分离形状作为线分割 canvas + Select 点以从数组绘制形状
Separate shapes as lines split the canvas + Select points to draw shapes from arrays
我想检测单独的形状，因为随机生成的线将 canvas 分开。我在 x 和 y 位置（相同顺序）的单独数组中保存了线交点，但不知道如何连接
完成多个形状的点。
- 是否有任何方法可以检测附近的点以闭合可能的最小形状，无论它是三角形、矩形还是多边形（例如，通过使用 beginShape 和 endShape）？
- 如果1）太复杂，有什么方法可以select从一个数组中随机抽取3个或更多点吗？
这是一个示例图片，其中有 4 条线将 canvas 分开，它们的交点用红色标记。我还保存了每个随机生成的线的顶部和底部点（标记为黑色），加上 canvas 的四个角分别在相同的数组中用于 x 和 y 位置（px，py）。
多行拆分 canvas.
如何在 Processing 中按线分割形状？
我能够得到所有的交点，但在将它们连接成单独的形状时遇到了问题。这是我正在处理的处理代码：
//Run in Processing.
//Press r to refresh.
//Top and bottom points are added to px and py when refreshed (filled in black).
//Intersection points are added to px and py when detected (filled in red).
int l = 4; //set number of lines
float[] r1 = new float[l];
float[] r2 = new float[l];
float[] px = {}; //array to save x positions of all possible points
float[] py = {}; //array to save y positions of all possible points
boolean added = false;
void setup(){
size(800, 800);
background(255);
refresh();
}
void draw(){
background(255);
stroke(0, 150, 255, 150);
strokeWeight(1);
for(int i=0; i < r1.length; i++){
for(int j=0; j < r1.length; j++){
if(i>j){
boolean hit = lineLine(r1[i], 0, r2[i], height, r1[j], 0, r2[j], height);
if (hit) stroke(255, 150, 0, 150);
else stroke(0, 150, 255, 150);
}
line(r1[i], 0, r2[i], height);
}
}
added = true;
print(px.length);
}
//source: http://jeffreythompson.org/collision-detection/line-line.php
boolean lineLine(float x1, float y1, float x2, float y2, float x3, float y3, float x4, float y4) {
// calculate the distance to intersection point
float uA = ((x4-x3)*(y1-y3) - (y4-y3)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
float uB = ((x2-x1)*(y1-y3) - (y2-y1)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
// if uA and uB are between 0-1, lines are colliding
if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
// optionally, draw a circle where the lines meet
float intersectionX = x1 + (uA * (x2-x1));
float intersectionY = y1 + (uA * (y2-y1));
fill(255,0,0);
noStroke();
ellipse(intersectionX,intersectionY, 20,20);
if(added==false){
px = append(px, intersectionX);
py = append(py, intersectionY);
}
return true;
}
return false;
}
void refresh(){
added = false;
px = new float[0];
py = new float[0];
r1 = new float[l];
r2 = new float[l];
px = append(px, 0);
py = append(py, 0);
px = append(px, 0);
py = append(py, height);
px = append(px, width);
py = append(py, 0);
px = append(px, width);
py = append(py, height);
for(int i=0; i< r1.length; i++){
r1[i] = random(800);
}
for(int i=0; i< r2.length; i++){
r2[i] = random(800);
}
for(int i=0; i < r1.length; i++){
stroke(0);
line(r1[i], 0, r2[i], height);
px = append(px, r1[i]);
py = append(py, 0);
px = append(px, r2[i]);
py = append(py, height);
}
}
void keyReleased() {
if (key == 'r') refresh();
}
我不清楚你的目标。您可以以任意顺序连接任意一组点并将其称为形状。你的标准是什么？
如果你想找到连接给定子集所有点的最短路径，我建议寻找旅行商问题。
我可以看到您的代码不是 javascript，但由于您没有指定语言，我假设您只需要一个方法并可以转换为您的语言。
我处理这个问题的方法是为每一行分配一个行号。如果我可以在一条线上识别出 2 个相邻点，那么我将通过检查它们不共享的线的交叉处是否有一个点来知道第三个点是否存在。
示例：
有 3 行（第 1、2、3 行）
我在 3 号线和 1 号线之间有一个交点，现在我沿着 3 号线走到相邻的点。我找到了一个，它的交点是 3 和 2。我唯一能得到三角形的方法是 1 和 2 线穿过某处。所以我们可以通过编程方式检查。
请记住，我实际上从未为此使用过角度。我确实在函数中计算了它们，但决定不使用它们，因为我使用了上面解释的方法。我使用 0.1 的 alpha 值给三角形上色，这样您就可以看到重叠的地方。
这只是检查三角形
let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");
canvas.width = 400;
canvas.height = 400;
let lines = []; //holds each line
let points = []; //all intersection point are pushed here [{x: num, y: num}, {x: num, y: num},...]
let sortedPts = []; //all points sorted bu first number are pushed here in 2d array.
let lineNum = 15;
class Lines {
constructor(num) {
this.x = Math.round(Math.random() * canvas.width);
this.x2 = Math.round(Math.random() * canvas.width);
this.pt1 = {
x: this.x,
y: 0
};
this.pt2 = {
x: this.x2,
y: canvas.height
};
this.num = num;
this.rads = Math.atan2(this.pt2.y - this.pt1.y, this.pt2.x - this.pt1.x);
this.angle = this.rads * (180 / Math.PI);
}
draw() {
ctx.beginPath();
ctx.moveTo(this.pt1.x, this.pt1.y);
ctx.lineTo(this.pt2.x, this.pt2.y);
ctx.stroke();
}
}
//creates the lines. I also use this function to prepare the 2d array by pushing an empty array for each line into sortedPts.
