行进平方算法中的模棱两可案例 [英] Ambiguous cases in Marching square algorithm
问题描述
如果我们考虑行军广场上的Wikipedia文章，则会发现该案例#5和case#10被认为是模棱两可的案例.
If we take Wikipedia article on Marching square into account, we see that case#5 and case#10 are said to be ambiguous cases.
我已经按照以下方式实施了游行广场，但我不了解如何产生模棱两可的情况:
I have implemented Marching Square as follows and I am not understanding how an ambiguous case can arise:
public class LinesRectangle
{
public Graphics Graphics { get; set; }
public Color Color { get; set; }
public Pen Pen { get; set; }
public int Thickness { get; set; }
public LinesRectangle()
{
Color = Color.Blue;
Thickness = 2;
Pen = new Pen(Color, Thickness);
}
public void DrawLines(int x, int y, int width, int code)
{
int height = width;
Graphics.DrawRectangle(Pen, new System.Drawing.Rectangle(x, y, width, height));
int x1 = 0, y1 = 0;
int x2 = 0, y2 = 0;
switch (code)
{
case 0:
case 15:
break;
case 1:
case 14:
x1 = x; y1 = y + height/2;
x2 = x + width/2; y2 = y + height;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 2:
case 13:
x1 = x + width/2; y1 = y + height;
x2 = x + width; y2 = y + height/2;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 3:
case 12:
x1 = x; y1 = y + height / 2;
x2 = x + width; y2 = y + height / 2;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 4:
case 11:
x1 = x+width/2; y1 = y;
x2 = x + width; y2 = y + height / 2;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 5:
x1 = x ; y1 = y + height/2;
x2 = x + width/2; y2 = y;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
x1 = x + width / 2; y1 = y + height;
x2 = x + width; y2 = y + height / 2;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 6:
case 9:
x1 = x + width / 2; y1 = y;
x2 = x + width/2; y2 = y + height;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 7:
case 8:
x1 = x; y1 = y + height / 2;
x2 = x + width / 2; y2 = y;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
case 10:
x1 = x + width / 2; y1 = y;
x2 = x + width; y2 = y + height / 2;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
x1 = x; y1 = y + height / 2;
x2 = x + width / 2; y2 = y + height;
Graphics.DrawLine(Pen, x1, y1, x2, y2);
break;
}
}
}
您可以在这里看到每种情况都是单独处理的.
You can see here each of the cases are taken care of individually.
输出:
谁能告诉我我想念的东西吗?
Can anyone tell me what I am missing?
驱动程序:
public enum What
{
lines, surface, both
}
public partial class DrawingForm : System.Windows.Forms.Form
{
public int [,] Data { get; set; }
public void Print(int[,] data, int xn, int yn)
{
for (int j = 0; j < yn; j++)
{
for (int i = 0; i < xn; i++)
{
Console.Write(data[i, j] + ", ");
}
Console.WriteLine();
}
}
public int[,] normalize(int[,] data, int xn, int yn)
{
for (int j = 0; j < yn; j++)
{
for (int i = 0; i < xn; i++)
{
if (data[i, j] > 1)
{
data[i, j] = 0;
}
else
{
data[i, j] = 1;
}
}
}
return data;
}
public int[,] marching_square(int x, int y, int[,] data, int isovalue, What what)
{
int xn = x;
int yn = y;
data = normalize(data, xn, yn);
int[,] bitMask = new int[xn - 1, yn - 1];
for (int j = 0; j < yn - 1; j++)
{
for (int i = 0; i < xn - 1; i++)
{
StringBuilder sb = new StringBuilder();
sb.Append(data[i, j]);
sb.Append(data[i + 1, j]);
sb.Append(data[i + 1, j + 1]);
sb.Append(data[i, j + 1]);
bitMask[i, j] = Convert.ToInt32(sb.ToString(), 2);
}
}
return bitMask;
}
public DrawingForm()
{
InitializeComponent();
}
private void MainForm_Paint(object sender, System.Windows.Forms.PaintEventArgs e)
{
int[,] data = new int[,] {
{ 1,1,1,1,1 },
{ 1,2,3,2,1 },
{ 1,3,1,3,1 },
{ 1,2,3,2,1 },
{ 1,1,1,1,1 }
};
int[,] bitMask = marching_square(5, 5, data, 0, What.lines);
Graphics g = this.CreateGraphics();
LinesRectangle rect = new LinesRectangle();
rect.Graphics = g;
for (int j = 0; j < 4; j++)
{
for (int i = 0; i < 4; i++)
{
rect.DrawLines(i*50, j*50, 50, bitMask[i,j]);
}
}
}
}
编辑:如果有以下数据( @JeremyLakeman 指出):
In case of the following data (as pointed out by @JeremyLakeman):
{ 2,1,2,1,2 },
{ 1,2,1,2,1 },
{ 2,1,2,1,2 },
{ 1,2,1,2,1 },
{ 2,1,2,1,2 }
我的程序产生了以下输出:
my program produced the following output:
推荐答案
哦，我理解你.令人惊讶的是，这是一个好问题！
Oh man, I understand you. Surprisingly thats a good question!
当您确定值高于等值"时，就会清楚地看到歧义.是黑色或白色，等值线以下等值线"相反.
Ambiguity is seen clearly in a moment when you decide if "value above the isovalue" is black or white and opposite for the "value below the isovalue".
让我解释一下我的意思. 如果您手动执行算法，则可以获得以下结果.在遵循Wiki中描述的算法时，您要做的唯一选择-决定在绘制节点时使用哪种颜色.
Let me explain what I mean. If you do algorithm by hand you can get following results. The only choice you do while following algorithm described on wiki - is to decide what color to use when painting nodes.
{ 1, 1, 1 },
{ 1, 2, 1 },
{ 1, 1, 1 }
没有无歧义的情况，因此 选择无关紧要 -无论"1"是";黑点"或白点".
has no ambiguous cases so the choice does not matter - resulting image will be the same no matter if '1' is a "black dot" or a "white dot".
但是，请查看示例具有不明确情况的情况:
{ 1, 2, 1 },
{ 2, 1, 2 },
{ 1, 2, 1 }
如果"1"为白色，则算法将在中间点周围提供一个圆，如果将"1"选择为黑色，则相同的算法将在中间点附近提供4条圆弧.
algorith would provide a circle around the middle point if '1's are white, and same algorithm would provide 4 arcs near the middle points if '1's are chosen to be black.
我认为选择时刻在normalize函数中
I think moment of choice is in normalize function at
if (data[i, j] > 1)
如果您更改"> "，到"< "如果情况不明，您将可以更改图片.对于无歧义的情况，它什么也不会改变.如果您看方法思想而不是算法，那么歧义更容易理解.查看鞍点-在绘制轮廓时存在歧义，因为从一方面来看，鞍点是一个最小值，而最大值是最大值-取决于测量方向.
If you change ">" to "<" you will get change of image for ambigous cases. And it would change nothing for non-ambigous cases. Ambiguity is easier to understand if you look at methods idea not algorithm. Look at saddle point - there is ambiguity in drawing contour because from the one hand saddle point is a minimum and from the other its a maximum - depends on direction of measurements.
希望能消除混乱.
编辑:我在评论中进行了详细说明，但为了便于查看，我将其复制在此处
I elaborated in comments but for visibility I'm duplicating it here
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