人工智能归结原理实验

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文章目录

• 前言
• 一、流程
• 二、步骤
• • 0.声明
  • 1.消去蕴含词和等值词
  • 2.使否定词仅作用于原子公式
  • 3.使量词间不含同名指导变元
  • 4.化为前束范式
  • 5.消去存在量词
  • 6.消去全称量词
  • 7.化公式为合（&）取范式
  • 8.消去合取词，得到子句集
  • 9.合一算法
  • 10.归结原理
• 运行结果
• • • 1.
    • 2.
    • 3.
    • • • 代码简陋，还请包涵
      • 当时做的时候没想那么多，所以前面化简和后面归结存在一些问题。

---

前言

考察 命题逻辑归结推理
要求 熟悉C++迭代器使用，有一定离散数学基础
复杂式子无法实现

---

一、流程

二、步骤

0.声明

typedef string Formula;
typedef vector<string> SubsentenceSet;
typedef string::iterator FormulaIter;
typedef string::reverse_iterator FormulaRevIter;
// 公式符号定义
const char EQ = '#';          // 存在量词符号
const char UQ = '@';          // 全称量词符号
const char IMPLICATION = '>'; // 蕴含符号
const char NEGATION = '~';    // 否定符号
const char CONJUNCTION = '&'; // 合取符号
const char DISJUNCTION = '|'; // 析取符号
const char CONSTANT_ALPHA[] = {'a', 'b', 'c', 'd', 'e', 'i', 'j', 'k'};
const char FUNC_ALPHA[] = {'f', 'g', 'h', 'l', 'm', 'n', 'o'};

1.消去蕴含词和等值词

//消去蕴含连接词
Formula &RemoveImplication(Formula &f)
{
        FormulaIter iter;
        while ((iter = find(f.begin(), f.end(), IMPLICATION)) != f.end())//寻找蕴含符号
        {
                *iter = DISJUNCTION; // 将蕴含符号替换为析取符号
                FormulaRevIter revIter(iter);
                revIter = GetBeginOfFormula(revIter, f.rend()); // 查找蕴含前件
                iter = revIter.base();                          // 反向迭代器到正向迭代器
                f.insert(iter, NEGATION); // 在前件前面插入否定
        }
        return f;
}

2.使否定词仅作用于原子公式

//将否定符号移到紧靠谓词的位置
Formula &MoveNegation(Formula &f)
{
        FormulaIter iter = find(f.begin(), f.end(), NEGATION);
        while (iter != f.end())
        {
                if (*(iter + 1) == '(')
                { // 否定不是直接修饰谓词公式，需要内移
                        // 否定符号修饰着带量词的谓词公式
                        if (*(iter + 2) == EQ || *(iter + 2) == UQ)
                        {
                                // 量词取反
                                *(iter + 2) == EQ ? *(iter + 2) = UQ : *(iter + 2) = EQ;

                                string leftDonePart(f.begin(), iter + 5);
                                // 移除否定符号

                                leftDonePart.erase(find(leftDonePart.begin(), leftDonePart.end(), NEGATION));

                                string rightPart(iter + 5, f.end());
                                // 否定内移
                                rightPart.insert(rightPart.begin(), NEGATION);
                                // 递归处理右部分
                                MoveNegation(rightPart);
                                string(leftDonePart + rightPart).swap(f);
                                return f;
                        }
                        else
                        {                                            // 修饰着多个公式，形如~(P(x)|Q(x))
                                iter = f.insert(iter + 2, NEGATION); // 内移否定符号
                                while (1)
                                {
                                        iter = FindFormula(iter, f.end());
                                        //assert(iter != f.end() && "No Predicate Formula!");
                                        FormulaIter iter2 = FindPairChar(
                                            iter, f.end(), '(', ')');
                                        ++iter2;
                                        if (IsConnector(*iter2))
                                        {
                                                *iter2 == DISJUNCTION ? *iter2 = CONJUNCTION
                                                                      : *iter2 = DISJUNCTION;
                                                iter = f.insert(iter2 + 1, NEGATION);
                                        }
                                        else
                                                break;
                                }
                                f.erase(find(f.begin(), f.end(),
                                             NEGATION)); // 清除原否定符号
                                return MoveNegation(f);
                        }
                }
                else if (*(iter + 1) == NEGATION)
                { // 两个否定，直接相消
                        f.erase(iter, iter + 2);
                        return MoveNegation(f); // 重新处理
                }
                else
                {
                        string leftDonePart(f.begin(), iter + 1);
                        string rightPart(iter + 1, f.end());
                        MoveNegation(rightPart);
                        string(leftDonePart + rightPart).swap(f);
                        return f;
                        //iter = find(iter + 1, f.end(), NEGATION);
                }
        }
        return f;
}