function createLines() {
for (let i = 0; i < lineNum; i++) {
lines.push(new Lines(i + 1));
sortedPts.push([])
}
}
createLines();
//Visually draws lines on screen
function drawLines() {
for (let i = 0; i < lines.length; i++) {
lines[i].draw();
}
}
drawLines();
//intersecting formula
function lineSegmentsIntersect(line1, line2) {
let a_dx = line1.pt2.x - line1.pt1.x;
let a_dy = line1.pt2.y - line1.pt1.y;
let b_dx = line2.pt2.x - line2.pt1.x;
let b_dy = line2.pt2.y - line2.pt1.y;
let s =
(-a_dy * (line1.pt1.x - line2.pt1.x) + a_dx * (line1.pt1.y - line2.pt1.y)) /
(-b_dx * a_dy + a_dx * b_dy);
let t =
(+b_dx * (line1.pt1.y - line2.pt1.y) - b_dy * (line1.pt1.x - line2.pt1.x)) /
(-b_dx * a_dy + a_dx * b_dy);
if (s >= 0 && s <= 1 && t >= 0 && t <= 1) {
//this is where we create our array but we also add the line number of where each point intersects. I also add the angle but have not used it throughout the rest of this...yet.
points.push({
x: Math.round(line1.pt1.x + t * (line1.pt2.x - line1.pt1.x)),
y: Math.round(line1.pt1.y + t * (line1.pt2.y - line1.pt1.y)),
num: {
first: line1.num,
second: line2.num
},
angle: {
a1: line1.angle,
a2: line2.angle
}
});
}
}
//just checks each line against the others by passing to lineSegmentsIntersect() function
function callIntersect() {
for (let i = 0; i < lines.length; i++) {
for (let j = i + 1; j < lines.length; j++) {
lineSegmentsIntersect(lines[i], lines[j]);
}
}
}
callIntersect();
function drawPoints() {
//just draws the black points for reference
for (let i = 0; i < points.length; i++) {
ctx.beginPath();
ctx.arc(points[i].x, points[i].y, 2, 0, Math.PI * 2);
ctx.fill();
}
}
drawPoints();
function createSortedArray() {
//Now we take the points array and sort the points by the first number to make using i and j below possible
points.sort((a, b) => a.num.first - b.num.first)
//We push each group of points into an array inside sortedPts creating the 2d array
for (let i = 0; i < lineNum; i++) {
for (let j = 0; j < points.length; j++) {
if (points[j].num.first == (i + 1)) {
sortedPts[i].push(points[j]);
}
}
}
//now sort the 2d arrays by y value. This allows or next check to go in order from point to point per line.
sortedPts.forEach(arr => arr.sort((a, b) => a.y - b.y));
fillTriangles();
}
createSortedArray();
/*
The last step iterates through each point in the original points array
and check to see if either the first or second number matches the second
number of a point in our sortedPts array AND do the first or second number
match the next points in the sortedPtsd array. If so then we must have a
triangle.
Quick breakdown. If we have 3 lines (line 1, 2, 3) and I have a points on lines
2 & 3. I also have another point on lines 2 & 1. Then in order to have a triangle
the last point must be on lines 1 & 3.
That's all this is doing.
*/
function fillTriangles() {
//iterate through each array inside sortedPts array
for (let i = 0; i < sortedPts.length; i++) {
//iterate through all points inside each array of points inside the sortedPts array
for (let j = 0; j < sortedPts[i].length - 1; j++) {
//iterate over the original points and compare
for (let k = 0; k < points.length; k++) {
if (
(points[k].num.first == sortedPts[i][j].num.second ||
points[k].num.second == sortedPts[i][j].num.second) &&
(points[k].num.first == sortedPts[i][j + 1].num.second ||
points[k].num.second == sortedPts[i][j + 1].num.second)
) {
ctx.fillStyle = "rgba(200, 100, 0, 0.1)";
ctx.beginPath();
ctx.moveTo(sortedPts[i][j].x, sortedPts[i][j].y);
ctx.lineTo(sortedPts[i][j + 1].x, sortedPts[i][j + 1].y);
ctx.lineTo(points[k].x, points[k].y);
ctx.closePath();
ctx.fill();
}
}
}
}
}
<canvas id="canvas"></canvas>
我还认为有一种很好的方法可以利用交叉线的角度来做到这一点，并且正在努力以这种方式做到这一点。我希望我可以根据边数确定形状的类型，但我不认为这是一个快速的项目。
如果你只想绘制一个由交点组成的形状，你就在 beginShape()/endShape() 的正确轨道上。
目前看起来您正在放置 所有 px 中的点，py：交点以及定义所用线的点首先计算交点。
您可能希望将两者分开，例如，一对仅用于定义线的点的数组和另一对仅用于交点的 x,y 数组。您只需要遍历相交坐标以在 beginShape()/endShape() 之间放置 vertex(x, y) 调用。这是您的代码的修改版本来说明这个想法：
//Run in Processing.
