Calculate circle coordinates in a sequential fashi

问题:

I need to have all point coordinates for a given circle one after another, so I can make an object go in circles by hopping from one point to the next. I tried the Midpoint circle algorithm, but the version I found is meant to draw and the coordinates are not sequential. They are produced simultaneously for 8 quadrants and in opposing directions on top of that. If at least they were in the same direction, I could make a separate array for every quadrant and append them to one another at the end. This is the JavaScript adapted code I have now:

  function calcCircle(centerCoordinates, radius) {
    var coordinatesArray = new Array();
    // Translate coordinates
    var x0 = centerCoordinates.left;
    var y0 = centerCoordinates.top;
    // Define variables
    var f = 1 - radius;
    var ddFx = 1;
    var ddFy = -radius << 1;
    var x = 0;
    var y = radius;
    coordinatesArray.push(new Coordinates(x0, y0 + radius));
    coordinatesArray.push(new Coordinates(x0, y0 - radius));
    coordinatesArray.push(new Coordinates(x0 + radius, y0));
    coordinatesArray.push(new Coordinates(x0 - radius, y0));
    // Main loop
    while (x < y) {
      if (f >= 0) {
        y--;
        ddFy += 2;
        f += ddFy;
      }
      x++;
      ddFx += 2;
      f += ddFx;
      coordinatesArray.push(new Coordinates(x0 + x, y0 + y));
      coordinatesArray.push(new Coordinates(x0 - x, y0 + y));
      coordinatesArray.push(new Coordinates(x0 + x, y0 - y));
      coordinatesArray.push(new Coordinates(x0 - x, y0 - y));
      coordinatesArray.push(new Coordinates(x0 + y, y0 + x));
      coordinatesArray.push(new Coordinates(x0 - y, y0 + x));
      coordinatesArray.push(new Coordinates(x0 + y, y0 - x));
      coordinatesArray.push(new Coordinates(x0 - y, y0 - x));
    }
    // Return the result
    return coordinatesArray;
  }

I prefer some fast algorithm without trigonometry, but any help is appreciated!

EDIT

This is the final solution. Thanks everybody!

  function calcCircle(centerCoordinates, radius) {
    var coordinatesArray = new Array();
    var octantArrays =
      {oct1: new Array(), oct2: new Array(), oct3: new Array(), oct4: new Array(),
       oct5: new Array(), oct6: new Array(), oct7: new Array(), oct8: new Array()};
    // Translate coordinates
    var xp = centerCoordinates.left;
    var yp = centerCoordinates.top;
    // Define add coordinates to array
    var setCrd =
      function (targetArray, xC, yC) {
        targetArray.push(new Coordinates(yC, xC));
      };
    // Define variables
    var xoff = 0;
    var yoff = radius;
    var balance = -radius;
    // Main loop
    while (xoff <= yoff) {
      // Quadrant 7 - Reverse
      setCrd(octantArrays.oct7, xp + xoff, yp + yoff);
      // Quadrant 6 - Straight
      setCrd(octantArrays.oct6, xp - xoff, yp + yoff);
      // Quadrant 3 - Reverse
      setCrd(octantArrays.oct3, xp - xoff, yp - yoff);
      // Quadrant 2 - Straight
      setCrd(octantArrays.oct2, xp + xoff, yp - yoff);
      // Avoid duplicates
      if (xoff != yoff) {
        // Quadrant 8 - Straight
        setCrd(octantArrays.oct8, xp + yoff, yp + xoff);
        // Quadrant 5 - Reverse
        setCrd(octantArrays.oct5, xp - yoff, yp + xoff);
        // Quadrant 4 - Straight
        setCrd(octantArrays.oct4, xp - yoff, yp - xoff);
        // Quadrant 1 - Reverse
        setCrd(octantArrays.oct1, xp + yoff, yp - xoff);
      }
      // Some weird stuff
      balance += xoff++ + xoff;
      if (balance >= 0) {
        balance -= --yoff + yoff;
      }
    }
    // Reverse counter clockwise octant arrays
    octantArrays.oct7.reverse();
    octantArrays.oct3.reverse();
    octantArrays.oct5.reverse();
    octantArrays.oct1.reverse();
    // Remove counter clockwise octant arrays last element (avoid duplicates)
    octantArrays.oct7.pop();
    octantArrays.oct3.pop();
    octantArrays.oct5.pop();
    octantArrays.oct1.pop();
    // Append all arrays together
    coordinatesArray =
      octantArrays.oct4.concat(octantArrays.oct3).concat(octantArrays.oct2).concat(octantArrays.oct1).
        concat(octantArrays.oct8).concat(octantArrays.oct7).concat(octantArrays.oct6).concat(octantArrays.oct5);
    // Return the result
    return coordinatesArray;
  }

回答1:

Use can try the following approach: use the algorithm you gave but push your coordinates to eight different coordinateArrays. Afterwards you have to reverse half of them (those with (x0+x,y0-y), (x0-x,y0+y), (x0+y,y0+x), (x0-y,y0-x)) and afterwards append all arrays in the correct order. Take care that you add the first four points to the correct arrays.

回答2:

As far as I know, you cannot do it without trigonometry, but it works pretty fast for me. Sorry I'm not familiar with Java, so I write the code in VB:

Dim PointList As New List(Of PointF)
For angle = 0 To Math.PI * 2 Step 0.01
    'the smaller the step, the more points you get
    PointList.Add(New PointF(Math.Cos(angle) * r + x0, Math.Sin(angle) * r + y0))
Next

x0 and y0 are the center coordinates of the circle, r is the radius.

Hope I answered your question.

回答3:

Here is a javascript implementation based on Dave's answer. A bit over-engineered, I wanted to avoid calling sin and cos more than necessary. Ironically making use of Dave's first answer without the radius :)

function calculateCircle(x,y,radius) {

  var basicPoints = getBasicCircle();
  var i = basicPoints.length;
  var points = []; // don't change basicPoints: that would spoil the cache.
  while (i--) {
    points[i] = {
      x: x + (radius * basicPoints[i].x),
      y: y + (radius * basicPoints[i].y)
    };
  }
  return points;
}

function getBasicCircle() {
  if (!arguments.callee.points) {
    var points = arguments.callee.points = [];
    var end = Math.PI * 2;
    for (var angle=0; angle < end; angle += 0.1) {
      points.push({x: Math.sin(angle), 
                   y: Math.cos(angle)
                  });
    }
  }
  return arguments.callee.points
}