如何解决行进平方算法中的插值
Here是源代码的GitRepository。
在“线性插值”部分中，本文讨论了当直线倾斜时如何插值。
例如，对于案例2，它具有以下计算：
我实现了如下插值：
public class Square
{
public Point A { get; set; }//bottom left point
public Point B { get; set; }//bottom right point
public Point C { get; set; }//top right point
public Point D { get; set; }//top left point
public double A_data { get; set; }//bottom left data
public double B_data { get; set; }//bottom right data
public double C_data { get; set; }//top roght data
public double D_data { get; set; }//top left data
public Square()
{
A = new Point();
B = new Point();
C = new Point();
D = new Point();
}
private double GetCaseId(double threshold)
{
int caseId = 0;
if (A_data >= threshold)
{
caseId |= 1;
}
if (B_data >= threshold)
{
caseId |= 2;
}
if (C_data >= threshold)
{
caseId |= 4;
}
if (D_data >= threshold)
{
caseId |= 8;
}
return caseId;
}
public List<Line> GetLines(double Threshold)
{
List<Line> linesList = new List<Line>();
double caseId = GetCaseId(Threshold);
if (caseId == 0) {/*do nothing*/ }
if (caseId == 15) {/*do nothing*/ }
if ((caseId == 1) || (caseId == 14))
{
double pX = B.X + (A.X - B.X) * ((1 - B_data) / (A_data - B_data));
double pY = B.Y;
Point p = new Point(pX,pY);
double qX = D.X;
double qY = D.Y + (A.Y - D.Y) * ((1 - D_data) / (A_data - D_data));
Point q = new Point(qX,qY);
Line line = new Line(p,q);
linesList.Add(line);
}
/*2==13*/
if ((caseId == 2) || (caseId == 13))//B
{
double pX = A.X + (B.X - A.X) * ((1 - A_data) / (B_data - A_data));
double pY = A.Y;
Point p = new Point(pX,pY);
double qX = C.X;
double qY = C.Y + (B.Y - C.Y) * ((1 - C_data) / (B_data - C_data));
Point q = new Point(qX,q);
linesList.Add(line);
}
/*3==12*/
if ((caseId == 3) || (caseId == 12))
{
double pX = A.X;
double pY = A.Y + (D.Y - A.Y) * ((1 - A_data) / (D_data - A_data));
Point p = new Point(pX,q);
linesList.Add(line);
}
/*4==11*/
if ((caseId == 4) || (caseId == 11))
{
double pX = D.X + (C.X - D.X) * ((1 - D_data) / (C_data - D_data));
double pY = D.Y;
Point p = new Point(pX,pY);
double qX = B.X;
double qY = B.Y + (C.Y - B.Y) * ((1 - B_data) / (C_data - B_data));
Point q = new Point(qX,q);
linesList.Add(line);
}
/*6==9*/
if ((caseId == 6) || (caseId == 9))
{
double pX = A.X + (B.X - A.X) * ((1 - A_data) / (B_data - A_data));
double pY = A.Y;
Point p = new Point(pX,pY);
double qX = C.X + (D.X - C.X) * ((1 - C_data) / (D_data - C_data));
double qY = C.Y;
Point q = new Point(qX,q);
linesList.Add(line);
}
/*7==8*/
if ((caseId == 7) || (caseId == 8))
{
double pX = C.X + (D.X - C.X) * ((1 - C_data) / (D_data - C_data));
double pY = C.Y;
Point p = new Point(pX,pY);
double qX = A.X;
double qY = A.Y + (D.Y - A.Y) * ((1 - A_data) / (D_data - A_data));
Point q = new Point(qX,q);
linesList.Add(line);
}
/*ambiguous case*/
if (caseId == 5)
{
double pX1 = A.X + (B.X - A.X) * ((1 - A_data) / (B_data - A_data));
double pY1 = A.Y;
Point p1 = new Point(pX1,pY1);
double qX1 = C.X;
double qY1 = C.Y + (B.Y - C.Y) * ((1 - C_data) / (B_data - C_data));
Point q1 = new Point(qX1,qY1);
Line line1 = new Line(p1,q1);
double pX2 = C.X + (D.X - C.X) * ((1 - C_data) / (D_data - C_data));
double pY2 = C.Y;
Point p2 = new Point(pX2,pY2);
double qX2 = A.X;
double qY2 = A.Y + (D.Y - A.Y) * ((1 - A_data) / (D_data - A_data));
Point q2 = new Point(qX2,qY2);
Line line2 = new Line(p2,q2);
linesList.Add(line1);
linesList.Add(line2);
}
if (caseId == 10)
{
double pX1 = B.X + (A.X - B.X) * ((1 - B_data) / (A_data - B_data));
double pY1 = B.Y;
Point p1 = new Point(pX1,pY1);
