Code:
/ 4.0 / 4.0 / DEVDIV_TFS / Dev10 / Releases / RTMRel / wpf / src / Core / CSharp / MS / Internal / Media3D / LineUtil.cs / 1305600 / LineUtil.cs
//----------------------------------------------------------------------------
//
// Copyright (C) Microsoft Corporation. All rights reserved.
//
//---------------------------------------------------------------------------
// GLOSSARY
// Kernel of matrix M : Space of vectors that vanish (go to 0) when multiplied by M
// Null-space of matrix M : Same thing
// Line matrix : Representation of a line using a matrix whose kernel is points on the line
// L : Matrix representing a line
// LT : Transpose of L
//
// This file offers a small line transform utility function. Given a line (lin) defined by Point3D
// origin & Vector3D direction and a model matrix M it returns (in-place) a line (lout) in "model
// space" so that any point on the line when transformed by M is on the original line.
//
// In other words x in lout ===> x M in lin
//
// This works even if M is rank-deficient, but if M is rank 2 or less then lout is not uniquely
// determined.
//
// The basic technique is as follows, we represent lin as a matrix where points on the line are in
// the null-space (kernel) of lin (this is straightforward.) Letting the symbols lin & lout refer
// to both the line & the corresponding matrix this means:
//
// x in lin ===> x lin = [0,0,0,0]
//
// Then we can transform a line matrix into model space by left multiplication with M
//
// lout = M lin
//
// That way,
//
// x in lout ===> x lout = [0,0,0,0] ===> x M lin = [0,0,0,0] ===> x M in lin
//
// Which is what we want. The hard part is going back from the matrix lout to a point and a vector.
// The smallest two eigenvalues of the l matrix are zero. The corresponding eigenvectors are two
// different points on the line.
//
// I find the eigenvectors and eigenvalues of a line matrix L by first computing the normal matrix
// N = L LT which has symmetric eigenvectors equal to the left eigenvectors of L. I find the
// eigenvectors of N using a Jacobi method for symmetric eigenvalue problems.
//
// The method is described in Chapter 8.4 of Golub & Van Loan (Matrix Computation), but here's a
// brief summary.
//
// We apply a series of 2D rotations (A1...An) to the matrix N that each make the matrix more
// diagonal. So if the whole sequence is A = A1 ... An then the final matrix E = AT N A is
// diagonal.
//
// Because A is orthonormal its columns are the eigenvectors of N and the diagonal elements of E are
// the eigenvalues.
//
// NOTES
//
// Forming the normal matrix N involves fourth powers of the input values. I mitigate this by
// scaling the matrix so that its largest value is 1 before squaring it. There may be a better
// method (perhaps even a modified Jacobi method) that would work directly on L and perhaps be more
// stable.
//
// None of this code understands rays. This is all in terms of lines, lines, lines, lines!
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Windows;
using System.Windows.Media.Media3D;
using MS.Utility;
namespace MS.Internal.Media3D
{
[Flags]
internal enum FaceType
{
None = 0,
Front = 1 << 0,
Back = 1 << 1,
};
internal static class LineUtil
{
// Coordinates of elements above the diagonal.
readonly static int[,] s_pairs = new int[,]{ {0,1}, {0,2}, {0,3}, {1,2}, {1,3}, {2,3} };
const int s_pairsCount = 6;
public static void Transform(Matrix3D modelMatrix,
ref Point3D origin, ref Vector3D direction, out bool isRay)
{
if (modelMatrix.InvertCore())
{
Point4D o = new Point4D(origin.X,origin.Y,origin.Z,1);
Point4D d = new Point4D(direction.X,direction.Y,direction.Z,0);
modelMatrix.MultiplyPoint(ref o);
modelMatrix.MultiplyPoint(ref d);
if (o.W == 1 && d.W == 0)
{
// Affine transformation
origin = new Point3D(o.X, o.Y, o.Z);
direction = new Vector3D(d.X, d.Y, d.Z);
isRay = true;
}
else
{
// Non-affine transformation (likely projection)
// Form 4x2 matrix with two points on line in two columns.
double[,] linepoints = new double[,]{{o.X,d.X},{o.Y,d.Y},{o.Z,d.Z},{o.W,d.W}};
ColumnsToAffinePointVector(linepoints,0,1,out origin, out direction);
isRay = false;
}
}
else
{
TransformSingular(ref modelMatrix, ref origin, ref direction);
isRay = false;
}
}
// modelMatrix is passed by reference for efficiency only. It is not modified.
