1 / 36

G eneral P roblem S olver

G eneral P roblem S olver. Monkey and Banana. 2013 Artificial Intelligence Lecture no. 14. G eneral P roblem S olver. 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.

zoltin
Download Presentation

G eneral P roblem S olver

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. GeneralProblemSolver Monkey and Banana 2013 Artificial Intelligence Lecture no. 14

  2. 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

  3. Monkeyand Banana How the monkey gets banana？ Needs to climb up on the box

  4. 状態表現(State Representaiton)

  5. 初期状態(initial state),目的状態(target state) AT(monkey, a) AT(box, b) EMPTY HOLD (monkey holds banana) Initial State Target State

  6. Operators

  7. 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 Reduce difference=G-S

  8. Importance of Differences D3 = distance between monkey and banana Most important D2= distance between the box and banana Necessary to solve D3 D1 = distance between monkey and banana Necessary to solve D2 importance D3 > D2 > D1

  9. Operators to reduce distances operators distances

  10. GPS algorithm LOOP1: if S satisfy then exit(S) find D=most important difference between G & S put all the operators to reduce D into oplist LOOP2: if empty(oplist) then return(null) 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 if (pc = null) then goto APPLY S1 = GPS(S, pc) If S1=fail then goto LOOP2 APPLY : S := operator (S1) goto LOOP1

  11. GPS algorithm solves“Monkey and Banana”

  12. Introduction AT(monkey, a) AT(box, b) EMPTY HOLD Monkey gets banana Initial state S0 Final state G0 GPS(S0 , G0)

  13. 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| D3is most important!

  14. GPS(S0, G0) Step 3 Operator to reduce D3is GRASP → GRASP is put into oplist Step 4 oplist is not empty Continue GPS

  15. GPS(S0, G0) Step 5 operator = first(oplist), first(oplist) is removed from oplist, pc GRASP solves D !

  16. GPS(S0, G0) Step 6 Continue to Step 7 if pc exists (if not, jump to Step 9) Step 7 GPS(S0, pc)

  17. GPS(S0, G1) Step 1 Continue GPS, unless S0=G1 Step 2 Distances between S0& G1 D2 =|box - banana| D1 = |monkey – box| D2is most important

  18. GPS(S0, G1) Step 3 D is solved by MOVEBOX, Put MOVEBOX into oplist Step 4 ContinueGPS unless oplist is empty

  19. 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

  20. GPS(S0, G1) Step 6 Continue to Step 7 since pc is not empty (if not, jump to Step9) Step 7 Run GPS(S0, pc)

  21. GPS(S0, G2) Step 1 Continue GPS unless S0=G2 Step 2 Distance between S0and G2 D1 =|monkey - box| D1is most important

  22. GPS(S0, G2) Step 3 Find operators to reduce D1すand put them into oplist GOTO and CLIMB are put into oplist Step 4 Continue GPS unless oplistis empty

  23. GPS(S0, G2) Step 5 oplist = {CLIMB} , operator =GOTO Prerequisit conditions of the operator are substituted to pc pc = (none) GOTOreduces D1 Step 6 No more pc exist Jump to Step 9

  24. GPS(S0, G2) → GPS(S1, G2) Step 9 Apply operator on S0 operator =GOTO(b) New state S1

  25. GPS(S1, G2) → GPS(S1, G1) Step 10 Go to Step 1 Step 1 S0=G1 End GPS(S1, G2) Go back to GPS(S1, G1)

  26. GPS(S1, G1) → GPS(S2, G1) Step 8 Continue if S1 is not fail Step 9 Apply operator on S1 operator =MOVEBOX(c) New state S2

  27. GPS(S2, G1) Step10 Go to Step1 Step 1 Continue GPS unless S0=G1 Step 2 Difference between S2and G1 D1 =|monkey – box| D1is most important

  28. GPS(S2, G1) Step3 Operators to solve D1 are GOTO and CLIMB Add GOTO and CLIMB to oplist Step 4 Continue GPS unless oplist is empty

  29. GPS(S2, G1) Step5 oplist = {CLIMB} , operator = GOTO Substitute prerequisite conditions to pc pc = (none) Go to Step 4, since No operator can reduce D1

  30. 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 since CLIMB reduces D1

  31. GPS(S2, G1) Step 6 Continue since pc exists Step 7 Run GPS(S2, pc)

  32. GPS(S2, G3) → GPS(S2, G1) Step1 S2=G3 GPS ends. Go upwards

  33. GPS(S2, G1) → GPS(S3, G1) Step 8 Continue since S2 is not fail Step 9 Apply operator on S1 operator =CLIMB New state S3

  34. GPS(S3, G1) → GPS(S3, G0) Step 10 Go to Step 1 Step 1 S3=G1 Finish GPS Go upwards.

  35. GPS(S3, G0) → GPS(S4, G0) Step8 Continue since S2is not fail Step 9 Apply operator on S1 operator =DRASP New state S4

  36. GPS(S4, G0) Step10 Go to Step 1 Step 1 S4=G0 Finish GPS End

More Related