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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 矮化砧嫁接的苹果树树冠体积小于乔化砧嫁接的苹果树树冠体积,矮化砧苹果树单株产量低于乔化砧苹果树,所以,栽植矮化苹果树必须根据不同的矮化砧木和不同类型的短枝型品种适当加大栽培密度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
2、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,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,
3、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(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(
4、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
5、,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(monkey)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)B
6、ANANA(banana)AT(monkey,b)HOLD(monkey,box)ON(banana,ceiling)CLEAR(b)CLEAR(c)BOX(box)BANANA(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
7、,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;*/ i
8、nt box; /*-1:box at A;0:box at B;1:box at C;*/ int banana; /*Banana at B,Banana=0*/ int 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)rout
9、esavei=Monkey go to A;Statesi+1=Statesi;Statesi+1.monkey=-1;else if(a=0)routesavei=Monkey 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 m
10、ove the box to the other place*/void movebox(int a,int i) int B;B=a;if(B=-1)routesavei=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 b
11、ox to C;Statesi+1=Statesi;Statesi+1.monkey=1;Statesi+1.box=1;elseprintf(parameter is 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
12、 from the box*/void climbdown(int i) routesavei=Monkey climb down from the box;Statesi+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*
13、/void showSolution(int i) int c; printf (%s n, Result to problem:); for(c=0; c=150) 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.m