人工智能导论实验一 基于图搜索技术的八数码问题求解

广州大学学生实验报告开课学院及实验室：计算机科学与工程实验室 2020年10月14日(***报告只能为文字和图片,老师评语将添加到此处,学生请勿作答***)一、实验内容1. 分别用广度优先搜索策略、深度优先搜索策略和启发式搜索算法（至少两种）求解八数码问题；分析估价函数对启发式搜索算法的影响；探究讨论各个搜索算法的特点。

二、实验设备1. 实验设备：计算机；2. 平台：Windows操作系统，Visual C++ 6.0 / Python Anaconda三、实验步骤1. 随机生成一个八数码问题分布，设计一个可解的目标状态（要求棋盘9个位置都不同）2. 分别用广度优先搜索策略、深度优先搜索策略和至少两种启发式搜索算法求解八数码问题3. 分析估价函数对启发式搜索算法的影响4. 探究讨论各个搜索算法的特点四、分析说明（包括核心代码及解释）广度优先搜索：首先创建一个结构体node，来记录节点移动方向和扩展的节点。

struct node{int ab[3][3];//节点int direction;//方向};struct node sh[102], end;int count = 1;然后创建一个init函数来初始化棋盘起始状态和目标状态，使用for语句填写棋盘数字用loction函数确定0节点的位置，通过for语句和if语句判断sh[num].ab[i / 3][i % 3] == 0，即可得到0节点的位置Sign函数用来获取棋盘状态，将当前棋盘数字顺序生成一个数，即可得知棋盘状态。

Mobile函数用来移动0节点，先用loction函数获取0节点的位置，再通过if语句来判断0节点位置和所能移动方向，然后进行移动。

Display函数使用for语句来打印当前棋盘。

Search函数使用display函数来打印从初始状态移动到目标状态的中间状态棋盘，在while（1）语句下利用mobile函数移动0节点，直到目标状态找到或者超过寻找次数。

#include<stdio.h>#include<iostream>struct node{int ab[3][3];//节点int direction;//方向};struct node sh[102], end;int count = 1;void init(){printf("输入起始棋盘的状态:\n");int i, j;for (i = 0; i < 3; i++)for (j = 0; j < 3; j++)scanf("%d", &sh[0].ab[i][j]);sh[0].direction = -1;printf("输入目标棋盘的状态:\n");for (i = 0; i < 3; i++)for (j = 0; j < 3; j++)scanf("%d", &sh[101].ab[i][j]);sh[101].direction = -1;printf("中间状态:\n");}//找出0的位置int loction(int num){int i;for (i = 0; i < 9; i++)if (sh[num].ab[i / 3][i % 3] == 0) return i;}//获取棋盘状态long long sign(int num){long long sum;sum = sh[num].ab[0][0] * 100000000 + sh[num].ab[0][1] * 10000000 + sh[num].ab[0][2] * 1000000 + sh[num].ab[1][0] * 100000 + sh[num].ab[1][1] * 10000 + sh[num].ab[1][2] * 1000 + sh[num].ab[2][0] * 100 + sh[num].ab[2][1] * 10 + sh[num].ab[2][2];return sum;}//移动0节点void mobile(int num){int temp;int loc;int up = 1, down = 1, left = 1, right = 1;loc = loction(num);int stand = sh[num].direction;//direction的0 1 2 3分别代表左上右下if (loc / 3 != 0 && stand != 1){sh[count] = sh[num];temp = sh[count].ab[loc / 3][loc % 3];sh[count].ab[loc / 3][loc % 3] = sh[count].ab[loc / 3 - 1][loc % 3];sh[count].ab[loc / 3 - 1][loc % 3] = temp;sh[count].direction = 3;count++;};if (loc / 3 != 2 && stand != 3){sh[count] = sh[num];temp = sh[count].ab[loc / 3][loc % 3];sh[count].ab[loc / 3][loc % 3] = sh[count].ab[loc / 3 + 1][loc % 3];sh[count].ab[loc / 3 + 1][loc % 3] = temp;sh[count].direction = 1;count++;}if (loc % 3 != 0 && stand != 0){sh[count] = sh[num];temp = sh[count].ab[loc / 3][loc % 3];sh[count].ab[loc / 3][loc % 3] = sh[count].ab[loc / 3][loc % 3 - 1];sh[count].ab[loc / 3][loc % 3 - 1] = temp;sh[count].direction = 2;count++;}if (loc % 3 != 2 && stand != 2){sh[count] = sh[num];temp = sh[count].ab[loc / 3][loc % 3];sh[count].ab[loc / 3][loc % 3] = sh[count].ab[loc / 3][loc % 3 + 1];sh[count].ab[loc / 3][loc % 3 + 1] = temp;sh[count].direction = 0;count++;}}//打印当前棋盘void display(int num){int i, j;for (i = 0; i < 3; i++){for (j = 0; j < 3; j++)printf("%d ", sh[num].ab[i][j]);printf("\n");}}//寻找目标状态int search(){int i = 0;while (1){printf("\n");display(i);printf("\n");if (i == 100){printf("超出了上限次数\n");return 0;}if (sign(i) == sign(101)){printf("在第%d次找到了\n", i);display(i);return i;}mobile(i);i++;}}int main(){init();search();system("pause");}宽度优先搜索：创建一个结构体Octal，其中的arr[4][4]用来存储八数码To_number函数将八数码棋盘中的数据转为数字存储。

