实心圆的中点圆算法

Question

Midpoint circle algorithm 可以用来光栅化圆的边框。但是，我希望填充圆圈，而不是多次绘制像素（这非常重要）。

这个答案提供了算法的修改，产生一个实心圆，但一些像素被访问了几次： fast algorithm for drawing filled circles?

问：如何在不多次绘制像素的情况下栅格化一个圆？请注意，RAM 非常有限！

更新：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace CircleTest
{
    class Program
    {
        static void Main(string[] args)
        {
            byte[,] buffer = new byte[50, 50];
            circle(buffer, 25, 25, 20);

            for (int y = 0; y < 50; ++y)
            {
                for (int x = 0; x < 50; ++x)
                    Console.Write(buffer[y, x].ToString());

                Console.WriteLine();
            }
        }

        // 'cx' and 'cy' denote the offset of the circle center from the origin.
        static void circle(byte[,] buffer, int cx, int cy, int radius)
        {
            int error = -radius;
            int x = radius;
            int y = 0;

            // The following while loop may altered to 'while (x > y)' for a
            // performance benefit, as long as a call to 'plot4points' follows
            // the body of the loop. This allows for the elimination of the
            // '(x != y)' test in 'plot8points', providing a further benefit.
            //
            // For the sake of clarity, this is not shown here.
            while (x >= y)
            {
                plot8points(buffer, cx, cy, x, y);

                error += y;
                ++y;
                error += y;

                // The following test may be implemented in assembly language in
                // most machines by testing the carry flag after adding 'y' to
                // the value of 'error' in the previous step, since 'error'
                // nominally has a negative value.
                if (error >= 0)
                {
                    error -= x;
                    --x;
                    error -= x;
                }
            }
        }

        static void plot8points(byte[,] buffer, int cx, int cy, int x, int y)
        {
            plot4points(buffer, cx, cy, x, y);
            if (x != y) plot4points(buffer, cx, cy, y, x);
        }

        // The '(x != 0 && y != 0)' test in the last line of this function
        // may be omitted for a performance benefit if the radius of the
        // circle is known to be non-zero.
        static void plot4points(byte[,] buffer, int cx, int cy, int x, int y)
        {
#if false // Outlined circle are indeed plotted correctly!
            setPixel(buffer, cx + x, cy + y);
            if (x != 0) setPixel(buffer, cx - x, cy + y);
            if (y != 0) setPixel(buffer, cx + x, cy - y);
            if (x != 0 && y != 0) setPixel(buffer, cx - x, cy - y);
#else // But the filled version plots some pixels multiple times...
            horizontalLine(buffer, cx - x, cy + y, cx + x);
            //if (x != 0) setPixel(buffer, cx - x, cy + y);
            //if (y != 0) setPixel(buffer, cx + x, cy - y);
            //if (x != 0 && y != 0) setPixel(buffer, cx - x, cy - y);
#endif
        }

        static void setPixel(byte[,] buffer, int x, int y)
        {
            buffer[y, x]++;
        }

        static void horizontalLine(byte[,] buffer, int x0, int y0, int x1)
        {
            for (int x = x0; x <= x1; ++x)
                setPixel(buffer, x, y0);
        }
    }
}

以下是相关结果：

00000111111111111111111111111111111111111111110000
00000111111111111111111111111111111111111111110000
00000111111111111111111111111111111111111111110000
00000111111111111111111111111111111111111111110000
00000111111111111111111111111111111111111111110000
00000011111111111111111111111111111111111111100000
00000011111111111111111111111111111111111111100000
00000011111111111111111111111111111111111111100000
00000001111111111111111111111111111111111111000000
00000001111111111111111111111111111111111111000000
00000000111111111111111111111111111111111110000000
00000000111111111111111111111111111111111110000000
00000000011111111111111111111111111111111100000000
00000000001111111111111111111111111111111000000000
00000000000111111111111111111111111111110000000000
00000000000011111111111111111111111111100000000000
00000000000001111111111111111111111111000000000000
00000000000000122222222222222222222210000000000000
00000000000000001222222222222222221000000000000000
00000000000000000012333333333332100000000000000000
00000000000000000000012345432100000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000

