猴子摘香蕉实验报告(含代码)

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1、华中科技大学人工智能与模式识别 实验报告 院 系： 电子与信息工程系 班 级： 姓 名： 同 组 人： 指导老师： 电 话： 邮 箱： 日 期： 2013年12月24日 1、 实验内容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（即下图所示状态），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。）2、 实验平台VC6.03、 实验分析1. 定义描述环境状态的谓词。AT(x,w)：x在t处，

2、个体域：xmonkey，wa,b,c,box；HOLD(x,t)：x手中拿着t，个体域：tbox,banana；EMPTY(x)：x手中是空的；ON(t,y)：t在y处，个体域：yb,c,ceiling；CLEAR(y)：y上是空的；BOX(u)：u是箱子，个体域：ubox；BANANA(v)：v是香蕉，个体域：vbanana；2. 使用谓词、连结词、量词来表示环境状态。问题的初始状态可表示为：So：AT(monkey,a)EMPTY(monkey)ON(box,c)ON(banana,ceiling)CLEAR(b)BOX(box)BANANA(banana)要达到的目标状态为：Sg：AT(

3、monkey,box)HOLD(monkey,banana)ON(box,b)CLEAR(ceiling)CLEAR(c)BOX(box)BANANA(banana)3. 从初始状态到目标状态的转化, 猴子需要完成一系列操作, 定义操作类谓词表示其动作。WALK(m,n)：猴子从m走到n处，个体域：m,na,b,c；CARRY(s,r)：猴子在r处拿到s，个体域：rc,ceiling，sbox,banana；CLIMB(u,b)：猴子在b处爬上u；这3个操作也可分别用条件和动作来表示。条件直接用谓词公式表示，是为完成相应操作所必须具备的条件；当条件中的事实使其均为真时，则可激活操作规则，于是可

4、执行该规则中的动作部分。动作通过前后状态的变化表示，即通过从动作前删除或增加谓词公式来描述动作后的状态。WALK(m,n)：猴子从m走到n处条件：AT(monkey,m)动作：CARRY(s,r)：猴子在r处拿到s条件：AT(monkey,r)EMPTY(monkey)ON(s,r)BOX(box)BANANA(banana)动作：CLIMB(u,b)：猴子在b处爬上u条件：AT(monkey,b)HOLD(monkey,u)CLEAR(b)BOX(box)BANANA(banana)动作：4. 按照行动计划, 一步步进行状态替换, 直至目标状态。AT(monkey,a)EMPTY(monke

5、y)ON(box,c)ON(banana,ceiling)CLEAR(b)BOX(box)BANANA(banana)AT(monkey,c)EMPTY(monkey)ON(box,c)ON(banana,ceiling)CLEAR(b)BOX(box)BANANA(banana)AT(monkey,c)HOLD(monkey,box)ON(banana,ceiling)CLEAR(b)CLEAR(c)BOX(box)BANANA(banana)AT(monkey,b)HOLD(monkey,box)ON(banana,ceiling)CLEAR(b)CLEAR(c)BOX(box)BANAN

6、A(banana)AT(monkey,box)EMPTY(monkey)ON(box,b)ON(banana,ceiling)CLEAR(c)BOX(box)BANANA(banana)AT(monkey,box)HOLD(monkey,banana)ON(box,b)CLEAR(ceiling)CLEAR(c)BOX(box)BANANA(banana)（目标得解）猴子行动的规则序列是：WALK(a,c)CARRY(c,box)WALK(c,b)CLIMB(box,b)CARRY(banana,ceiling)在上述过程中，我们应该注意，当猴子执行某一个操作之前，需要检查当前状态是否可使所要

7、求的条件得到满足，即证明当前状态是否蕴涵操作所要求的状态的过程。在行动过程中， 检查条件的满足性后才进行变量的代换。代入新条件后的新状态如果是目标状态，则问题解决；否则看是否满足下面的操作，如果不满足或即使满足却又回到了原来的状态，那么代入无效。4、 源代码#include struct Stateint monkey; /*-1:Monkey at A;0: Monkey at B;1:Monkey at C;*/ int box; /*-1:box at A;0:box at B;1:box at C;*/ int banana; /*Banana at B,Banana=0*/ int

