我已经在 C# 中实现了行进立方体、双行进立方体和自适应行进立方体，结果发现我需要双重轮廓来实现我的目的。
我已经阅读了所有关于双等高线的作品，除了双等高线本身的核心之外，我获得了所有内容:最小化二次误差函数 (QEF)。
现在，我只是通过找到共享该单个顶点(3 到 6 个边)的所有边点之间的平均值来计算内部体素的顶点位置，并且效果很好，但显然它没有在正确的位置创建内部顶点。
这是我正在尝试创建的一段代码。任何帮助将不胜感激
/// <summary>
/// ORIGINAL WORK: Dual Contouring of Hermite Data by Tao Ju (remember me of a MechCommander 2 character)
/// 2.3 Representing and minimizing QEFs
/// The function E[x] can be expressed as the inner
/// product (Ax-b)T (Ax-b) where A is a matrix whose rows are the
/// normals ni and b is a vector whose entries are ni*pi. <------------ (dot product?)>
/// Typically, the quadratic function E[x] is expanded into the form
/// E[x] = xT AT Ax - 2xT AT b + bT b (2)
/// where the matrix AT A is a symmetric 3x3 matrix, AT b is a column
/// vector of length three and bT b is a scalar. The advantage of this expansion
/// is that only the matrices AT A, AT b and bT b need be stored
/// (10 floats), as opposed to storing the matrices A and b. Furthermore,
/// a minimizing value ˆ x for E[x] can be computed by solving
/// the normal equations AT Aˆ x = AT b.
/// </summary>
public Vector3 GetMinimumError(Vector3 p0, Vector3 p1, Vector3 p2, Vector3 n0, Vector3 n1, Vector3 n2)
{
//so, here we are. I'm creating a vector to store the final value.
Vector3 position = Vector3.Zero;
//Values of b are supposed to b (:P) three floats. The only way i know to find a float value
//by multiplying 2 vectors is to use dot product.
Vector3 b = new Vector3(
Vector3.Dot(p0, n0),
Vector3.Dot(p1, n1),
Vector3.Dot(p2, n2));
//What the transpose of a vector is supposed to be?
//I don't know, but i think should be the vector itself :)
float bTb = Vector3.Dot(b, b);
//i create a square matrix 3x3, so i can use c# matrix transformation libraries.
//i know i will probably have to build bigger matrix later on, but it should fit for now
Matrix A = new Matrix(
n0.X, n0.Y, n0.Z, 0,
n1.X, n1.Y, n1.Z, 0,
n2.X, n2.Y, n2.Z, 0,
0, 0, 0, 0);
//easy
Matrix AT = Matrix.Transpose(A);
//EASY
Matrix ATA = Matrix.Multiply(AT, A);
//Another intuition. Hope makes sense...
Vector3 ATb = Vector3.Transform(b, AT);
//...
// some cool stuff about solving
// the normal equations AT Aˆ x = AT b
//...
return position; //profit!
}
最佳答案
QEF 比较难理解。希望我能帮上忙。双重轮廓法计算每个交叉点处的“Hermite”数据，或者换句话说，在体素边缘上创建的每个点处，表面的法线是已知的。用一个点和一个法线可以计算平面的方程。
QEF是从体素内部点到与体素关联的每个平面的距离的平方和。下面是一些用于计算 QEF 的伪代码。
double get_QEF(Point3d point, Voxel3d voxel)
{
double QEF = 0.0;
foreach(plane in voxel.planes)
{
double dist_to_plane = plane.distance(point);
QEF += dist_to_plane*dist_to_plane;
}
return(QEF);
}
目标是在体素内选择一个最小化 QEF 的点。文献建议使用 Grahm-Schmidt 过程来定位最佳点，但这可能很复杂，也可能导致点位于体素之外。
另一种选择(hack-ish)是在体素内部创建一个点网格并计算每个点的QEF，然后选择具有最低点的QEF，网格越精细，越接近最佳点，但越长计算。
关于matrix-multiplication - 双重轮廓和二次误差函数，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16734792/