我试图在Javascript中实现一种分而治之的算法来计算Delanay三角剖分，以获得Voronoi图。编码基于Graphic Gems IV中描述的C ++实现，而后者又基于1985年着名的Guibas和Stolfi论文。它使用了四边数据结构。
算法将点集分成两半;然后，它分别对这些部分进行三角测量;最后它继续“结婚”两半。
问题在于，它不起作用。当“结合”两半时，它会陷入无限递归。我认为问题可能出在四边形结构本身，因为第一个边缘和它的目标边缘具有相同的坐标，其双重和反向旋转边缘指向（0,0）。
这是代码（C ++的翻译不是100％，我请求帮助，因为我还是JS菜鸟）：
//The data structure starts here.
var Edge = function (num, qe) {
var _this = this;
this.num = num;
this.qe = qe;
this.coord = new paper.Point();
this.sym = function () {
return (_this.num < 2) ? _this.qe[_this.num + 2] : this.qe[_this.num - 2];
};
this.oNext = function () {
return this.next;
};
this.setNext = function (next) {
this.next = next
};
this.rot = function () {
return (this.num < 3) ? _this.qe[_this.num + 1] : _this.qe[_this.num - 3];
};
this.invRot = function () {
return (_this.num > 0) ? _this.qe[_this.num - 1] : _this.qe[_this.num + 3];
};
this.oPrev = function () {
return _this.rot().oNext().rot();
};
this.dNext = function () {
return _this.sym().next.sym();
};
this.dPrev = function () {
return _this.invRot().oNext().invRot();
};
this.lNext = function () {
return _this.invRot().oNext().rot();
};
this.lPrev = function () {
return _this.oNext().sym();
};
this.rNext = function () {
return _this.rot().oNext().invRot();
};
this.rPrev = function () {
return _this.sym().oNext();
};
this.dest = function () {
return _this.sym().coord;
};
this.endPoints = function (or, de) {
this.coord = or;
this.sym().coord = de;
};
};
//The first edge always points to itself
//as in edge.oNext().coord === edge.coord
var QuadEdge = function () {
this.edges = new Array(4);
for (var i = 0; i < 4; i++) {
this.edges[i] = new Edge(i, this.edges);
}
this.edges[0].setNext(this.edges[0]);
this.edges[1].setNext(this.edges[3]);
this.edges[2].setNext(this.edges[2]);
this.edges[3].setNext(this.edges[1]);
};
//Constructs and returns a new QuadEdge.
function makeEdge() {
var qe = new QuadEdge();
return qe.edges[0];
}
//Helper function, returns twice of the area of the triangle formed
//by a, b and c. Negative if the triangle is clockwise.
function triArea(a, b, c) {
return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}
//Tests if the point d is inside the circumcircle of the triangle formed by
//a, b, and c. As described by Guibas and Stolfi in their paper.
function inCircle(a, b, c, d) {
return (Math.pow(a.x, 2) + Math.pow(a.y, 2)) * triArea(b, c, d) -
(Math.pow(b.x, 2) + Math.pow(b.y, 2)) * triArea(a, c, d) +
(Math.pow(c.x, 2) + Math.pow(c.y, 2)) * triArea(a, b, d) -
(Math.pow(d.x, 2) + Math.pow(d.y, 2)) * triArea(a, b, d) > 0;
}
//Tests if triangle is in counterclockwise order.
function cClockwise(a, b, c) {
return triArea(a, b, c) > 0;
}
//Tests if point is left of the provided edge.
function leftOf(point, edge) {
return cClockwise(point, edge.coord, edge.dest());
}
//Tests if point is right of the provided edge.
function rightOf(point, edge) {
return cClockwise(point, edge.dest(), edge.coord);
}
//If a and b are distinct, splice will combine them; if not, it will
//separate them in two.