底部像素被绘制了太多次。我在这里错过了什么？

更新 #2：此解决方案有效：

static void circle(byte[,] buffer, int cx, int cy, int radius)
{
    int error = -radius;
    int x = radius;
    int y = 0;

    while (x >= y)
    {
        int lastY = y;

        error += y;
        ++y;
        error += y;

        plot4points(buffer, cx, cy, x, lastY);

        if (error >= 0)
        {
            if (x != lastY)
                plot4points(buffer, cx, cy, lastY, x);

            error -= x;
            --x;
            error -= x;
        }
    }
}

static void plot4points(byte[,] buffer, int cx, int cy, int x, int y)
{
    horizontalLine(buffer, cx - x, cy + y, cx + x);
    if (y != 0)
        horizontalLine(buffer, cx - x, cy - y, cx + x);
}    

Answer 1

另一个问题的答案很好。不过，由于它造成了混乱，我将稍微解释一下。

你在 Wikipedia 中看到的算法基本上找到了 1/8 个圆的 x 和 y（角度 0 到 pi/4），然后绘制 8 个点作为它的镜像。例如：

    (o-y,o+x) x         x (o+y,o+x)

(o-x,o+y) x                  x (o+x,o+y) <-- compute x,y

                   o

(o-x,o-y) x                  x (o+x,o-y)

    (o-y,o-x) x         x (o+y,o-x)

另一种解决方案的建议是，如果您仔细观察这张图片，这将是非常有意义的，而不是绘制 8 个点，而是绘制 4 条水平线：

    (o-y,o+x) x---------x (o+y,o+x)

(o-x,o+y) x-----------------x (o+x,o+y) <-- compute x,y

                   o

(o-x,o-y) x-----------------x (o+x,o-y)

    (o-y,o-x) x---------x (o+y,o-x)

现在，如果您计算 (x,y) 中的角度 [0, pi/4] 并为每个计算点绘制这 4 条线，您将绘制许多水平线填充一个圆圈，而没有任何线重叠。

更新

在圆圈底部出现重叠线的原因是 (x,y) 坐标是四舍五入的，因此在这些位置，(x,y) 会自行水平移动。

如果你看看this wikipedia图片：

您会注意到，在圆圈的顶部，一些像素水平对齐。从这些点开始绘制水平线重叠。

如果您不想这样，解决方案很简单。您必须保留之前绘制的x（因为顶部和底部是原始(x,y) 的镜像，您应该保留表示这些线的y 的之前的x）并且只绘制水平线，如果那样的话值变化。如果不是，说明你们在同一行。

考虑到您将首先遇到最里面的点，只有当新点具有不同的x 时，您才应该为前一个点画线（当然，总是画最后一条线）。或者，您可以从角度 PI/4 开始绘制到 0 而不是 0 到 PI/4，并且您将首先遇到外部点，因此每次看到新的x 时都会绘制线条。

Answer 2

我想出了一个算法来绘制已经填充的圆圈。
它遍历将绘制圆的像素，仅此而已。
从这里开始，关于绘制像素函数的速度。

Here's a *.gif that demonstrates what the algorithm does !

至于算法，代码如下：

    //The center of the circle and its radius.
    int x = 100;
    int y = 100;
    int r = 50;
    //This here is sin(45) but i just hard-coded it.
    float sinus = 0.70710678118;
    //This is the distance on the axis from sin(90) to sin(45). 
    int range = r/(2*sinus);
    for(int i = r ; i >= range ; --i)
    {
        int j = sqrt(r*r - i*i);
        for(int k = -j ; k <= j ; k++)
        {
            //We draw all the 4 sides at the same time.
            PutPixel(x-k,y+i);
            PutPixel(x-k,y-i);
            PutPixel(x+i,y+k);
            PutPixel(x-i,y-k);
        }
    }
    //To fill the circle we draw the circumscribed square.
    range = r*sinus;
    for(int i = x - range + 1 ; i < x + range ; i++)
    {
        for(int j = y - range + 1 ; j < y + range ; j++)
        {
            PutPixel(i,j);
        }
    }

