实心圆的中点圆算法

时间：2012-06-04 08:12:11

标签： c# algorithm graphics geometry

可以使用Midpoint circle algorithm栅格化圆的边框。但是，我想要填充圆圈，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，产生了一个实心圆，但有几个像素被访问了几次： fast algorithm for drawing filled circles?

问：如何在不多次绘制像素的情况下栅格化圆圈？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

底部像素绘制的次数太多。我在这里缺少什么？

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);
}

5 个答案:

答案 0 :(得分：15)

另一个问题的答案非常好。然而，由于它造成了混乱，我将稍微解释一下。

你在维基百科中看到的算法基本上找到1/8圆的x和y（角度0到pi/4），然后绘制8个作为其镜像的点。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)

另一个解决方案建议，如果你仔细观察这张照片，这是完全合理的，而不是绘制8个点，画出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，如果您为(x,y)中的角度计算[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，这些线条填充一个圆而没有任何线与另一条线重叠。

更新

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

如果你看一下this wikipedia图片：

enter image description here

您会注意到，在圆圈顶部，某些像素水平对齐。绘制源自这些点的水平线重叠。

如果您不想这样，解决方案非常简单。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，您应该保留前面的x代表这些行的y）并且只绘制水平线，如果该值更改。如果没有，则意味着您在同一条线上。

鉴于您将首先遇到最里面的点，您应该为前一点绘制线条，只有新点具有不同的x（当然，最后一行总是被绘制）。或者，您可以从角度PI / 4向下绘制到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。

答案 1 :(得分：3)

我需要这样做，这是我提出的代码。此处的可视图像显示绘制的像素，其中数字是遍历像素的顺序，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。

enter image description here 

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);
        }

    }

}

答案 2 :(得分：3)

我想出了一个算法来绘制已填充的圆圈 它会迭代圆圈将被绘制的像素，而不是其他任何东西 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，这里是代码：

    //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);
        }
    }

希望这会有所帮助...一些新用户...抱歉发帖 〜Shmiggy

答案 3 :(得分：1)

我想评论你的更新＃2：这个解决方案有效:(但我想我首先需要更多的声誉......）解决方案中存在一个小错误，巧合的是在绘制小圆圈时。如果将半径设置为1，则得到

00000
00000
01110
00100
00000

要解决这个问题，您需要做的就是从

更改plot4points中的条件检查 
if (x != 0 && y != 0)

到

if (y != 0)

我已经在小圆圈和大圆圈上对此进行了测试，以确保每个像素仍然只分配一次。似乎工作得很好。我认为不需要x！= 0。节省一点性能。

答案 4 :(得分：0)

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;
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