double qX1 = D.X;
double qY1 = D.Y + (A.Y - D.Y) * ((1 - D_data) / (A_data - D_data));
Point q1 = new Point(qX1,q1);
double pX2 = D.X + (C.X - D.X) * ((1 - D_data) / (C_data - D_data));
double pY2 = D.Y;
Point p2 = new Point(pX2,pY2);
double qX2 = B.X;
double qY2 = B.Y + (C.Y - B.Y) * ((1 - B_data) / (C_data - B_data));
Point q2 = new Point(qX2,q2);
linesList.Add(line1);
linesList.Add(line2);
}
return linesList;
}
但是，这不能正常工作。
谁能检查插值部分并告诉我哪里出了问题？
解决方法
我检查了您的仓库，错误不在您发布的代码中：|
实际错误在于您如何在MarchingSquare.cs中初始化数据
替换以下内容：
double a = Data[j + 1,i];
double b = Data[j + 1,i + 1];
double c = Data[j,i + 1];
double d = Data[j,i];
Point A = new Point(xVector[i],yVector[j + 1]);//A
Point B = new Point(xVector[i + 1],yVector[j + 1]);//B
Point C = new Point(xVector[i + 1],yVector[j]);//C
Point D = new Point(xVector[i],yVector[j]);//D
具有：
double a = Data[j,i];
double b = Data[j,i + 1];
double c = Data[j+1,i + 1];
double d = Data[j+1,i];
Point A = new Point(j,i);//A
Point B = new Point(j,i+1);//B
Point C = new Point(j+1,i+1);//C
Point D = new Point(j+1,i);//D
您将获得所需的结果（也许是转置的结果）。
编辑：我想我把它调换了：Plot[Sin[x/100.0] * Cos[y/100.0]] = 0.9,x = 0 to 250,y = 0 to 500
稍微看一下插值代码：
- 使用枚举代替大小写常量
- 请勿将caseId转换为两倍
- 将插值提取为辅助方法！
using System;
using System.Collections.Generic;
namespace G__Marching_Sqaure
{
public class Square
{
public Point A { get; set; } //bottom left point
public Point B { get; set; } //bottom right point
public Point C { get; set; } //top right point
public Point D { get; set; } //top left point
public double A_data { get; set; } //bottom left data
public double B_data { get; set; } //bottom right data
public double C_data { get; set; } //top roght data
public double D_data { get; set; } //top left data
public Square()
{
A = new Point();
B = new Point();
C = new Point();
D = new Point();
}
private LineShapes GetCaseId(double threshold)
{
int caseId = 0;
if (A_data >= threshold)
{
caseId |= 1;
}
if (B_data >= threshold)
{
caseId |= 2;
}
if (C_data >= threshold)
{
caseId |= 4;
}
if (D_data >= threshold)
{
caseId |= 8;
}
return (LineShapes)caseId;
}
public List<Line> GetLines(double Threshold)
{
List<Line> linesList = new List<Line>();
LineShapes caseId = GetCaseId(Threshold);
if (caseId == LineShapes.Empty)
{
/*do nothing*/
}
if (caseId == LineShapes.All)
{
/*do nothing*/
}
if ((caseId == LineShapes.BottomLeft) || (caseId == LineShapes.AllButButtomLeft))
{
var p = InterpolateHorizonal(B,A,B_data,A_data);
var q = InterpolateVertical(D,D_data,A_data);
Line line = new Line(p,q);
linesList.Add(line);
}
/*2==13*/
if ((caseId == LineShapes.BottomRight) || (caseId == LineShapes.AllButButtomRight)) //B
{
var p = InterpolateHorizonal(A,B,A_data,B_data);
var q = InterpolateVertical(C,C_data,B_data);
Line line = new Line(p,q);
linesList.Add(line);
}
/*3==12*/
if ((caseId == LineShapes.Bottom) || (caseId == LineShapes.Top))
{
// interpolate vertical
var p = InterpolateVertical(A,D,D_data);
var q = InterpolateVertical(C,q);
linesList.Add(line);
}
/*4==11*/
if ((caseId == LineShapes.TopRight) || (caseId == LineShapes.AllButTopRight))
{
var p = InterpolateHorizonal(D,C,C_data);
var q = InterpolateVertical(B,C_data);
Line line = new Line(p,q);
linesList.Add(line);
}
/*6==9*/
if ((caseId == LineShapes.Right) || (caseId == LineShapes.Left))
{
var p = InterpolateHorizonal(A,B_data);
var q = InterpolateHorizonal(C,D_data);
Line line = new Line(p,q);
linesList.Add(line);
}
/*7==8*/
if ((caseId == LineShapes.AllButTopLeft) || (caseId == LineShapes.TopLeft))
{
