本文实例讲述了C#中矩阵运算方法。分享给大家供大家参考。具体分析如下：
一、测试环境：
主机:XP
开发环境:VS2008
二、功能:
在C#中实现矩阵运算
三、源代码：
using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Windows.Forms; //矩阵数据结构 //二维矩阵 class _Matrix { public int m; public int n; public float[] arr; //初始化 public _Matrix() { m = 0; n = 0; } public _Matrix(int mm,int nn) { m = mm; n = nn; } //设置m public void set_mn(int mm,int nn) { m = mm; n = nn; } //设置m public void set_m(int mm) { m = mm; } //设置n public void set_n(int nn) { n = nn; } //初始化 public void init_matrix() { arr = new float[m * n]; } //释放 public void free_matrix() { //delete [] arr; } //读取i,j坐标的数据 //失败返回-31415,成功返回值 public float read(int i,int j) { if (i >= m || j >= n) { return -31415; } //return *(arr + i * n + j); return arr[i * n + j]; } //写入i,j坐标的数据 //失败返回-1,成功返回1 public int write(int i,int j,float val) { if (i >= m || j >= n) { return -1; } arr[i * n + j] = val; return 1; } }; //二维运算类 class _Matrix_Calc { //初始化 public _Matrix_Calc() { } //C = A + B //成功返回1,失败返回-1 public int add(ref _Matrix A,ref _Matrix B,ref _Matrix C) { int i = 0; int j = 0; //判断是否可以运算 if (A.m != B.m || A.n != B.n || A.m != C.m || A.n != C.n) { return -1; } //运算 for (i = 0;i < C.m;i++) { for (j = 0;j < C.n;j++) { C.write(i,j,A.read(i,j) + B.read(i,j)); } } return 1; } //C = A - B //成功返回1,失败返回-1 public int subtract(ref _Matrix A,ref _Matrix B, ref _Matrix C) { int i = 0; int j = 0; //判断是否可以运算 if (A.m != B.m || A.n != B.n || A.m != C.m || A.n != C.n) { return -1; } //运算 for (i = 0;i < C.m;i++) { for (j = 0;j < C.n;j++) { C.write(i,j,A.read(i,j) - B.read(i,j)); } } return 1; } //C = A * B //成功返回1,失败返回-1 public int multiply(ref _Matrix A, ref _Matrix B, ref _Matrix C) { int i = 0; int j = 0; int k = 0; float temp = 0; //判断是否可以运算 if (A.m != C.m || B.n != C.n || A.n != B.m) { return -1; } //运算 for (i = 0;i < C.m;i++) { for (j = 0;j < C.n;j++) { temp = 0; for (k = 0;k < A.n;k++) { temp += A.read(i,k) * B.read(k,j); } C.write(i,j,temp); } } return 1; } //行列式的值,只能计算2 * 2,3 * 3 //失败返回-31415,成功返回值 public float det(ref _Matrix A) { float value = 0; //判断是否可以运算 if (A.m != A.n || (A.m != 2 && A.m != 3)) { return -31415; } //运算 if (A.m == 2) { value = A.read(0,0) * A.read(1,1) - A.read(0,1) * A.read(1,0); } else { value = A.read(0,0) * A.read(1,1) * A.read(2,2) + A.read(0,1) * A.read(1,2) * A.read(2,0) + A.read(0,2) * A.read(1,0) * A.read(2,1) - A.read(0,0) * A.read(1,2) * A.read(2,1) - A.read(0,1) * A.read(1,0) * A.read(2,2) - A.read(0,2) * A.read(1,1) * A.read(2,0); } return value; } //求转置矩阵,B = AT //成功返回1,失败返回-1 public int transpos(ref _Matrix A,ref _Matrix B) { int i = 0; int j = 0; //判断是否可以运算 if (A.m != B.n || A.n != B.m) { return -1; } //运算 for (i = 0;i < B.m;i++) { for (j = 0;j < B.n;j++) { B.write(i,j,A.read(j,i)); } } return 1; } //求逆矩阵,B = A^(-1) //成功返回1,失败返回-1 public int inverse(ref _Matrix A, ref _Matrix B) { int i = 0; int j = 0; int k = 0; _Matrix m = new _Matrix(A.m,2 * A.m); float temp = 0; float b = 0; //判断是否可以运算 if (A.m != A.n || B.m != B.n || A.m != B.m) { return -1; } /* //如果是2维或者3维求行列式判断是否可逆 if (A.m == 2 || A.m == 3) { if (det(A) == 0) { return -1; } } */ //增广矩阵m = A | B初始化 m.init_matrix(); for (i = 0;i < m.m;i++) { for (j = 0;j < m.n;j++) { if (j <= A.n - 1) { m.write(i,j,A.read(i,j)); } else { if (i == j - A.n) { m.write(i,j,1); } else { m.write(i,j,0); } } } } //高斯消元 //变换下三角 for (k = 0;k < m.m - 1;k++) { //如果坐标为k,k的数为0,则行变换 if (m.read(k,k) == 0) { for (i = k + 1;i < m.m;i++) { if (m.read(i,k) != 0) { break; } } if (i >= m.m) { return -1; } else { //交换行 for (j = 0;j < m.n;j++) { temp = m.read(k,j); m.write(k,j,m.read(k + 1,j)); m.write(k + 1,j,temp); } } } //消元 for (i = k + 1;i < m.m;i++) { //获得倍数 b = m.read(i,k) / m.read(k,k); //行变换 for (j = 0;j < m.n;j++) { temp = m.read(i,j) - b * m.read(k,j); m.write(i,j,temp); } } } //变换上三角 for (k = m.m - 1;k > 0;k--) { //如果坐标为k,k的数为0,则行变换 if (m.read(k,k) == 0) { for (i = k + 1;i < m.m;i++) { if (m.read(i,k) != 0) { break; } } if (i >= m.m) { return -1; } else { //交换行 for (j = 0;j < m.n;j++) { temp = m.read(k,j); m.write(k,j,m.read(k + 1,j)); m.write(k + 1,j,temp); } } } //消元 for (i = k - 1;i >= 0;i--) { //获得倍数 b = m.read(i,k) / m.read(k,k); //行变换 for (j = 0;j < m.n;j++) { temp = m.read(i,j) - b * m.read(k,j); m.write(i,j,temp); } } } //将左边方阵化为单位矩阵 for (i = 0;i < m.m;i++) { if (m.read(i,i) != 1) { //获得倍数 b = 1 / m.read(i,i); //行变换 for (j = 0;j < m.n;j++) { temp = m.read(i,j) * b; m.write(i,j,temp); } } } //求得逆矩阵 for (i = 0;i < B.m;i++) { for (j = 0;j < B.m;j++) { B.write(i,j,m.read(i,j + m.m)); } } //释放增广矩阵 m.free_matrix(); return 1; } }; namespace test { public partial class Form1 : Form { double zk; double xkg, pkg, kk, xk, pk, q, r; public Form1() { InitializeComponent(); xk = 0; pk = 0; q = 0.00001; r = 0.0001; int i = 0; int j = 0; int k = 0; _Matrix_Calc m_c = new _Matrix_Calc(); //_Matrix m1 = new _Matrix(3,3); //_Matrix m2 = new _Matrix(3,3); //_Matrix m3 = new _Matrix(3,3); _Matrix m1 = new _Matrix(2, 2); _Matrix m2 = new _Matrix(2, 2); _Matrix m3 = new _Matrix(2, 2); //初始化内存 m1.init_matrix(); m2.init_matrix(); m3.init_matrix(); //初始化数据 k = 1; for (i = 0;i < m1.m;i++) { for (j = 0;j < m1.n;j++) { m1.write(i,j,k++); } } for (i = 0;i < m2.m;i++) { for (j = 0;j < m2.n;j++) { m2.write(i,j,k++); } } m_c.multiply(ref m1,ref m2, ref m3); //output.Text = Convert.ToString(m3.read(1,1)); output.Text = Convert.ToString(m_c.det(ref m1)); } /* private void button1_Click(object sender, EventArgs e) { zk = Convert.ToDouble(input.Text); //时间方程 xkg = xk; pkg = pk + q; //状态方程 kk = pkg / (pkg + r); xk = xkg + kk * (zk - xkg); pk = (1 - kk) * pkg; //输出 output.Text = Convert.ToString(xk); } private void textBox1_TextChanged(object sender, EventArgs e) { } * */ } }
希望本文所述对大家的C#程序设计有所帮助。
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