private static void TransformSingular(ref Matrix3D modelMatrix,
ref Point3D origin, ref Vector3D direction)
{
double [,] matrix = TransformedLineMatrix(ref modelMatrix, ref origin, ref direction);
matrix = Square(matrix);
double[,] eigen = new double[,]{ {1,0,0,0}, {0,1,0,0}, {0,0,1,0}, {0,0,0,1} };
// We'll just do 5 iterations with each pair because according to my results & Golub &
// Van Loan this process converges quickly.
int iterations = 5 * s_pairsCount;
for (int iter = 0; iter < iterations; ++iter)
{
int pair = iter % s_pairsCount;
JacobiRotation jrot = new JacobiRotation(s_pairs[pair,0],s_pairs[pair,1],matrix);
matrix = jrot.LeftRightMultiply(matrix);
eigen = jrot.RightMultiply(eigen);
}
// That was it as far as finding eigenvectors
int evec1,evec2;
FindSmallestTwoDiagonal(matrix, out evec1, out evec2);
// The eigenvectors corresponding to the two smallest eigenvalues are columns evec1 &
// evec2. These, in homogeneous space, are two different points on our line. We are
// going to convert them to an affine point & vector.
ColumnsToAffinePointVector(eigen, evec1, evec2, out origin, out direction);
}
private static void ColumnsToAffinePointVector(double[,] matrix, int col1, int col2, out Point3D origin, out Vector3D direction)
{
// The col1 & col2 columsn of matrix are two different homogeneous points on a line. We
// are going to convert them to an affine point & vector using the following procedure.
// 1. Pick the eigenvector with the largest W, call it big
// 2. Scale it by 1/W to form an affine point, still call it big
// 3. Add a weighted multiple of big to small to make small.W zero.
//
// If both w are zero then we have a projective line that intersects no affine points
// (i.e. it's a line at infinity.) This will cause overflow but that's fine because the
// line at infinity doesn't intersect anything & the caller of this function needs be
// able to handle results that come back with inf or nan by "doing nothing."
// Step 1.
if (matrix[3,col1]*matrix[3,col1] < matrix[3,col2]*matrix[3,col2])
{
int temp = col1;
col1 = col2;
col2 = temp;
}
// Step 2.
double s = 1/matrix[3,col1];
origin = new Point3D(s*matrix[0,col1],
s*matrix[1,col1],
s*matrix[2,col1]);
// Step 3.
s = -matrix[3,col2];
direction = new Vector3D(matrix[0,col2]+s*origin.X,
matrix[1,col2]+s*origin.Y,
matrix[2,col2]+s*origin.Z);
}
// Returns the indices of the smallest two diagonal elements of matrix
private static void FindSmallestTwoDiagonal(double[,] matrix, out int evec1, out int evec2)
{
evec1 = 0;
evec2 = 1;
// And corresponding squared eigenvalues.
double eval1 = matrix[0,0]*matrix[0,0];
double eval2 = matrix[1,1]*matrix[1,1];
for (int i = 2; i < 4; ++i)
{
// Replace second smallest if necessary.
double val = matrix[i,i]*matrix[i,i];
if (val < eval1)
{
if (eval1 < eval2)
{
eval2 = val;
evec2 = i;
}
else
{
eval1 = val;
evec1 = i;
}
}
else if (val < eval2)
{
eval2 = val;
evec2 = i;
}
}
}
// Returns the "line matrix" corresponding to the line (origin,direction) transformed by the
// inverse *transform* of modelMatrix. (To transform a line by the inverse transform
// requires multiplying by the non-inverted matrix.)
private static double[,] TransformedLineMatrix(ref Matrix3D modelMatrix,
ref Point3D origin, ref Vector3D direction)
{
double x1 = origin.X;
double y1 = origin.Y;
double z1 = origin.Z;
// w1 = 1
double x2 = direction.X;
double y2 = direction.Y;
double z2 = direction.Z;
// w2 = 0
// To prove to yourself that this matrix is correct just multiply by the two
// (homogeneous) points on the line. Any other homogeneous point on the line is a
// linear combination of them.
double a = y2*z1-y1*z2;
double b = x1*z2-x2*z1;
double c = x2*y1-x1*y2;
Matrix3D m = modelMatrix *
new Matrix3D(a, y2, z2, 0,
b, -x2, 0, z2,
c, 0, -x2, -y2,
0, c, -b, a);
double[,] matrix = new double[4,4];
matrix[0,0] = m.M11;
matrix[0,1] = m.M12;
matrix[0,2] = m.M13;
matrix[0,3] = m.M14;
matrix[1,0] = m.M21;
matrix[1,1] = m.M22;
matrix[1,2] = m.M23;
matrix[1,3] = m.M24;
matrix[2,0] = m.M31;
matrix[2,1] = m.M32;
matrix[2,2] = m.M33;
matrix[2,3] = m.M34;
matrix[3,0] = m.OffsetX;
matrix[3,1] = m.OffsetY;
matrix[3,2] = m.OffsetZ;
matrix[3,3] = m.M44;
return matrix;
}
// Scales M so that its largest element is 1 and then returns M MT
// (MT=transpose(M))
private static double [,] Square(double[,] m)
{
double[,] o = new double[4,4];
// Scale the matrix so that its largest element is 1.