//Press r to refresh.
//Top and bottom points are added to px and py when refreshed (filled in black).
//Intersection points are added to px and py when detected (filled in red).
int l = 4; //set number of lines
float[] r1 = new float[l];
float[] r2 = new float[l];
float[] px = {}; //array to save x positions of all possible points
float[] py = {}; //array to save y positions of all possible points
float[] ipx = {}; // array to save x for intersections only
float[] ipy = {}; // array to save y for intersections only
boolean added = false;
void setup(){
size(800, 800);
background(255);
refresh();
}
void draw(){
background(255);
stroke(0, 150, 255, 150);
strokeWeight(1);
for(int i=0; i < r1.length; i++){
for(int j=0; j < r1.length; j++){
if(i>j){
boolean hit = lineLine(r1[i], 0, r2[i], height, r1[j], 0, r2[j], height);
if (hit) stroke(255, 150, 0, 150);
else stroke(0, 150, 255, 150);
}
line(r1[i], 0, r2[i], height);
}
}
added = true;
// draw intersections
beginShape();
for(int i = 0 ; i < ipx.length; i++){
vertex(ipx[i], ipy[i]);
}
endShape();
//print(px.length);
//println(px.length, py.length);
}
//source: http://jeffreythompson.org/collision-detection/line-line.php
boolean lineLine(float x1, float y1, float x2, float y2, float x3, float y3, float x4, float y4) {
// calculate the distance to intersection point
float uA = ((x4-x3)*(y1-y3) - (y4-y3)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
float uB = ((x2-x1)*(y1-y3) - (y2-y1)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
// if uA and uB are between 0-1, lines are colliding
if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
// optionally, draw a circle where the lines meet
float intersectionX = x1 + (uA * (x2-x1));
float intersectionY = y1 + (uA * (y2-y1));
fill(255,0,0);
noStroke();
ellipse(intersectionX,intersectionY, 20,20);
if(added==false){
px = append(px, intersectionX);
py = append(py, intersectionY);
// store intersections
ipx = append(ipx, intersectionX);
ipy = append(ipy, intersectionY);
}
return true;
}
return false;
}
void refresh(){
added = false;
px = new float[0];
py = new float[0];
ipx = new float[0];
ipy = new float[0];
r1 = new float[l];
r2 = new float[l];
px = append(px, 0);
py = append(py, 0);
px = append(px, 0);
py = append(py, height);
px = append(px, width);
py = append(py, 0);
px = append(px, width);
py = append(py, height);
for(int i=0; i< r1.length; i++){
r1[i] = random(800);
}
for(int i=0; i< r2.length; i++){
r2[i] = random(800);
}
for(int i=0; i < r1.length; i++){
stroke(0);
line(r1[i], 0, r2[i], height);
px = append(px, r1[i]);
py = append(py, 0);
px = append(px, r2[i]);
py = append(py, height);
}
}
void keyReleased() {
if (key == 'r') refresh();
}
请记住，这个 simlpy 按照计算交点的顺序绘制点。在美好的一天，你会得到这样的东西：
不排除多边形顶点顺序错误（缠绕）的可能性：
你也可能得到凸多边形。
如果您只需要这些交点的外部 'shell'，您可能需要 convex hull algorithm
至少在视觉上分割形状的一个选项可能是将 beginShape(TRIANGLES); 与 endShape(CLOSE); 一起使用，其中 应该 遍历点并为每个坐标三元组绘制一个三角形，但是给定随机点和交叉点数，您可能最终会缺少一个或两个三角形（例如，6 个点 = 2 个三角形，7 个点 = 2 个三角形和 1 个点，没有缺失对）
我唯一的其他注意事项是关于语法：数组可以开始使用，但您可能需要研究 ArrayList and PVector。这将允许您使用具有 x、y 属性的 PVector 实例的单个动态数组。
更新
总体上代码可以简化。如果我们取出与线相交相关的代码，我们就可以摆脱这样的情况：
int l = 4; //set number of random lines
float[] r1 = new float[l]; // random x top
float[] r2 = new float[l]; // random x bottom
void setup() {
size(800, 800);
strokeWeight(3);
stroke(0, 150, 255, 150);
refresh();
}
void draw() {
background(255);
// random lines
for (int i=0; i < r1.length; i++) {
line(r1[i], 0, r2[i], height);
}
// borders
line(0, 0, width, 0);
line(width, 0, width - 1, height - 1);
line(0, height - 1, width - 1, height - 1);
line(0, 0, 0, height - 1);
}
void refresh() {
r1 = new float[l];
r2 = new float[l];
for (int i=0; i< r1.length; i++) {
r1[i] = random(800);
r2[i] = random(800);
}
}
void keyReleased() {
if (key == 'r') refresh();
}
如果我们要使用基本的 Line class 并利用 PVector 和 ArrayList 我们可以将上面的代码重写为：
int numRandomLines = 4;
ArrayList<PVector> points = new ArrayList<PVector>();
void setup() {
size(800, 800);
stroke(0, 150, 255, 150);
strokeWeight(3);
refresh();
}
void refresh(){
// remove previous points
points.clear();
//add borders
points.add(new PVector(0, 0)); points.add(new PVector(width, 0));
points.add(new PVector(width, 0));points.add(new PVector(width - 1, height - 1));
points.add(new PVector(0, height - 1));points.add(new PVector(width - 1, height - 1));
points.add(new PVector(0, 0)); points.add(new PVector(0, height - 1));
// add random lines
for (int i=0; i< numRandomLines; i++) {
points.add(new PVector(random(800), 0)); points.add(new PVector(random(800), height));
}
}
void draw(){
background(255);
beginShape(LINES);
for(PVector point : points) vertex(point.x, point.y);
endShape();
}
void keyReleased() {
