对偶轮廓和二次误差函数

Question

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，只是为了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对偶轮廓的作品，除了对偶轮廓本身的核心:最小化二次误差函数(QEF)之外，我得到了所有作品。

现在，我正在计算内部体素的顶点位置，只是通过找到共享单个顶点(3到6条边)的所有边缘点之间的平均值，它工作得很好，但它显然没有在正确的地方创建内部顶点。

这是我要创建的代码。如有任何帮助，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!
    }

Answer 1

QEF相当难以理解。希望我能帮上忙。双重轮廓法计算每个交叉点的"赫米特"数据，或者换句话说，在体素边缘创建的每个点上，表面的法线是已知的。用点和法线可以计算平面的方程。

QEF是从体素的内部点到与体素相关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使QEF最小的点。文献建议使用格雷厄姆-施密特过程来定位最佳点，但这可能很复杂，也可能导致点位于体素之外。

另一种选择是在体素内创建一个点网格，计算每个点的QEF，并选择最低的点，网格越细越接近最佳点，但计算时间越长。

Answer 2

在我目前的双重轮廓实现中，我使用一种非常简单的方法来解决QEF。由于QEF本质上是最小二乘近似，因此我发现计算QEF的最简单方法是计算伪逆。这个伪逆可以用你的语言中的任何代数库来计算。

这是我使用的代码:

    public static Vector<float> CalculateCubeQEF(Vector3[] normals, Vector3[] positions, Vector3 meanPoint)
    {
        var A = DenseMatrix.OfRowArrays(normals.Select(e => new[] { e.X, e.Y, e.Z }).ToArray());
        var b = DenseVector.OfArray(normals.Zip(positions.Select(p => p - meanPoint), Vector3.Dot).ToArray());
        var pseudo = PseudoInverse(A);
        var leastsquares = pseudo.Multiply(b);
        return leastsquares + DenseVector.OfArray(new[] { meanPoint.X, meanPoint.Y, meanPoint.Z });
    }

函数的输入是交点和法线，meanPoint是给定交点的平均值。

总结数学:这个函数计算所有平面的交点和法线的交点。因为它没有精确解，所以计算最小二乘近似，找到"最小错误"的点。此外，交点被"移动"，使平均点成为原点。这确保了当QEF有多个解时，最接近均值点的解被选择。