3.使量词间不含同名指导变元

//适当改名使量词间不含同名指导变元，对变元标准化。
Formula &StandardizeValues(Formula &f)
{
        set<char> checkedAlpha;
        FormulaIter iter = FindQuantifier(f.begin(), f.end());
        while (iter != f.end())
        {
                char varName = *++iter; // 获取变量名
                if (checkedAlpha.find(varName) == checkedAlpha.end())
                {
                        checkedAlpha.insert(varName);
                }
                else
                { // 变量名冲突了，需要改名
                        // 获取新名子
                        char newName = FindNewLowerAlpha(checkedAlpha);

                        checkedAlpha.insert(newName);
                        // 查找替换右边界
                        FormulaIter rightBorder = FindPairChar(
                            iter + 2, f.end(), '(', ')');
                        // 将冲突变量名替换为新的名子
                        *iter = newName;
                        replace(iter, rightBorder, varName, newName);
                        iter = rightBorder; // 移动到新的开始
                }
                iter = FindQuantifier(iter, f.end());
        }
        return f;
}

4.化为前束范式

//化为前束范式。
Formula &TransformToPNF(Formula &f)
{
        FormulaIter iter = FindQuantifier(f.begin(), f.end());
        if (iter == f.end())
                return f;
        else if (iter - 1 == f.begin())
        { // 量词已经在最前面
                iter += 3;
                string leftPart(f.begin(), iter);
                string rightPart(iter, f.end());
                TransformToPNF(rightPart); // 递归处理右部分
                (leftPart + rightPart).swap(f);
        }
        else
        {                                          // 量词在内部，需要提到前面
                string quantf(iter - 1, iter + 3); // 保存量词
                f.erase(iter - 1, iter + 3);       // 移除量词
                f.insert(f.begin(), quantf.begin(), quantf.end());
                return TransformToPNF(f); // 继续处理
        }
        return f;
}

5.消去存在量词

//消去存在量词。
Formula &RemoveEQ(Formula &f)
{
        set<char> checkedAlpha;
        FormulaIter eqIter = find(f.begin(), f.end(), EQ);
        if (eqIter == f.end())
                return f;
        FormulaRevIter left = leftleft(FormulaRevIter(eqIter), f.rend());
        FormulaRevIter uqIter = findRev(left, f.rend(), UQ);
        if (uqIter == f.rend())
        { // 该存在量词前没有任意量词
                char varName = *(eqIter + 1);
                char newName = GetNewConstantAlha(f);
                auto rightBound = FindPairChar(eqIter + 3, f.end(), '(', ')');
                //assert(rightBound != f.end());
                replace(eqIter + 3, rightBound, varName, newName); // 常量化
                f.erase(eqIter - 1, eqIter + 3);                   // 移除存在量词
        }
        else
        {
                // 记录公式中已经存在的字母
                copy_if(f.begin(), f.end(), checkedAlpha);
                const char oldName = *(eqIter + 1);
                // 准备任意量词的函数来替换该存在量词
                const char funcName = FindFuncAlpha(checkedAlpha);
                string funcFormula;
                funcFormula = funcFormula + funcName + '(' + *(uqIter - 1) + ')';
                f.erase(eqIter - 1, eqIter + 3); // 移除存在量词
                ReplaceAlphaWithString(f, oldName, funcFormula);
        }
        //RemoveOuterBracket(f, f.begin());
        return RemoveEQ(f); // 递归处理
}

