Constraint Satisfaction Problems (CSP)

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Constraint Satisfaction Problems

(CSP)

CSP

atau Constraint Satisfaction Problem adalah

permasalahan yang tujuannya adalah mendapatkan suatu

kombinasi variabel-variabel tertentu yang memenuhi

aturan-aturan (constraints) tertentu.

State

didefinisikan dengan variables X

_i

yang mempunyai

values dari domain D

_i

Goal Test

adalah sebuah himpunan constraints yang

memberikan kombinasi yang diijinkan untuk mengisi variabel

Batasan CSP dalam perkuliahan ini: diskrit (solusi deterministik),

absolut (solusi pasti tersedia dalam domain), unair atau biner (satu

atau dua variabel yang harus diisi).

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CSP Example: Map-Coloring

Variables: WA, NT, Q, NSW, V, SA, T

Domains: D_i = {red, green, blue}

Constraints: adjacent regions must have different colors

e.g.: WA ≠ NT, WA ≠ SA, NT ≠ SA, ... (if the language allow this), or (WA,NT) Є { (red, green), (red, blue), (green, red),...}

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CSP Example: Map-Coloring

Solutions

are complete and consistent assignments, e.g.

{WA=red, NT=green, Q=red, NSW=green,

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Varieties of Constraints

Unary

: Constraints involve a single variable

e.g.: SA ≠ green

Binary

: Constraints involve pairs of variables

e.g.: SA ≠ WA

Higher-Order

: Constraints involve 3 or more variables

e.g.: cryptarithmetic column constraints

Preferences (soft constraints)

:

e.g.: blue is better than green.

(Often representable by a cost for each variable assignment



Constrained Optimization Problems

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Standard Search Formulation for

CSP

States ditentukan dengan nilai yang sudah dialokasikan sekarang

Initial State

: { }

Successor Function

: assign value ke variable yang belum terisi

nilai tidak boleh melanggar constraint

Goal Test

: bila assignment selesai dilakukan

Catatan:

1. Hal ini berlaku untuk setiap masalah CSP

2. Karena variable terbatas maka setiap solusi akan muncul pada kedalaman n dengan n variable  gunakan depth-first Search

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Backtracking Example: Map

Coloring

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Backtracking Example: Map

Coloring

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Backtracking Example: Map

Coloring

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Backtracking Example: Map

Coloring

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Memperbaiki backtracking

1. Variable yang mana yang harus di assign terlebih dahulu ?

2. Bagaimana urutan nilai dicoba ?

3. Bisakah kita mendeteksi kegagalan lebih awal

4. Dapatkah kita menggunakan struktur problem untuk membantu

kita ?

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Forward Checking

Idenya :

• Simpan nilai valid untuk variable yang belum diassign

• Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid maka pencarian dihentikan

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Idenya :

• Simpan nilai valid untuk variable yang belum di-assign

• Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid maka search dihentikan

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Forward Checking

Idenya :

• Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid maka search dihentikan

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Forward Checking

Idenya :

• Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid search dihentikan

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Constraint Propagation

Forward checking memberikan informasi dari variabel yang dialokasi, namun tidak dapat mendeteksi kegagalan sebelumnya.

NT dan SA tidak boleh diberikan warna biru !

Constraint Propagation secara berulang mengevaluasi alokasi variabel dalam skala lokal (solusi sementara)

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Isikan bidang (R1..R7) di atas dengan warna:

merah

,

kuning

,

hijau

,

biru

.

Bidang bertetangga tidak boleh memiliki warna yang sama.

1. Apakah variabel yang Anda gunakan? 2. Apakah domain yang tersedia?

3. Bagaimana Anda mengevaluasi constraints-nya?

R1

R2 R3 R4 R5 R7

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1. Variabel yang harus diisi: R1, .. R7

2. Domain yang tersedia: warna (

merah

,

kuning

,

hijau

,

biru)

3. Constraints:

1. R1 <> R2, …, R7,

2. R2 <> R3,

3. R3 <> R4,

4. R4 <> R5,

5. R5 <> R6,

6. R6 <> R7

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Most Constrained Variable Heuristic

Pilih variabel dengan kemungkinan nilai legal paling

sedikit, constraint terbesar. (Cari variabel yang paling

susah untuk diisi)

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Least Constraining Value

Diberikan sebuah variabel, pilihlah yang memiliki nilai

constraint paling sedikit (legalitas terbesar, variabel

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Degree Heuristic

Pilih variabel dengan constraint paling besar diantara

variabel yang belum terisi (kumpulkan variabel-variabel

yang paling sulit diisi)

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

Why game playing ?

