not(or) |not(=>) |not |not(and) |not |and |not(=>) |not(xor) |false |and |not(xor) |xor |and |not(xor) |false |=> |not(=>) |=> |not |=> |xor |=> |not |or |false |not(xor) |or |xor |not(xor) |or |not(=>) |true |xor |and |true |xor |or |true |
整理一下
not(or)not(and)not,ornot,andnot,=>not,not(=>)not(=>),=>not(=>),true not(=>),not(xor) =>,false=>,xorxor,and,not(xor)xor,or,not(xor)xor,and,true xor,or,truenot(xor),and,falsenot(xor),or,false xor功能比false更丰富，因为false = a xor a，所以一切出现false的地方都可以用xor取代
I got this result by a violent method:If you run this program ,you can see more things.
#include#include #include #include using namespace std;const int N = 1 << 16;//There are 65536 probability.bool a[16][4]; //A 10 B 12const char *desc[16] = { "false", "not (a or b)","not (a => b)", "not b","not (b => a)", "not a","a xor b", "not (a and b)","a and b", "not (a xor b)","a", "b => a","b", "a => b","a or b", "true" };const char*desc_simple[16] = { "false", "not(or)", "not(=>)", "not", "not(=>)", "not", "xor", "not(and)", "and", "not(xor)", "a", "=>", "b", "=>", "or", "true" };/*上表中，只有条件运算符和非条件运算符是不满足交换律，但是它们是等价的也就是说：2，4等价，11，13等价,3,5等价*/int eq[] = { 2, 4, 11, 13, 3, 5 }, eq_size = 3;void init(){//初始化16个逻辑词 for (int i = 0; i < 16; i++){ int x = i; for (int j = 0; j < 4; j++){ a[i][j] = x & 1; x >>= 1; } }}void show(int n){ printf("\t%-16s", desc[n]); for (int i = 0; i < 4; i++)printf("%2d", a[n][i]); puts("");}void table(){ show(10), show(12);//10和12代表着原始的a和b puts("\t---------------------------"); for (int i = 0; i < 16; i++)show(i);}int ans[N], ai = 0;//存储全部的答案int book[N][16][3];//对于N中逻辑词组，16个逻辑词来源于三个人：a，b，操作 //判断集合set是否是完备的逻辑词组bool ok(int n){ int temp = n; n |= (1 << 10) | (1 << 12);again: for (int i = 0; i < 16; i++) if (n&(1 << i)) for (int j = 0; j < 16; j++) if (n&(1 << j)) for (int k = 0; k < 16; k++) if (n&(1 << k)){ //i,j分别操作数，k表示操作符，也就是k表示运算法则 int t = 0;//t表示运算结果 for (int l = 0; l < 4; l++)//将四位分别进行运算 t |= (a[k][a[i][l] + a[j][l] * 2] << l); if ((n&(1 << t)) == 0){ n |= (1 << t); book[temp][t][0] = i, book[temp][t][1] = j, book[temp][t][2] = k; goto again; } } return n == N - 1;}void print(bool detail){ for (int i = 0; i < ai; i++){ int n = ans[i]; for (int j = 0; j < 16; j++){ if (n&(1 << j)){ const char*des = detail ? desc[j] : desc_simple[j]; printf("%-15s|", des); } } puts(""); if (detail){ for (int i = 0; i < 16; i++){ if ((n&(1 << i)) || i == 10 || i == 12)continue; printf("\t\t%-15s = %-15s [%-15s] %-15s\n", desc[i], desc[book[n][i][0]], desc[book[n][i][2]], desc[book[n][i][1]]); } puts(""); } }}//根据eq数组去掉等价的元素void simple(){ for (int i = 0; i < ai; i++){ int n = ans[i]; for (int j = 0; j < eq_size; j++){ if (n&(1 << eq[j << 1 | 1])){ n &= ~(1 << eq[j << 1 | 1]); n |= (1 << eq[j << 1]); } } ans[i] = n; for (int j = 0; j < i; j++) if (ans[i] == ans[j]) ans[i] = -1; } int ind = 0; for (int i = 0; i < ai; i++) if (ans[i] != -1)ans[ind++] = ans[i]; ai = ind;}bool visited(int x){ for (int i = 0; i < ai; i++) if ((ans[i] & x) == ans[i]) return true; return false;}int main(){ init(); table();//打印表，说清逻辑运算及其对应的编码 memset(book, -1, sizeof(book)); for (int i = 0; i < N; i++){ //先判断i的某个子集是否已经是最小逻辑词组，如果是，则i必然不是最小逻辑词组 if (visited(i))continue; if (ok(i)) ans[ai++] = i; } simple(); print(false); return 0;}