C# 矩阵预算和一些基本的几何运算
Posted 季末的寂寞
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以前工作中写的，这里备个份，有可能用到
基本的矩阵运算类，测试20阶以内应该没啥问题，超过20阶不好使。。。
/// <summary> /// 矩阵 异常 512索引 1024无解 2046矩阵行列 /// </summary> public class Matrix { private int m_row;//行 private int m_col;//列 private double[,] m_data;//数据 /// <summary>元素 /// </summary> /// <param name="ro"></param> /// <param name="co"></param> /// <returns></returns> public double this[int ro, int co] { get { if (ro >= m_row || co >= m_col || ro < 0 || co < 0) throw new Exception("512"); return m_data[ro, co]; } set { if (ro >= m_row || co >= m_col || ro < 0 || co < 0) throw new Exception("512"); m_data[ro, co] = value; } } /// <summary>行数 /// </summary> public int Row { get { return m_row; } } /// <summary>列数 /// </summary> public int Column { get { return m_col; } } public Matrix() { m_row = 0; m_col = 0; m_data = new double[0, 0]; } public Matrix(double[,] matrix) { m_row = matrix.GetLength(0); m_col = matrix.GetLength(1); m_data = matrix; } public Matrix(int ro, int co) { if (ro < 0 || co < 0) throw new Exception("512"); m_row = ro; m_col = co; m_data = new double[ro, co]; } public static Matrix operator *(Matrix left, Matrix right) { if (left.Column != right.Row) throw new Exception("2046"); Matrix re = new Matrix(left.Row, right.Column); for (int i = 0; i < left.Row; i++) { for (int j = 0; j < right.Column; j++) { for (int k = 0; k < left.Column; k++) { re[i, j] += left[i, k] * right[k, j]; } } } return re; } public static Matrix operator +(Matrix left, Matrix right) { if (left.Row != right.Row || left.Column != right.Column) throw new Exception("2046"); Matrix re = new Matrix(left.Row, left.Column); for (int i = 0; i < left.Row; i++) { for (int j = 0; j < left.Column; j++) { re[i, j] = left[i, j] + right[i, j]; } } return re; } public static Matrix operator -(Matrix left, Matrix right) { if (left.Row != right.Row || left.Column != right.Column) throw new Exception("2046"); Matrix re = new Matrix(left.Row, left.Column); for (int i = 0; i < left.Row; i++) { for (int j = 0; j < left.Column; j++) { re[i, j] = left[i, j] - right[i, j]; } } return re; } public static Matrix operator *(double factor, Matrix right) { Matrix re = new Matrix(right.Row, right.Column); for (int i = 0; i < right.Row; i++) { for (int j = 0; j < right.Column; j++) { re[i, j] = right[i, j] * factor; } } return re; } public static Matrix operator *(Matrix left, double factor) { return factor * left; } /// <summary>转置 /// </summary> /// <returns></returns> public Matrix Matrixtran() { Matrix re = new Matrix(this.m_col, this.m_row); for (int i = 0; i < this.m_row; i++) { for (int j = 0; j < this.m_col; j++) { re[j, i] = this[i, j]; } } return re; } /// <summary>行列式 //加边法 /// </summary> /// <param name="Matrix"></param> /// <returns></returns> public double Matrixvalue() { if (this.m_row != this.m_col) { throw new Exception("2046"); } int n = this.m_row; if (n == 2) return this[0, 0] * this[1, 1] - this[0, 1] * this[1, 0]; double dsum = 0, dSign = 1; for (int i = 0; i < n; i++) { Matrix tempa = new Matrix(n - 1, n - 1); for (int j = 0; j < n - 1; j++) { for (int k = 0; k < n - 1; k++) { tempa[j, k] = this[j + 1, k >= i ? k + 1 : k]; } } dsum += this[0, i] * dSign * tempa.Matrixvalue(); dSign = dSign * -1; } return dsum; } /// <summary>求逆 /// </summary> /// <param name="Matrix"></param> /// <returns></returns> public Matrix InverseMatrix() { int row = this.m_row; int col = this.m_col; if (row != col) throw new Exception("2046"); Matrix re = new Matrix(row, col); double val = this.Matrixvalue(); if (Math.Abs(val) <= 1E-6) { throw new Exception("1024"); } re = this.AdjointMatrix(); for (int i = 0; i < row; i++) { for (int j = 0; j < row; j++) { re[i, j] = re[i, j] / val; } } return re; } /// <summary>求伴随矩阵 /// </summary> /// <param name="Matrix"></param> /// <returns></returns> public Matrix AdjointMatrix() { int row = this.m_row; Matrix re = new Matrix(row, row); for (int i = 0; i < row; i++) { for (int j = 0; j < row; j++) { Matrix temp = new Matrix(row - 1, row - 1); for (int x = 0; x < row - 1; x++) { for (int y = 0; y < row - 1; y++) { temp[x, y] = this[x < i ? x : x + 1, y < j ? y : y + 1]; } } re[j, i] = ((i + j) % 2 == 0 ? 