有没有一种简单的方法来确定一个点是否在三角形内?是2D的,不是3D的。


当前回答

因为没有JS的答案, 顺时针和逆时针解决方案:

function triangleContains(ax, ay, bx, by, cx, cy, x, y) {

    let det = (bx - ax) * (cy - ay) - (by - ay) * (cx - ax)

    return  det * ((bx - ax) * (y - ay) - (by - ay) * (x - ax)) >= 0 &&
            det * ((cx - bx) * (y - by) - (cy - by) * (x - bx)) >= 0 &&
            det * ((ax - cx) * (y - cy) - (ay - cy) * (x - cx)) >= 0    

}

编辑:修正了两个拼写错误(关于符号和比较)。

https://jsfiddle.net/jniac/rctb3gfL/

function triangleContains(ax, ay, bx, by, cx, cy, x, y) { let det = (bx - ax) * (cy - ay) - (by - ay) * (cx - ax) return det * ((bx - ax) * (y - ay) - (by - ay) * (x - ax)) > 0 && det * ((cx - bx) * (y - by) - (cy - by) * (x - bx)) > 0 && det * ((ax - cx) * (y - cy) - (ay - cy) * (x - cx)) > 0 } let width = 500, height = 500 // clockwise let triangle1 = { A : { x: 10, y: -10 }, C : { x: 20, y: 100 }, B : { x: -90, y: 10 }, color: '#f00', } // counter clockwise let triangle2 = { A : { x: 20, y: -60 }, B : { x: 90, y: 20 }, C : { x: 20, y: 60 }, color: '#00f', } let scale = 2 let mouse = { x: 0, y: 0 } // DRAW > let wrapper = document.querySelector('div.wrapper') wrapper.onmousemove = ({ layerX:x, layerY:y }) => { x -= width / 2 y -= height / 2 x /= scale y /= scale mouse.x = x mouse.y = y drawInteractive() } function drawArrow(ctx, A, B) { let v = normalize(sub(B, A), 3) let I = center(A, B) let p p = add(I, rotate(v, 90), v) ctx.moveTo(p.x, p.y) ctx.lineTo(I.x, I .y) p = add(I, rotate(v, -90), v) ctx.lineTo(p.x, p.y) } function drawTriangle(ctx, { A, B, C, color }) { ctx.beginPath() ctx.moveTo(A.x, A.y) ctx.lineTo(B.x, B.y) ctx.lineTo(C.x, C.y) ctx.closePath() ctx.fillStyle = color + '6' ctx.strokeStyle = color ctx.fill() drawArrow(ctx, A, B) drawArrow(ctx, B, C) drawArrow(ctx, C, A) ctx.stroke() } function contains({ A, B, C }, P) { return triangleContains(A.x, A.y, B.x, B.y, C.x, C.y, P.x, P.y) } function resetCanvas(canvas) { canvas.width = width canvas.height = height let ctx = canvas.getContext('2d') ctx.resetTransform() ctx.clearRect(0, 0, width, height) ctx.setTransform(scale, 0, 0, scale, width/2, height/2) } function drawDots() { let canvas = document.querySelector('canvas#dots') let ctx = canvas.getContext('2d') resetCanvas(canvas) let count = 1000 for (let i = 0; i < count; i++) { let x = width * (Math.random() - .5) let y = width * (Math.random() - .5) ctx.beginPath() ctx.ellipse(x, y, 1, 1, 0, 0, 2 * Math.PI) if (contains(triangle1, { x, y })) { ctx.fillStyle = '#f00' } else if (contains(triangle2, { x, y })) { ctx.fillStyle = '#00f' } else { ctx.fillStyle = '#0003' } ctx.fill() } } function drawInteractive() { let canvas = document.querySelector('canvas#interactive') let ctx = canvas.getContext('2d') resetCanvas(canvas) ctx.beginPath() ctx.moveTo(0, -height/2) ctx.lineTo(0, height/2) ctx.moveTo(-width/2, 0) ctx.lineTo(width/2, 0) ctx.strokeStyle = '#0003' ctx.stroke() drawTriangle(ctx, triangle1) drawTriangle(ctx, triangle2) ctx.beginPath() ctx.ellipse(mouse.x, mouse.y, 4, 4, 0, 0, 2 * Math.PI) if (contains(triangle1, mouse)) { ctx.fillStyle = triangle1.color + 'a' ctx.fill() } else if (contains(triangle2, mouse)) { ctx.fillStyle = triangle2.color + 'a' ctx.fill() } else { ctx.strokeStyle = 'black' ctx.stroke() } } drawDots() drawInteractive() // trigo function add(...points) { let x = 0, y = 0 for (let point of points) { x += point.x y += point.y } return { x, y } } function center(...points) { let x = 0, y = 0 for (let point of points) { x += point.x y += point.y } x /= points.length y /= points.length return { x, y } } function sub(A, B) { let x = A.x - B.x let y = A.y - B.y return { x, y } } function normalize({ x, y }, length = 10) { let r = length / Math.sqrt(x * x + y * y) x *= r y *= r return { x, y } } function rotate({ x, y }, angle = 90) { let length = Math.sqrt(x * x + y * y) angle *= Math.PI / 180 angle += Math.atan2(y, x) x = length * Math.cos(angle) y = length * Math.sin(angle) return { x, y } } * { margin: 0; } html { font-family: monospace; } body { padding: 32px; } span.red { color: #f00; } span.blue { color: #00f; } canvas { position: absolute; border: solid 1px #ddd; } <p><span class="red">red triangle</span> is clockwise</p> <p><span class="blue">blue triangle</span> is couter clockwise</p> <br> <div class="wrapper"> <canvas id="dots"></canvas> <canvas id="interactive"></canvas> </div>