double maxvalue = 0;
for (int i = 0; i < 4; ++i)
{
for (int j = 0; j < 4; ++j)
{
maxvalue = Math.Max(maxvalue,m[i,j]*m[i,j]);
}
}
maxvalue = Math.Sqrt(maxvalue);
for (int i = 0; i < 4; ++i)
{
for (int j = 0; j < 4; ++j)
{
m[i,j] /= maxvalue;
}
}
// Compute its square.
for (int i = 0; i < 4; ++i)
{
for (int j = 0; j < 4; ++j)
{
double d = 0;
for (int k = 0; k < 4; ++k)
{
d += m[i,k]*m[j,k];
}
o[i,j] = d;
}
}
return o;
}
// See section 8.4 of Golub & Van Loan "Matrix Computation"
// Remember, J is this Jacobi rotation. JT is the transpose.
// Also section 5.1.8 shows the formula for multiplying by the Jacobi rotation
// Though as they helpfully point out a Jacobi rotation is the same as a Givens rotation
private struct JacobiRotation
{
public JacobiRotation(int p, int q, double[,] a)
{
// Constructs a 2D rotation matrix M as follows
// Start with identity.
// Zero the p & q rows & columns
// Then set the intersections of these rows & columns as follows
// [ M_pp M_pq ] = [ c s]
// [ M_qp M_qq ] [-s c]
//
// Where c & s are a cosine-sine pair calculated so that multiplying a by this will
// descrease the the off-diagonal elements.
_p = p;
_q = q;
double tau = (a[q,q] - a[p,p])/(2*a[p,q]);
if (tau < Double.MaxValue && tau > -Double.MaxValue)
{
double root = Math.Sqrt(1+tau*tau);
// Choose the smaller of -tau +/- root
double t = -tau < 0 ? -tau + root : -tau - root;
_c = 1/Math.Sqrt(1+t*t);
_s = t * _c;
}
else
{
_c = 1;
_s = 0;
}
}
// These functions overwrite & return their argument.
// returns JT a J
public double[,] LeftRightMultiply(double [,] a)
{
return RightMultiply(LeftMultiplyTranspose(a));
}
// returns a J
public double[,] RightMultiply(double [,] a)
{
for (int j = 0; j < 4; ++j)
{
double tau1 = a[j,_p];
double tau2 = a[j,_q];
a[j,_p] = _c * tau1 - _s * tau2;
a[j,_q] = _s * tau1 + _c * tau2;
}
return a;
}
// returns JT a
public double[,] LeftMultiplyTranspose(double [,] a)
{
for (int j = 0; j < 4; ++j)
{
double tau1 = a[_p,j];
double tau2 = a[_q,j];
a[_p,j] = _c * tau1 - _s * tau2;
a[_q,j] = _s * tau1 + _c * tau2;
}
return a;
}
private int _p, _q;
private double _c, _s;
}
// This method determines if the line/ray intersects the triangle.
// If "origin" and "direction" truely represent a line, "type" should be front and back because
// we don't have any true direction.
//
// origin/direction define the line/ray
// v0/v1/v2 define the triangle
//
// origin, direction, v0, v1, v2 are passed by ref for perf. They are NOT MODIFIED
//
// If this method returns false, ignore the values of hitCoord and dist.
