我一直在谷歌上搜索，但找不到任何基本的东西。在最基本的形式中，双轮廓(对于体素地形)是如何实现的？我知道它的作用和原因，但不明白如何去做。 JS或C#(最好)都好。有没有人用过Dual contouring，能简单介绍一下吗？

最佳答案

好的。所以今晚我很无聊，决定尝试一下自己实现双重轮廓。正如我在评论中所说，所有相关 Material 都在以下论文的第 2 节中:

• 原始版本:http://www.frankpetterson.com/publications/dualcontour/dualcontour.pdf
• 存档版本:https://web.archive.org/web/20170713094715if_/http://www.frankpetterson.com/publications/dualcontour/dualcontour.pdf

特别是，网格的拓扑在以下部分的 2.2 部分中进行了描述，引用:

1. For each cube that exhibits a sign change, generate a vertex positioned at the minimizer of the quadratic function of equation 1.

2. For each edge that exhibits a sign change, generate a quad connecting the minimizing vertices of the four cubes containing the edge.

这就是它的全部!您解决线性最小二乘问题以获得每个立方体的顶点，然后将相邻顶点与四边形连接。因此，使用这个基本思想，我使用 numpy 在 python 中编写了一个双等高线等值面提取器(部分原因是为了满足我自己对它如何工作的病态好奇心)。这是代码:

import numpy as np
import numpy.linalg as la
import scipy.optimize as opt
import itertools as it

#Cardinal directions
dirs = [ [1,0,0], [0,1,0], [0,0,1] ]

#Vertices of cube
cube_verts = [ np.array([x, y, z])
    for x in range(2)
    for y in range(2)
    for z in range(2) ]

#Edges of cube
cube_edges = [ 
    [ k for (k,v) in enumerate(cube_verts) if v[i] == a and v[j] == b ]
    for a in range(2)
    for b in range(2)
    for i in range(3) 
    for j in range(3) if i != j ]

#Use non-linear root finding to compute intersection point
def estimate_hermite(f, df, v0, v1):
    t0 = opt.brentq(lambda t : f((1.-t)*v0 + t*v1), 0, 1)
    x0 = (1.-t0)*v0 + t0*v1
    return (x0, df(x0))

#Input:
# f = implicit function
# df = gradient of f
# nc = resolution
def dual_contour(f, df, nc):

    #Compute vertices
    dc_verts = []
    vindex   = {}
    for x,y,z in it.product(range(nc), range(nc), range(nc)):
        o = np.array([x,y,z])

        #Get signs for cube
        cube_signs = [ f(o+v)>0 for v in cube_verts ]

        if all(cube_signs) or not any(cube_signs):
            continue

        #Estimate hermite data
        h_data = [ estimate_hermite(f, df, o+cube_verts[e[0]], o+cube_verts[e[1]]) 
            for e in cube_edges if cube_signs[e[0]] != cube_signs[e[1]] ]

        #Solve qef to get vertex
        A = [ n for p,n in h_data ]
        b = [ np.dot(p,n) for p,n in h_data ]
        v, residue, rank, s = la.lstsq(A, b)

        #Throw out failed solutions
        if la.norm(v-o) > 2:
            continue

        #Emit one vertex per every cube that crosses
        vindex[ tuple(o) ] = len(dc_verts)
        dc_verts.append(v)

    #Construct faces
    dc_faces = []
    for x,y,z in it.product(range(nc), range(nc), range(nc)):
        if not (x,y,z) in vindex:
            continue

        #Emit one face per each edge that crosses
        o = np.array([x,y,z])   
        for i in range(3):
            for j in range(i):
                if tuple(o + dirs[i]) in vindex and tuple(o + dirs[j]) in vindex and tuple(o + dirs[i] + dirs[j]) in vindex:
                    dc_faces.append( [vindex[tuple(o)], vindex[tuple(o+dirs[i])], vindex[tuple(o+dirs[j])]] )
                    dc_faces.append( [vindex[tuple(o+dirs[i]+dirs[j])], vindex[tuple(o+dirs[j])], vindex[tuple(o+dirs[i])]] )

    return dc_verts, dc_faces

它不是很快，因为它使用 SciPy 的通用非线性求根方法来查找等值面上的边缘点。但是，它似乎确实以相当通用的方式工作得相当好。为了测试它，我使用以下测试用例(在 Mayavi2 可视化工具包中)运行它:

import enthought.mayavi.mlab as mlab

center = np.array([16,16,16])
radius = 10

def test_f(x):
    d = x-center
    return np.dot(d,d) - radius**2

def test_df(x):
    d = x-center
    return d / sqrt(np.dot(d,d))

verts, tris = dual_contour(f, df, n)

mlab.triangular_mesh( 
            [ v[0] for v in verts ],
            [ v[1] for v in verts ],
            [ v[2] for v in verts ],
            tris)

这将球体定义为隐式方程，并通过双等高线求解 0 等值面。当我在工具包中运行它时，结果是这样的:

总之，它似乎有效。

关于c# - 基本双重轮廓理论，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/6485908/