人工智能实验报告完整版八数码+验证

1、把数码问题就是把一串数字变成下面这个样子：1 2 38 0 47 6 5实现方法1.过程表示源代码：#includestatic int style9=1,2,3,6,7,4,5,8,0;/输入的数码/2,5,4,3,0,7,1,8,63,2,1,8,0,4,7,6,5 1,0,4,2,7,3,8,5,61,0,3,8,2,4,7,6,5static int arrayStep416=5,4,3,6,7,8;/第四步和第六步共用的数组，所以设为全局量 static int arrayStep714=3,6,7,4;static int local;/空 格的位置int i,j;/所用到的变量i

2、nt number=0;/记录移动步数void print();void step1();void step2();void step3();void step4();void step5();void step6();void step7();void step8();void step9();void exchange(int x,int y);void judge();void judge()/判断空格位置number = 0;for(i=0;i9;i+)if(stylei=0)local=i;return;void exchange(int x,int y)/交换两个数int temp

3、;print();temp=stylex;stylex=styley;styley=temp;local=y;number+;void step1()int arrayStep115=3,0,1,2,5;int arrayStep126=6,7,8,5,2,1;if(style2!=0)&style2!=1)return;elseif(local=2)if(style1=1)exchange(2,5);elseexchange(2,1);return;elseif(local=4)exchange(4,1);i=2;while(local!=5)exchange(arrayStep11i,ar

4、rayStep11i+1);i+;return;for(i=0;i3;i+)if(arrayStep11i=local)while(local!=5)exchange(arrayStep11i,arrayStep11i+1);i+；return;for(i=0;i4;i+)if(arrayStep12i=local)while(local!=1)exchange(arrayStep12i,arrayStep12i+1);i+;return;return;void step2()int arrayStep218=0,3,6,7,8,5,4,1;for(i=0;i8;i+)if(arrayStep

5、21i=local)while(style0!=1)exchange(arrayStep21i%8,arrayStep21(i+1)%8);i+;break;void step3()int arrayStep318=2,1,4,3,6,7,8,5;for(i=0;i8;i+)if(arrayStep31i=local)while(style1!=2)exchange(arrayStep31i%8,arrayStep31(i+1)%8);i+；break;void step4()for(i=0;i6;i+)if(arrayStep41i=local)while(style4!=3)exchang

6、e(arrayStep41i%6,arrayStep41(i+1)%6);i=(i+1)%6;while(local!=3)exchange(arrayStep41i%6,arrayStep41(i+5)%6);i=(i+5)%6;break;void step5()int arrayStep519=3,0,1,4,5,2,1,0,3;i=0;doexchange(arrayStep51i,arrayStep51i+1);i+;while(local!=3);void step6()for(i=0;i6;i+)if(arrayStep41i=local)while(style5!=4)exch

7、ange(arrayStep41i%6,arrayStep41(i+1)%6);i+;if(local=8)exchange(8,7);break;return;void step7()for(i=0;i4;i+)if(arrayStep71i=local)while(style4!=5)exchange(arrayStep71i%4,arrayStep71(i+1)%4);i=(i+1)%4;while(local!=3)exchange(arrayStep71i%4,arrayStep71(i+3)%4);i=(i+3)%4;break;void step8()int arrayStep8

8、113=3,0,1,2,5,4,7,8,5,2,1,0,3;i=0;doexchange(arrayStep81i,arrayStep81i+1);i+;while(local!=3);void step9()for(i=0;i4;i+)if(arrayStep71i=local)while(style7!=6)exchange(arrayStep71i%4,arrayStep71(i+1)%4);i=(i+1)%4;while(local!=4)exchange(arrayStep71i%4,arrayStep71(i+3)%4);i=(i+3)%4;break;void print()fo

9、r(j=0;j9;j+)if(stylej=0)printf( t);elseprintf(%dt,stylej);if(j+1)%3=0)printf(n);printf(* %d *n,number);void loop()printf(请输入数码：n);for(i=0;i9;i+)scanf(%d”,&stylei);judge();step1();step2();step3();if(style !=3)step4();step5();step6();if(style8!=5)step7();step8();step9();print();if(!(style3=8)&(style6=