6.消去全称量词

//消去全称量词。
Formula &RemoveUQ(Formula &f)
{
        FormulaIter uqIter = find(f.begin(), f.end(), UQ);
        while (uqIter != f.end())
        {
                uqIter = f.erase(uqIter - 1, uqIter + 3); // 直接移除全称量词
                uqIter = find(uqIter, f.end(), UQ);       // 继续扫描
        }
        //RemoveOuterBracket(f, f.begin());
        return f;
}

7.化公式为合（&）取范式

//化为合取范式。
Formula &TransformToConjunction(Formula &f)
{
        FormulaIter dis = find(f.begin(), f.end(), DISJUNCTION);
        FormulaRevIter left = leftfind((FormulaRevIter)dis, f.rend());
        FormulaIter leftt = left.base() - 1;
        FormulaIter right = rightfind(dis + 1, f.end());
        FormulaRevIter leftcon = findRev((FormulaRevIter)dis, left, CONJUNCTION);
        FormulaIter lefttcon = leftcon.base() - 1;
        FormulaIter rightcon = find(dis, right, CONJUNCTION);
        if (leftt != lefttcon && rightcon == right)
        {
                cout << string(dis + 1, right) << endl;
                cout << string(leftt + 1, lefttcon) << endl;
                cout << string(lefttcon + 1, dis) << endl;
                string temp = "(" + string(dis + 1, right) + "|" + string(leftt + 1, lefttcon) + ")&(" + string(dis + 1, right) + "|" + string(lefttcon + 1, dis - 1) + ")";
                // cout << temp << endl;
                f.replace(leftt, right, temp);
        }
        else if (leftt == lefttcon && right != rightcon)
        {
                // cout << string(leftt + 1, dis) << endl;
                // cout << string(dis + 1, rightcon) << endl;
                // cout << string(rightcon + 1, right - 1) << endl;
                string temp = "(" + string(leftt + 1, dis) + "|" + string(dis + 1, rightcon) + ")&(" + string(leftt + 1, dis) + "|" + string(rightcon + 1, right - 1) + ")";
                // cout << temp << endl;
                f.replace(leftt, right, temp);
        }
        else if (leftt != lefttcon && right != rightcon)
        {
                string str1 = string(leftt + 1, lefttcon);
                string str2 = string(lefttcon + 1, dis);
                string str3 = string(dis + 1, rightcon);
                string str4 = string(rightcon + 1, right);
                cout << str1 << endl;
                cout << str2 << endl;
                cout << str3 << endl;
                cout << str4 << endl;
                string temp;
                if (str1 == str3)
                {
                        temp = str1 + "&(" + str2 + "|" + str4 + ")";
                }
                else if (str1 == str4)
                {
                        temp = str1 + "&(" + str2 + "|" + str3 + ")";
                }
                else if (str2 == str3)
                {
                        temp = str2 + "&(" + str1 + "|" + str4 + ")";
                }
                else if (str2 == str4)
                {
                        temp = str2 + "&(" + str1 + "|" + str3 + ")";
                }
                f.replace(leftt, right + 1, temp);
        }
        return f;
}

8.消去合取词，得到子句集

这里没有对子句做标准化，即不同的子句用不同的变元，和前面的标准化大同小异。

//消去合取词，以子句为元素组成一个集合S。
void ExtractSubsentence(SubsentenceSet &subset, Formula &f)
{
        FormulaIter leftIter = f.begin();
        FormulaIter middleIter = find(f.begin(), f.end(), CONJUNCTION);
        while (middleIter != f.end())
        {
                if (*leftIter == '(' && *(middleIter - 1) == ')')
                        subset.push_back(string(leftIter + 1, middleIter - 1));
                else
                        subset.push_back(string(leftIter, middleIter));
                leftIter = middleIter + 1;
                middleIter = find(middleIter + 1, f.end(), CONJUNCTION);
        }
        if (*leftIter == '(' && *(middleIter - 1) == ')')
                subset.push_back(string(leftIter + 1, middleIter - 1));
        else
                subset.push_back(string(leftIter, middleIter));
}

