如何检查点是否在四面体中？

时间：2014-08-07 10:07:09

标签： c# opengl tetrahedra

我知道四面体的所有坐标和我想确定的点。那么有谁知道怎么做？我试图确定该点属于四面体的每个三角形，如果它对所有三角形都是真的那么该点在四面体中。但这绝对是错误的。

6 个答案:

答案 0 :(得分：13)

对于四面体的每个平面，检查该点是否与剩余的顶点在同一侧：

bool SameSide(v1, v2, v3, v4, p)
{
    normal := cross(v2 - v1, v3 - v1)
    dotV4 := dot(normal, v4 - v1)
    dotP := dot(normal, p - v1)
    return Math.Sign(dotV4) == Math.Sign(dotP);
}

你需要检查每架飞机：

bool PointInTetrahedron(v1, v2, v3, v4, p)
{
    return SameSide(v1, v2, v3, v4, p) &&
           SameSide(v2, v3, v4, v1, p) &&
           SameSide(v3, v4, v1, v2, p) &&
           SameSide(v4, v1, v2, v3, p);               
}

答案 1 :(得分：2)

您可以通过四个顶点定义四面体。 A B C和D. 因此，您还可以使用4个三角形来定义四面体的表面。

你知道，如果P点位于飞机的另一侧，那就好了。每个平面的法线指向远离四面体的中心。 所以你只需要测试4架飞机。

您的平面方程如下所示：a*x+b*y+c*z+d=0只需填写点值（x y z）。如果结果的符号> 0，则该点与法线相同，结果== 0，则点位于平面中，在您的情况下，您需要第三个选项。 ＆lt; 0表示它位于飞机的背面。 如果所有4个飞机都已满，则您的点位于四面体内

答案 2 :(得分：1)

从Hugues' solution开始，这是一个更简单，甚至更有效的方法：

import numpy as np

def tetraCoord(A,B,C,D):
  # Almost the same as Hugues' function, 
  # except it does not involve the homogeneous coordinates.
  v1 = B-A ; v2 = C-A ; v3 = D-A
  mat = np.array((v1,v2,v3)).T
  # mat is 3x3 here
  M1 = np.linalg.inv(mat)
  return(M1)

def pointInside(v1,v2,v3,v4,p):
  # Find the transform matrix from orthogonal to tetrahedron system
  M1=tetraCoord(v1,v2,v3,v4)
  # apply the transform to P (v1 is the origin)
  newp = M1.dot(p-v1)
  # perform test
  return (np.all(newp>=0) and np.all(newp <=1) and np.sum(newp)<=1)

在与四面体关联的坐标系中，与原点相对的面（在此表示为v1）的特征是x + y + z = 1。因此，从该面起与v1相同侧的半空间满足x + y + z ＜1。

作为比较，这是用于比较Nico，Hugues和我提出的方法的完整代码：

import numpy as np
import time

def sameside(v1,v2,v3,v4,p):
    normal = np.cross(v2-v1, v3-v1)
    return (np.dot(normal, v4-v1) * np.dot(normal, p-v1) > 0)

# Nico's solution
def pointInside_Nico(v1,v2,v3,v4,p):   
    return sameside(v1, v2, v3, v4, p) and sameside(v2, v3, v4, v1, p) and sameside(v3, v4, v1, v2, p) and sameside(v4, v1, v2, v3, p)      

# Hugues' solution
def tetraCoord(A,B,C,D):
    v1 = B-A ; v2 = C-A ; v3 = D-A
    # mat defines an affine transform from the tetrahedron to the orthogonal system
    mat = np.concatenate((np.array((v1,v2,v3,A)).T, np.array([[0,0,0,1]])))
    # The inverse matrix does the opposite (from orthogonal to tetrahedron)
    M1 = np.linalg.inv(mat)
    return(M1)

def pointInside_Hugues(v1,v2,v3,v4,p):
    # Find the transform matrix from orthogonal to tetrahedron system
    M1=tetraCoord(v1,v2,v3,v4)
    # apply the transform to P
    p1 = np.append(p,1)
    newp = M1.dot(p1)
    # perform test
    return(np.all(newp>=0) and np.all(newp <=1) and sameside(v2,v3,v4,v1,p))    