10、7)printf(-用书上所给算法来看此数码错误！ n”);void main()while(1)loop();2.深度优先实现/* 说明 *“ 个个个个个个个个个个个个个个个个个个个个个个用宽度优先搜索算法实现八数码问题*/ #include#include#include#include#include string.h#include assert.h#include windows.h using namespace std;int wholeStyle9 = 2,8,3,1,6,4,7,0,5;int standard19 = 1,2,3,8,0,4,7,6,5;int local,

11、i,j;int startKey = 0,endKey = 0,equalKey = 1,tempSpace;struct node *openHead,*open; /open 表struct node *closedHead,*closed; /closed 表struct node *tempNode; 临时节点struct node *answer; /找到的路径 int num = 0;struct nodeint style9;struct node *next;struct node *father;void updateData() /更新要判断数据int i;printf(请

12、输入八数码原始状态：n);for(i = 0;i 9;i+)scanf(%d”,&wholeStylei);printf(请输入八数码最终状态：n);for(i = 0;i 9;i+)scanf(%d”,&standard1i);void judge1(struct node *head)/判 断空格位置for(i = 0;i stylei = 0)local = i;return;int judge2(struct node *head) /判断是否与标准八数码相等，不相等返回值为0for(i = 0;i stylei != standard1i)if(i = 3)&(head-style3

13、 = standard16);else if(i = 6)&(head-style6 = standard13);elsereturn 0;return 1;void judge3() /判断新生成的八数码是否就是最终状态或者在open closed表中出现 if(judge2(tempNode)endKey = 1;elsewhile(openHead-next-next-style0 != 9)for(i = 0;i next-next-stylei != tempNode-stylei)equalKey = 1;break;elseequalKey = 0;if(equalKey)/不

14、相等openHead = openHead-next;elsebreak;openHead = open-next;if(equalKey)/不 相等while(closedHead-next-style0 != 9)for(i = 0;i next-stylei != tempNode-stylei)equalKey = 1;break;elseequalKey = 0;if(!equalKey)/相 等break;elseclosedHead = closedHead-next; closedHead = closed-next;if(equalKey)/不 相等open-next = t

15、empNode;tempNode-next = openHead;tempNode-father = openHead-next; open = open-next;void print(struct node *temp) 输出八数码表for(j = 0;j stylej = 0) printf( t);elseprintf(%dt,temp-stylej);if(j + 1) % 3 = 0) printf(n);void write2txt()ofstream out(F:out_details.txt”,ios:app);if( out.fail()cerr未找到文件endl;out第

16、 +numn;outnext-style0 != 9)for(i = 0;i next-styleivvt”;if(i+1) % 3 = 0) outn;outnext;openHead = openHead-next;out*n;outnext-style0 != 9)for(i = 0;i 9;i +)outnext-styleit;if(i + 1) % 3 =0)outn;outnext;closedHead = closedHead-next;outn;out.close();void main()/updateData();/输入八数码数据for(i = 0;i style0 =

17、9;openHead-next = open;for(i = 0;i stylei = wholeStylei;open-next = openHead;open-father = openHead;closedHead = new node();closed = new node();closedHead-style0 = 9;closedHead-next = closedHead;closed = closedHead;while(open-style0 != 9)/当 open 表不为空时一直循环judge1(openHead-next);if(local % 3 0)/右移equal

18、Key = 1;tempNode = new node();for(i = 0;i stylei = openHead-next-stylei;tempSpace = tempNode-stylelocal - 1;tempNode-stylelocal - 1 = tempNode-stylelocal; tempNode-stylelocal = tempSpace;judge3();if(endKey)break;if(local 2)/下移equalKey = 1;tempNode = new node();for(i = 0;i stylei = openHead-next-styl

19、ei;tempSpace = tempNode-stylelocal - 3;tempNode-stylelocal - 3 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;judge3();if(endKey)break;if(local % 3 2)/左移equalKey = 1;tempNode = new node();for(i = 0;i stylei = openHead-next-stylei;tempSpace = tempNode-stylelocal + 1;tempNode-stylelocal + 1 = t

