c#实现二元关系性质的判定
逻辑层代码如下：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace 二元关系判定
{
class ClassDuality
{
private int age;
public int Age
{
get { return age; }
set { age = value; }
}
/// <summary>
/// 判断自反和反自反
/// </summary>
/// <param name="a"></param>
/// <returns></returns>
public int zifan( double[,] a, int n)
{
int flag1 = 0,flag2=0;
for( int i=0;i<n;i++)
for ( int j = 0; j < n; j++)
{
if (i == j)
{
if (a[i, j] == 1)
{
flag1++;
}
else if (a[i, j] == 0)
{
flag2++;
}
}
}
if (flag1 == n)
{
return 1; //若为1表示自反性
}
else if (flag2 == n)
{
return 0; //若为1表示反自反性
}
else
{
return -1; //若为-1表示既不自反也不反自反
}
}
/// <summary>
/// 判断对称和反对陈
/// </summary>
/// <param name="a"></param>
/// <returns></returns>
public int duichen( double[,] a, int n)
{
int flag = 0, flag1 = 0,flag2=0;
for( int i=0;i<n;i++)
for ( int j = 0; j < n; j++)
{
if (i != j&&a[i, j] != 0)
{
flag = 1;
}
else if (i != j && a[i, j] != a[j, i])
{
flag1 = 1;
}
else if (a[i, j] == 1 && a[j, i] == 0)
{
flag2++;
}
}
if (flag == 0)
{
return 1; //表示既对称有反对称
}
else if (flag1 == 0)
{
return 2; //表示对称
}
else if (flag2 == n)
{
return 3; //表示反对陈
}
else
{
return 0;
}
}
/// <summary>
/// 判断传递
/// </summary>
/// <param name="a"></param>
/// <returns></returns>
public int chuandi( double[,] a, int n)
{
bool t = true;
int i = 0,j=0,k=0;
while (i < n)
{
while (j < n)
{
if (a[i, j] == 1)
{
while (k < n)
{
if (a[j, k] == 1 && a[i, k] != 1)
{
t = false; break;
}
else
{
k++;
}
}
}
j++;
}
i++;
}
if (t == true)
{
return 1; //传递
}
else
{
return 0; //非传递
}
}
}
}
实现代码如下：
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;
namespace 二元关系判定
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button2_Click( object sender, EventArgs e)
{
textBox1.Text = "";
}
private int arr()
{
string text1 = textBox1.Text;
string[] result = System.Text.RegularExpressions.Regex.Split(text1, "\r\n");
string[] ss = System.Text.RegularExpressions.Regex.Split(result[0], ",");
return result.Length;
}
private double[,] read()
{
string text1 = textBox1.Text;
double[,] resultArray;
string[] result = System.Text.RegularExpressions.Regex.Split(text1, "\r\n");
string[] ss = System.Text.RegularExpressions.Regex.Split(result[0], ",");
resultArray= new double[result.Length,ss.Length];
if (result.Length != ss.Length)
{
return resultArray;
}
int i = 0, j = 0;
foreach ( string str in result)
{
string[] res = System.Text.RegularExpressions.Regex.Split(str, ",");
foreach ( string re in res)
{
resultArray[i, j] = Convert.ToDouble(re);
j++;
}
i++;
j = 0;
}
return resultArray;
}
private void button1_Click( object sender, EventArgs e)
{
try
{
double[,] result = read();
if (result.Length == 0)
{
MessageBox.Show( "输入的矩阵错误！");
}
foreach ( double i in result)
{
if (i != 1 && i != 0)
{
MessageBox.Show( "输入数字有误！", "提示",MessageBoxButtons.OK,MessageBoxIcon.Error);
return;
}
}
ClassDuality cd = new ClassDuality();
string res = "该矩阵有以下性质:\n";
if (cd.zifan(result,arr()) == 1)
{
res = res + "1、自反的\n";
}
else if (cd.zifan(result,arr()) == 0)
{
res = res + "1、反自反的\n";
}
else if (cd.zifan(result,arr()) == -1)
{
res = res + "1、既不是自反的也不是反自反的\n";
}
if (cd.duichen(result,arr()) == 1)
{
res = res + "2、既是对称又是反对陈的\n";
}
else if (cd.duichen(result,arr()) == 2)
{
res = res + "2、对称的\n";
}
else if (cd.duichen(result,arr()) == 3)
{
res = res + "2、反对称的\n";
}
else if (cd.duichen(result,arr()) == 0)
{
res = res + "2、既不是对称的又不是反对称的\n";
}
if (cd.chuandi(result,arr()) == 1)
{
res = res + "3、传递的\n";
}
else if (cd.chuandi(result,arr()) == 0)
{
res = res + "4、非传递的";
}
MessageBox.Show(res, "二元关系性质",MessageBoxButtons.OK,MessageBoxIcon.Information);
}
catch
{
MessageBox.Show( "输入错误字符！", "提示",MessageBoxButtons.OK,MessageBoxIcon.Error);
}
}
}
}
整个源程序在附件