//
// Ported from dxg\d3dx9\mesh\intersect.cpp (12/04/03)
// which is an implementation of "Fast, Minimum Storage Ray-Triangle Intersection" by Moller + Trumbore
internal static bool ComputeLineTriangleIntersection(
FaceType type,
ref Point3D origin,
ref Vector3D direction,
ref Point3D v0,
ref Point3D v1,
ref Point3D v2,
out Point hitCoord,
out double dist)
{
Vector3D e1;
Point3D.Subtract(ref v1, ref v0, out e1);
Vector3D e2;
Point3D.Subtract(ref v2, ref v0, out e2);
Vector3D r;
Vector3D.CrossProduct(ref direction, ref e2, out r);
double a = Vector3D.DotProduct(ref e1, ref r);
Vector3D s;
if (a > 0 && (type & FaceType.Front) != 0)
{
Point3D.Subtract(ref origin, ref v0, out s);
}
else if (a < 0 && (type & FaceType.Back) != 0)
{
Point3D.Subtract(ref v0, ref origin, out s);
a = -a;
}
else
{
hitCoord = new Point();
dist = 0;
return false;
}
double u = Vector3D.DotProduct(ref s, ref r);
if ((u < 0) || (a < u))
{
hitCoord = new Point();
dist = 0;
return false;
}
Vector3D q;
Vector3D.CrossProduct(ref s, ref e1, out q);
double v = Vector3D.DotProduct(ref direction, ref q);
if ((v < 0) || (a < (u + v)))
{
hitCoord = new Point();
dist = 0;
return false;
}
double t = Vector3D.DotProduct(ref e2, ref q);
double f = 1 / a;
t = t * f;
u = u * f;
v = v * f;
hitCoord = new Point(u, v);
dist = t;
return true;
}
// This function returns true if the probe line intersects the bbox volume (not
// just the surface of the box). Does LINE and RAY intersection tests.
//
// Based on Woo's method presented in Gems I, p. 395. See also "Real-Time
// Rendering", Haines, sec 10.4.2.
//
// origin/direction define the non-oriented line or ray
// box is the volume to intersect
//
// origin, direction, and box are passed by ref for perf. They are NOT MODIFIED
//
// Ported from dxg\d3dx9\mesh\intersect.cpp (12/04/03)
internal static bool ComputeLineBoxIntersection(ref Point3D origin, ref Vector3D direction, ref Rect3D box, bool isRay)
{
// Reject empty bounding boxes.
if (box.IsEmpty)
{
return false;
}
bool inside = true;
bool[] middle = new bool[3]; // True if ray origin in middle for coord i.
double[] plane = new double[3]; // Candidate BBox Planes
int i; // General Loop Counter
// Find all candidate planes; select the plane nearest to the ray origin
// for each coordinate.
double[] rgfMin = new double[] { box.X, box.Y, box.Z };
double[] rgfMax = new double[] { box.X + box.SizeX, box.Y + box.SizeY, box.Z + box.SizeZ };
double[] rgfRayPos = new double[] { origin.X, origin.Y, origin.Z };
double[] rgfRayDir = new double[] { direction.X, direction.Y, direction.Z };
for (i = 0; i < 3; ++i)
{
if (rgfRayPos[i] < rgfMin[i])
{
middle[i] = false;
plane[i] = rgfMin[i];
inside = false;
}
else if (rgfRayPos[i] > rgfMax[i])
{
middle[i] = false;
plane[i] = rgfMax[i];
inside = false;
}
else
{
middle[i] = true;
}
}
// If the ray origin is inside the box, then it must intersect the volume
// of the bounding box.
if (inside)
{
return true;
}
double rayt;
if (isRay)
{
// If we never end up finding the furthest plane, the box will be
// rejected since rayt is negative
rayt = -1;
}
else
{
// Can't use -1 in the line case because rayt^2 is 1 and we
// would miss valid ts in the furthest plane search
rayt = 0;
}
int maxPlane = 0;
for (i = 0; i < 3; ++i)
{
if (!middle[i] && (rgfRayDir[i] != 0))
{
double t = (plane[i] - rgfRayPos[i]) / rgfRayDir[i];
if (isRay)
{
if (t > rayt)
{
rayt = t;
maxPlane = i;
}
}
else
{
// In the original ray algorithm this test to find the furthest plane from the
// origin was t > rayt which only considered planes in the positive direction.
// I changed it to compare squared values so that we look for the farthest
// plane in either direction.
// Note that if the line intersects the box then all of the planes considered
// in this loop must be on the same side of the origin (because we are finding
// the intersection of the line with the space formed by the intersection of
// the half-spaces formed by the planes -- which incidentally point away from
// the origin.)
if (t * t > rayt * rayt)
{
rayt = t;
maxPlane = i;
}
}
}
}
// If the box is behind the ray, or if the box is beyond the extent of the
// ray, then return no-intersect.
if (isRay && rayt < 0)
{
return false;
}
// The intersection candidate point is within acceptible range; test each
// coordinate here to ensure that it actually hits the box.
for (i = 0; i < 3; ++i)
{
if (i != maxPlane)
{
double c = rgfRayPos[i] + (rayt * rgfRayDir[i]);
if ((c < rgfMin[i]) || (rgfMax[i] < c))
return false;
}
}
return true;
}
}
}
// File provided for Reference Use Only by Microsoft Corporation (c) 2007.