Up函数用来移动0节点，同时判断0节点是否能向上移动Down函数用来判断0节点是否能向下移动Left函数用来判断0节点是否能向左移动Right函数用来判断0节点是否能向右移动主函数中用while（！open.empty（））语句来循环搜索目标状态，用if语句判断当前状态若为目标状态则break 跳出循环，然后打印寻找目标状态过程的中间状态#include<iostream>#include<vector>using namespace std;struct Octal{int arr[4][4]; // 存储八数码int parent; // 父亲的值int current; // 当前的值int depth; // 当前深度};int to_number(int[4][4]);bool up(int[4][4]);bool down(int[4][4]);bool left(int[4][4]);bool right(int[4][4]);int main(){int arr[4][4] = { 0 }; // 存储当前八数码vector<Octal> open; // open表vector<Octal> closed; // closed表int depth; // 深度Octal octal; // 八数码cout <<"请输入起始棋盘状态："<< endl;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){cin >> arr[i][j];octal.arr[i][j] = arr[i][j];}}cout <<"请输入深度："<< endl;cin >> depth;cout <<"中间状态："<< endl;octal.parent = 0;octal.current = to_number(arr);octal.depth = 0;open.push_back(octal);int goal = 123804765; // 最终的八数码值bool flag = false; // 成功解决为true，失败为falseOctal temp;int count = 1;while (!open.empty()){temp.current = open[open.size() - 1].current;temp.parent = open[open.size() - 1].parent;temp.depth = open[open.size() - 1].depth;for (int i = 1; i <= 3; i++)for (int j = 1; j <= 3; j++)temp.arr[i][j] = open[open.size() - 1].arr[i][j];open.pop_back();closed.push_back(temp);if (temp.current == goal){flag = true;break;}if (temp.depth >= depth){continue;}Octal produce[4]; // 四个可能生成的八数码数组for (int i = 0; i < 4; i++){produce[i].parent = temp.current;for (int j = 1; j <= 3; j++)for (int k = 1; k <= 3; k++)produce[i].arr[j][k] = temp.arr[j][k];produce[i].depth = temp.depth + 1;}bool flagProduce[4][2]; // 四个八数码对应的标志，0为open中是否含有，1位closed中是否含有for (int i = 0; i < 4; i++)for (int j = 0; j < 2; j++)flagProduce[i][j] = true;if (up(produce[0].arr)){produce[0].current = to_number(produce[0].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[0].current){flagProduce[0][0] = false;break;}if (flagProduce[0][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[0].current){flagProduce[0][1] = false;break;}if (flagProduce[0][1])open.push_back(produce[0]);}}if (down(produce[1].arr)){produce[1].current = to_number(produce[1].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[1].current){flagProduce[1][0] = false;break;}if (flagProduce[1][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[1].current){flagProduce[1][1] = false;break;}if (flagProduce[1][1])open.push_back(produce[1]);}}if (left(produce[2].arr)){produce[2].current = to_number(produce[2].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[2].current){flagProduce[2][0] = false;break;}if (flagProduce[2][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[2].current){flagProduce[2][1] = false;break;}if (flagProduce[2][1])open.push_back(produce[2]);}}if (right(produce[3].arr)){produce[3].current = to_number(produce[3].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[3].current){flagProduce[3][0] = false;break;}if (flagProduce[3][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[3].current){flagProduce[3][1] = false;break;}if (flagProduce[3][1])open.push_back(produce[3]);}}}if (flag){vector<Octal> answer;answer.push_back(temp);int parent = temp.parent;while (parent != 0){for (int j = 0; j < closed.size(); j++){if (closed[j].current == parent){temp.current = closed[j].current;temp.depth = closed[j].depth;temp.parent = closed[j].parent;for (int k = 1; k <= 3; k++){for (int m = 1; m <= 3; m++){temp.arr[k][m] = closed[j].arr[k][m];}}answer.push_back(temp);parent = closed[j].parent;break;}}}for (int i = answer.size() - 1; i >= 0; i--){if(i<=0){cout <<"第"<< answer.size() - i - 1 <<"次找到了"<< endl; }for (int j = 1; j <= 3; j++){for (int k = 1; k <= 3; k++){cout << answer[i].arr[j][k] <<" ";}cout << endl;}cout << endl;}}else{cout <<"在深度限制内的所有结果搜索完毕，没有答案！"<< endl;}system("pause");}int to_number(int arr[4][4]){// 将八数码矩阵中的数据转换为数字，方便存储和比较// 从左到右，从上至下，按照从1到9递增，1为亿位，……，9为个位int index = 100000000;long long result = 0;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){result += arr[i][j] * index;index /= 10;}}return result;}bool up(int arr[4][4]){// 空格向上int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (i == 1)return false; // 此时无法向上else{arr[i][j] = arr[i - 1][j];arr[i - 1][j] = 0;}return true;}bool down(int arr[4][4]){// 空格向下int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (i == 3)return false; // 此时无法向下else{arr[i][j] = arr[i + 1][j];arr[i + 1][j] = 0;}return true;}bool left(int arr[4][4]){// 空格向左int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (j == 1)return false; // 此时无法向左else{arr[i][j] = arr[i][j - 1];arr[i][j - 1] = 0;}return true;}bool right(int arr[4][4]){// 空格向右int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (j == 3)return false; // 此时无法向右else{arr[i][j] = arr[i][j + 1];arr[i][j + 1] = 0;}return true;}启发式搜索：首先创建一个结构体Node。