猴子摘香蕉问题python一阶谓词 谓词逻辑猴子摘香蕉

利用一阶谓词逻辑求解猴子摘香蕉问题：房内有一个猴子，一个箱子，天花板上挂了一串香蕉，其位置如图1所示，猴子为了拿到香蕉，它必须把箱子搬到香蕉下面，然后再爬到箱子上。请定义必要的谓词，列出问题的初始化状态（可变），目标状态（猴子拿到了香蕉，站在箱子上，箱子位于位置b）。（附加：从初始状态到目标状态的谓词演算过程。）

香蕉位置在B即0

思路：
用数字表示猴子和香蕉的位置，-1为A，0为B，1为C；当猴子站在箱子上用1表示，反之则用-1表示。用一个结构体表示整个场景的状态，
其中包括猴子和箱子的位置以及猴子是否在箱子上，猴子每次进行下一次行为记为下一个状态。所有状态用整个结构体数组存储，每一次猴子行为
用另一字符指针数组存储，最后打印输出猴子摘香蕉过程。
代码：

#include<iostream>
#include <stdio.h>
using namespace std;

struct State
{
    int monkey; 
    int box;
    int banana;//香蕉在B
    int monbox; 
};
struct State States[10];
const char* actionsave[10];

void monkeygoto(int b, int i)//猴子走向位置
{
    int a;
    a = b;
    if (a == -1)
    {
        actionsave[i] = "Monkey go to A";
        States[i + 1] = States[i];
        States[i + 1].monkey = -1;
    }
    else if (a == 0)
    {
        actionsave[i] = "Monkey go to B";
        States[i + 1] = States[i];
        States[i + 1].monkey = 0;
    }
    else if (a == 1)
    {
        actionsave[i] = "Monkey go to C";
        States[i + 1] = States[i];
        States[i + 1].monkey = 1;
    }
   
}

void movebox(int a, int i)//猴子推箱子移动位置
{
    int B;
    B = a;
    if (B == -1)
    {
        actionsave[i] = "monkey move box to A";
        States[i + 1] = States[i];
        States[i + 1].monkey = -1;
        States[i + 1].box = -1;
    }
    else if (B == 0)
    {
        actionsave[i] = "monkey move box to B";
        States[i + 1] = States[i];
        States[i + 1].monkey = 0;
        States[i + 1].box = 0;
    }
    else if (B == 1)
    {
        actionsave[i] = "monkey move box to C";
        States[i + 1] = States[i];
        States[i + 1].monkey = 1;
        States[i + 1].box = 1;
    }
}

void climbonto(int i)//猴子爬上箱子
{
    actionsave[i] = "Monkey climb onto the box";
    States[i + 1] = States[i];
    States[i + 1].monbox = 1;
}

void climbdown(int i)//如果初始状态猴子在箱子上，则需要爬下来
{
    actionsave[i] = "Monkey climb down from the box";
    States[i + 1] = States[i];
    States[i + 1].monbox = -1;
}

void reach(int i)//猴子拿到香蕉
{
    actionsave[i] = "Monkey reach the banana";
}

void showSolution(int i)//打印猴子过程操作
{
    int c;
    cout<<"Result to problem:"<<endl;
    for (c = 0; c < i + 1; c++)
    {
        cout<<actionsave[c]<<endl;
    }
    cout << endl;
}

void nextStep(int i)//猴子下一步动作
{
    int c;
    int j;
    //超过一定步数，判断为有问题
    if (i >= 10)
    {
        cout << "steplength reached 10,have problem！ " << endl;
        return;
    }
    //成功拿到香蕉
    if (States[i].monbox == 1 && States[i].monkey == 0 && States[i].banana == 0 && States[i].box == 0)
    {
        if (i == 0)
        {
            reach(i);
        }
        showSolution(i);
        return;
    }

    j = i + 1;//进行数据更新，用来标记当前是第几个状态
    if (States[i].monkey == 0)//猴子位置B
    {
        if (States[i].box == 0)//箱子位置B
        {
            if (States[i].monbox == -1)//猴子没爬上箱子
            {
                climbonto(i);
                reach(i + 1);
                nextStep(j);
            }
        }
        else if(States[i].box == 1)//箱子位置C
        {
            monkeygoto(States[i].box, i);
            nextStep(j);
        }
        else//箱子位置A
        {
            monkeygoto(States[i].box, i);
            nextStep(j);
        }
    }
   
    if (States[i].monkey == -1)//猴子位置A
    {
        if (States[i].box == -1)//同上
        {
            if (States[i].monbox == -1)
            {
                movebox(0, i);
                nextStep(j);
            }
            else
            {
                climbdown(i);
                nextStep(j);
            }
        }
        else if (States[i].box == 0)
        {
            monkeygoto(0, i);
            nextStep(j);
        }
        else
        {
            monkeygoto(1, i);
            nextStep(j);
        }
    }
    
    if (States[i].monkey == 1)//猴子位置C
    {
        if (States[i].box == 1)
        {
            if (States[i].monbox == -1)
            {
                movebox(0, i);
                nextStep(j);
            }
            else
            {
                climbdown(i);
                nextStep(j);
            }
        }
        else if (States[i].box == -1)
        {
            monkeygoto(-1, i);
            nextStep(j);
        }
        else
        {
            monkeygoto(0, i);
            nextStep(j);
        }
    }
}
int main()
{
    while (1)   //测试
    {
        States[0].banana = 0;
        cout << "请输入初始位置：\n";
        cout << "monkey(-1 or 0 or 1)：";
        cin >> States[0].monkey;
        cout << "box(-1 or 0 or 1)：";
        cin >> States[0].box;
        cout << "monbox(-1 or 1)：";
        cin >> States[0].monbox;   
        //判断输入是否符合要求，错误则提示并且重新输入
        if (States[0].monkey == -1 || States[0].monkey == 0 || States[0].monkey== 1)
        {
            if (States[0].box == -1 || States[0].box == 0 || States[0].box == 1)
            {
                if (States[0].monbox == -1 || States[0].monbox == 1)
                {
                    if ((States[0].monkey != States[0].box) && States[0].monbox == 1)
                    {
                        cout << "input wrong" << endl<<endl;
                    }
                    else  nextStep(0);
                }
                else cout << "input wrong!" << endl<<endl;
            }
            else cout << "input wrong!" << endl<<endl;
        }
        else cout << "input wrong!" << endl<<endl;
    }
}

运行展示：

初始猴子在C，箱子在B；然后猴子走向B，爬上箱子摘到香蕉。

初始猴子和箱子在C并且在箱子上面，这时猴子需要 箱子上爬下来，然后把箱子推到B，再爬上箱子摘到香蕉。