// Copyright (c) Microsoft Corporation. All rights reserved.
//----------------------------------------------------------------------------
//
// Copyright (C) Microsoft Corporation. All rights reserved.
//
//---------------------------------------------------------------------------
// GLOSSARY
// Kernel of matrix M : Space of vectors that vanish (go to 0) when multiplied by M
// Null-space of matrix M : Same thing
// Line matrix : Representation of a line using a matrix whose kernel is points on the line
// L : Matrix representing a line
// LT : Transpose of L
//
// This file offers a small line transform utility function. Given a line (lin) defined by Point3D
// origin & Vector3D direction and a model matrix M it returns (in-place) a line (lout) in "model
// space" so that any point on the line when transformed by M is on the original line.
//
// In other words x in lout ===> x M in lin
//
// This works even if M is rank-deficient, but if M is rank 2 or less then lout is not uniquely
// determined.
//
// The basic technique is as follows, we represent lin as a matrix where points on the line are in
// the null-space (kernel) of lin (this is straightforward.) Letting the symbols lin & lout refer
// to both the line & the corresponding matrix this means:
//
// x in lin ===> x lin = [0,0,0,0]
//
// Then we can transform a line matrix into model space by left multiplication with M
//
// lout = M lin
//
// That way,
//
// x in lout ===> x lout = [0,0,0,0] ===> x M lin = [0,0,0,0] ===> x M in lin
//
// Which is what we want. The hard part is going back from the matrix lout to a point and a vector.
// The smallest two eigenvalues of the l matrix are zero. The corresponding eigenvectors are two
// different points on the line.
//
// I find the eigenvectors and eigenvalues of a line matrix L by first computing the normal matrix
// N = L LT which has symmetric eigenvectors equal to the left eigenvectors of L. I find the
// eigenvectors of N using a Jacobi method for symmetric eigenvalue problems.
//
// The method is described in Chapter 8.4 of Golub & Van Loan (Matrix Computation), but here's a
// brief summary.
//
// We apply a series of 2D rotations (A1...An) to the matrix N that each make the matrix more
// diagonal. So if the whole sequence is A = A1 ... An then the final matrix E = AT N A is
// diagonal.
//
// Because A is orthonormal its columns are the eigenvectors of N and the diagonal elements of E are
// the eigenvalues.
//
// NOTES
//
// Forming the normal matrix N involves fourth powers of the input values. I mitigate this by
// scaling the matrix so that its largest value is 1 before squaring it. There may be a better
// method (perhaps even a modified Jacobi method) that would work directly on L and perhaps be more
// stable.
//
// None of this code understands rays. This is all in terms of lines, lines, lines, lines!
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Windows;
using System.Windows.Media.Media3D;
using MS.Utility;
namespace MS.Internal.Media3D
{
[Flags]
internal enum FaceType
{
None = 0,
Front = 1 << 0,
Back = 1 << 1,
};
internal static class LineUtil
{
// Coordinates of elements above the diagonal.
readonly static int[,] s_pairs = new int[,]{ {0,1}, {0,2}, {0,3}, {1,2}, {1,3}, {2,3} };
const int s_pairsCount = 6;
public static void Transform(Matrix3D modelMatrix,
ref Point3D origin, ref Vector3D direction, out bool isRay)
{
if (modelMatrix.InvertCore())
{
Point4D o = new Point4D(origin.X,origin.Y,origin.Z,1);
Point4D d = new Point4D(direction.X,direction.Y,direction.Z,0);
modelMatrix.MultiplyPoint(ref o);
modelMatrix.MultiplyPoint(ref d);
if (o.W == 1 && d.W == 0)
{
// Affine transformation
origin = new Point3D(o.X, o.Y, o.Z);
direction = new Vector3D(d.X, d.Y, d.Z);
isRay = true;
}
else
{
// Non-affine transformation (likely projection)
// Form 4x2 matrix with two points on line in two columns.
double[,] linepoints = new double[,]{{o.X,d.X},{o.Y,d.Y},{o.Z,d.Z},{o.W,d.W}};
ColumnsToAffinePointVector(linepoints,0,1,out origin, out direction);
isRay = false;
}
}
else
{
TransformSingular(ref modelMatrix, ref origin, ref direction);
isRay = false;
}
}
// modelMatrix is passed by reference for efficiency only. It is not modified.