8、monbox; /*-1: monkey on the box;1: monkey the box;*/;struct State States 150;char* routesave150; /*function monkeygoto,it makes the monkey goto the other place*/void monkeygoto(int b,int i) int a;a=b;if (a=-1)routesavei=Monkey go to A;Statesi+1=Statesi;Statesi+1.monkey=-1;else if(a=0)routesavei=Monk

9、ey go to B;Statesi+1=Statesi;Statesi+1.monkey=0;else if(a=1)routesavei=Monkey go to C;Statesi+1=Statesi;Statesi+1.monkey=1;else printf(parameter is wrong);/*end function monkeyygoto*/ /*function movebox,the monkey move the box to the other place*/void movebox(int a,int i) int B;B=a;if(B=-1)routesave

10、i=monkey move box to A;Statesi+1=Statesi;Statesi+1.monkey=-1;Statesi+1.box=-1;else if(B=0)routesavei = monkey move box to B;Statesi+1=Statesi;Statesi+1.monkey=0;Statesi+1.box=0;else if(B=1)routesavei = monkey move box to C;Statesi+1=Statesi;Statesi+1.monkey=1;Statesi+1.box=1;elseprintf(parameter is

11、wrong);/*end function movebox*/ /*function climbonto,the monkey climb onto the box*/void climbonto(int i)routesavei=Monkey climb onto the box;Statesi+1=Statesi;Statesi+1.monbox=1;/*function climbdown,monkey climb down from the box*/void climbdown(int i) routesavei=Monkey climb down from the box;Stat

12、esi+1=Statesi;Statesi+1.monbox=-1;/*function reach,if the monkey,box,and banana are at the same place,the monkey reach banana*/void reach(int i) routesavei=Monkey reach the banana;/*output the solution to the problem*/void showSolution(int i) int c; printf (%s n, Result to problem:); for(c=0; c=150)

13、 printf(%s n, steplength reached 150,have problem ); return; for (c=0; ci; c+) /*if the current state is same to previous,retrospect*/ if(Statesc.monkey=Statesi.monkey&Statesc.box=Statesi.box&Statesc.banana=Statesi.banana&Statesc.monbox=Statesi.monbox) return; if(Statesi.monbox=1&Statesi.monkey=0&St

14、atesi.banana=0&Statesi.box=0) showSolution(i);printf(Press any key to continue n);getchar();/*to save screen for user,press any key to continue*/return; j=i+1; if(Statesi.monkey=0) if(Statesi.box=0)if(Statesi.monbox=-1)climbonto(i);reach(i+1);nextStep(j);/*monkeygoto(-1,i);nextStep(j);monkeygoto(0,i

15、);nextStep(j);movebox(-1,i);nextStep(j);movebox(0,i);nextStep(j);*/elsereach(i+1);nextStep(j);/*climbdown(i);nextStep(j);*/ else if(Statesi.box=1)/*monkeygoto(-1,i);nextStep(j);*/monkeygoto(1,i);nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);else /*box=-1*/monkeygoto(-1,i);

16、nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j); /*end if*/if(Statesi.monkey=-1) if(Statesi.box=-1)if(Statesi.monbox=-1) movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);elseclimbdown(i);nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);els

17、e if(Statesi.box=0) monkeygoto(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);elsemonkeygoto(1,i);nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);/*end if*/if(Statesi.monkey=1) if (Statesi.box=1)if(Statesi.monbox=-1) movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nex

18、tStep(j);elseclimbdown(i);nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j); else if(Statesi.box=-1) monkeygoto(-1,i);nextStep(j);movebox(0,i);nextStep(j); movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);elsemonkeygoto(0,i);nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j); /*end if*/*end nextStep*/ int main() States0.monkey=-1; States0.box=1; States0.banana=0; States0.monbox=-1; nextStep(0);5、 实验截图