function splice(a, b) {
var alpha = a.oNext().rot();
var beta = b.oNext().rot();
var t1 = b.oNext();
var t2 = a.oNext();
var t3 = beta.oNext();
var t4 = alpha.oNext();
a.setNext(t1);
b.setNext(t2);
alpha.setNext(t3);
beta.setNext(t4);
}
function deleteEdge(e) {
splice(e, e.oPrev());
splice(e.sym(), e.sym().oPrev());
}
function connect(e1, e2) {
var e = makeEdge();
e.endPoints(e1.dest(), e2.coord);
splice(e, e1.lNext());
splice(e.sym(), e2);
return e;
}
//Fixed thanks to: http://www.rpenalva.com/blog/?p=74
function valid(e, basel) {
return rightOf(e.dest(), basel);
}
//This is the actual algorithm.
function divideAndConquer(vertices) {
if (vertices.length === 2) {
var a = makeEdge();
a.coord = vertices[0];
a.sym().coord = vertices[1];
return {right: a, left: a.sym()};
}
else if (vertices.length === 3) {
var a = makeEdge(), b = makeEdge();
splice(a.sym(), b);
a.coord = vertices[0];
b.coord = vertices[1];
a.sym().coord = b.coord;
b.sym().coord = vertices[2];
if (cClockwise(vertices[0], vertices[1], vertices[2])) {
connect(b, a);
return {right: a, left: b.sym()};
}
else if (cClockwise(vertices[0], vertices[2], vertices[1])) {
var c = connect(b, a);
return {right: c.sym(), left: c};
}
else
return {right: a, left: b.sym()};
}
else if (vertices.length >= 4) {
var half = Math.floor(vertices.length / 2);
var lObjects = divideAndConquer(vertices.slice(0, half));
var rObjects = divideAndConquer(vertices.slice(half, vertices.length));
var ldo = lObjects.right, ldi = lObjects.left;
var rdi = rObjects.right, rdo = rObjects.left;
while (true) {
if (leftOf(rdi.coord, ldi))
ldi = ldi.lNext();
else if (rightOf(ldi.coord, rdi))
rdi = rdi.rPrev();
else
break;
}
var basel = connect(rdi.sym(), ldi);
if (ldi.coord === ldo.coord)
ldo = basel.sym();
if (rdi.coord === rdo.coord)
rdo = basel;
while (true) {
var lcand = basel.sym().oNext();
if (valid(lcand, basel)) {
while (inCircle(basel.dest(), basel.coord, lcand.dest(), lcand.oPrev().dest())) {
var t = lcand.oNext();
deleteEdge(lcand);
lcand = t;
}
}
var rcand = basel.oPrev();
if (valid(rcand, basel)) {
while (inCircle(basel.dest(), basel.coord, rcand.dest(), rcand.oPrev().dest())) {
var t = rcand.oPrev();
deleteEdge(rcand);
rcand = t;
}
}
//This is the part where it gets stuck,
//it never reaches this condition.
if (!valid(lcand, basel) && !valid(rcand, basel)) {
break;
}
if (!valid(lcand, basel) || (valid(rcand, basel) && inCircle(lcand.dest(), lcand.coord, rcand.coord, rcand.dest()))) {
basel = connect(rcand, basel.sym());
}
else {
basel = connect(basel.sym(), lcand.sym());
}
}
return {right: ldo, left: rdo};
}
}
点类来自PaperJS。
编辑：设置Codepen说明问题;单击输出窗口以添加点。添加第四个点时会发生递归。
EDIT 2：修复了由有效函数引起的无限递归，这要归功于Ruben Penalva blog上的代码。但是，指向原点的双边问题仍然存在。那就是：
edge.rot().coord === (0, 0)
这让我更加想到问题在于QuadEdge结构本身。
编辑3：第n次仔细阅读代码，我意识到旋转和反向旋转边缘的坐标永远不会设置。所以我尝试通过旋转原始边缘来设置它们。输出现在绘制某些东西，虽然我很确定这不是Delaunay三角剖分。我用进度更新了Codepen示例。