private static void TransformSingular(ref Matrix3D modelMatrix,
ref Point3D origin, ref Vector3D direction)
{
double [,] matrix = TransformedLineMatrix(ref modelMatrix, ref origin, ref direction);
matrix = Square(matrix);
double[,] eigen = new double[,]{ {1,0,0,0}, {0,1,0,0}, {0,0,1,0}, {0,0,0,1} };
// We'll just do 5 iterations with each pair because according to my results & Golub &
// Van Loan this process converges quickly.
int iterations = 5 * s_pairsCount;
for (int iter = 0; iter < iterations; ++iter)
{
int pair = iter % s_pairsCount;
JacobiRotation jrot = new JacobiRotation(s_pairs[pair,0],s_pairs[pair,1],matrix);
matrix = jrot.LeftRightMultiply(matrix);
eigen = jrot.RightMultiply(eigen);
}
// That was it as far as finding eigenvectors
int evec1,evec2;
FindSmallestTwoDiagonal(matrix, out evec1, out evec2);
// The eigenvectors corresponding to the two smallest eigenvalues are columns evec1 &
// evec2. These, in homogeneous space, are two different points on our line. We are
// going to convert them to an affine point & vector.
ColumnsToAffinePointVector(eigen, evec1, evec2, out origin, out direction);
}
private static void ColumnsToAffinePointVector(double[,] matrix, int col1, int col2, out Point3D origin, out Vector3D direction)
{
// The col1 & col2 columsn of matrix are two different homogeneous points on a line. We
// are going to convert them to an affine point & vector using the following procedure.
// 1. Pick the eigenvector with the largest W, call it big
// 2. Scale it by 1/W to form an affine point, still call it big
// 3. Add a weighted multiple of big to small to make small.W zero.
//
// If both w are zero then we have a projective line that intersects no affine points
// (i.e. it's a line at infinity.) This will cause overflow but that's fine because the
// line at infinity doesn't intersect anything & the caller of this function needs be
// able to handle results that come back with inf or nan by "doing nothing."
// Step 1.
if (matrix[3,col1]*matrix[3,col1] < matrix[3,col2]*matrix[3,col2])
{
int temp = col1;
col1 = col2;
col2 = temp;
}
// Step 2.
double s = 1/matrix[3,col1];
origin = new Point3D(s*matrix[0,col1],
s*matrix[1,col1],
s*matrix[2,col1]);
// Step 3.
s = -matrix[3,col2];
direction = new Vector3D(matrix[0,col2]+s*origin.X,
matrix[1,col2]+s*origin.Y,
matrix[2,col2]+s*origin.Z);
}
// Returns the indices of the smallest two diagonal elements of matrix
private static void FindSmallestTwoDiagonal(double[,] matrix, out int evec1, out int evec2)
{
evec1 = 0;
evec2 = 1;
// And corresponding squared eigenvalues.
double eval1 = matrix[0,0]*matrix[0,0];
double eval2 = matrix[1,1]*matrix[1,1];
for (int i = 2; i < 4; ++i)
{
// Replace second smallest if necessary.
double val = matrix[i,i]*matrix[i,i];
if (val < eval1)
{
if (eval1 < eval2)
{
eval2 = val;
evec2 = i;
}
else
{
eval1 = val;
evec1 = i;
}
}
else if (val < eval2)
{
eval2 = val;
evec2 = i;
}
}
}
// Returns the "line matrix" corresponding to the line (origin,direction) transformed by the
// inverse *transform* of modelMatrix. (To transform a line by the inverse transform
// requires multiplying by the non-inverted matrix.)
private static double[,] TransformedLineMatrix(ref Matrix3D modelMatrix,
ref Point3D origin, ref Vector3D direction)
{
double x1 = origin.X;
double y1 = origin.Y;
double z1 = origin.Z;
// w1 = 1
double x2 = direction.X;
double y2 = direction.Y;
double z2 = direction.Z;
// w2 = 0
// To prove to yourself that this matrix is correct just multiply by the two
// (homogeneous) points on the line. Any other homogeneous point on the line is a
// linear combination of them.
double a = y2*z1-y1*z2;
double b = x1*z2-x2*z1;
double c = x2*y1-x1*y2;
Matrix3D m = modelMatrix *
new Matrix3D(a, y2, z2, 0,
b, -x2, 0, z2,
c, 0, -x2, -y2,
0, c, -b, a);
double[,] matrix = new double[4,4];
matrix[0,0] = m.M11;
matrix[0,1] = m.M12;
matrix[0,2] = m.M13;
matrix[0,3] = m.M14;
matrix[1,0] = m.M21;
matrix[1,1] = m.M22;
matrix[1,2] = m.M23;
matrix[1,3] = m.M24;
matrix[2,0] = m.M31;
matrix[2,1] = m.M32;
matrix[2,2] = m.M33;
matrix[2,3] = m.M34;
matrix[3,0] = m.OffsetX;
matrix[3,1] = m.OffsetY;
matrix[3,2] = m.OffsetZ;
matrix[3,3] = m.M44;
return matrix;
}
// Scales M so that its largest element is 1 and then returns M MT
// (MT=transpose(M))
private static double [,] Square(double[,] m)
{
double[,] o = new double[4,4];
// Scale the matrix so that its largest element is 1.