20、empNode-stylelocal;tempNode-stylelocal = tempSpace;judge3();if(endKey)break;if(local 6)/上移equalKey = 1;tempNode = new node();/tempNode = malloc(sizeof(struct node);for(i = 0;i stylei = openHead-next-stylei;tempSpace = tempNode-stylelocal + 3;tempNode-stylelocal + 3 = tempNode-stylelocal;tempNode-sty

21、lelocal = tempSpace;judge3();if(endKey)break;closed-next = openHead-next; /把 open 的标头添加到 closed 表中 openHead-next = openHead-next-next;closed = closed-next;closed-next = closedHead;write2txt();open-next = tempNode;/把找到的新节点添加到open表中 tempNode-next = openHead;tempNode-father = openHead-next;open = open-

22、next;closed-next = openHead-next; /把 open 的标头添加到 closed 表中 openHead-next = openHead-next-next;closed = closed-next;closed-next = closedHead;write2txt();answer = new node();tempNode = new node();tempNode = open;while(tempNode-style0 != 9)/将结果路径存于 answer answer = tempNode;tempNode = tempNode-father;te

23、mpNode-next = answer;num = 0;while(answer-next-style0 != 9)/输出 answerprintf(*第d 步*n,num+);print(answer);answer = answer-next;printf(n);printf(*第d 步*n,num+);print(answer);if(answer-style3 != standard13)printf(n!输入的八数码不合法，不能从初始状态到最终状态nnnn); return;3.宽度优先:/* 说明 *“ 个个个个个个个个个个个个个个个个个个个个个个用宽度优先搜索算法实现八数码问题

24、*/ #include#include#include#include#include string.h#include assert.h#include windows.husing namespace std;int wholeStyle9 = 2,8,3,1,6,4,7,0,5;int standard19 = 1,2,3,8,0,4,7,6,5;int local,i,j;int startKey = 0,endKey = 0,equalKey = 1,tempSpace;struct node *openHead,*open; /open 表struct node *closedHe

25、ad,*closed; /closed 表struct node *tempNode; 临时节点struct node *answer; /找到的路径 int num = 0;struct nodeint style9;struct node *next;struct node *father;void updateData() /更新要判断数据int i;printf(请输入八数码原始状态：n);for(i = 0;i 9;i+)scanf(%d,&wholeStylei);printf(请输入八数码最终状态：n);for(i = 0;i 9;i+)scanf(%d,&standard1i)

26、;void judge1(struct node *head)/判 断空格位置for(i = 0;i stylei = 0)local = i;return;int judge2(struct node *head) /判断是否与标准八数码相等，不相等返回值为0for(i = 0;i stylei != standard1i)if(i = 3)&(head-style3 = standard16);else if(i = 6)&(head-style6 = standard13);elsereturn 0;return 1;void judge3() /判断新生成的八数码是否就是最终状态或者在

27、open closed表中出现if(judge2(tempNode)endKey = 1;elsewhile(openHead-next-next-style0 != 9)for(i = 0;i next-next-stylei != tempNode-stylei)equalKey = 1;break;elseequalKey = 0;if(equalKey)/不 相等openHead = openHead-next;elsebreak;openHead = open-next;if(equalKey)/不 相等while(closedHead-next-style0 != 9)for(i

28、= 0;i next-stylei != tempNode-stylei)equalKey = 1;break;elseequalKey = 0;if(!equalKey)/相 等break;elseclosedHead = closedHead-next;closedHead = closed-next;if(equalKey)/不 相等open-next = tempNode;tempNode-next = openHead;tempNode-father = openHead-next;open = open-next;void print(struct node *temp) 输出八数

29、码表for(j = 0;j stylej = 0)printf( t);elseprintf(%dt,temp-stylej);if(j + 1) % 3 = 0)printf(n);void write2txt()ofstream out(F:out_details.txt”,ios:app);if( out.fail()cerr未找到文件endl;out第 +numn;outnext-style0 != 9)for(i = 0;i next-styleivvt”; if(i+1) % 3 = 0) outn;outnext;openHead = openHead-next;out*n;ou