if (key == 'r') refresh();
}
并将一对点 (PVector) 分组为 Line class:
int numRandomLines = 4;
ArrayList<Line> lines = new ArrayList<Line>();
void setup() {
size(800, 800);
stroke(0, 150, 255, 150);
strokeWeight(3);
refresh();
}
void refresh(){
// remove previous points
lines.clear();
//add borders
lines.add(new Line(new PVector(0, 0), new PVector(width, 0)));
lines.add(new Line(new PVector(width, 0), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, height - 1), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, 0), new PVector(0, height - 1)));
// add random lines
for (int i=0; i< numRandomLines; i++) {
lines.add(new Line(new PVector(random(800), 0), new PVector(random(800), height)));
}
}
void draw(){
background(255);
for(Line line : lines) line.draw();
}
void keyReleased() {
if (key == 'r') refresh();
}
class Line{
PVector start;
PVector end;
Line(PVector start, PVector end){
this.start = start;
this.end = end;
}
void draw(){
line(start.x, start.y, end.x, end.y);
}
}
在这个阶段，为了获得图表所描述的各个形状，我们可以作弊并使用像 OpenCV. This is if course overkill (as we'd get() a PImage copy of the drawing, convert that to an OpenCV image) then simply use findContours() 这样的计算机视觉库来获得每个 shape/contour。
回到原来的方法，线与线相交功能可以集成到Line class:
int numRandomLines = 4;
ArrayList<Line> lines = new ArrayList<Line>();
ArrayList<PVector> intersections = new ArrayList<PVector>();
void setup() {
size(800, 800);
strokeWeight(3);
refresh();
}
void refresh(){
// remove previous points
lines.clear();
intersections.clear();
//add borders
lines.add(new Line(new PVector(0, 0), new PVector(width, 0)));
lines.add(new Line(new PVector(width, 0), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, height - 1), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, 0), new PVector(0, height - 1)));
// add random lines
for (int i=0; i< numRandomLines; i++) {
lines.add(new Line(new PVector(random(800), 0), new PVector(random(800), height)));
}
// compute intersections
int numLines = lines.size();
// when looping only check if lineA intersects lineB but not also if lineB intersects lineA (redundant)
for (int i = 0; i < numLines - 1; i++){
Line lineA = lines.get(i);
for (int j = i + 1; j < numLines; j++){
Line lineB = lines.get(j);
// check intersection
PVector intersection = lineA.intersect(lineB);
// if there is one, append the intersection point to the list
if(intersection != null){
intersections.add(intersection);
}
}
}
}
void draw(){
background(255);
stroke(0, 150, 255, 150);
// draw lines
for(Line line : lines) line.draw();
stroke(255, 0, 0, 150);
// draw intersections
for(PVector intersection : intersections) ellipse(intersection.x, intersection.y, 9, 9);
}
void keyReleased() {
if (key == 'r') refresh();
}
class Line{
PVector start;
PVector end;
Line(PVector start, PVector end){
this.start = start;
this.end = end;
}
void draw(){
line(start.x, start.y, end.x, end.y);
}
//source: http://jeffreythompson.org/collision-detection/line-line.php
//boolean lineLine(float this.start.x, float this.start.y, float this.end.x, float this.end.y,
//float other.start.x, float other.start.y, float other.end.x, float other.end.y) {
PVector intersect(Line other) {
// calculate the distance to intersection point
float uA = ((other.end.x-other.start.x)*(this.start.y-other.start.y) - (other.end.y-other.start.y)*(this.start.x-other.start.x)) / ((other.end.y-other.start.y)*(this.end.x-this.start.x) - (other.end.x-other.start.x)*(this.end.y-this.start.y));
float uB = ((this.end.x-this.start.x)*(this.start.y-other.start.y) - (this.end.y-this.start.y)*(this.start.x-other.start.x)) / ((other.end.y-other.start.y)*(this.end.x-this.start.x) - (other.end.x-other.start.x)*(this.end.y-this.start.y));
// if uA and uB are between 0-1, lines are colliding
if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
// optionally, draw a circle where the lines meet
float intersectionX = this.start.x + (uA * (this.end.x-this.start.x));
float intersectionY = this.start.y + (uA * (this.end.y-this.start.y));
return new PVector(intersectionX, intersectionY);
}
return null;
}
}
下一步将是一个更复杂的算法，根据 x、y 位置（例如从上到下、从左到右）对点进行排序，遍历通过距离和角度比较第一个点与其余点的点并尝试计算距离和角度变化最小的连续点是否连接。
在网上快速浏览一下，我可以看到这样的算法，例如：
- Polygon Detection from a Set of Lines
- Bentley Ottman 算法（上述论文中提到的算法之一）实际上是在 CGAL. (While there are CGAL Java bindings 中实现的，构建这些算法并为处理连接或制作包装器并非易事。
我想检测单独的形状，因为随机生成的线将 canvas 分开。我在 x 和 y 位置（相同顺序）的单独数组中保存了线交点，但不知道如何连接 完成多个形状的点。
- 是否有任何方法可以检测附近的点以闭合可能的最小形状，无论它是三角形、矩形还是多边形（例如，通过使用 beginShape 和 endShape）？
- 如果1）太复杂，有什么方法可以select从一个数组中随机抽取3个或更多点吗？
这是一个示例图片，其中有 4 条线将 canvas 分开，它们的交点用红色标记。我还保存了每个随机生成的线的顶部和底部点（标记为黑色），加上 canvas 的四个角分别在相同的数组中用于 x 和 y 位置（px，py）。
我能够得到所有的交点，但在将它们连接成单独的形状时遇到了问题。这是我正在处理的处理代码：
//Run in Processing.
