to undo our gaussian elimination from earlier on */
/* add r0 * first col to last col */
/* add r1 * 2nd col to last col */
/* add r2 * 3rd col to last col */
for (i = 0; i < 4; i++) {
inv[i][3] += r[0] * inv[i][0] + r[1] * inv[i][1] + r[2] * inv[i][2];
inv[i][3] *= s;
}
if (swap >= 0)
for (i = 0; i < 4; i++) SWAP(inv[i][swap], inv[i][3], temp);
}
MAT3copy(result_mat, inv);
return(TRUE);
}
View git blame Copy permalink