30、tnext-style0 != 9)for(i = 0;i 9;i +)outnext-styleit;if(i + 1) % 3 =0)outn;outnext;closedHead = closedHead-next;outn;out.close();void main()/updateData();/输入八数码数据for(i = 0;i style0 = 9;openHead-next = open;for(i = 0;i stylei = wholeStylei;open-next = openHead;open-father = openHead;closedHead = new n

31、ode();closed = new node();closedHead-style0 = 9;closedHead-next = closedHead;closed = closedHead;while(open-style0 != 9)/当 open 表不为空时一直循环judge1(openHead-next);if(local % 3 0)/右移equalKey = 1;tempNode = new node();for(i = 0;i stylei = openHead-next-stylei; tempSpace = tempNode-stylelocal - 1;tempNode-

32、stylelocal - 1 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;judge3();if(endKey)break;if(local 2)/下移equalKey = 1;tempNode = new node();for(i = 0;i stylei = openHead-next-stylei;tempSpace = tempNode-stylelocal - 3;tempNode-stylelocal - 3 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;j

33、udge3();if(endKey)break;if(local % 3 2)/左移equalKey = 1;tempNode = new node();for(i = 0;i stylei = openHead-next-stylei;tempSpace = tempNode-stylelocal + 1;tempNode-stylelocal + 1 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;judge3();if(endKey)break;if(local 6)/上移equalKey = 1;tempNode = new

34、node();/tempNode = malloc(sizeof(struct node);for(i = 0;i stylei = openHead-next-stylei; tempSpace = tempNode-stylelocal + 3;tempNode-stylelocal + 3 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;judge3();if(endKey)break;closed-next = openHead-next; /把 open 的标头添加到 closed 表中 openHead-next = op

35、enHead-next-next;closed = closed-next;closed-next = closedHead;write2txt();open-next = tempNode;/把找到的新节点添加到open表中tempNode-next = openHead;tempNode-father = openHead-next;open = open-next;closed-next = openHead-next; /把 open 的标头添加到 closed 表中 openHead-next = openHead-next-next;closed = closed-next;clo

36、sed-next = closedHead;write2txt();answer = new node();tempNode = new node();tempNode = open;while(tempNode-style0 != 9)/将结果路径存于 answeranswer = tempNode;tempNode = tempNode-father;tempNode-next = answer;num = 0;while(answer-next-style0 != 9)/输出 answerprintf(*第d 步*n,num+);print(answer);answer = answer

37、-next; printf(n);printf(*第(!步*n,num+);print(answer);if(answer-style3 != standard13)printf(n!输入的八数码不合法，不能从初始状态到最终状态nnnn); return;4.A算法/* 说明 *“ 个个个个个个个个个个个个个个个个个个个个个个* A算法实现八数码问题 */ #include #include#include using namespace std;void print(struct node *temp);int wholeStyle9 = 2,8,3,1,6,4,7,0,5;int stan

38、dard19 = 1,2,3,8,0,4,7,6,5;int local,i,j;int tempSpace;struct node *openHead,*open; /open 表struct node *closedHead,*closed; /closed 表 struct node *tempNode; 临时节点 struct node *answer; /找到的路径 int num = 0;bool endKey;struct nodeint depth;int judgement_based;int style9;struct node *next;struct node *fat

39、her;void updateData() /更新要判断数据int i;cout”请输入八数码原始状态：endl;for(i = 0;i wholeStylei;cout请输入八数码最终状态：endl;for(i = 0;i standard1i;int judge1(struct node *head)/判 断空格位置for(i = 0;i stylei = 0)return i;int evaluation_function(struct node *head) 计算不在正确位置的点的个数int not_correct_position = 0;for(i = 0;i stylei !=

