c++ 离散数学 群的相关判断及求解

  1 #include<iostream>
  2 
  3 using namespace std;
  4 //返回元素下标
  5 //n为集合元素个数，set[]为存放元素集合，s为要求下标的元素
  6 int sub(int n, char set[], char s)
  7 {
  8     for (int i = 0; i < n; i++)
  9     {
 10         if (set[i] == s)
 11             return i;
 12     }
 13 }
 14 
 15 //判断群是不是循环群
 16 //n为群的阶数，，set[]为存放集合元素的数组，ele[6][6]为存放群运算表的二维数组
 17 int Iscyc(int n, char set[], char ele[][6])
 18 {
 19     int x, count;
 20     int cnt;
 21     for (int i = 0; i < n; i++)
 22     {
 23         count = 0;//计量---集合中的元素是否由set[i]的幂组成
 24 
 25         for (int j = 0; j < n; j++)
 26         {
 27             x = i;
 28             cnt = 2;//防止死循环
 29             if (set[j] == set[i])
 30                 count++;
 31             if (set[j] == ele[x][i])
 32                 count++;
 33 
 34             while (set[j] != ele[x][i] && cnt < n)
 35             {
 36                 x = sub(n, set, ele[x][i]);
 37                 cnt++;
 38                 if (set[j] == ele[x][i])
 39                     count++;
 40             }
 41 
 42 
 43         }
 44         if (count == n)
 45             return 1;
 46     }
 47     return 0;
 48 
 49 }
 50 
 51 //输出循环群的生成元
 52 //n为群的阶数，，set[]为存放集合元素的数组，ele[6][6]为存放群运算表的二维数组
 53 void gen(int n, char set[], char ele[][6])
 54 {
 55     int x, count;
 56     int cnt;
 57     for (int i = 0; i < n; i++)
 58     {
 59         count = 0;
 60         for (int j = 0; j < n; j++)
 61         {
 62             x = i;
 63             cnt = 2;
 64             if (set[j] == set[i])
 65                 count++;
 66             if (set[j] == ele[x][i])
 67                 count++;
 68             while (set[j] != ele[x][i] && cnt < n)
 69             {
 70                 x = sub(n, set, ele[x][i]);
 71                 cnt++;
 72                 if (set[j] == ele[x][i])
 73                     count++;
 74             }
 75         }
 76         if (count == n)
 77             cout << "循环群的生成元是：" << set[i] << endl;
 78     }
 79 }
 80 
 81 //给出群中每个元素的阶
 82 //n为群的阶数，，set[]为存放集合元素的数组，ele[6][6]为存放群运算表的二维数组,ie为幺元
 83 void order(int n, char set[], char ele[][6], char ie)
 84 {
 85     int  x, cnt;
 86     for (int i = 0; i < n; i++)
 87     {
 88         if (set[i] != ie)
 89         {
 90             x = i;
 91             cnt = 2;
 92             if (ele[x][i] == ie)
 93                 cout << set[i] << "的阶是" << cnt << endl;
 94             while (ele[x][i] != ie)
 95             {
 96                 x = sub(n, set, ele[x][i]);
 97                 cnt++;
 98                 if (ele[x][i] == ie)
 99                 {
100                     cout << set[i] << "的阶是" << cnt << endl;
101                     break;
102                 }
103             }
104         }
105         else
106             cout << set[i] << "的阶是1" << endl;
107     }
108 
109 
110 }
111 
112 //输出群中每个元素的逆元
113 //n为群的阶数，，set[]为存放集合元素的数组，ele[6][6]为存放群运算表的二维数组
114 //ie为幺元
115 void inver(int n, char set[], char ele[][6], char ie)
116 {
117 
118     for (int i = 0; i < n; i++)
119         for (int j = 0; j < n; j++)
120         {
121             if (ele[i][j] == ie)
122             {
123                 cout << set[i] << "的逆元是" << set[j] << endl;
124                 break;
125             }
126         }
127 
128 
129 }
130 
131 
132 int main()
133 {
134     cout << "请输入元素的个数(要求3到6之间)" << endl;
135     int inputnum;
136     cin >> inputnum;
137     cout << "请输入元素" << endl;
138     char Y[4];
139     for (int i = 0; i < inputnum; i++)
140     {
141         cin >> Y[i];
142     }
143     cout << "请输入群的幺元" << endl;
144     char e;
145     cin >> e;
146     char outputform[7][7];
147     char out1[6][6];
148     outputform[0][0] = ‘*‘;
149     for (int i = 1; i <= inputnum; i++)
150     {
151         outputform[0][i] = Y[i - 1];
152         outputform[i][0] = Y[i - 1];
153     }
154     int m;
155     for (int i = 0; i < inputnum; i++)
156     {
157         if (Y[i] == e)
158         {
159             m = i;
160             break;
161         }
162     }
163     int n = m;
164     int r = m;
165     for (int i = 0; i < inputnum; i++)
166     {
167         out1[n%inputnum][m] = Y[n%inputnum];
168         n = n + 1;
169     }
170     int q;
171     for (int i = 0; i < inputnum; i++)
172     {
173         q = 0;
174         for (int y = 0; y < inputnum; y++)
175         {
176             if (out1[i][m] == Y[y])
177             {
178                 q = y;
179                 break;
180             }
181         }
182         for (int j = m; j < inputnum + m; j++)
183         {
184             out1[i][j%inputnum] = Y[q%inputnum];
185             q++;
186         }
187     }
188     for (int i = 1; i <= inputnum; i++)
189     {
190         for (int j = 1; j <= inputnum; j++)
191         {
192             outputform[i][j] = out1[i - 1][j - 1];
193         }
194     }
195     for (int i = 0; i <= inputnum; i++)
196     {
197         for (int j = 0; j <= inputnum; j++)
198         {
199             cout << outputform[i][j] << ‘ ‘;
200         }
201         cout << endl;
202     }
203 
204 
205     cout << "---------------------------" << endl;
206     if (Iscyc(inputnum, Y, out1))
207     {
208         cout << "<G,*>是循环群" << endl;
209         gen(inputnum, Y, out1);
210     }
211     else
212         cout << "<G,*>不是循环群" << endl;
213     cout << "---------------------------" << endl;
214     order(inputnum, Y, out1, e);
215     cout << "---------------------------" << endl;
216     inver(inputnum, Y, out1, e);
217 
218 
219     return 0;
220 }