人工智能实验 传教士野人

return (pLift->m == pLift->c); }
void PrintLift(struct Lift *pLift) {
int rm=3-(pLift->m), rc=3-(pLift->c); printf("(%d, %d, %d) (%d, %d)\n", pLift->m, pLift->c, pLift->b,rm, rc); }
四、实验描述及要求：
（1） 设计该问题的状态。例如：((左岸牧师数，左岸野人数),(右岸牧师数，右岸野人 数),船的位置)。
（2） 定义目标状态。这里是：((0,0),(3,3),1) （3） 描述可能的动作。船上所能够载人的状态就是可能的操作。用谓词 move/2 表示。 （4） 判断合法状态 （5） 深度优先搜索
struct BOAT *CopyBoat(struct BOAT *pBoat) {
return NewBoat(pBoat->m, pBoat->c); }
struct Lift *In(struct Lift *pLift, struct Lift *pList) {
if (pList == NULL) return NULL; if (Equal(pLift, pList)) return pList; return In(pLift, pList->pNext); }
if (!Safe(pRData)) free(pRData);
{ continue;
}
pRDataList = ConsLift(pRData, pDataList); pPath = Backtrack1(pRDataList, bound);
if (pPath == FAIL) {
free(pRData); }
人工智能实验报告
题目：求解传教士和野人问题 姓名： 班级： 学号： 学院：计算机科学与信息 专业：计算机科学与技术 指导教师：
日期：2011 年 12 月 6 日
一、实验目的：
复习经典谓词演算中的归结原理，掌握人工智能程序设计语言 Prolog，理解通过搜索求 解问题实现人工智能的思想。
二、实验原理：
pData = pDataList; if (In(pData, pDataList->pNext)) {
return FAIL; } if (IsGoal(pData)) {
FreeLiftList(pDataList); return NULL; } if (!Safe(pData)) { return FAIL; } if (Length(pDataList) > bound) { return FAIL; }
else return 0;
}
int Safe(struct Lift *pLift) {
if (pLift->m < 0 || pLift->c < 0 || pLift->m > M || pLift->c > C || pLift->c > M) return 0;
if (pLift->m == M || pLift->m == 0) return 1;
void FreeLiftList(struct Lift *pLiftList) {
struct Lift *pLift = NULL; while (pLiftList) {
pLift = pLiftList; pLiftList = pLiftList->pNext; free(pLift); } }
continue;
pBoat = CopyBoat(pBoat); if (pBoat == NULL) return FAIL; return ConsBoat(pBoat, pPath); } return FAIL; }
void main(void) {
struct Lift *pData = NULL; struct BOAT *pPath = NULL; g_pBoatSet = InitBoats(); cout<<"3个?传ä?教¨¬士º?和¨ª3个?野°¡ã人¨?过y河¨®问¨º题¬a的Ì?答äe案ã?为a：êo\n 左Á¨®岸ã? 右®¨°岸ã?"<<endl; pData = NewLift(3, 3, 1); pPath = Backtrack1(pData, 20);
if (pLift1->m == pLift2->m && pLift1->c == pLift2->c && pLift1->b == pLift2->b) return 1;
else return 0;
}
struct Lift *NewLift(int m, int c, int b) {
struct Lift *pLift = NULL; pLift = (Lift*)malloc(sizeof(struct Lift)); if (pLift == NULL) return NULL; pLift->m = m; pLift->c = c; pLift->b = b; pLift->pNext = NULL; return pLift; }
PrintBoatList(pPath);
if (pPath == FAIL) {
free(pData); } FreeBoatList(pPath); FreeBoatList(g_pBoatSet); }
}
int Length(struct Lift *pList) {
if (pList == NULL) return 0; return Length(pList->pNext) + 1; }
struct Lift *ConsLift(struct Lift *pLift, struct Lift *pList) {
五、实验结果：
六：实验代码
#include "stdafx.h"
#include <cstdio> #include <cstdlib> #include <iostream> using namespace std;
#define #define #define #define
FAIL ((struct BOAT *)-1) M3 C3 K2
pLift->pNext = pList; return pLift; }
struct BOAT *ConsBoat(struct BOAT *pBoat, struct BOAT *pList) {
pBoat->pNext = pList; return pBoat; }
struct Lift *Gen(struct BOAT *pBoat, struct Lift *pData) {
if (pBoat == NULL) return NULL;
pBoat->pNext = pBoats;
pBoats = pBoat;
}
}
}
return pBoats;
}
int IsGoal(struct Lift *pLift) {
if (pLift->m == 0 && pLift->c == 0 && pLift->b == 0) return 1;
struct BOAT *NewBoat(int m, int c) {
struct BOAT *pBoat = NULL; pBoat = (BOAT*)malloc(sizeof(struct BOAT));
if (pBoat == NULL) return NULL; pBoat->m = m; pBoat->c = c; pBoat->pNext = NULL; return pBoat; }
struct Lift {
int m; int c;
int b; struct Lift *pNext;
};
struct BOAT {
int m; int c; struct BOAT *pNext; };
struct BOAT *g_pBoatSet = NULL;
int Equal(struct Lift *pLift1, struct Lift *pLift2) {
谓词演算中的消解法。
三、实验内容：
设有 3 个传教士和 3 个野人同在河的左岸，他们都要到对岸去；河里只有一条船，他们 都会划船，但每次渡船至多只能乘两人；如果在任何一边河岸上，野人的数量超过传教士， 野人就要吃掉传教士，问怎样才能用船将 3 个传教士和 3 个野人从左岸都渡到右岸，又不会 发生传教士被吃的事件呢？通过 Prolog 程序，给出乘船过河的动作序列。
if (pData->b == 1) {
return NewLift(pData->m - pBoat->m, pData->c - pBoatLeabharlann >c, 0); } else {
return NewLift(pData->m + pBoat->m, pData->c + pBoat->c, 1); } }
struct BOAT *InitBoats(void) {
struct BOAT *pBoats = NULL, *pBoat = NULL; int i, j;
for (i = 0; i <= 2; i++)