public static int computeMaximumOnCircle(double[] a, double[] normal) {
double db0, db1, db2, db3, db4;
double val1;
double[] zeros = new double[4];
double u, v, maxval, maxu = -1, maxv = -1, inc, arg, polyval;
int index, nroots;
index = -1;
nroots = -1;
db0 = a[2] - a[3];
db1 = 4 * a[1] - 2 * a[4] - 4 * a[0];
db2 = -6 * a[2];
db3 = -4 * a[1] - 2 * a[4] + 4 * a[0];
db4 = a[2] + a[3];
/** polynomial is constant on circle, pick (0,1) as a solution */
if (Math.abs(db0) < zerotol
&& Math.abs(db1) < zerotol
&& Math.abs(db2) < zerotol
&& Math.abs(db3) < zerotol) {
normal[0] = 0.0;
normal[1] = 1.0;
return 1;
}
if (db0 != 0) {
double[] c = new double[] {db4, db3, db2, db1, db0};
nroots = GraphicsGems.SolveQuartic(c, zeros);
} else if (db1 != 0) {
double[] c = new double[] {db4, db3, db2, db1};
nroots = GraphicsGems.SolveCubic(c, zeros);
} else
/** TODO case where db2 is zero */
{
double[] c = new double[] {db4, db3, db2};
nroots = GraphicsGems.SolveQuadric(c, zeros);
}
if (nroots <= 0) {
return -1;
}
switch (nroots) {
case 1:
index = 0;
break;
default:
double[] vals = new double[nroots];
for (int i = 0; i < vals.length; i++) {
vals[i] = evalPoly(a, zeros[i]);
}
index = Utils.indexOfMax(vals);
}
/**
* I noticed that the fact that the pont (0,-1) on the circle can only be attained in the limit
* causes it to be missed in case it really is the maximum. Hence it is necessary to investigate
* a neighboring region to find the potential maximum there, we look at the segment from 260
* degress to 280 degrees (270 degrees being the limit point).
*/
normal[0] = 2 * zeros[index] / (1 + zeros[index] * zeros[index]);
normal[1] = (1 - zeros[index] * zeros[index]) / (1 + zeros[index] * zeros[index]);
/** test the correctness of solution: */
maxval = -1000;
for (int k = 0; k <= 20; k++) {
inc = (1 / 9.0) / 20 * k;
arg = Math.PI * (26.0 / 18.0 + inc);
u = Math.cos(arg);
v = Math.sin(arg);
polyval = a[0] * u * u + a[1] * v * v + a[2] * u * v + a[3] * u + a[4] * v + a[5];
if (maxval < polyval) {
maxval = polyval;
maxu = u;
maxv = v;
}
}
val1 = evalPoly(a, zeros[index]);
if (maxval > val1) {
normal[0] = maxu;
normal[1] = maxv;
}
return 1;
}
public static Vec3f[] getNormals(int[][] coeff) {
Vec3f[] normals = new Vec3f[coeff.length];
double[] a;
double length2d;
double disc;
int stat;
int maxfound;
for (int i = 0; i < normals.length; i++) {
double[] normal = new double[3];
a = new double[6];
a[0] = (double) coeff[i][0];
a[1] = (double) coeff[i][1];
a[2] = (double) coeff[i][2];
a[3] = (double) coeff[i][3];
a[4] = (double) coeff[i][4];
a[5] = (double) coeff[i][5];
/* now remove factor of 256 in scaling values */
a[0] /= 256;
a[1] /= 256;
a[2] /= 256;
a[3] /= 256;
a[4] /= 256;
a[5] /= 256;
/*
* These coordinates are where the first deriviative of the
* polynominal is zero
*/
/* Derivation in lab notebook 2951-43 */
/*
* normal[0] = (2 * a[1] * a[3] - (a[4] / a[2]) ) / (1.0 - 4 * a[0] *
* a[1]);
*/
/** zero denominator in upcoming computations */
if (Math.abs(a[2] * a[2] - 4 * a[1] * a[0]) < zerotol) {
normal[0] = 0;
normal[1] = 0;
} else {
if (Math.abs(a[2]) < zerotol) {
normal[0] = -1.0 * a[3] / (2.0 * a[0]);
normal[1] = -1.0 * a[4] / (2.0 * a[1]);
} else {
normal[0] = (2.0 * a[1] * a[3] - a[2] * a[4]) / (a[2] * a[2] - 4.0 * a[1] * a[0]);
normal[1] = (-2.0 * a[0] * normal[0] - a[3]) / a[2];
}
}
/** polynomial is constant we are done, set normal to be at 0,0,1 */
if (Math.abs(a[0]) < zerotol
&& Math.abs(a[1]) < zerotol
&& Math.abs(a[2]) < zerotol
&& Math.abs(a[3]) < zerotol
&& Math.abs(a[4]) < zerotol) {
normal[0] = 0.0;
normal[1] = 0.0;
normal[2] = 1.0;
}
/**
* clip normal[0], normal[1] values - these can both be unbounded theoretically. first check
* if the vector (normal[0],normal[1]) is greater than 1
*/
else {
length2d = normal[0] * normal[0] + normal[1] * normal[0];
/*
* Add check for saddle or minimum. if p_uu >0 and/or
* p_uu*p_vv-p_uv*p_uv <0 -> saddle or minimum if this is the
* case then we should always look at boundary
*/
if (4 * a[0] * a[1] - a[2] * a[2] > eps && a[0] < -eps) maxfound = 1;
else maxfound = 0;
/**
* Changed by Hans J. Wolters 12-11-99: instead of clipping, we attempt to find the minimum
* value on the circle. we know that since the polynomial is monotonous in the interior the
* extrema must be attained at the boundary. The circle will be parametrized by x =
* 2t/(1+t^2), y = (1-t^2)/(1+t^2). by using Maple we derived the polynomial in t and
* subsequently its derivative wrt t, a quartic. The task is to find the real root(s) of
* this polynomial. We denote the 3 coefficients as db0,db1,db2,db3, db4 For precise
* derivation see Hans Wolters' lab book
*/
if (length2d > 1 - eps || maxfound == 0) {
stat = computeMaximumOnCircle(a, normal);
if (stat == -1) // failed
{
length2d = Math.sqrt(length2d);
if (length2d > zerotol) {
normal[0] /= length2d;
normal[1] /= length2d;
}
}
}
disc = 1.0 - normal[0] * normal[0] - normal[1] * normal[1];
if (disc < 0.0) {
normal[2] = 0;
} else {
normal[2] = (float) Math.sqrt(disc);
}
}
normal = Utils.normalize3(normal);
normals[i] = new Vec3f((float) normal[0], (float) normal[1], (float) normal[2]);
}
return normals;
}