# Proposed solution
def tetraCoord_Dorian(A,B,C,D):
    v1 = B-A ; v2 = C-A ; v3 = D-A
    # mat defines an affine transform from the tetrahedron to the orthogonal system
    mat = np.array((v1,v2,v3)).T
    # The inverse matrix does the opposite (from orthogonal to tetrahedron)
    M1 = np.linalg.inv(mat)
    return(M1) 

def pointInside_Dorian(v1,v2,v3,v4,p):
    # Find the transform matrix from orthogonal to tetrahedron system
    M1=tetraCoord_Dorian(v1,v2,v3,v4)
    # apply the transform to P
    newp = M1.dot(p-v1)
    # perform test
    return (np.all(newp>=0) and np.all(newp <=1) and np.sum(newp)<=1)

npt=100000    
Pt=np.random.rand(npt,3)
A=np.array([0.1, 0.1, 0.1])
B=np.array([0.9, 0.2, 0.1])
C=np.array([0.1, 0.9, 0.2])
D=np.array([0.3, 0.3, 0.9])

inTet_Nico=np.zeros(shape=(npt,1),dtype=bool)
inTet_Hugues=inTet_Nico
inTet_Dorian=inTet_Nico

start_time = time.time()
for i in range(0,npt):
    inTet_Nico[i]=pointInside_Nico(A,B,C,D,Pt[i,:])
print("--- %s seconds ---" % (time.time() - start_time)) # https://stackoverflow.com/questions/1557571/how-do-i-get-time-of-a-python-programs-execution

start_time = time.time()
for i in range(0,npt):
    inTet_Hugues[i]=pointInside_Hugues(A,B,C,D,Pt[i,:])
print("--- %s seconds ---" % (time.time() - start_time))   

start_time = time.time()    
for i in range(0,npt):
    inTet_Dorian[i]=pointInside_Dorian(A,B,C,D,Pt[i,:])
print("--- %s seconds ---" % (time.time() - start_time))

这是关于运行时间的结果：

--- 15.621951341629028 seconds ---
--- 8.97989797592163 seconds ---
--- 4.597853660583496 seconds ---

答案 3 :(得分：1)

我已经矢量化了Dorian和Hughes解决方案，以将整个点数组作为输入。我也将tetraCoord函数移到pointsInside函数之外，并重命名了这两个函数，因为没有必要为每个点调用它。

在我的计算机上，@ Dorian的解决方案和示例在2.5秒内运行。在相同的数据上，我的运行速度为0.003秒，快了近千倍。如果出于某种原因甚至需要更高的速度，则将GPU cupy软件包导入为“ np”会将其推入100微秒的范围。

import time
# alternatively, import cupy as np if len(points)>1e7 and GPU
import numpy as np 

def Tetrahedron(vertices):
    """
    Given a list of the xyz coordinates of the vertices of a tetrahedron, 
    return tetrahedron coordinate system
    """
    origin, *rest = vertices
    mat = (np.array(rest) - origin).T
    tetra = np.linalg.inv(mat)
    return tetra, origin

def pointInside(point, tetra, origin):
    """
    Takes a single point or array of points, as well as tetra and origin objects returned by 
    the Tetrahedron function.
    Returns a boolean or boolean array indicating whether the point is inside the tetrahedron.
    """
    newp = np.matmul(tetra, (point-origin).T).T
    return np.all(newp>=0, axis=-1) & np.all(newp <=1, axis=-1) & (np.sum(newp, axis=-1) <=1)

npt=10000000
points = np.random.rand(npt,3)
# Coordinates of vertices A, B, C and D
A=np.array([0.1, 0.1, 0.1])
B=np.array([0.9, 0.2, 0.1])
C=np.array([0.1, 0.9, 0.2])
D=np.array([0.3, 0.3, 0.9])

start_time = time.time()
vertices = [A, B, C, D]
tetra, origin = Tetrahedron(vertices)
inTet = pointInside(points, tetra, origin)
print("--- %s seconds ---" % (time.time() - start_time))

