问题
If we take Wikipedia article on Marching square into account, we see that case#5 and case#10 are said to be ambiguous cases.
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;
}
}
}
You can see here each of the cases are taken care of individually.
Output:
Can anyone tell me what I am missing?
Driver Program:
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]);
}
}
}
}
Edit: 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 }
my program produced the following output:
回答1:
Oh man, I understand you. Surprisingly thats a good question!
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".
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 }
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".
BUT lets see example with ambiguous cases:
{ 1, 2, 1 },
{ 2, 1, 2 },
{ 1, 2, 1 }
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.
I think moment of choice is in normalize function at
if (data[i, j] > 1)
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.
Hope that helps to clear the confusion.
Edit: I elaborated in comments but for visibility I'm duplicating it here
回答2:
Your example doesn't include any ambiguous cases. What output would you expect with the following input;
{ 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 }
Circles around the 1's? Circles around the 2's? Diagonal lines?
Edit;
From your code;
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 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;
You could swap those case labels. You could pick one, delete the other, and merge the cases. You could look at more surrounding pixels to pick one. You could roll a random number to pick which way to draw it.
But you didn't. You arbitrarily decided that you would always draw those cases this way.
来源：https://stackoverflow.com/questions/62868990/ambiguous-cases-in-marching-square-algorithm