//Press r to refresh.
//Top and bottom points are added to px and py when refreshed (filled in black).
//Intersection points are added to px and py when detected (filled in red).
int l = 4; //set number of lines
float[] r1 = new float[l];
float[] r2 = new float[l];
float[] px = {}; //array to save x positions of all possible points
float[] py = {}; //array to save y positions of all possible points
boolean added = false;
void setup(){
size(800, 800);
background(255);
refresh();
}
void draw(){
background(255);
stroke(0, 150, 255, 150);
strokeWeight(1);
for(int i=0; i < r1.length; i++){
for(int j=0; j < r1.length; j++){
if(i>j){
boolean hit = lineLine(r1[i], 0, r2[i], height, r1[j], 0, r2[j], height);
if (hit) stroke(255, 150, 0, 150);
else stroke(0, 150, 255, 150);
}
line(r1[i], 0, r2[i], height);
}
}
added = true;
print(px.length);
}
//source: http://jeffreythompson.org/collision-detection/line-line.php
boolean lineLine(float x1, float y1, float x2, float y2, float x3, float y3, float x4, float y4) {
// calculate the distance to intersection point
float uA = ((x4-x3)*(y1-y3) - (y4-y3)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
float uB = ((x2-x1)*(y1-y3) - (y2-y1)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
// if uA and uB are between 0-1, lines are colliding
if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
// optionally, draw a circle where the lines meet
float intersectionX = x1 + (uA * (x2-x1));
float intersectionY = y1 + (uA * (y2-y1));
fill(255,0,0);
noStroke();
ellipse(intersectionX,intersectionY, 20,20);
if(added==false){
px = append(px, intersectionX);
py = append(py, intersectionY);
}
return true;
}
return false;
}
void refresh(){
added = false;
px = new float[0];
py = new float[0];
r1 = new float[l];
r2 = new float[l];
px = append(px, 0);
py = append(py, 0);
px = append(px, 0);
py = append(py, height);
px = append(px, width);
py = append(py, 0);
px = append(px, width);
py = append(py, height);
for(int i=0; i< r1.length; i++){
r1[i] = random(800);
}
for(int i=0; i< r2.length; i++){
r2[i] = random(800);
}
for(int i=0; i < r1.length; i++){
stroke(0);
line(r1[i], 0, r2[i], height);
px = append(px, r1[i]);
py = append(py, 0);
px = append(px, r2[i]);
py = append(py, height);
}
}
void keyReleased() {
if (key == 'r') refresh();
}
我不清楚你的目标。您可以以任意顺序连接任意一组点并将其称为形状。你的标准是什么？
如果你想找到连接给定子集所有点的最短路径，我建议寻找旅行商问题。
我可以看到您的代码不是 javascript，但由于您没有指定语言，我假设您只需要一个方法并可以转换为您的语言。
我处理这个问题的方法是为每一行分配一个行号。如果我可以在一条线上识别出 2 个相邻点，那么我将通过检查它们不共享的线的交叉处是否有一个点来知道第三个点是否存在。
示例： 有 3 行（第 1、2、3 行）
我在 3 号线和 1 号线之间有一个交点，现在我沿着 3 号线走到相邻的点。我找到了一个，它的交点是 3 和 2。我唯一能得到三角形的方法是 1 和 2 线穿过某处。所以我们可以通过编程方式检查。
请记住，我实际上从未为此使用过角度。我确实在函数中计算了它们，但决定不使用它们，因为我使用了上面解释的方法。我使用 0.1 的 alpha 值给三角形上色，这样您就可以看到重叠的地方。
这只是检查三角形
let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");
canvas.width = 400;
canvas.height = 400;
let lines = []; //holds each line
let points = []; //all intersection point are pushed here [{x: num, y: num}, {x: num, y: num},...]
let sortedPts = []; //all points sorted bu first number are pushed here in 2d array.
let lineNum = 15;
class Lines {
constructor(num) {
this.x = Math.round(Math.random() * canvas.width);
this.x2 = Math.round(Math.random() * canvas.width);
this.pt1 = {
x: this.x,
y: 0
};
this.pt2 = {
x: this.x2,
y: canvas.height
};
this.num = num;
this.rads = Math.atan2(this.pt2.y - this.pt1.y, this.pt2.x - this.pt1.x);
this.angle = this.rads * (180 / Math.PI);
}
draw() {
ctx.beginPath();
ctx.moveTo(this.pt1.x, this.pt1.y);
ctx.lineTo(this.pt2.x, this.pt2.y);
ctx.stroke();
}
}
//creates the lines. I also use this function to prepare the 2d array by pushing an empty array for each line into sortedPts.