答案 4 :(得分：0)

给出4个点A，B，C，D定义一个非退化的四面体，并指定一个点P进行测试，一种方法是将P的坐标转换为四面体坐标系，例如以A为原点，向量BA，CA，DA为单位向量。

在此坐标系中，如果P的坐标在P内，则它们都在0到1之间，但是它也可以在由原点和3个单位向量定义的变换立方体中的任何位置。 断言P在内部（A，B，C，D）的一种方法是依次将点（A，B，C和D）和其他三个点定义为新坐标系。此测试重复4次是有效的，但可以改进。

仅转换一次坐标并重新使用之前提出的SameSide函数是最有效的，例如，以A为原点，转换为（A，B，C，D）坐标系，P和A必须位于在（B，C，D）平面的同一侧。

以下是该测试的numpy / python实现。测试表明，该方法比Planes方法快2-3倍。

import numpy as np

def sameside(v1,v2,v3,v4,p):
    normal = np.cross(v2-v1, v3-v1)
    return ((np.dot(normal, v4-v1)*p.dot(normal, p-v1) > 0)

def tetraCoord(A,B,C,D):
    v1 = B-A ; v2 = C-A ; v3 = D-A
    # mat defines an affine transform from the tetrahedron to the orthogonal system
    mat = np.concatenate((np.array((v1,v2,v3,A)).T, np.array([[0,0,0,1]])))
    # The inverse matrix does the opposite (from orthogonal to tetrahedron)
    M1 = np.linalg.inv(mat)
    return(M1)

def pointInsideT(v1,v2,v3,v4,p):
    # Find the transform matrix from orthogonal to tetrahedron system
    M1=tetraCoord(v1,v2,v3,v4)
    # apply the transform to P
    p1 = np.append(p,1)
    newp = M1.dot(p1)
    # perform test
    return(np.all(newp>=0) and np.all(newp <=1) and sameside(v2,v3,v4,v1,p))

答案 5 :(得分：0)

借助Dorian的测试用例脚本，我可以开发另一种解决方案，并将其与迄今为止的解决方案进行快速比较。

直觉

对于三角形ABC和点P，如果将P连接到角以获得矢量PA，PB，PC并比较被PA，PC和PB，PC跨越的两个三角形X和Y，则该点如果X和Y重叠，则P位于三角形ABC内。

换句话说，如果P在三角形ABC中，则仅通过正系数线性组合PC和PB来构建向量PA是不可能的。

从那以后，我试图将其转移到四面体情况，并读到here，可以通过检查从向量构成的矩阵的行列式作为非列来检查向量是否线性独立-零。 我尝试了使用行列式的各种方法，我偶然发现了这个方法：

让PA，PB，PC，PD是P到四面体点ABCD的连接（即PA = A-P等）。计算决定因素detA = det（PB PC PD），detB，detC和detD（如detA）。

如果满足以下条件，则点P处在ABCD所跨越的四面体之内：

detA> 0且detB <0且detC> 0且detD <0

或

detA <0且detB> 0且detC <0且detD> 0

所以行列式从负数开始或从正数开始切换符号。

可以吗？显然。为什么行得通？我不知道，或者至少我无法证明这一点。也许其他具有更好数学技能的人可以在这里为我们提供帮助。

（编辑：实际上，重心坐标可以使用这些行列式定义，最后，重心坐标需要累加一个。类似于将P的组合与点组成的四面体的体积进行比较A，B，C，D和四面体ABCD本身的体积。观察到的行列式符号表示的情况目前尚不清楚，它是否可以正常使用，我不推荐使用）

