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
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”
var ctx = $("canvas")[0].getContext("2d"); var W = 500; var H = 500; var point = { x: W / 2, y: H / 2 }; var triangle = randomTriangle(); $("canvas").click(function(evt) { point.x = evt.pageX - $(this).offset().left; point.y = evt.pageY - $(this).offset().top; test(); }); $("canvas").dblclick(function(evt) { triangle = randomTriangle(); test(); }); document.querySelector('#performance').addEventListener('click', _testPerformance); test(); function test() { var result = checkClockwise(triangle.a, triangle.b, triangle.c) ? pointInTriangle(point, triangle.a, triangle.c, triangle.b) : pointInTriangle(point, triangle.a, triangle.b, triangle.c); var info = "point = (" + point.x + "," + point.y + ")\n"; info += "triangle.a = (" + triangle.a.x + "," + triangle.a.y + ")\n"; info += "triangle.b = (" + triangle.b.x + "," + triangle.b.y + ")\n"; info += "triangle.c = (" + triangle.c.x + "," + triangle.c.y + ")\n"; info += "result = " + (result ? "true" : "false"); $("#result").text(info); render(); } function _testPerformance () { var px = [], py = [], p0x = [], p0y = [], p1x = [], p1y = [], p2x = [], p2y = [], p = [], p0 = [], p1 = [], p2 = []; for(var i = 0; i < 1000000; i++) { p[i] = {x: Math.random() * 100, y: Math.random() * 100}; p0[i] = {x: Math.random() * 100, y: Math.random() * 100}; p1[i] = {x: Math.random() * 100, y: Math.random() * 100}; p2[i] = {x: Math.random() * 100, y: Math.random() * 100}; } console.time('optimal: pointInTriangle'); for(var i = 0; i < 1000000; i++) { pointInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('optimal: pointInTriangle'); console.time('original: ptInTriangle'); for(var i = 0; i < 1000000; i++) { ptInTriangle(p[i], p0[i], p1[i], p2[i]); } console.timeEnd('original: ptInTriangle'); } 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; } function ptInTriangle(p, p0, p1, p2) { var s = (p0.y * p2.x - p0.x * p2.y + (p2.y - p0.y) * p.x + (p0.x - p2.x) * p.y); var t = (p0.x * p1.y - p0.y * p1.x + (p0.y - p1.y) * p.x + (p1.x - p0.x) * p.y); if (s <= 0 || t <= 0) return false; var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return (s + t) < A; } function render() { ctx.fillStyle = "#CCC"; ctx.fillRect(0, 0, 500, 500); drawTriangle(triangle.a, triangle.b, triangle.c); drawPoint(point); } function checkClockwise(p0, p1, p2) { var A = (-p1.y * p2.x + p0.y * (-p1.x + p2.x) + p0.x * (p1.y - p2.y) + p1.x * p2.y); return A > 0; } function drawTriangle(p0, p1, p2) { ctx.fillStyle = "#999"; ctx.beginPath(); ctx.moveTo(p0.x, p0.y); ctx.lineTo(p1.x, p1.y); ctx.lineTo(p2.x, p2.y); ctx.closePath(); ctx.fill(); ctx.fillStyle = "#000"; ctx.font = "12px monospace"; ctx.fillText("1", p0.x, p0.y); ctx.fillText("2", p1.x, p1.y); ctx.fillText("3", p2.x, p2.y); } function drawPoint(p) { ctx.fillStyle = "#F00"; ctx.beginPath(); ctx.arc(p.x, p.y, 5, 0, 2 * Math.PI); ctx.fill(); } function rand(min, max) { return Math.floor(Math.random() * (max - min + 1)) + min; } function randomTriangle() { return { a: { x: rand(0, W), y: rand(0, H) }, b: { x: rand(0, W), y: rand(0, H) }, c: { x: rand(0, W), y: rand(0, H) } }; } <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script> <button id="performance">Run performance test (open console)</button> <pre>Click: place the point. Double click: random triangle.</pre> <pre id="result"></pre> <canvas width="500" height="500"></canvas>
受此启发: http://www.phatcode.net/articles.php?id=459
其他回答
我同意Andreas Brinck的观点,重心坐标对于这项任务来说非常方便。注意,不需要每次都求解一个方程组:只需计算解析解。使用Andreas的符号,解是:
s = 1/(2*Area)*(p0y*p2x - p0x*p2y + (p2y - p0y)*px + (p0x - p2x)*py);
t = 1/(2*Area)*(p0x*p1y - p0y*p1x + (p0y - p1y)*px + (p1x - p0x)*py);
其中Area是三角形的(带符号的)面积:
Area = 0.5 *(-p1y*p2x + p0y*(-p1x + p2x) + p0x*(p1y - p2y) + p1x*p2y);
只计算st和1-s-t。点p在三角形内当且仅当它们都是正的。
编辑:请注意,上面的区域表达式假设三角形节点编号是逆时针方向的。如果编号是顺时针的,这个表达式将返回一个负的面积(但大小正确)。然而,测试本身(s>0 && t>0 && 1-s-t>0)并不依赖于编号的方向,因为如果三角形节点的方向改变,上面乘以1/(2*Area)的表达式也会改变符号。
编辑2:为了获得更好的计算效率,请参阅下面的coproc注释(其中指出,如果三角形节点的方向(顺时针或逆时针)事先已知,则可以避免在s和t的表达式中除以2*Area)。在Andreas Brinck的回答下面的评论中也可以看到Perro Azul的jsfiddle-code。
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。
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
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]。
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。