function createLines() {
for (let i = 0; i < lineNum; i++) {
lines.push(new Lines(i + 1));
sortedPts.push([])
}
}
createLines();
//Visually draws lines on screen
function drawLines() {
for (let i = 0; i < lines.length; i++) {
lines[i].draw();
}
}
drawLines();
//intersecting formula
function lineSegmentsIntersect(line1, line2) {
let a_dx = line1.pt2.x - line1.pt1.x;
let a_dy = line1.pt2.y - line1.pt1.y;
let b_dx = line2.pt2.x - line2.pt1.x;
let b_dy = line2.pt2.y - line2.pt1.y;
let s =
(-a_dy * (line1.pt1.x - line2.pt1.x) + a_dx * (line1.pt1.y - line2.pt1.y)) /
(-b_dx * a_dy + a_dx * b_dy);
let t =
(+b_dx * (line1.pt1.y - line2.pt1.y) - b_dy * (line1.pt1.x - line2.pt1.x)) /
(-b_dx * a_dy + a_dx * b_dy);
if (s >= 0 && s <= 1 && t >= 0 && t <= 1) {
//this is where we create our array but we also add the line number of where each point intersects. I also add the angle but have not used it throughout the rest of this...yet.
points.push({
x: Math.round(line1.pt1.x + t * (line1.pt2.x - line1.pt1.x)),
y: Math.round(line1.pt1.y + t * (line1.pt2.y - line1.pt1.y)),
num: {
first: line1.num,
second: line2.num
},
angle: {
a1: line1.angle,
a2: line2.angle
}
});
}
}
//just checks each line against the others by passing to lineSegmentsIntersect() function
function callIntersect() {
for (let i = 0; i < lines.length; i++) {
for (let j = i + 1; j < lines.length; j++) {
lineSegmentsIntersect(lines[i], lines[j]);
}
}
}
callIntersect();
function drawPoints() {
//just draws the black points for reference
for (let i = 0; i < points.length; i++) {
ctx.beginPath();
ctx.arc(points[i].x, points[i].y, 2, 0, Math.PI * 2);
ctx.fill();
}
}
drawPoints();
function createSortedArray() {
//Now we take the points array and sort the points by the first number to make using i and j below possible
points.sort((a, b) => a.num.first - b.num.first)
//We push each group of points into an array inside sortedPts creating the 2d array
for (let i = 0; i < lineNum; i++) {
for (let j = 0; j < points.length; j++) {
if (points[j].num.first == (i + 1)) {
sortedPts[i].push(points[j]);
}
}
}
//now sort the 2d arrays by y value. This allows or next check to go in order from point to point per line.
sortedPts.forEach(arr => arr.sort((a, b) => a.y - b.y));
fillTriangles();
}
createSortedArray();
/*
The last step iterates through each point in the original points array
and check to see if either the first or second number matches the second
number of a point in our sortedPts array AND do the first or second number
match the next points in the sortedPtsd array. If so then we must have a
triangle.
Quick breakdown. If we have 3 lines (line 1, 2, 3) and I have a points on lines
2 & 3. I also have another point on lines 2 & 1. Then in order to have a triangle
the last point must be on lines 1 & 3.
That's all this is doing.
*/
function fillTriangles() {
//iterate through each array inside sortedPts array
for (let i = 0; i < sortedPts.length; i++) {
//iterate through all points inside each array of points inside the sortedPts array
for (let j = 0; j < sortedPts[i].length - 1; j++) {
//iterate over the original points and compare
for (let k = 0; k < points.length; k++) {
if (
(points[k].num.first == sortedPts[i][j].num.second ||
points[k].num.second == sortedPts[i][j].num.second) &&
(points[k].num.first == sortedPts[i][j + 1].num.second ||
points[k].num.second == sortedPts[i][j + 1].num.second)
) {
ctx.fillStyle = "rgba(200, 100, 0, 0.1)";
ctx.beginPath();
ctx.moveTo(sortedPts[i][j].x, sortedPts[i][j].y);
ctx.lineTo(sortedPts[i][j + 1].x, sortedPts[i][j + 1].y);
ctx.lineTo(points[k].x, points[k].y);
ctx.closePath();
ctx.fill();
}
}
}
}
}
<canvas id="canvas"></canvas>
我还认为有一种很好的方法可以利用交叉线的角度来做到这一点，并且正在努力以这种方式做到这一点。我希望我可以根据边数确定形状的类型，但我不认为这是一个快速的项目。
如果你只想绘制一个由交点组成的形状，你就在 beginShape()/endShape() 的正确轨道上。
目前看起来您正在放置 所有 px 中的点，py：交点以及定义所用线的点首先计算交点。
您可能希望将两者分开，例如，一对仅用于定义线的点的数组和另一对仅用于交点的 x,y 数组。您只需要遍历相交坐标以在 beginShape()/endShape() 之间放置 vertex(x, y) 调用。这是您的代码的修改版本来说明这个想法：
//Run in Processing.
//Press r to refresh.
//Top and bottom points are added to px and py when refreshed (filled in black).