我将测试用例更改为不针对一个四面体T检查n个点Pi，而是针对n个四面体Ti检查n个点Pi。所有答案仍然会给出正确的结果。我认为这种方法更快的原因是它不需要矩阵求逆。 我离开了用一个四面体实现的TomNorway的方法，由于对python和numpy不太熟悉，所以将这种新方法的向量化留给了其他人。

import numpy as np
import time

def sameside(v1,v2,v3,v4,p):
    normal = np.cross(v2-v1, v3-v1)
    return (np.dot(normal, v4-v1) * np.dot(normal, p-v1) > 0)

# Nico's solution
def pointInside_Nico(v1,v2,v3,v4,p):   
    return sameside(v1, v2, v3, v4, p) and sameside(v2, v3, v4, v1, p) and sameside(v3, v4, v1, v2, p) and sameside(v4, v1, v2, v3, p)      

# Hugues' solution
def tetraCoord(A,B,C,D):
    v1 = B-A ; v2 = C-A ; v3 = D-A
    # mat defines an affine transform from the tetrahedron to the orthogonal system
    mat = np.concatenate((np.array((v1,v2,v3,A)).T, np.array([[0,0,0,1]])))
    # The inverse matrix does the opposite (from orthogonal to tetrahedron)
    M1 = np.linalg.inv(mat)
    return(M1)

def pointInside_Hugues(v1,v2,v3,v4,p):
    # Find the transform matrix from orthogonal to tetrahedron system
    M1=tetraCoord(v1,v2,v3,v4)
    # apply the transform to P
    p1 = np.append(p,1)
    newp = M1.dot(p1)
    # perform test
    return(np.all(newp>=0) and np.all(newp <=1) and sameside(v2,v3,v4,v1,p))    

#Dorian's solution
def tetraCoord_Dorian(A,B,C,D):
    v1 = B-A ; v2 = C-A ; v3 = D-A
    # mat defines an affine transform from the tetrahedron to the orthogonal system
    mat = np.array((v1,v2,v3)).T
    # The inverse matrix does the opposite (from orthogonal to tetrahedron)
    M1 = np.linalg.inv(mat)
    return(M1) 

def pointInside_Dorian(v1,v2,v3,v4,p):
    # Find the transform matrix from orthogonal to tetrahedron system
    M1=tetraCoord_Dorian(v1,v2,v3,v4)
    # apply the transform to P
    newp = M1.dot(p-v1)
    # perform test
    return (np.all(newp>=0) and np.all(newp <=1) and np.sum(newp)<=1)

#TomNorway's solution adapted to cope with n tetrahedrons

def Tetrahedron(vertices):
    """
    Given a list of the xyz coordinates of the vertices of a tetrahedron, 
    return tetrahedron coordinate system
    """
    origin, *rest = vertices
    mat = (np.array(rest) - origin).T
    tetra = np.linalg.inv(mat)
    return tetra, origin

def pointInside(point, tetra, origin):
    """
    Takes a single point or array of points, as well as tetra and origin objects returned by 
    the Tetrahedron function.
    Returns a boolean or boolean array indicating whether the point is inside the tetrahedron.
    """
    newp = np.matmul(tetra, (point-origin).T).T
    return np.all(newp>=0, axis=-1) & np.all(newp <=1, axis=-1) & (np.sum(newp, axis=-1) <=1)