希望这对您有所帮助...一些新用户...抱歉发布了 necro-posting。
~Shmiggy

Answer 3

我需要这样做，这是我想出的代码。此处的可视图像显示了绘制的像素，其中数字是像素遍历的顺序，绿色数字表示使用对称性的列完成反射绘制的像素，如代码所示。

void drawFilledMidpointCircleSinglePixelVisit( int centerX, int centerY, int radius )
{
    int x = radius;
    int y = 0;
    int radiusError = 1 - x;

    while (x >= y)  // iterate to the circle diagonal
    {

        // use symmetry to draw the two horizontal lines at this Y with a special case to draw
        // only one line at the centerY where y == 0
        int startX = -x + centerX;
        int endX = x + centerX;
        drawHorizontalLine( startX, endX, y + centerY );
        if (y != 0)
        {
            drawHorizontalLine( startX, endX, -y + centerY );
        }

        // move Y one line
        y++;

        // calculate or maintain new x
        if (radiusError<0)
        {
            radiusError += 2 * y + 1;
        }
        else 
        {
            // we're about to move x over one, this means we completed a column of X values, use
            // symmetry to draw those complete columns as horizontal lines at the top and bottom of the circle
            // beyond the diagonal of the main loop
            if (x >= y)
            {
                startX = -y + 1 + centerX;
                endX = y - 1 + centerX;
                drawHorizontalLine( startX, endX,  x + centerY );
                drawHorizontalLine( startX, endX, -x + centerY );
            }
            x--;
            radiusError += 2 * (y - x + 1);
        }

    }

}

Answer 4

我想对您的更新 #2 发表评论：此解决方案有效：（但我想我首先需要更多的声誉......）解决方案中有一个小错误，巧合的是在绘制小圆圈时。如果将半径设置为 1，则会得到

00000
00000
01110
00100
00000

要解决这个问题，您需要做的就是更改 plot4points 中的条件检查

if (x != 0 && y != 0)

到

if (y != 0)

我已经在大大小小的圆圈上进行了测试，以确保每个像素仍然只分配一次。似乎工作得很好。我认为 x != 0 是不需要的。也节省了一点性能。

Answer 5

更新 #2

 if (error >= 0)
 {
    if (x != lastY) 
       plot4points(buffer, cx, cy, lastY, x);

到

 if (error >= 0)
 {     
    plot4points(buffer, cx, cy, lastY, x);

Circle 和 FillCircle 版本：

Const
  Vypln13:Boolean=False;  // Fill Object


//Draw a circle at (cx,cy)
Procedure Circle(cx: integer; cy: integer; radius: integer );
Var
   error,x,y: integer;
Begin  
   error := -radius;
   x := radius;
   y := 0;

   while (x >= y) do
   Begin

     Draw4Pixel(cx,cy, x, y);
     if ( Not Vypln13 And ( x <> y) ) Then Draw4Pixel(cx,cy, y, x);

     error := error + y;
     y := y + 1;
     error := error + y;

     if (error >= 0) Then
     Begin

       if ( Vypln13) then Draw4Pixel(cx, cy, y - 1, x);

       error := error - x;
       x := x - 1;
       error := error - x;
     End;
   End;
End;


Procedure Draw4Pixel(cx,cy,dx,dy: integer);
Begin
  if ( (dx = 0) And (dy = 0) ) then
  begin
    PutPixel (cx , cy , Color13);
    exit;
  End;

  IF Vypln13 Then
  Begin
    HorizontLine (cx - dx,  cx + dx, cy + dy, Color13);
    if ( dy = 0 ) then exit;
    HorizontLine (cx - dx,  cx + dx, cy - dy, Color13);
    exit;
  end;

  PutPixel (cx + dx, cy + dy, Color13);
  if ( dx <> 0 ) then
  begin
    PutPixel (cx - dx, cy + dy, Color13);
    if ( dy = 0 ) then exit;
    PutPixel (cx + dx, cy - dy, Color13);
  End;
  PutPixel (cx - dx, cy - dy, Color13);

End;