//Intersection points are added to px and py when detected (filled in red).
int l = 4; //set number of lines
float[] r1 = new float[l];
float[] r2 = new float[l];
float[] px = {}; //array to save x positions of all possible points
float[] py = {}; //array to save y positions of all possible points
float[] ipx = {}; // array to save x for intersections only
float[] ipy = {}; // array to save y for intersections only
boolean added = false;
void setup(){
size(800, 800);
background(255);
refresh();
}
void draw(){
background(255);
stroke(0, 150, 255, 150);
strokeWeight(1);
for(int i=0; i < r1.length; i++){
for(int j=0; j < r1.length; j++){
if(i>j){
boolean hit = lineLine(r1[i], 0, r2[i], height, r1[j], 0, r2[j], height);
if (hit) stroke(255, 150, 0, 150);
else stroke(0, 150, 255, 150);
}
line(r1[i], 0, r2[i], height);
}
}
added = true;
// draw intersections
beginShape();
for(int i = 0 ; i < ipx.length; i++){
vertex(ipx[i], ipy[i]);
}
endShape();
//print(px.length);
//println(px.length, py.length);
}
//source: http://jeffreythompson.org/collision-detection/line-line.php
boolean lineLine(float x1, float y1, float x2, float y2, float x3, float y3, float x4, float y4) {
// calculate the distance to intersection point
float uA = ((x4-x3)*(y1-y3) - (y4-y3)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
float uB = ((x2-x1)*(y1-y3) - (y2-y1)*(x1-x3)) / ((y4-y3)*(x2-x1) - (x4-x3)*(y2-y1));
// if uA and uB are between 0-1, lines are colliding
if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
// optionally, draw a circle where the lines meet
float intersectionX = x1 + (uA * (x2-x1));
float intersectionY = y1 + (uA * (y2-y1));
fill(255,0,0);
noStroke();
ellipse(intersectionX,intersectionY, 20,20);
if(added==false){
px = append(px, intersectionX);
py = append(py, intersectionY);
// store intersections
ipx = append(ipx, intersectionX);
ipy = append(ipy, intersectionY);
}
return true;
}
return false;
}
void refresh(){
added = false;
px = new float[0];
py = new float[0];
ipx = new float[0];
ipy = new float[0];
r1 = new float[l];
r2 = new float[l];
px = append(px, 0);
py = append(py, 0);
px = append(px, 0);
py = append(py, height);
px = append(px, width);
py = append(py, 0);
px = append(px, width);
py = append(py, height);
for(int i=0; i< r1.length; i++){
r1[i] = random(800);
}
for(int i=0; i< r2.length; i++){
r2[i] = random(800);
}
for(int i=0; i < r1.length; i++){
stroke(0);
line(r1[i], 0, r2[i], height);
px = append(px, r1[i]);
py = append(py, 0);
px = append(px, r2[i]);
py = append(py, height);
}
}
void keyReleased() {
if (key == 'r') refresh();
}
请记住，这个 simlpy 按照计算交点的顺序绘制点。在美好的一天，你会得到这样的东西：
不排除多边形顶点顺序错误（缠绕）的可能性：
你也可能得到凸多边形。
如果您只需要这些交点的外部 'shell'，您可能需要 convex hull algorithm
至少在视觉上分割形状的一个选项可能是将 beginShape(TRIANGLES); 与 endShape(CLOSE); 一起使用，其中 应该 遍历点并为每个坐标三元组绘制一个三角形，但是给定随机点和交叉点数，您可能最终会缺少一个或两个三角形（例如，6 个点 = 2 个三角形，7 个点 = 2 个三角形和 1 个点，没有缺失对）
我唯一的其他注意事项是关于语法：数组可以开始使用，但您可能需要研究 ArrayList and PVector。这将允许您使用具有 x、y 属性的 PVector 实例的单个动态数组。
更新
总体上代码可以简化。如果我们取出与线相交相关的代码，我们就可以摆脱这样的情况：
int l = 4; //set number of random lines
float[] r1 = new float[l]; // random x top
float[] r2 = new float[l]; // random x bottom
void setup() {
size(800, 800);
strokeWeight(3);
stroke(0, 150, 255, 150);
refresh();
}
void draw() {
background(255);
// random lines
for (int i=0; i < r1.length; i++) {
line(r1[i], 0, r2[i], height);
}
// borders
line(0, 0, width, 0);
line(width, 0, width - 1, height - 1);
line(0, height - 1, width - 1, height - 1);
line(0, 0, 0, height - 1);
}
void refresh() {
r1 = new float[l];
r2 = new float[l];
for (int i=0; i< r1.length; i++) {
r1[i] = random(800);
r2[i] = random(800);
}
}
void keyReleased() {
if (key == 'r') refresh();
}
如果我们要使用基本的 Line class 并利用 PVector 和 ArrayList 我们可以将上面的代码重写为：
int numRandomLines = 4;
ArrayList<PVector> points = new ArrayList<PVector>();
void setup() {
size(800, 800);
stroke(0, 150, 255, 150);
strokeWeight(3);
refresh();
}
void refresh(){
// remove previous points
points.clear();
//add borders
points.add(new PVector(0, 0)); points.add(new PVector(width, 0));
points.add(new PVector(width, 0));points.add(new PVector(width - 1, height - 1));
points.add(new PVector(0, height - 1));points.add(new PVector(width - 1, height - 1));
points.add(new PVector(0, 0)); points.add(new PVector(0, height - 1));
// add random lines
for (int i=0; i< numRandomLines; i++) {
points.add(new PVector(random(800), 0)); points.add(new PVector(random(800), height));
}
}
void draw(){