# Proposed solution
def det3x3_Philipp(b,c,d):
    return b[0]*c[1]*d[2] + c[0]*d[1]*b[2] + d[0]*b[1]*c[2] - d[0]*c[1]*b[2] - c[0]*b[1]*d[2] - b[0]*d[1]*c[2]

def pointInside_Philipp(v0,v1,v2,v3,p):
    a = v0 - p
    b = v1 - p
    c = v2 - p
    d = v3 - p
    detA = det3x3_Philipp(b,c,d)
    detB = det3x3_Philipp(a,c,d)
    detC = det3x3_Philipp(a,b,d)
    detD = det3x3_Philipp(a,b,c)
    ret0 = detA > 0.0 and detB < 0.0 and detC > 0.0 and detD < 0.0
    ret1 = detA < 0.0 and detB > 0.0 and detC < 0.0 and detD > 0.0
    return ret0 or ret1


npt=100000
Pt= np.array([ np.array([p[0]-0.5,p[1]-0.5,p[2]-0.5]) for p in np.random.rand(npt,3)])
A=np.array([ np.array([p[0]-0.5,p[1]-0.5,p[2]-0.5]) for p in np.random.rand(npt,3)])
B=np.array([ np.array([p[0]-0.5,p[1]-0.5,p[2]-0.5]) for p in np.random.rand(npt,3)])
C=np.array([ np.array([p[0]-0.5,p[1]-0.5,p[2]-0.5]) for p in np.random.rand(npt,3)])
D=np.array([ np.array([p[0]-0.5,p[1]-0.5,p[2]-0.5]) for p in np.random.rand(npt,3)])

inTet_Nico=np.zeros(shape=(npt,1),dtype=bool)
inTet_Hugues=np.copy(inTet_Nico)
inTet_Dorian=np.copy(inTet_Nico)
inTet_Philipp=np.copy(inTet_Nico)



print("non vectorized, n points, different tetrahedrons:")

start_time = time.time()
for i in range(0,npt):
    inTet_Nico[i]=pointInside_Nico(A[i,:],B[i,:],C[i,:],D[i,:],Pt[i,:])
print("Nico's:   --- %s seconds ---" % (time.time() - start_time)) # https://stackoverflow.com/questions/1557571/how-do-i-get-time-of-a-python-programs-execution

start_time = time.time()
for i in range(0,npt):
    inTet_Hugues[i]=pointInside_Hugues(A[i,:],B[i,:],C[i,:],D[i,:],Pt[i,:])
print("Hugues':  --- %s seconds ---" % (time.time() - start_time))   

start_time = time.time()    
for i in range(0,npt):
    inTet_Dorian[i]=pointInside_Dorian(A[i,:],B[i,:],C[i,:],D[i,:],Pt[i,:])
print("Dorian's: --- %s seconds ---" % (time.time() - start_time))

start_time = time.time()
for i in range(0,npt):
    inTet_Philipp[i]=pointInside_Philipp(A[i,:],B[i,:],C[i,:],D[i,:],Pt[i,:])
print("Philipp's:--- %s seconds ---" % (time.time() - start_time))   

print("vectorized, n points, 1 tetrahedron:")

start_time = time.time()
vertices = [A[0], B[0], C[0], D[0]]
tetra, origin = Tetrahedron(vertices)
inTet_Tom = pointInside(Pt, tetra, origin)
print("TomNorway's: --- %s seconds ---" % (time.time() - start_time)) 


for i in range(0,npt):
    assert inTet_Hugues[i] == inTet_Nico[i]
    assert inTet_Dorian[i] == inTet_Hugues[i]
    #assert inTet_Tom[i] == inTet_Dorian[i] can not compare because Tom implements 1 tetra instead of n
    assert inTet_Philipp[i] == inTet_Dorian[i]

'''errors = 0
for i in range(0,npt):
    if ( inTet_Philipp[i] != inTet_Dorian[i]):
        errors = errors + 1 
print("errors " + str(errors))'''

结果：

non vectorized, n points, different tetrahedrons:
Nico's:   --- 25.439453125 seconds ---
Hugues':  --- 28.724457263946533 seconds ---
Dorian's: --- 15.006574153900146 seconds ---
Philipp's:--- 4.389788389205933 seconds ---
vectorized, n points, 1 tetrahedron:
TomNorway's: --- 0.008165121078491211 seconds ---
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