1 : -1) * temp.Matrixvalue(); } } return re; } }
一些基本的几何运算
public class CalcCls { /// <summary> /// 根据两点计算距离两点连线为R的两点,默认垂足为A /// </summary> /// <param name="pointa">A 已知点</param> /// <param name="pointb">B 已知点</param> /// <param name="R">距离</param> /// <param name="pointc">C 待求点</param> /// <param name="pointd">D 待求点</param> /// <returns>AB两点重合返回false 正常为true</returns> protected bool Vertical(PointXY pointa, PointXY pointb, double R, ref PointXY pointc, ref PointXY pointd) { try { //（X-xa)^2+(Y-ya)^2=R*R 距离公式 //(X-xa)*(xb-xa)+(Y-ya)*(yb-ya)=0 垂直 //解方程得两点即为所求点 pointc.X = pointa.X - (pointb.Y - pointa.Y) * R / Distance(pointa, pointb); pointc.Y = pointa.Y + (pointb.X - pointa.X) * R / Distance(pointa, pointb); pointd.X = pointa.X + (pointb.Y - pointa.Y) * R / Distance(pointa, pointb); pointd.Y = pointa.Y - (pointb.X - pointa.X) * R / Distance(pointa, pointb); return true; } catch { //如果A,B两点重合会报错，那样就返回false return false; } } /// <summary> 判断pt在pa到pb的左侧</summary> /// <returns></returns> protected bool IsLeft(PointXY pa, PointXY pb, PointXY pt) { //ax+by+c=0 double a = pb.Y - pa.Y; double b = pa.X - pb.X; double c = pb.X * pa.Y - pa.X * pb.Y; double d = a * pt.X + b * pt.Y + c; if (d < 0) { return false; } else { return true; } } /// <summary> 计算P1P2和P3P4两条线段所在直线的交点 /// 如果两条线段在同一直线，返回较短的线段的端点,p2或P3</summary> /// <param name="pt">输出交点</param> /// <returns>有交点返回true 否则返回false</returns> protected void calcIntersectionPoint(PointXY pt1, PointXY pt2, PointXY pt3, PointXY pt4, ref PointXY pt) { //ax+by=c double a1, b1, c1, a2, b2, c2; a1 = pt1.Y - pt2.Y; b1 = pt2.X - pt1.X; c1 = pt1.Y * pt2.X - pt2.Y * pt1.X; a2 = pt3.Y - pt4.Y; b2 = pt4.X - pt3.X; c2 = pt3.Y * pt4.X - pt4.Y * pt3.X; double db = a1 * b2 - a2 * b1; if (db == 0)//平行或共线 { if (a1 * pt3.X + b1 * pt3.Y == c1)//共线 { pt = (Distance(pt1, pt2) > Distance(pt3, pt4)) ? pt3 : pt2; } } else { pt.X = (b2 * c1 - b1 * c2) / db; pt.Y = (a1 * c2 - a2 * c1) / db; } } /// <summary>判断P1P2和P3P4线段是否相交</summary> protected bool IsIntersect(PointXY pt1, PointXY pt2, PointXY pt3, PointXY pt4) { //return ((max(u.s.x, u.e.x) >= min(v.s.x, v.e.x)) && //排斥实验 //(max(v.s.x, v.e.x) >= min(u.s.x, u.e.x)) && //(max(u.s.y, u.e.y) >= min(v.s.y, v.e.y)) && //(max(v.s.y, v.e.y) >= min(u.s.y, u.e.y)) && //(multiply(v.s, u.e, u.s) * multiply(u.e, v.e, u.s) >= 0) && //跨立实验 //(multiply(u.s, v.e, v.s) * multiply(v.e, u.e, v.s) >= 0)); //判断矩形是否相交 if (Math.Max(pt1.X, pt2.X) >= Math.Min(pt3.X, pt4.X) && Math.Max(pt3.X, pt4.X) >= Math.Min(pt3.X, pt4.X) && Math.Max(pt1.Y, pt2.Y) >= Math.Min(pt3.Y, pt4.Y) && Math.Max(pt3.Y, pt4.Y) >= Math.Min(pt1.Y, pt2.Y)) { //线段跨立 if (multiply(pt3, pt2, pt1) * multiply(pt2, pt4, pt1) >= 0 && multiply(pt1, pt4, pt3) * multiply(pt4, pt2, pt3) >= 0) { return true; } } return false; } /****************************************************************************** r=multiply(sp,ep,op),得到(sp-op)*(ep-op)的叉积 r>0:ep在矢量opsp的逆时针方向； r=0：opspep三点共线； r<0:ep在矢量opsp的顺时针方向 *******************************************************************************/ /// <summary> 计算p1p0和p2p0叉积 </summary> /// <returns></returns> protected double multiply(PointXY