40、0)&(head-stylei != standard1i)not_correct_position+;return not_correct_position;void print(struct node *temp) 输出八数码表for(j = 0;j stylej = 0)cout t;elsecoutstylejt;if(j + 1) % 3 = 0)coutendl;void write2txt()/将过程记录到文本文档中ofstream out(F:out_details4.txt”,ios:app);if( out.fail()cerr未找到文件endl;out第 +numn;ou

41、tnext-style0 != 9)for(i = 0;i 9;i +)outnext-styleit; if(i+1) % 3 = 0) outn;out估值：”；outnext-judgement_basednext;openHead = openHead-next;outvv*n;outnext-style0 != 9)for(i = 0;i next-styleivvt”;if(i + 1) % 3 =0) outvvn;outvv 深 度 ： next-depth 估 值 ： next-judgement_basednext;closedHead = closedHead-next;

42、outvvn;out.close();void basic_exchange_oper() 每一次基本的交换都要进行的动作if(judge1(tempNode) != judge1(closed-father)/当新生成节点的空格位置与 closed 父亲 节点的空格位置不相等tempNode-father = closed; /新生成节点的指向它的父亲节点tempNode-depth = closed-depth + 1;/深度加 1tempNode-judgement_based = evaluation_function(tempNode) + tempNode-depth;/ 计算估值

43、while(openHead-next-style0 != 9) if(tempNode-judgement_based openHead-next-judgement_based)openHead = openHead-next;elsetempNode-next = openHead-next;openHead-next = tempNode;openHead = open-next;return;/插入的值比最后一个还大进行下面的操作openHead = open-next;/先恢复表头 否则Head指向为最后一个 open-next = tempNode;tempNode-next =

44、 openHead;open = open-next;return;void main()/updateData();/输入八数码数据openHead = new node(); /初始化 open 表open = new node();openHead-style0 = 9;openHead-next = open;for(i = 0;i stylei = wholeStylei;open-next = openHead;open-father = openHead;open-depth = 0;open-judgement_based = evaluation_function(open)

45、 + open-depth;/这行可以不写closedHead = new node(); /初始化 closed 表closed = new node();closedHead-style0 = 9;closedHead-next = closedHead;closed = closedHead;int min_judgement1 = open-judgement_based;while(open-style0 != 9)/当 open 表不为空时一直循环/updateData();/输入八数码数据write2txt();closed-next = openHead-next;/把 ope

46、n 表中第一组八数码移到 closed 表中openHead-next = openHead-next-next;closed = closed-next;closed-next = closedHead;if(openHead-next-style0 = 9)/如果是第一次 则 open 暂时指向 openHead open = openHead;if(closed-judgement_based = closed-depth)/判断新移进 closed 表中的八数码的 f 值是否等于它的层数coutsucceed!style0 != 9)answer = closed;closed = c

47、losed-father;closed-next = answer;num = 0;while(answer-style0 != 9)cout* 第num+步 *n;print(answer);coutnext;return;else /如果closed表末的八数码是最终状态则拓展新的八数码local = judge1(closed);if(local % 3 0)/右移tempNode = new node();for(i = 0;i stylei = closed-stylei;tempSpace = tempNode-stylelocal - 1;tempNode-stylelocal

48、- 1 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;basic_exchange_oper();if(local 6)/上移tempNode = new node();for(i = 0;i stylei = closed-stylei;tempSpace = tempNode-stylelocal + 3;tempNode-stylelocal + 3 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;basic_exchange_oper();if(local 2)/下

49、移tempNode = new node();for(i = 0;i stylei = closed-stylei;tempSpace = tempNode-stylelocal - 3;tempNode-stylelocal - 3 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;basic_exchange_oper();if(local % 3 2)/左移tempNode = new node();for(i = 0;i stylei = closed-stylei;tempSpace = tempNode-stylelocal

50、 + 1;tempNode-stylelocal + 1 = tempNode-stylelocal;tempNode-stylelocal = tempSpace;basic_exchange_oper();coutvvfalse!”vvendl;神经网络实验群神经网络实验一、实验目的：理解反向传播网络的结构和原理，掌握反向传播算法对神经元的训练过程，了解反向传播公式。通过构建BP网络实例，熟悉前馈网络的原理及结构。二、 实验原理：反向传播（BP）算法是一种计算单个权值变化引起网络性能变化值的较 为简单的方法。BP算法过程从输出节点开始，反向地向第一隐含层（即最接近输入层的隐含 层）传播由总

