离散数学实验(六)判断代数系统是半群,幺半群,群。
作者:互联网
给出一个代数系统<G,* >,其中:G={1,2,…,n},* 运算由运算表矩阵 给出,要判断: (1)<G,* >是否为半群; (2)<G,*>是否为含幺半群; (3)<G,*>是否为群。
编程要求 |
通过编程判断给定代数系统是否为群。
#include<iostream>
using namespace std;
int jiehe(int n,char g[100][100])//判断结合律
{
int i, j, k;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
for (k = 0; k < n; k++)
{
if (g[g[i][j] - 'a'][k] != g[i][g[j][k] - 'a']) // 存在一个不满足条件,即没有结合律
{
return 0;
}
}
}
}
return 1;
}
int zuoyao(int n, char g[100][100], char a[100])//判断左幺元
{
int i, j, k, s = 0;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (g[i][j] == a[i])
{
s++;
}
}
if (s == n)
{
return 1;
}
}
}
int youyao(int n, char g[100][100], char a[100])//判断右幺元
{
int i, j, k, s = 0;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (g[j][i] == a[i])
{
s++;
}
}
if (s == n)
{
return 1;
}
}
}
int niyuan(int n, char g[100][100], char a[100])//判断逆元
{
int i, j, k, s = 0;
char e;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (g[i][j] == a[j])
{
s++;
}
}
if (s == n)
{
e = a[i];
break;
}
}
s = 0;
int c = 0;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
if (g[i][j] == e)
{
s++;
}
}
if (s == n)
{
c++;
}
}
if (c == n)
{
return 0;
}
}
int main()
{
char g[100][100];
char a[100];
int i, j, k;
int n;
cout << "输入元素个数:" << endl;
cin >> n;
cout << "输入元素:" << endl;
for (i = 0; i < n; i++)
{
cin >> a[i];
}
cout << "输入代数系统的运算表:" << endl;
cout << "*";
for (i = 0; i < n; i++)
{
cout << " " << a[i];
}
cout << endl;
for (i = 0; i < n; i++)
{
cout << a[i] << " ";
for (j = 0; j < n; j++)
{
cin >> g[i][j];
}
}
int s = 0;
int c = 0;
for (i = 0; i < n; i++)
{
for (j = 0; j < n; j++)
{
for (k = 0; k < n; k++)
{
if (g[i][j] == a[k])
{
s++;
}
}
}
}
if (s==n*n&&jiehe(n,g))
{
cout << "是半群" << endl;
if (zuoyao(n, g, a) && youyao(n, g, a))
{
cout << "是幺半群" << endl;
if (niyuan(n, g, a))
{
cout << "是群" << endl;
}
}
}
}
标签:char,return,int,离散数学,++,半群,100,代数 来源: https://blog.csdn.net/qq_57173265/article/details/117441578