GeneralProblemSolver

Monkey and Banana

2013 Artificial Intelligence

Lecture no. 14

GeneralProblemSolver

人間の問題解決過程のモデル与えられた知識 → 問題を解決

• 初期状態 (Initial State)

• 目的状態 (target State)

• オペレータ(operator)

• 状態表現 (state representation)

与えられた知識Given Knowledge

など

Monkey and Banana

How the monkey gets banana ？Needs to climb up on the box

状態表現 (State Representaiton)

AT(A, X) A is at XON(A, X) A is on XEMPTY Monkey’s HAND is EMPTYHOLD Monkey HOLDS banana

初期状態 (initial state), 目的状態 (target state)

AT(monkey, a)

AT(box, b)

EMPTY

Initial State

HOLD

Target State

(monkey holds banana)

Operators

GOTO(a) Monkey goes to “a” CLIMB Monkey climbs up on the Box

PC 前提条件 (none) PC 前提条件 AT(monkey, y),AT(box, y)

D 削除リスト

AT(monkey, x) D 削除リスト AT(monkey, y)

A 追加リスト

AT(monkey, a) A 追加リスト ON(monkey, box)

MOVEBOX(a)

Monkey moves the box to “a”

GRASP バナナを掴む

PC 前提条件 AT(monkey, y) ,AT(box, y)

PC 前提条件 AT(box, c), ON(monkey,box)

D 削除リスト

AT(monkey, y) , AT(box, y)

D 削除リスト EMPTY

A 追加リスト

AT(monkey, a) AT(box, a)

A 追加リスト HOLD

問題解決 How to solve the problem ？

現在の状態 (S) と目的状態 (G) を近づける

　　　　　　　　　　　　 S 　→　 G

S=G ： 同じ状態

difference 　 = 　 G - S

Reduce

Importance of Differences

D3 = distance between monkey and banana

D2= distance between the box and banana

D1 = distance between monkey and banana

Most important

Necessary to solve D3

Necessary to solve D2

importance D3 > D2 > D1

Operators to reduce distances

GOTOMOVEBO

XCLIMB GRASP

D1(monkey) ○ ○ ○

D2(box) ○

D3(monkey’s hand) ○

operatorsdistance

s

GPS algorithm1.LOOP1: if S satisfy then exit(S)2.find D=most important difference between G

& S3.put all the operators to reduce D into

oplist4.LOOP2: if empty(oplist) then

return(null)5. operator:=first(oplist), remove(operator, oplist), pc := operator の前提条件 , if not effective(operator) then goto LOOP2 : give up, if operator fails to reduce d 6. if (pc = null) then goto APPLY 7. S1 = GPS(S, pc)8. If S1=fail then goto LOOP29. APPLY : S := operator (S1)10.goto LOOP1

GPS algorithm solves“Monkey and Banana”

Introduction

AT(monkey, a)AT(box, b)EMPTYInitial state S0

HOLD

Final state G0

Monkey gets banana

GPS(S0 , G0)

GPS(S0, G0)Step 1: Continue GPS unless S0=G0

Step 2: Find differences between S0 and G0

D3 = |monkey – banana| D2 = |box – banana| D1 = |box – monkey|

D3 is most important!

GPS(S0, G0)

Step 3Operator to reduce D3 is GRASP→GRASP is put into oplist

Step 4 oplist is not empty

Continue GPS

GPS(S0, G0)

Step 5

operator = first(oplist), first(oplist) is removed from oplist,pc

GRASP solves D !

GPS(S0, G0)

Step 6Continue to Step 7 if pc exists (if not, jump to Step 9)

Step 7GPS(S0, pc)

GPS(S0, G1)Step 1

Continue GPS, unless S0=G1

Step 2Distances between S0& G1

D2 =|box - banana| 　 D1 = |monkey – box|

D2 is most important

GPS(S0, G1)Step 3

D is solved by MOVEBOX,Put MOVEBOX into oplist

Step 4Continue GPS unless oplist is empty

GPS(S0, G1)Step 5

Substitute prerequisit conditions into pc

AT(box, y), AT(monkey, y)}where y=b is necessary to solve D2

pc = {AT(box, b), AT(monkey, b)}

Continue, since MOVEBOX reduces D2

GPS(S0, G1)Step 6

Continue to Step 7 since pc is not empty(if not, jump to Step9)

Step 7Run GPS(S0, pc)

GPS(S0, G2)Step 1

Continue GPS unless S0=G2

Step 2Distance between S0 and G2

D1 =|monkey - box|

D1 is most important

GPS(S0, G2)

Step 3Find operators to reduce D1 す and put them into oplist GOTO and CLIMB are put into oplist

Step 4Continue GPS unless oplist is empty

GPS(S0, G2)

Step 5oplist = {CLIMB} , operator = GOTO

Prerequisit conditions of the operator are substituted to pc

pc = (none)

GOTO reduces D1

Step 6No more pc existJump to Step 9

GPS(S0, G2) → GPS(S1, G2)Step 9

Apply operator on S0

operator = GOTO(b)

New state S1

GPS(S1, G2) → GPS(S1, G1) Step 10

Go to Step 1

Step 1S0=G1

End GPS(S1, G2)

Go back to GPS(S1, G1)

GPS(S1, G1) → GPS(S2, G1) Step 8

Continue if S1 is not fail

Step 9Apply operator on S1

operator = MOVEBOX(c)

New state S2

GPS(S2, G1)Step10

Go to Step1

Step 1Continue GPS unless S0=G1

Step 2Difference between S2and G1

D1 =|monkey – box|

D1 is most important

GPS(S2, G1)Step3

Operators to solve D1 are GOTO and CLIMBAdd GOTO and CLIMB to oplist

Step 4Continue GPS unless oplist is empty

GPS(S2, G1)Step5

oplist = {CLIMB} , operator = GOTO

Substitute prerequisite conditions to pc

pc = (none)

Go to Step 4, sinceNo operator can reduce D1

GPS(S2, G1)Step 5

oplist = {} , operator = CLIMB

Substitute prerequisite conditions of operators into pc

AT(box, y), AT(monkey, y)}

where y=c reduces D1 pc= {AT(box, c), AT(monkey, c)}

Continue sinceCLIMB reduces D1

GPS(S2, G1)Step 6

Continue since pc exists

Step 7Run GPS(S2, pc)

GPS(S2, G3) → GPS(S2, G1) Step1

S2=G3

GPS ends.

Go upwards

GPS(S2, G1) → GPS(S3, G1) Step 8

Continue since S2 is not fail

Step 9Apply operator on S1

operator = CLIMB

New state S3

GPS(S3, G1) → GPS(S3, G0) Step 10

Go to Step 1

Step 1S3=G1

Finish GPS

Go upwards.

GPS(S3, G0) → GPS(S4, G0) Step8

Continue since S2 is not fail

Step 9Apply operator on S1

operator = DRASP

New state S4

GPS(S4, G0)Step10

Go to Step 1

Step 1S4=G0

Finish GPS

End