pt1, PointXY pt2, PointXY p0) { //return ((sp.x - op.x) * (ep.y - op.y) - (ep.x - op.x) * (sp.y - op.y)); double mult = (pt1.X - p0.X) * (pt2.Y - p0.Y) - (pt2.X - p0.X) * (pt1.Y - p0.Y); return mult; } /// <summary>按照距离将交点排序 </summary> /// <param name="temline"></param> /// <param name="ps">起点</param> /// <returns></returns> protected PointXY[] sortPoint(PointXY[] temline, PointXY ps) { List<PointXY> lp = new List<PointXY>(); List<PointXY> sortlp = new List<PointXY>(); lp.AddRange(temline); if (temline.Length < 1) return sortlp.ToArray(); for (; lp.Count > 1; ) { PointXY pd = lp[0]; for (int i = 0; i < lp.Count - 1; i++) { if (Distance(ps, pd) > Distance(ps, lp[i + 1])) { pd = lp[i + 1]; } } lp.Remove(pd); sortlp.Add(pd); } sortlp.Add(lp[0]); return sortlp.ToArray(); } /// <summary> /// 根据两点计算两点连线上距离A点L的点 输出C点为距离B近的点上的点 D为距离B远的点 /// </summary> /// <param name="pointa">A点 默认为计算距离A点L的点</param> /// <param name="pointb">B点</param> /// <param name="L">距离</param> /// <param name="pointc">返回点C</param> /// <param name="pointd">返回点D 有时候没用</param> /// <returns></returns> protected bool GapP(PointXY pointa, PointXY pointb, double L, ref PointXY pointc, ref PointXY pointd) { try { PointXY pc = new PointXY(); PointXY pd = new PointXY(); //(north-xa)^2+(east-ya)^2=L 两点距离公式 //(north-xa)*(ya-yb)-(east-ya)(xa-xb)=0 直线平行条件，注意，是CA和AB平行（实际是C点在AB上） pc.X = pointa.X + (pointa.X - pointb.X) * L / Distance(pointa, pointb); pc.Y = pointa.Y + (pointa.Y - pointb.Y) * L / Distance(pointa, pointb); pd.X = pointa.X - (pointa.X - pointb.X) * L / Distance(pointa, pointb); pd.Y = pointa.Y - (pointa.Y - pointb.Y) * L / Distance(pointa, pointb); pointc = Distance(pc, pointb) > Distance(pd, pointb) ? pd : pc; //取距离B近的点 pointd = Distance(pc, pointb) > Distance(pd, pointb) ? pc : pd;//取距离B远的点 return true; } catch { //如果A,B两点重合会报错，那样就返回false return false; } } /// <summary> 返回两点的距离 </summary> /// <param name="pa"></param> /// <param name="pb"></param> /// <returns></returns> public double Distance(double xa, double ya, double xb, double yb) { double L; L = Math.Sqrt(Math.Pow(xa - xb, 2) + Math.Pow(ya - yb, 2)); return L; } public double Distance(PointXY pa, PointXY pb) { return Distance(pa.X, pa.Y, pb.X, pb.Y); } /// <summary> 用VS自带算法判断点是否在区域内 </summary> /// <param name="pt"></param> /// <param name="pointsArray"></param> /// <returns></returns> public bool PtInPolygon(PointXY pd, PointXY[] pdsArray) { PointF pt = new PointF(); pt.X = (float)pd.X; pt.Y = (float)pd.Y; List<Point> pl = new List<Point>(); for (int i = 0; i < pdsArray.Length; i++) { Point p = new Point(); p.X = (int)pdsArray[i].X; p.Y = (int)pdsArray[i].Y; pl.Add(p); } if (pl.Count < 3) return false; using (System.Drawing.Drawing2D.GraphicsPath gp = new System.Drawing.Drawing2D.GraphicsPath()) { gp.AddPolygon(pl.ToArray()); return gp.IsVisible(pt); } } /// <summary> /// 求线段的方位角0-360 /// </summary> /// <param name="pa">线段起点</param> /// <param name="pb">线段终点</param> /// <returns>从0度顺时针偏转到该线段所划过的角度</returns> protected double Angle(PointXY pa, PointXY pb) { double Ang = 0; double a; try { a = Math.Acos((pb.X - pa.X) / Distance(pa, pb)); if (pb.Y - pa.Y < 0) { a = 2 * Math.PI - a; } Ang = a * 180d / Math.PI; return Ang; } catch { Ang = 0; return Ang; } } /// <summary>角度到弧度</summary> /// <param name="value"></param> /// <returns></returns> private double AngleToRadian(double value) { return value * Math.PI / 180d; } /// <summary> 弧度到角度</summary> /// <param name="value"></param> /// <returns></returns> private double RadianToAngle(double value) { return value * 180d / Math.PI; } } public struct PointXY { public double Y; public double X; }
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