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C#中矩阵运算方法 C#中矩阵运算方法实例分析

jdh99 人气:7
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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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