9.合一算法

合一代码，直接调用syncretism函数。

bool syncretism(const string tf1, const string tf2, vector<transform> &mgu) //合一方法，判断是否可进行合一
{
        string f1 = tf1, f2 = tf2;
        while (!same(f1, f2)) //f1与f2中的符号不完全相同时才进入while循环
        {
                transform t = dif(f1, f2); //得到f1和f2的一个差异集，并把它赋给t
                int flag = legal(t);
                if (flag == 0)
                        return false;
                else
                {
                        mgu.push_back(t);
                        f1 = change(f1, mgu.back());
                        f2 = change(f2, mgu.back());
                        cout << "after change:" << endl;
                        cout << "f1:" << f1 << endl;
                        cout << "f2:" << f2 << endl;
                        if (same(f1, f2))
                                return true; //f1和f2相同时就停止循环
                }
        }
        return false;
}
bool same(const string f1, const string f2) //判断两个谓词f1和f2是否相同
{
        if (f1.length() == f2.length())
        {
                int i = 0;
                while (i < f1.length() && f1.at(i) == f2.at(i))
                        i++;
                if (i == f1.length())
                        return true;
                else
                {
                        return false;
                }
        }
        else
                return false;
}
transform dif(const string f1, const string f2) //求解f1和f2的差异集
{
        int i = 0;
        transform t;
        while (f1.at(i) == f2.at(i)) //第i个相等时就转向比较i+1，直到遇到不相等时就跳出while循环
                i++;
        int j1 = i;
        while (j1 < f1.length() - 1 && f1.at(j1) != ',') //从不相等的部分开始，直到遇到‘，’或到达结尾时跳出while循环
                j1++;
        if (j1 - i == 0)
                return t;
        t.t_f1 = f1.substr(i, j1 - i); //将f1中的不相同的项截取出来放入变量t.t_f1中
        int j2 = i;
        while (j2 < f2.length() - 1 && f2.at(j2) != ',')
                j2++;
        if (j2 - i == 0)
                return t;
        t.t_f2 = f2.substr(i, j2 - i);                   //将f2中的不相同的项截取出来放入变量t.t_f2中
        while (t.t_f1[j1 - i - 1] == t.t_f2[j2 - i - 1]) //去除相同的部分
        {
                t.t_f1.erase(j1 - 1 - i);
                t.t_f2.erase(j2 - i - 1);
                j1--;
                j2--;
        }
        return t;
}
int legal(transform &t) //判断置换t是否合法
{
        if (t.t_f1.length() == 0 || t.t_f2.length() == 0)
                return 0;
        if (var(t.t_f2))
        {
                if (var(t.t_f1) && (varData(t.t_f1) == varData(t.t_f2))) //不能代换合一
                        return 0;
                else
                        return 2;
        }
        if (!var(t.t_f1)) //若t_f1和t_f2都不是变量，也不能合一
                return 0;
        string temp;
        temp = t.t_f1;
        t.t_f1 = t.t_f2;
        t.t_f2 = temp; //在t_f1是变量而t_f2不是变量的情况下，交换t_f1和t_f2
        return 1;
}
string varData(string s) //该函数是剥去外层括号
{
        if (s.length() == 1 || s.length() == 0)
                return s;
        if (s.length() > 1)
        {
                int i = 0;
                while (i < s.length() && s.at(i) != '(')
                        i++;
                int j = i;
                while (j < s.length() && s.at(j) != ')')
                        j++;
                string ss = s.substr(i + 1, j - i - 1);
                return varData(ss);
        }
}
bool var(const string s)
{
        if (s.length() == 0)
                return false;
        if (s.length() == 1 && s[0] >= 'A' && s[0] <= 'Z')
                return false;
        if (s.length() > 1)
        {
                int i = 0;
                while (i < s.length() && s.at(i) != '(') //略过不是'('的字符
                        i++;
                int j = i;
                while (j < s.length() && s.at(j) != ')') //略过')'前的字符
                        j++;
                string ss = s.substr(i + 1, j - i - 1); //取出'('和')'之间的串
                return var(ss);                         //递归调用var
        }
        else
        {
                return true;
        }
}
string change(string f, transform q) //该函数查找t_f2在f中的位置并用t_f1替代f中相应的t_f2
{
        int i = f.find(q.t_f2);
        while (i < f.length())
        {
                i = f.find(q.t_f2);
                if (i < f.length())
                        f = f.replace(i, q.t_f2.length(), q.t_f1);
        }
        return f;
}