double maxvalue = 0;
for (int i = 0; i < 4; ++i)
{
for (int j = 0; j < 4; ++j)
{
maxvalue = Math.Max(maxvalue,m[i,j]*m[i,j]);
}
}
maxvalue = Math.Sqrt(maxvalue);
for (int i = 0; i < 4; ++i)
{
for (int j = 0; j < 4; ++j)
{
m[i,j] /= maxvalue;
}
}
// Compute its square.
for (int i = 0; i < 4; ++i)
{
for (int j = 0; j < 4; ++j)
{
double d = 0;
for (int k = 0; k < 4; ++k)
{
d += m[i,k]*m[j,k];
}
o[i,j] = d;
}
}
return o;
}
// See section 8.4 of Golub & Van Loan "Matrix Computation"
// Remember, J is this Jacobi rotation. JT is the transpose.
// Also section 5.1.8 shows the formula for multiplying by the Jacobi rotation
// Though as they helpfully point out a Jacobi rotation is the same as a Givens rotation
private struct JacobiRotation
{
public JacobiRotation(int p, int q, double[,] a)
{
// Constructs a 2D rotation matrix M as follows
// Start with identity.
// Zero the p & q rows & columns
// Then set the intersections of these rows & columns as follows
// [ M_pp M_pq ] = [ c s]
// [ M_qp M_qq ] [-s c]
//
// Where c & s are a cosine-sine pair calculated so that multiplying a by this will
// descrease the the off-diagonal elements.
_p = p;
_q = q;
double tau = (a[q,q] - a[p,p])/(2*a[p,q]);
if (tau < Double.MaxValue && tau > -Double.MaxValue)
{
double root = Math.Sqrt(1+tau*tau);
// Choose the smaller of -tau +/- root
double t = -tau < 0 ? -tau + root : -tau - root;
_c = 1/Math.Sqrt(1+t*t);
_s = t * _c;
}
else
{
_c = 1;
_s = 0;
}
}
// These functions overwrite & return their argument.
// returns JT a J
public double[,] LeftRightMultiply(double [,] a)
{
return RightMultiply(LeftMultiplyTranspose(a));
}
// returns a J
public double[,] RightMultiply(double [,] a)
{
for (int j = 0; j < 4; ++j)
{
double tau1 = a[j,_p];
double tau2 = a[j,_q];
a[j,_p] = _c * tau1 - _s * tau2;
a[j,_q] = _s * tau1 + _c * tau2;
}
return a;
}
// returns JT a
public double[,] LeftMultiplyTranspose(double [,] a)
{
for (int j = 0; j < 4; ++j)
{
double tau1 = a[_p,j];
double tau2 = a[_q,j];
a[_p,j] = _c * tau1 - _s * tau2;
a[_q,j] = _s * tau1 + _c * tau2;
}
return a;
}
private int _p, _q;
private double _c, _s;
}
// This method determines if the line/ray intersects the triangle.
// If "origin" and "direction" truely represent a line, "type" should be front and back because
// we don't have any true direction.
//
// origin/direction define the line/ray
// v0/v1/v2 define the triangle
//
// origin, direction, v0, v1, v2 are passed by ref for perf. They are NOT MODIFIED
//
// If this method returns false, ignore the values of hitCoord and dist.
