如何确定一个点是否在二维三角形中?

下面是一个python解决方案，它是高效的，文档化的，包含三个单元测试。它具有专业级的质量，并且可以以模块的形式放入您的项目中。

import unittest

###############################################################################
def point_in_triangle(point, triangle):
    """Returns True if the point is inside the triangle
    and returns False if it falls outside.
    - The argument *point* is a tuple with two elements
    containing the X,Y coordinates respectively.
    - The argument *triangle* is a tuple with three elements each
    element consisting of a tuple of X,Y coordinates.

    It works like this:
    Walk clockwise or counterclockwise around the triangle
    and project the point onto the segment we are crossing
    by using the dot product.
    Finally, check that the vector created is on the same side
    for each of the triangle's segments.
    """
    # Unpack arguments
    x, y = point
    ax, ay = triangle[0]
    bx, by = triangle[1]
    cx, cy = triangle[2]
    # Segment A to B
    side_1 = (x - bx) * (ay - by) - (ax - bx) * (y - by)
    # Segment B to C
    side_2 = (x - cx) * (by - cy) - (bx - cx) * (y - cy)
    # Segment C to A
    side_3 = (x - ax) * (cy - ay) - (cx - ax) * (y - ay)
    # All the signs must be positive or all negative
    return (side_1 < 0.0) == (side_2 < 0.0) == (side_3 < 0.0)

###############################################################################
class TestPointInTriangle(unittest.TestCase):

    triangle = ((22 , 8),
                (12 , 55),
                (7 , 19))

    def test_inside(self):
        point = (15, 20)
        self.assertTrue(point_in_triangle(point, self.triangle))

    def test_outside(self):
        point = (1, 7)
        self.assertFalse(point_in_triangle(point, self.triangle))

    def test_border_case(self):
        """If the point is exactly on one of the triangle's edges,
        we consider it is inside."""
        point = (7, 19)
        self.assertTrue(point_in_triangle(point, self.triangle))

###############################################################################
if __name__ == "__main__":
    suite = unittest.defaultTestLoader.loadTestsFromTestCase(TestPointInTriangle)
    unittest.TextTestRunner().run(suite)

上面的算法有一个额外的可选图形测试，以确认其有效性:

import random
from matplotlib import pyplot
from triangle_test import point_in_triangle

###############################################################################
# The area #
size_x = 64
size_y = 64

# The triangle #
triangle = ((22 , 8),
            (12 , 55),
            (7 , 19))

# Number of random points #
count_points = 10000

# Prepare the figure #
figure = pyplot.figure()
axes = figure.add_subplot(111, aspect='equal')
axes.set_title("Test the 'point_in_triangle' function")
axes.set_xlim(0, size_x)
axes.set_ylim(0, size_y)

# Plot the triangle #
from matplotlib.patches import Polygon
axes.add_patch(Polygon(triangle, linewidth=1, edgecolor='k', facecolor='none'))

# Plot the points #
for i in range(count_points):
    x = random.uniform(0, size_x)
    y = random.uniform(0, size_y)
    if point_in_triangle((x,y), triangle): pyplot.plot(x, y, '.g')
    else:                                  pyplot.plot(x, y, '.b')

# Save it #
figure.savefig("point_in_triangle.pdf")

制作以下图表:

2018-07-23 12:57:41

下面是一个高效的Python实现:

def PointInsideTriangle2(pt,tri):
    '''checks if point pt(2) is inside triangle tri(3x2). @Developer'''
    a = 1/(-tri[1,1]*tri[2,0]+tri[0,1]*(-tri[1,0]+tri[2,0])+ \
        tri[0,0]*(tri[1,1]-tri[2,1])+tri[1,0]*tri[2,1])
    s = a*(tri[2,0]*tri[0,1]-tri[0,0]*tri[2,1]+(tri[2,1]-tri[0,1])*pt[0]+ \
        (tri[0,0]-tri[2,0])*pt[1])
    if s<0: return False
    else: t = a*(tri[0,0]*tri[1,1]-tri[1,0]*tri[0,1]+(tri[0,1]-tri[1,1])*pt[0]+ \
              (tri[1,0]-tri[0,0])*pt[1])
    return ((t>0) and (1-s-t>0))

和一个示例输出:

2014-01-06 11:32:43

我在JavaScript中改编的高性能代码(文章如下):

function pointInTriangle (p, p0, p1, p2) {
  return (((p1.y - p0.y) * (p.x - p0.x) - (p1.x - p0.x) * (p.y - p0.y)) | ((p2.y - p1.y) * (p.x - p1.x) - (p2.x - p1.x) * (p.y - p1.y)) | ((p0.y - p2.y) * (p.x - p2.x) - (p0.x - p2.x) * (p.y - p2.y))) >= 0;
}

pointInTriangle(p, p0, p1, p2) -用于逆时针方向的三角形 pointInTriangle(p, p0, p1, p2) -用于顺时针三角形

在jsFiddle(包括性能测试)中，在一个单独的函数中也有缠绕检查。或按下面的“运行代码片段”

受此启发: http://www.phatcode.net/articles.php?id=459

2016-05-13 23:34:09

bool point2Dtriangle(double e,double f, double a,double b,double c, double g,double h,double i, double v, double w){
    /* inputs: e=point.x, f=point.y
               a=triangle.Ax, b=triangle.Bx, c=triangle.Cx 
               g=triangle.Ay, h=triangle.By, i=triangle.Cy */
    v = 1 - (f * (b - c) + h * (c - e) + i * (e - b)) / (g * (b - c) + h * (c - a) + i * (a - b));
    w = (f * (a - b) + g * (b - e) + h * (e - a)) / (g * (b - c) + h * (c - a) + i * (a - b));
    if (*v > -0.0 && *v < 1.0000001 && *w > -0.0 && *w < *v) return true;//is inside
    else return false;//is outside
    return 0;
}

从质心转换而来的几乎完美的笛卡尔坐标在*v (x)和*w (y)双精度内导出。在每种情况下，两个导出双精度对象前面都应该有一个*字符，可能是*v和*w 代码也可以用于四边形的另一个三角形。特此签名只写三角形abc从顺时针abcd的四边形。

A---B
|..\\.o|  
|....\\.| 
D---C

o点在ABC三角形内对于带有第二个三角形的测试，将此函数称为CDA方向，*v=1-*v后的结果应正确;* w = 1 - * w;为了四合院

2018-02-12 13:00:15

通过使用重心坐标的解析解(由Andreas Brinck指出)和:

不是把乘法分布在括号里的项上通过存储相同的项来避免多次计算还原比较(如coproc和Thomas Eding所指出的)

可以最小化“昂贵”操作的数量:

function ptInTriangle(p, p0, p1, p2) {
    var dX = p.x-p2.x;
    var dY = p.y-p2.y;
    var dX21 = p2.x-p1.x;
    var dY12 = p1.y-p2.y;
    var D = dY12*(p0.x-p2.x) + dX21*(p0.y-p2.y);
    var s = dY12*dX + dX21*dY;
    var t = (p2.y-p0.y)*dX + (p0.x-p2.x)*dY;
    if (D<0) return s<=0 && t<=0 && s+t>=D;
    return s>=0 && t>=0 && s+t<=D;
}

代码可以粘贴在Perro Azul jsfiddle中，或者通过点击下面的“运行代码片段”来尝试

导致:

变量“召回”:30 可变存储:7 补充:4 减法:8 乘法:6 部门:没有比较:4

这与Kornel Kisielewicz解决方案(25次召回，1次存储，15次减法，6次乘法，5次比较)相比非常好，如果需要顺时针/逆时针检测(它本身需要6次召回，1次加法，2次减法，2次乘法和1次比较，使用解析解行列式，如rhgb所指出的)，可能会更好。

2015-12-04 17:08:58

一般来说，最简单(也是最优)的算法是检查由边创建的半平面的哪一边是点。

以下是关于GameDev的一些高质量信息，包括性能问题。

这里有一些代码让你开始:

float sign (fPoint p1, fPoint p2, fPoint p3)
{
    return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}

bool PointInTriangle (fPoint pt, fPoint v1, fPoint v2, fPoint v3)
{
    float d1, d2, d3;
    bool has_neg, has_pos;

    d1 = sign(pt, v1, v2);
    d2 = sign(pt, v2, v3);
    d3 = sign(pt, v3, v1);

    has_neg = (d1 < 0) || (d2 < 0) || (d3 < 0);
    has_pos = (d1 > 0) || (d2 > 0) || (d3 > 0);

    return !(has_neg && has_pos);
}

2010-01-12 14:27:02

如何确定一个点是否在二维三角形中?

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