猴子摘香蕉實(shí)驗(yàn)報(bào)告(含代碼)

1、華中科技大學(xué)人工智能與模式識(shí)別 實(shí)驗(yàn)報(bào)告 院 系： 電子與信息工程系 班 級(jí)： 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?jī)?nèi)有一個(gè)猴子，一個(gè)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(qǐng)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(tái)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(gè)體域：xmonkey，wa,b,c,box；HOLD(x,t)：x手中拿著t，個(gè)體域：tbox,banana；EMPTY(x)：x手中是空的；ON(t,y)：t在y處，個(gè)體域：yb,c,ceiling；CLEAR(y)：y上是空的；BOX(u)：u是箱子，個(gè)體域：ubox；BANANA(v)：v是香蕉，個(gè)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(dòng)作。WALK(m,n)：猴子從m走到n處，個(gè)體域：m,na,b,c；CARRY(s,r)：猴子在r處拿到s，個(gè)體域：rc,ceiling，sbox,banana；CLIMB(u,b)：猴子在b處爬上u；這3個(gè)操作也可分別用條件和動(dòng)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(shí)，則可激活操作規(guī)則，于是可

4、執(zhí)行該規(guī)則中的動(dòng)作部分。動(dòng)作通過前后狀態(tài)的變化表示，即通過從動(dòng)作前刪除或增加謂詞公式來描述動(dòng)作后的狀態(tài)。WALK(m,n)：猴子從m走到n處條件：AT(monkey,m)動(dòng)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(dòng)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(dòng)作：4. 按照行動(dòng)計(jì)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(dòng)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(gè)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(dòng)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 <stdio.h>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,

8、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)routesavei="Monkey go to A"Statesi+1=Statesi;Statesi+1.monkey=-1;el

9、se 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 move the box to the other pla

10、ce*/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

12、 climbdown,monkey climb down 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

13、 banana"/*output the solution to the problem*/void showSolution(int i) int c; printf ("%s n", "Result to problem:"); for(c=0; c<i+1; c+) printf ("Step %d : %s n",c+1,routesavec); printf("n");/*perform next step*/void nextStep(int i)int c; int j; if(i&g

14、t;=150) printf("%s n", "steplength reached 150,have problem "); return; for (c=0; c<i; 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=S

15、tatesi.monbox) return; if(Statesi.monbox=1&&Statesi.monkey=0&&Statesi.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)i

16、f(Statesi.monbox=-1)climbonto(i);reach(i+1);nextStep(j);/*monkeygoto(-1,i);nextStep(j);monkeygoto(0,i);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);next

17、Step(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);else /*box=-1*/monkeygoto(-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);next

18、Step(j);elseclimbdown(i);nextStep(j);movebox(0,i);nextStep(j);climbonto(i);reach(i+1);nextStep(j);else 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