猴子摘香蕉實(shí)驗(yàn)報告(含代碼)

1、華中科技大學(xué)人工智能與模式識別 實(shí)驗(yàn)報告 院 系： 電子與信息工程系 班 級： 姓 名： 同 組 人： 指導(dǎo)老師： 電 話： 郵 箱： 日 期： 2013年12月24日 1、 實(shí)驗(yàn)內(nèi)容利用一階謂詞邏輯求解猴子摘香蕉問題：房內(nèi)有一個猴子，一個箱子，天花板上掛了一串香蕉，其位置如圖所示，猴子為了拿到香蕉，它必須把箱子搬到香蕉下面，然后再爬到箱子上。請定義必要的謂詞，列出問題的初始化狀態(tài)（即下圖所示狀態(tài)），目標(biāo)狀態(tài)（猴子拿到了香蕉，站在箱子上，箱子位于位置b）。（附加：從初始狀態(tài)到目標(biāo)狀態(tài)的謂詞演算過程。）2、 實(shí)驗(yàn)平臺vc6.03、 實(shí)驗(yàn)分析1. 定義描述環(huán)境狀態(tài)的謂詞。at(x,w)：x在t處，

2、個體域：xmonkey，wa,b,c,box；hold(x,t)：x手中拿著t，個體域：tbox,banana；empty(x)：x手中是空的；on(t,y)：t在y處，個體域：yb,c,ceiling；clear(y)：y上是空的；box(u)：u是箱子，個體域：ubox；banana(v)：v是香蕉，個體域：vbanana；2. 使用謂詞、連結(jié)詞、量詞來表示環(huán)境狀態(tài)。問題的初始狀態(tài)可表示為：so：at(monkey,a)empty(monkey)on(box,c)on(banana,ceiling)clear(b)box(box)banana(banana)要達(dá)到的目標(biāo)狀態(tài)為：sg：at(

3、monkey,box)hold(monkey,banana)on(box,b)clear(ceiling)clear(c)box(box)banana(banana)3. 從初始狀態(tài)到目標(biāo)狀態(tài)的轉(zhuǎn)化, 猴子需要完成一系列操作, 定義操作類謂詞表示其動作。walk(m,n)：猴子從m走到n處，個體域：m,na,b,c；carry(s,r)：猴子在r處拿到s，個體域：rc,ceiling，sbox,banana；climb(u,b)：猴子在b處爬上u；這3個操作也可分別用條件和動作來表示。條件直接用謂詞公式表示，是為完成相應(yīng)操作所必須具備的條件；當(dāng)條件中的事實(shí)使其均為真時，則可激活操作規(guī)則，于是可

4、執(zhí)行該規(guī)則中的動作部分。動作通過前后狀態(tài)的變化表示，即通過從動作前刪除或增加謂詞公式來描述動作后的狀態(tài)。walk(m,n)：猴子從m走到n處條件：at(monkey,m)動作：carry(s,r)：猴子在r處拿到s條件：at(monkey,r)empty(monkey)on(s,r)box(box)banana(banana)動作：climb(u,b)：猴子在b處爬上u條件：at(monkey,b)hold(monkey,u)clear(b)box(box)banana(banana)動作：4. 按照行動計劃, 一步步進(jìn)行狀態(tài)替換, 直至目標(biāo)狀態(tài)。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)（目標(biāo)得解）猴子行動的規(guī)則序列是：walk(a,c)carry(c,box)walk(c,b)climb(box,b)carry(banana,ceiling)在上述過程中，我們應(yīng)該注意，當(dāng)猴子執(zhí)行某一個操作之前，需要檢查當(dāng)前狀態(tài)是否可使所要

7、求的條件得到滿足，即證明當(dāng)前狀態(tài)是否蘊(yùn)涵操作所要求的狀態(tài)的過程。在行動過程中， 檢查條件的滿足性后才進(jìn)行變量的代換。代入新條件后的新狀態(tài)如果是目標(biāo)狀態(tài)，則問題解決；否則看是否滿足下面的操作，如果不滿足或即使?jié)M足卻又回到了原來的狀態(tài)，那么代入無效。4、 源代碼#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);nextstep(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); move