10.归结原理

int finditer(string s, char a, char b)//*iter.find("~P")的时候莫名报错，于是手写查找。
{
        for (int i = 0; i < s.length(); i++)
        {
                if (s[i] == a && s[i + 1] == b)
                {
                        return i;
                }
        }
        return 0;
}
bool resolution(SubsentenceSet &subset) //归结
{
        for (vector<string>::reverse_iterator rbegin = subset.rbegin(); rbegin != subset.rend();) //从尾部，即目标公式开始归结
        {
                bool f = false;
                bool rev = false; //是否取反
                string s = *rbegin;
                string ss;
                FormulaIter left = s.begin();
                FormulaIter middle = find(left, s.end(), DISJUNCTION);
                string goal = string(left, middle); //归结元素
                string _goal = pdrev(goal, rev);    //归结元素的逆
                vector<string>::reverse_iterator iter;
                while (middle != s.end()) //寻找归结元素的逆，找到末尾结束
                {
                        iter = findreviter(subset, _goal, rev);
                        if (iter == subset.rend()) //找不到继续循环
                        {
                                left = middle + 1;
                                middle = find(left, s.end(), DISJUNCTION);
                                goal = string(left, middle);
                                _goal = pdrev(goal, rev);
                        }
                        else
                        {
                                f = true; //找到标记
                                break;
                        }
                }
                if (!f) //判断是否找到，前面最后一个left和middle没有处理，这里继续处理。
                {
                        iter = findreviter(subset, _goal, rev);
                        if (iter == subset.rend())
                        {
                                rbegin++;
                                continue;
                        }
                }
                ss = (*iter);
                char cc = goal[0];
                int len = goal.length();
                if (rev)
                {
                        cc = goal[1];
                        len -= 1;
                }
                dosyncretism(s.substr(s.find(cc), len), ss.substr(ss.find(cc), len), *iter); //合一
                //擦除归结了的式子
                (*rbegin).erase((*rbegin).find(goal), goal.length());
                string c = "";
                if (rev)
                {
                        c += _goal[0];
                        (*iter).erase((*iter).find(_goal[0]), _goal.length());
                }
                else
                {
                        c += _goal[0] + _goal[1];
                        (*iter).erase(finditer(*iter, _goal[0], _goal[1]), _goal.length());//这里~P找不到，于是手写寻找函数。
                }
				//(*iter).erase(*iter.find(c),_goal.length());找不到，
                //合并剩余式子
                string newstr = (*rbegin) + "|" + (*iter);
                //擦除多余析取符
                for (int i = 0; i < newstr.length(); i++)
                {
                        if ((i == 0 || i == newstr.length() - 1) && newstr[i] == DISJUNCTION)
                        {
                                newstr.erase(i, 1);
                                i -= 1;
                        }
                        else if (newstr[i] == DISJUNCTION && newstr[i + 1] == DISJUNCTION)
                        {
                                newstr.erase(i + 1, 1);
                        }
                }
                cout << endl;
                //删除旧的式子，添加新的式子。
                subset.erase(subset.rbegin().base());
                subset.erase(iter.base() - 1);
                subset.push_back(newstr);

                rbegin = subset.rbegin();
                count(subset);
                //归结出空，则结论为真，归结结束。
                if (newstr == "")
                {
                        return true;
                }
        }
        return false;
}

---

运行结果

1.

