如何绘制具有n个边的任意不规则多边形？

Question

我正在尝试找到一种算法，用于绘制简单的（不允许线相交）不规则多边形。

边数应由用户n>3定义。

这是一个初始代码，它仅绘制一个复杂的多边形（线相交）：

var ctx = document.getElementById('drawpolygon').getContext('2d');

var sides = 10;

ctx.fillStyle = '#f00';
ctx.beginPath();
ctx.moveTo(0, 0);

for(var i=0;i<sides;i++)
{
    var x = getRandomInt(0, 100);
    var y = getRandomInt(0, 100);
    ctx.lineTo(x, y);
}
ctx.closePath();
ctx.fill();

// https://stackoverflow.com/a/1527820/1066234
function getRandomInt(min, max) {
    min = Math.ceil(min);
    max = Math.floor(max);
    return Math.floor(Math.random() * (max - min + 1)) + min;
}

JSFiddle：https://jsfiddle.net/kai_noack/op2La1jy/6/

我不知道如何确定连接线的下一个点，这样它就不会切断任何其他线。

此外，最后一点必须关闭多边形。

这是一个结果多边形的外观示例：

编辑：我今天认为，一种可行的算法是将多边形点规则排列（例如，以矩形的形式），然后在xy方向上将它们重新定位为随机量，同时检查所生成的线是否被切割。

Answer 1

我将此solution移植到Javascript 1到1。代码看起来不是最优的，但是会产生随机的凸（但仍然是不规则的）多边形。

//shuffle array in place
function shuffle(arr) {
    for (let i = arr.length - 1; i > 0; i--) {
        const j = Math.floor(Math.random() * (i + 1));
        [arr[i], arr[j]] = [arr[j], arr[i]];
    }
    return arr;
}

/** Based on Sander Verdonschot java implementation **/

class Point {
  constructor(x, y) {
    this.x = x;
    this.y = y
  }
}

function generateRandomNumbersArray(len) {
  const result = new Array(len);
  for (let i = 0; i < len; ++i) {
    result[i] = Math.random();
  }
  return result;
}

function generateRandomConvexPolygon(vertexNumber) {
  const xPool = generateRandomNumbersArray(vertexNumber);
  const yPool = generateRandomNumbersArray(vertexNumber);

//   debugger;
  xPool.sort();
  yPool.sort();

  const minX = xPool[0];
  const maxX = xPool[xPool.length - 1];
  const minY = yPool[0];
  const maxY = yPool[yPool.length - 1];

  const xVec = []
  const yVec = [];

  let lastTop = minX;
  let lastBot = minX;

  xPool.forEach(x => {
    if (Math.random() >= 0.5) {
      xVec.push(x - lastTop);
      lastTop = x;
    } else {
      xVec.push(lastBot - x);
      lastBot = x;
    }
  });

  xVec.push(maxX - lastTop);
  xVec.push(lastBot - maxX);

  let lastLeft = minY;
  let lastRight = minY;

  yPool.forEach(y => {
    if (Math.random() >= 0.5) {
      yVec.push(y - lastLeft);
      lastLeft = y;
    } else {
      yVec.push(lastRight - y);
      lastRight = y;
    }
  });

  yVec.push(maxY - lastLeft);
  yVec.push(lastRight - maxY);

  shuffle(yVec);
  
  vectors = [];
  for (let i = 0; i < vertexNumber; ++i) {
    vectors.push(new Point(xVec[i], yVec[i]));
  }
  
  vectors.sort((v1, v2) => {
    if (Math.atan2(v1.y, v1.x) > Math.atan2(v2.y, v2.x)) {
      return 1;
    } else {
      return -1;
    }
  });
  
  let x = 0, y = 0;
  let minPolygonX = 0;
  let minPolygonY = 0;
  let points = [];
  
  for (let i = 0; i < vertexNumber; ++i) {
    points.push(new Point(x, y));
    x += vectors[i].x;
    y += vectors[i].y;
    
    minPolygonX = Math.min(minPolygonX, x);
    minPolygonY = Math.min(minPolygonY, y);
  }
  
          // Move the polygon to the original min and max coordinates
  let xShift = minX - minPolygonX;
  let yShift = minY - minPolygonY;

  for (let i = 0; i < vertexNumber; i++) {
    const p = points[i];
    points[i] = new Point(p.x + xShift, p.y + yShift);
  }
  
  return points;
}


function draw() {
  const vertices = 10;
  const _points = generateRandomConvexPolygon(vertices);
  
  //apply scale
  const points = _points.map(p => new Point(p.x * 300, p.y * 300));

  const ctx = document.getElementById('drawpolygon').getContext('2d');


  ctx.fillStyle = '#f00';
  ctx.beginPath();
  ctx.moveTo(points[0].x, points[0].y);

  for(let i = 1; i < vertices ; ++i)
  {
    let x = points[i].x;
    let y = points[i].y;
    ctx.lineTo(x, y);
  }
  ctx.closePath();
  ctx.fill();
}

draw();

<canvas id="drawpolygon"></canvas>

Answer 2

您可以生成随机点，然后将其与大概的旅行推销员游览联系起来。无法通过2-opt的移动来改善的游览不会有任何交叉。

Answer 3

如果不需要是随机的，这是一个快速的不规则n点多边形：

Points are:
(0,0)
((i mod 2)+1, i) for 0 <= i <= n-2

Lines are between (0,0) and the two extreme points from the second row, as well as points generated by adjacent values of i.