background(255);
beginShape(LINES);
for(PVector point : points) vertex(point.x, point.y);
endShape();
}
void keyReleased() {
if (key == 'r') refresh();
}
并将一对点 (PVector) 分组为 Line class:
int numRandomLines = 4;
ArrayList<Line> lines = new ArrayList<Line>();
void setup() {
size(800, 800);
stroke(0, 150, 255, 150);
strokeWeight(3);
refresh();
}
void refresh(){
// remove previous points
lines.clear();
//add borders
lines.add(new Line(new PVector(0, 0), new PVector(width, 0)));
lines.add(new Line(new PVector(width, 0), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, height - 1), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, 0), new PVector(0, height - 1)));
// add random lines
for (int i=0; i< numRandomLines; i++) {
lines.add(new Line(new PVector(random(800), 0), new PVector(random(800), height)));
}
}
void draw(){
background(255);
for(Line line : lines) line.draw();
}
void keyReleased() {
if (key == 'r') refresh();
}
class Line{
PVector start;
PVector end;
Line(PVector start, PVector end){
this.start = start;
this.end = end;
}
void draw(){
line(start.x, start.y, end.x, end.y);
}
}
在这个阶段，为了获得图表所描述的各个形状，我们可以作弊并使用像 OpenCV. This is if course overkill (as we'd get() a PImage copy of the drawing, convert that to an OpenCV image) then simply use findContours() 这样的计算机视觉库来获得每个 shape/contour。
回到原来的方法，线与线相交功能可以集成到Line class:
int numRandomLines = 4;
ArrayList<Line> lines = new ArrayList<Line>();
ArrayList<PVector> intersections = new ArrayList<PVector>();
void setup() {
size(800, 800);
strokeWeight(3);
refresh();
}
void refresh(){
// remove previous points
lines.clear();
intersections.clear();
//add borders
lines.add(new Line(new PVector(0, 0), new PVector(width, 0)));
lines.add(new Line(new PVector(width, 0), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, height - 1), new PVector(width - 1, height - 1)));
lines.add(new Line(new PVector(0, 0), new PVector(0, height - 1)));
// add random lines
for (int i=0; i< numRandomLines; i++) {
lines.add(new Line(new PVector(random(800), 0), new PVector(random(800), height)));
}
// compute intersections
int numLines = lines.size();
// when looping only check if lineA intersects lineB but not also if lineB intersects lineA (redundant)
for (int i = 0; i < numLines - 1; i++){
Line lineA = lines.get(i);
for (int j = i + 1; j < numLines; j++){
Line lineB = lines.get(j);
// check intersection
PVector intersection = lineA.intersect(lineB);
// if there is one, append the intersection point to the list
if(intersection != null){
intersections.add(intersection);
}
}
}
}
void draw(){
background(255);
stroke(0, 150, 255, 150);
// draw lines
for(Line line : lines) line.draw();
stroke(255, 0, 0, 150);
// draw intersections
for(PVector intersection : intersections) ellipse(intersection.x, intersection.y, 9, 9);
}
void keyReleased() {
if (key == 'r') refresh();
}
class Line{
PVector start;
PVector end;
Line(PVector start, PVector end){
this.start = start;
this.end = end;
}
void draw(){
line(start.x, start.y, end.x, end.y);
}
//source: http://jeffreythompson.org/collision-detection/line-line.php
//boolean lineLine(float this.start.x, float this.start.y, float this.end.x, float this.end.y,
//float other.start.x, float other.start.y, float other.end.x, float other.end.y) {
PVector intersect(Line other) {
// calculate the distance to intersection point
float uA = ((other.end.x-other.start.x)*(this.start.y-other.start.y) - (other.end.y-other.start.y)*(this.start.x-other.start.x)) / ((other.end.y-other.start.y)*(this.end.x-this.start.x) - (other.end.x-other.start.x)*(this.end.y-this.start.y));
float uB = ((this.end.x-this.start.x)*(this.start.y-other.start.y) - (this.end.y-this.start.y)*(this.start.x-other.start.x)) / ((other.end.y-other.start.y)*(this.end.x-this.start.x) - (other.end.x-other.start.x)*(this.end.y-this.start.y));
// if uA and uB are between 0-1, lines are colliding
if (uA >= 0 && uA <= 1 && uB >= 0 && uB <= 1) {
// optionally, draw a circle where the lines meet
float intersectionX = this.start.x + (uA * (this.end.x-this.start.x));
float intersectionY = this.start.y + (uA * (this.end.y-this.start.y));
return new PVector(intersectionX, intersectionY);
}
return null;
}
}
下一步将是一个更复杂的算法，根据 x、y 位置（例如从上到下、从左到右）对点进行排序，遍历通过距离和角度比较第一个点与其余点的点并尝试计算距离和角度变化最小的连续点是否连接。
在网上快速浏览一下，我可以看到这样的算法，例如：
- Polygon Detection from a Set of Lines
- Bentley Ottman 算法（上述论文中提到的算法之一）实际上是在 CGAL. (While there are CGAL Java bindings 中实现的，构建这些算法并为处理连接或制作包装器并非易事。