var p = InterpolateHorizonal(C,D_data);
var q = InterpolateVertical(A,q);
linesList.Add(line);
}
/*ambiguous case*/
if (caseId == LineShapes.TopRightBottomLeft)
{
var p1 = InterpolateHorizonal(A,B_data);
var q1 = InterpolateVertical(C,B_data);
Line line1 = new Line(p1,q1);
var p2 = InterpolateHorizonal(C,D_data);
var q2 = InterpolateVertical(A,D_data);
Line line2 = new Line(p2,q2);
linesList.Add(line1);
linesList.Add(line2);
}
if (caseId == LineShapes.TopLeftBottomRight)
{
var p1 = InterpolateHorizonal(B,A_data);
var q1 = InterpolateVertical(D,A_data);
Line line1 = new Line(p1,q1);
var p2 = InterpolateHorizonal(D,C_data);
var q2 = InterpolateVertical(B,C_data);
Line line2 = new Line(p2,q2);
linesList.Add(line1);
linesList.Add(line2);
}
return linesList;
}
private static Point InterpolateVertical(Point point,Point point1,double data,double data1)
{
double qX = point.X;
double qY = point.Y + (point1.Y - point.Y) * ((1 - data) / (data1 - data));
Point q = new Point(qX,qY);
return q;
}
private static Point InterpolateHorizonal(Point start,Point end,double startForce,double endForce)
{
double pX = start.X + (end.X - start.X) * ((1 - startForce) / (endForce - startForce));
double pY = start.Y;
Point p = new Point(pX,pY);
return p;
}
}
internal enum LineShapes
{
Empty = 0,// ○----○
// | |
// | |
// ○----○
BottomLeft = 1,// ○----○
// | |
// | |
// ●----○
BottomRight = 2,// ○----○
// | |
// | |
// ○----●
Bottom = 3,// ○----○
// | |
// | |
// ●----●
TopRight = 4,// ○----●
// | |
// | |
// ○----○
TopRightBottomLeft = 5,// ○----●
// | |
// | |
// ●----○
Right = 6,// ○----●
// | |
// | |
// ○----●
AllButTopLeft = 7,// ○----●
// | |
// | |
// ●----●
TopLeft = 8,// ●----○
// | |
// | |
// ○----○
Left = 9,// ●----○
// | |
// | |
// ●----○
TopLeftBottomRight = 10,// ●----○
// | |
// | |
// ○----●
AllButTopRight = 11,// ●----○
// | |
// | |
// ●----●
Top = 12,// ●----●
// | |
// | |
// ○----○
AllButButtomRight = 13,// ●----●
// | |
// | |
// ●----○
AllButButtomLeft = 14,// ●----●
// | |
// | |
// ○----●
All = 15,// ●----●
// | |
// | |
// ●----●
}
}
还要清理package.json
更新
如果您再次阅读该文章，您会发现插值公式中的值1是阈值的值
如果检查具有任意阈值的情况，但不修改插值公式（因此将阈值设置为1进行插值），则可能会得到位于给定平方之外的点！
更新的插值例程：
private static Point InterpolateVertical(Point start,double endForce,double threshold)
{
double a = ((threshold - startForce) / (endForce - startForce));
double qX = start.X;
double qY = start.Y + (end.Y - start.Y) * a;
Point q = new Point(qX,qY);
return q;
}
private static Point InterpolateHorizonal(Point start,double threshold)
{
double a = ((threshold - startForce) / (endForce - startForce));
double pX = start.X + (end.X - start.X) * a;
double pY = start.Y;
Point p = new Point(pX,pY);
return p;
}
您正确地注意到在最初的示例中，您的网格过于密集，因此对每个像素运行测试同样有效。
我检查了您的sin-cos示例，您仍在采样每个点（正方形的宽度和高度等于1），因此行进点算法没有真正的好处。
检查以下带有较大正方形网格的初始化代码（分辨率x分辨率）
int height = 1000;
int width = height;
int resolution = 20;
double threshold = 0.9;
int xLen = width / resolution;
int yLen = height / resolution;
int[] x = new int[xLen];
int[] y = new int[yLen];
for (int i = 0; i < xLen; i++)
{
x[i] = i * resolution;
}
for (int j = 0; j < yLen; j++)
{
y[j] = j * resolution;
}
/////////////////////SIN-COS/////////////////////////////
double[,] example = new double[xLen,yLen];
for (int j = 0; j < yLen; j++)
{
for (int i = 0; i < xLen; i++)
{
double example_l = Math.Sin(i * resolution /100.0 ) * Math.Cos(j * resolution /100.0);
example[j,i] = example_l;
}
}
以分辨率播放（例如，尝试使用resolution = 10）以获得更平滑的形状。
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