51、误差引起的权值修正BP网络不仅含有输入节点和输出节点，而且含有一层或 多层隐（层）节点。输入信号先向前传递到隐节点，经过作用后，再把隐节点的输出信息传递 到输出节点，最后给出输出结果。三、实验内容：1通过BP网络各项参数的不同设置，观察BP算法的学习效果。2观察比较BP网络拓朴结构及其它各项参数变化对于训练结果的影响。3观察并分析不同训练数据对相同拓朴结构的BP网络建模的影响.四、实验步骤：验证性实验：1、进入实验环境；2、选择相关的实验模块；3、设置相应地实验参数（如设置初始权值为随机值）；4、选择实验运行方式；5、观测运行过程；6、修改相应地参数重复第15步直到满意为止。六、实验结论：包括

52、做实验的目的、方法、过程等，具体要写成实验报告，如下图所示（见 下页）.BP网络的基本结构及BP算法的训练过程。试述阈值函数和权值变化对BP网络推理结果的影响。训练数据集变化对网络训练结果的影响。姓名实验目的理解反向传播网络的结构和原理，掌握反向传播算法对神经元的训练过程，了解反向传播公式。通 过构建BP网络实例，熟悉前馈网络的原理及结构。训练数据集Training Examples |15krimtrapShrTtKiiaL: L 0I. Da.o.DE. 0.01. D1.0D. 01.03- 0 0D Da.oD 0a. ai. L.Od. n宜1. 00.0& L 01. 0,o1 Q

53、ft L 0D D0.00 00.07: 0.00 0,oQ 00.0KhovdIk音5hortH匕Het L. D1.Q0 01 02： 0. D1.01.01.03 a. D0. Q0 0a o1： L.D0.0a.o1.05 L. D1.Q1 01 06： L.D0.0a.oi.a7 a. g0. Q1 0a.oTrairiiii gi间含如fLTltiri 131 idFbE-4_LlEa: 1 DQ.QD D-1 00.01 0OJJ0.030 00.01 0L 01 040 Q0.01 0Q 00.0Q.O0.00.(1L. 00.Q1.00.00.0L. Q.Q1.00.00.G

54、0. Q.QD 0D1 dL 0an9 aD1 dQ D0. 10J.OD1 dD 0DMJ 0OJJL 01 012D.Q口 D0 flQ 0IDI wfiing ExanwhH 4训练误差第1代误差 4.99 第46代误差0.99 第69代误差0.29 第83代误差0.19第117代误差0.09第1代误差 1.68第51代误差0.52 第101代误差0.11 第151代误差0.05 第201代误差0.03第1代误差 3.0第47代误差1.5第86代误差1.289第122代误差1.199第224代误差1.1模拟的问题或函数Y=f(x1,x2,x3,x4)Y=f(x1,x2,x3)(y1,y2

55、,y3)=f(x1,x2)观测结果经过200代的进化，误差极 大地降低经过200代的进化，误差以明 显的阶梯型降低学生结论经过训练的BP网络可以进行 邮件识别，解决信息识别的 难题，可以极大地提高生产 力神经计算能够实现“多数赞成 表决器”功能搜索策略实验群一、实验目的：熟悉和掌握启发式搜索的定义、估价函数和算法过程，并利用A*算法求解N数码难题， 理解求解流程和搜索顺序。二、实验原理：A*算法是一种有序搜索算法，其特点在于对估价函数的定义上。对于一般的有序搜索， 总是选择f值最小的节点作为扩展节点。因此，f是根据需要找到一条最小代价路径的观点 来估算节点的，所以，可考虑每个节点n的估价函数值为两个分量：从起始节点到节点n 的代价以及从节点n到达目标节点的代价。三、实验内容：1分别以各种