我在这里使用与上面描述的相同的方法:如果一个点分别位于AB, BC, CA的“同”边,则它在ABC内。

其他回答

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:

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);
}

python中的其他函数,比Developer的方法更快(至少对我来说),并受到Cédric Dufour解决方案的启发:

def ptInTriang(p_test, p0, p1, p2):       
     dX = p_test[0] - p0[0]
     dY = p_test[1] - p0[1]
     dX20 = p2[0] - p0[0]
     dY20 = p2[1] - p0[1]
     dX10 = p1[0] - p0[0]
     dY10 = p1[1] - p0[1]

     s_p = (dY20*dX) - (dX20*dY)
     t_p = (dX10*dY) - (dY10*dX)
     D = (dX10*dY20) - (dY10*dX20)

     if D > 0:
         return (  (s_p >= 0) and (t_p >= 0) and (s_p + t_p) <= D  )
     else:
         return (  (s_p <= 0) and (t_p <= 0) and (s_p + t_p) >= D  )

你可以用:

X_size = 64
Y_size = 64
ax_x = np.arange(X_size).astype(np.float32)
ax_y = np.arange(Y_size).astype(np.float32)
coords=np.meshgrid(ax_x,ax_y)
points_unif = (coords[0].reshape(X_size*Y_size,),coords[1].reshape(X_size*Y_size,))
p_test = np.array([0 , 0])
p0 = np.array([22 , 8]) 
p1 = np.array([12 , 55]) 
p2 = np.array([7 , 19]) 
fig = plt.figure(dpi=300)
for i in range(0,X_size*Y_size):
    p_test[0] = points_unif[0][i]
    p_test[1] = points_unif[1][i]
    if ptInTriang(p_test, p0, p1, p2):
        plt.plot(p_test[0], p_test[1], '.g')
    else:
        plt.plot(p_test[0], p_test[1], '.r')

绘制网格需要花费很多时间,但是该网格在0.0195319652557秒内测试,而开发人员代码为0.0844349861145秒。

最后是代码注释:

# Using barycentric coordintes, any point inside can be described as:
# X = p0.x * r + p1.x * s + p2.x * t
# Y = p0.y * r + p1.y * s + p2.y * t
# with:
# r + s + t = 1  and 0 < r,s,t < 1
# then: r = 1 - s - t
# and then:
# X = p0.x * (1 - s - t) + p1.x * s + p2.x * t
# Y = p0.y * (1 - s - t) + p1.y * s + p2.y * t
#
# X = p0.x + (p1.x-p0.x) * s + (p2.x-p0.x) * t
# Y = p0.y + (p1.y-p0.y) * s + (p2.y-p0.y) * t
#
# X - p0.x = (p1.x-p0.x) * s + (p2.x-p0.x) * t
# Y - p0.y = (p1.y-p0.y) * s + (p2.y-p0.y) * t
#
# we have to solve:
#
# [ X - p0.x ] = [(p1.x-p0.x)   (p2.x-p0.x)] * [ s ]
# [ Y - p0.Y ]   [(p1.y-p0.y)   (p2.y-p0.y)]   [ t ]
#
# ---> b = A*x ; ---> x = A^-1 * b
# 
# [ s ] =   A^-1  * [ X - p0.x ]
# [ t ]             [ Y - p0.Y ]
#
# A^-1 = 1/D * adj(A)
#
# The adjugate of A:
#
# adj(A)   =   [(p2.y-p0.y)   -(p2.x-p0.x)]
#              [-(p1.y-p0.y)   (p1.x-p0.x)]
#
# The determinant of A:
#
# D = (p1.x-p0.x)*(p2.y-p0.y) - (p1.y-p0.y)*(p2.x-p0.x)
#
# Then:
#
# s_p = { (p2.y-p0.y)*(X - p0.x) - (p2.x-p0.x)*(Y - p0.Y) }
# t_p = { (p1.x-p0.x)*(Y - p0.Y) - (p1.y-p0.y)*(X - p0.x) }
#
# s = s_p / D
# t = t_p / D
#
# Recovering r:
#
# r = 1 - (s_p + t_p)/D
#
# Since we only want to know if it is insidem not the barycentric coordinate:
#
# 0 < 1 - (s_p + t_p)/D < 1
# 0 < (s_p + t_p)/D < 1
# 0 < (s_p + t_p) < D
#
# The condition is:
# if D > 0:
#     s_p > 0 and t_p > 0 and (s_p + t_p) < D
# else:
#     s_p < 0 and t_p < 0 and (s_p + t_p) > D
#
# s_p = { dY20*dX - dX20*dY }
# t_p = { dX10*dY - dY10*dX }
# D = dX10*dY20 - dY10*dX20
bool isInside( float x, float y, float x1, float y1, float x2, float y2, float x3, float y3 ) {
  float l1 = (x-x1)*(y3-y1) - (x3-x1)*(y-y1), 
    l2 = (x-x2)*(y1-y2) - (x1-x2)*(y-y2), 
    l3 = (x-x3)*(y2-y3) - (x2-x3)*(y-y3);
  return (l1>0 && l2>0  && l3>0) || (l1<0 && l2<0 && l3<0);
}

没有比这更有效率的了!三角形的每边都可以有独立的位置和方向,因此需要进行l1、l2和l3三个计算,每个计算需要进行2次乘法。一旦l1, l2和l3是已知的,结果只是一些基本的比较和布尔运算。

我只是想用一些简单的向量数学来解释安德里亚斯给出的重心坐标解,它会更容易理解。

区域A定义为s * v02 + t * v01给出的任意向量,条件s >= 0, t >= 0。如果三角形v0 v1 v2内的任意一点,它一定在区域A内。

如果进一步限制s, t属于[0,1]。得到包含s * v02 + t * v01的所有向量的区域B,条件s, t属于[0,1]。值得注意的是,区域B的下部是三角形v0, v1, v2的镜像。问题来了,我们是否可以给定一定的s和t条件,来进一步排除区域B的低部分。

假设我们给出一个值s, t在[0,1]内变化。在下图中,点p位于v1v2的边缘。s * v02 + t * v01的所有向量沿着虚线通过简单向量和得到。在v1v2和虚线交点p处,我们有:

(1-S)|V0v2|/ |v0v2|= tp|v0v1|/ |v0v1|

得到1 - s = tp,然后1 = s + tp。如果任意t > tp,即1 < s + t where在双虚线上,则该向量在三角形外,任意t <= tp,即1 >= s + t where在单虚线上,则该向量在三角形内。

如果我们给出[0,1]中的任意s,对应的t必须满足1 >= s + t,对于三角形内的向量。

最后我们得到v = s * v02 +t * v01, v在三角形内,条件s, t, s+t属于[0,1]。然后翻译到点,我们有

P - p0 = s * (p1 - p0) + t * (p2 - p0), and s, t, s + t in [0,1]

和Andreas解方程组的解是一样的 P = p0 + s * (p1 - p0) + t * (p2 - p0),带s, t, s + t属于[0,1]。