//
// Ported from dxg\d3dx9\mesh\intersect.cpp (12/04/03)
// which is an implementation of "Fast, Minimum Storage Ray-Triangle Intersection" by Moller + Trumbore
internal static bool ComputeLineTriangleIntersection(
FaceType type,
ref Point3D origin,
ref Vector3D direction,
ref Point3D v0,
ref Point3D v1,
ref Point3D v2,
out Point hitCoord,
out double dist)
{
Vector3D e1;
Point3D.Subtract(ref v1, ref v0, out e1);
Vector3D e2;
Point3D.Subtract(ref v2, ref v0, out e2);
Vector3D r;
Vector3D.CrossProduct(ref direction, ref e2, out r);
double a = Vector3D.DotProduct(ref e1, ref r);
Vector3D s;
if (a > 0 && (type & FaceType.Front) != 0)
{
Point3D.Subtract(ref origin, ref v0, out s);
}
else if (a < 0 && (type & FaceType.Back) != 0)
{
Point3D.Subtract(ref v0, ref origin, out s);
a = -a;
}
else
{
hitCoord = new Point();
dist = 0;
return false;
}
double u = Vector3D.DotProduct(ref s, ref r);
if ((u < 0) || (a < u))
{
hitCoord = new Point();
dist = 0;
return false;
}
Vector3D q;
Vector3D.CrossProduct(ref s, ref e1, out q);
double v = Vector3D.DotProduct(ref direction, ref q);
if ((v < 0) || (a < (u + v)))
{
hitCoord = new Point();
dist = 0;
return false;
}
double t = Vector3D.DotProduct(ref e2, ref q);
double f = 1 / a;
t = t * f;
u = u * f;
v = v * f;
hitCoord = new Point(u, v);
dist = t;
return true;
}
// This function returns true if the probe line intersects the bbox volume (not
// just the surface of the box). Does LINE and RAY intersection tests.
//
// Based on Woo's method presented in Gems I, p. 395. See also "Real-Time
// Rendering", Haines, sec 10.4.2.
//
// origin/direction define the non-oriented line or ray
// box is the volume to intersect
//
// origin, direction, and box are passed by ref for perf. They are NOT MODIFIED
//
// Ported from dxg\d3dx9\mesh\intersect.cpp (12/04/03)
internal static bool ComputeLineBoxIntersection(ref Point3D origin, ref Vector3D direction, ref Rect3D box, bool isRay)
{
// Reject empty bounding boxes.
if (box.IsEmpty)
{
return false;
}
bool inside = true;
bool[] middle = new bool[3]; // True if ray origin in middle for coord i.
double[] plane = new double[3]; // Candidate BBox Planes
int i; // General Loop Counter
// Find all candidate planes; select the plane nearest to the ray origin
// for each coordinate.
double[] rgfMin = new double[] { box.X, box.Y, box.Z };
double[] rgfMax = new double[] { box.X + box.SizeX, box.Y + box.SizeY, box.Z + box.SizeZ };
double[] rgfRayPos = new double[] { origin.X, origin.Y, origin.Z };
double[] rgfRayDir = new double[] { direction.X, direction.Y, direction.Z };
for (i = 0; i < 3; ++i)
{
if (rgfRayPos[i] < rgfMin[i])
{
middle[i] = false;
plane[i] = rgfMin[i];
inside = false;
}
else if (rgfRayPos[i] > rgfMax[i])
{
middle[i] = false;
plane[i] = rgfMax[i];
inside = false;
}
else
{
middle[i] = true;
}
}
// If the ray origin is inside the box, then it must intersect the volume
// of the bounding box.
if (inside)
{
return true;
}
double rayt;
if (isRay)
{
// If we never end up finding the furthest plane, the box will be
// rejected since rayt is negative
rayt = -1;
}
else
{
// Can't use -1 in the line case because rayt^2 is 1 and we
// would miss valid ts in the furthest plane search
rayt = 0;
}
int maxPlane = 0;
for (i = 0; i < 3; ++i)
{
if (!middle[i] && (rgfRayDir[i] != 0))
{
double t = (plane[i] - rgfRayPos[i]) / rgfRayDir[i];
if (isRay)
{
if (t > rayt)
{
rayt = t;
maxPlane = i;
}
}
else
{
// In the original ray algorithm this test to find the furthest plane from the
// origin was t > rayt which only considered planes in the positive direction.
// I changed it to compare squared values so that we look for the farthest
// plane in either direction.
// Note that if the line intersects the box then all of the planes considered
// in this loop must be on the same side of the origin (because we are finding
// the intersection of the line with the space formed by the intersection of
// the half-spaces formed by the planes -- which incidentally point away from
// the origin.)
if (t * t > rayt * rayt)
{
rayt = t;
maxPlane = i;
}
}
}
}
// If the box is behind the ray, or if the box is beyond the extent of the
// ray, then return no-intersect.
if (isRay && rayt < 0)
{
return false;
}
// The intersection candidate point is within acceptible range; test each
// coordinate here to ensure that it actually hits the box.
for (i = 0; i < 3; ++i)
{
if (i != maxPlane)
{
double c = rgfRayPos[i] + (rayt * rgfRayDir[i]);
if ((c < rgfMin[i]) || (rgfMax[i] < c))
return false;
}
}
return true;
}
}
}
// File provided for Reference Use Only by Microsoft Corporation (c) 2007.
// Copyright (c) Microsoft Corporation. All rights reserved.
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