Input:
(@x)(D(x)>(B(x)&F(x)))&(@x)(F(x)>N(x))&(@x)(G(x)>D(x))&~(@x)(G(x)>N(x))

Elimination of implied symbols://消去蕴含
(@x)(~D(x)|(B(x)&F(x)))&(@x)(~F(x)|N(x))&(@x)(~G(x)|D(x))&~(@x)(~G(x)|N(x))

Move the negation sign to the front of each predicate://处理非符
(@x)(~D(x)|(B(x)&F(x)))&(@x)(~F(x)|N(x))&(@x)(~G(x)|D(x))&(#x)(G(x)&~N(x))

Standardization of variables://变元标准化
(@x)(~D(x)|(B(x)&F(x)))&(@y)(~F(y)|N(y))&(@z)(~G(z)|D(z))&(#q)(G(q)&~N(q))

Eliminate existential quantifiers://消去存在
(@x)(~D(x)|(B(x)&F(x)))&(@y)(~F(y)|N(y))&(@z)(~G(z)|D(z))&(G(a)&~N(a))

Convert to a front bundle://化为前束
(@x)(@y)(@z)(~D(x)|(B(x)&F(x)))&(~F(y)|N(y))&(~G(z)|D(z))&(G(a)&~N(a))

Omit universal quantifiers://略去全称
(~D(x)|(B(x)&F(x)))&(~F(y)|N(y))&(~G(z)|D(z))&(G(a)&~N(a))

Standardization of Conjunction://化为合取式
(~D(x)|B(x))&(~D(x)|F(x))&(~F(y)|N(y))&(~G(z)|D(z))&G(a)&~N(a)

Eliminate the conjunction://子集
~D(x)|B(x)
~D(x)|F(x)
~F(y)|N(y)
~G(z)|D(z)
G(a)
~N(a)

after change://合一
f1:N(a)
f2:N(a)
mgu={ a/y }

~D(x)|B(x)
~D(x)|F(x)
~G(z)|D(z)
G(a)
~F(a)
after change:
f1:F(a)
f2:F(a)
mgu={ a/x }

~D(x)|B(x)
~G(z)|D(z)
G(a)
~D(a)
after change:
f1:D(a)
f2:D(a)
mgu={ a/z }

~D(x)|B(x)
G(a)
~G(a)
cannot be syncretized//相同变元无法合一

~D(x)|B(x)
//归结出空集，为真
Success

2.

Input:
~((#x)(P(x)&(@y)D(y)))

Elimination of implied symbols:
~((#x)(P(x)&(@y)D(y)))

Move the negation sign to the front of each predicate:
((@x)(~P(x)|(#y)~D(y)))

Standardization of variables:
((@x)(~P(x)|(#y)~D(y)))

Eliminate existential quantifiers:
((@x)(~P(x)|~D(f(x))))

Convert to a front bundle:
(@x)((~P(x)|~D(f(x))))

Omit universal quantifiers:
((~P(x)|~D(f(x))))

Standardization of Skolem:
((~P(x)|~D(f(x))))

Eliminate the conjunction:
(~P(x)|~D(f(x)))

Failure

3.

Input:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Elimination of implied symbols:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Move the negation sign to the front of each predicate:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Standardization of variables:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Eliminate existential quantifiers:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Convert to a front bundle:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Omit universal quantifiers:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Standardization of Skolem:
(~P(x)|Q(x))&(~P(y)|R(y))&S(a)&(~S(z)|~R(z))&P(a)

Eliminate the conjunction:
~P(x)|Q(x)
~P(y)|R(y)
S(a)
~S(z)|~R(z)
P(a)

after change:
f1:P(a)
f2:P(a)
mgu={ a/y }

~P(x)|Q(x)
S(a)
~S(z)|~R(z)
R(a)
after change:
f1:R(a)
f2:R(a)
mgu={ a/z }

~P(x)|Q(x)
S(a)
~S(a)
cannot be syncretized

~P(x)|Q(x)

Success

代码简陋，还请包涵

链接：https://pan.baidu/s/10yz6Yh83lEIpecEYdvhhuw
提取码：qwer

当时做的时候没想那么多，所以前面化简和后面归结存在一些问题。

本文标签： 人工智能原理