我很难尝试编写这些函数。他们工作不正常,不知道我做错了什么。至于Transitive,我甚至无法开始,希望能给你提供任何帮助,以及我在功能上做错了什么。谢谢。
采样输入:如何测试一组是自反的,对称的,反对称的和/或传递的?
0 1 2 3 //elements (A)
0 0 //relations (B)
1 1
2 2
3 3
x y z //elements (A)
x y //relations (B)
y z
y y
z z
x y z //elements (A)
x x //relations (B)
y z
x y
z y
x z
y y
z x
y x
z z
1 2 3 4 5 6 7 8 //elements (A)
1 4 //relations (B)
1 7
2 5
2 8
3 6
4 7
5 8
6 6
1 1
2 2
代码:
bool reflexive(int a[], int sizeOfA, int b[], int sizeOfB)
{
bool hold = true;
for(int i=0; i+1<sizeOfB; i+=2)
{
int e = b[i];
int e1 = b[i];
if(pair_is_in_relation(e1, e, b, sizeOfB) == false)
{
if (hold)
{
return false;
break;
}
}
}
if (hold)
cout << "Reflexive - Yes" << endl;
else
cout << "Reflexive - No" << endl;
return hold;
}
bool symmetric(int a[], int sizeOfA, int b[], int sizeOfB)
{
bool hold = true; // set hold to true
for(int i=0; i+1<sizeOfB; i+=2) // for each pair (e,f) in b
{
int e = b[i];
int f = b[i+1];
if(is_in_relation(f, e, b, sizeOfB)) // if pair(e,f) is not in b
{
if(hold) // set hold to false
{
return false;
break;
}
}
}
if(hold) // if hold return true
cout << "Symmetric - Yes" << endl;
else // if hold is false return false
cout << "Symmetric - No" << endl;
}
void antiSymmetric(int b[], int sizeOfB)
{
bool hold = true; // set hold to true
for(int i = 0; i < sizeOfB;) // for each pair (e,f) in b
{
if(hold == false)
{
cout << "AntiSymmetric - No" << endl;
break; //Did not find (e,e) in b
}
for(int j = 0; j < sizeOfB;)
{
if(b[i] == b[j+1] && b[i+1] == b[j]) //If true, then pair(f,e) exists
{
if(b[i+1] != b[i]) //If true, relation is antisymmetric
{
hold = true;
break;
}
else
{
hold = false;
j = j + 2;
}
}
else
{
hold = false;
j = j + 2;
}
}
i = i + 2;
}
if(hold == true)
cout << "AntiSymmetric - Yes" << endl;
}
void transitive(int a[], int sizeOfA, int b[], int sizeOfB)
{
}
int main()
{
char keepGoing = 'y';
while (keepGoing=='y') {
int set1[4] = {0, 1, 2, 3};
int rel1[8] = {0, 0, 1, 1, 2, 2, 3, 3};
cout << "Set 1: " << endl;
reflexive(set1, 3, rel1, 4);
symmetric(set1, 3, rel1, 4);
antiSymmetric(set1, 3, rel1, 4);
cout << endl;
char set2[4] = {'x', 'y', 'z'};
char rel2[8] = {'x', 'y', 'y', 'z', 'y', 'y', 'z', 'z'};
cout << "Set 2: " << endl;
charReflexive(set2, 4, rel2, 8);
charSymmetric(set2, 4, rel2, 8);
charAntiSymmetric(set2, 4, rel2, 8);
cout << endl;
char set3[3] = {'x', 'y', 'z'};
char rel3[18] = {'x', 'x', 'y', 'z', 'x', 'y', 'z', 'y', 'x',
'z', 'y', 'y', 'z', 'x', 'y', 'x', 'z', 'z'};
cout << "Set 3: " << endl;
charReflexive(set3, 3, rel3, 18);
charSymmetric(set3, 3, rel3, 18);
charAntiSymmetric(set3, 3, rel3, 18);
cout << endl;
int set4[8] = {1, 2, 3, 4, 5, 6, 7, 8};
int rel4[20] = {1, 7, 2, 5, 2, 8, 3, 6, 4, 7, 5, 8, 6, 6, 1, 1,
2, 2};
cout << "Set 4: " << endl;
reflexive(set4, 8, rel4, 20);
symmetric(set4, 8, rel4, 20);
antiSymmetric(set4, 8, rel4, 20);
cout << endl << "Would you like to test it again? (y/n): ";
cin >> keepGoing;
}
return 0;
}
您的问题需要更具体。什么不行? (即你期望的产出是什么,你会得到什么?)。您是否尝试过在调试器中运行最小数据集? –
'他们工作不正常。他们在做什么?你期望他们做什么? –
另外,我不是集合论专家,但我认为像“反身”等属性适用于*关系*,而不是集合? –