›

It’s Fun

›

Game Playing is non trivial



Player need ‘human-like’ intelligence



Games can vary in complexity



Decision making should be done in limited time

›

Games are :



Well defined and repeatable



Limited and accessible

›

Games Can directly compare

human and computer

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Checkers:

– 1994: Chinook (U.of A.) beat world champion Marion Tinsley, ending 40-yr reign. Othello:

– 1997: Logistello (NEC research) beat the human world champion.

– Today: world champions refuse to play AI computer program (because it’s too good).

Chess:

 1997: Deep Blue (IBM) beat world champion Gary Kasparov

 2005: a team of computers (Hydra, Deep Junior and Fritz), wins 8.5-3.5

against a rather strong human team formed by Veselin Topalov, Ruslan Ponomariov and Sergey Karjakin, who had an average ELO rating of 2681.

 2006: The undisputed world champion, Vladimir Kramnik, is defeated 4-2 by

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Backgammon:

– TD-Gammon (IBM) is world champion amongst humans and computers

Go:

– Human champions refuse to play top AI player (because it’s too weak)

Bridge:

– Still out of reach for AI players. Why ?

Others :  ???

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Perfect vs. Imperfect information:

 Perfect: See the exact state of the game

›

e.g. chess, backgammon, checkers, go, othello

 Imperfect: Information is hidden

›

e.g. scrabble, bridge, most card games

Deterministic vs Stochastic:

 Deterministic: Change in state is fully determined by player

move.

›

e.g. chess, othello

 Stochastic: Change in state is partially determined by

chance.

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Deterministic Stochastic admissible, perfect info Checkers, Chess, Go, Othello Backgammon, Monopoly not admissible, imperfect info ??? Bridge, Poker, Scrabble

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

We can model game playing as a

search in a state space as we did with



In order to model a game into a search

problem we need to decide

›

the states,

›

operator,

›

initial state,

›

goal, and

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 Consider a two player board game: – e.g., chess, checkers, tic-tac-toe

– board configuration: unique arrangement of "pieces“  Representing board games as search problem:

– states: board configurations – operators: legal moves

– initial state: current board configuration

– terminal state: winning/terminal board configuration – utility function: values for terminal state

(win: +1, loss: -1, draw: 0)

 We want to find a strategy (i.e. way of picking moves) that wins the

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

Assume the opponent’s moves can be predicted given

the computer's moves



How complex would search be in this case?

–

Worst case:

O(b

m

) b

ranching factor,

m

ax depth

–

Tic-Tac-Toe: ~5 legal moves, max of 9 moves

 59 = 1,953,125 states

–

Chess: ~35 legal moves, ~100 moves per game

 bd ~ 35100 ~10154 states, “only” ~1040 legal states

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The Problem is that the enemy will not do exactly as we planned, in fact the enemy will try to do the best move for it and thus creating the worst move for the player

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

Search Tree ?

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• Expand complete search tree in DFS manner,

until terminal states have been reached and their utilities

computed.

• Computer favors high utility value and the opponent favors

low utility value.

Computer will choose the moves that maximize the utility

value.

• Go back up from leaves towards the current state of the game.

 At

each min

node: backup the worst value among the

children. (opponent’s move)

 At

each max

node: backup the best value among the

children. (computer’s move)

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1. Generate the complete game tree.

2. Apply the utility function to all the terminal

states.

3. Use the utility of the terminal states to

calculate a utility value for their parents

(either max or min) depending on depth.

4. Continue up to root node.

5. Choose move with highest value from root

node.

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The utility function is only applied to terminal nodes.

If max makes move A1 then Min should make move A11. Thus the result of making move A1 is a value of 3 for the utility function.

Similarly A2 leads to a value of 2 and A3 a value of 2.

Max wants to maximise the utility function and so A1 is the best move to make.

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Complete ?

Only if tree is finite.

NB a finite strategy can exist even in an infinite tree

Optimality ?

Yes, against an optimal opponent. (Otherwise we don’t

know)

Time Complexity ?

O(b

m

₎

Space Complexity ?

O(bm) (depth-first exploration)

Why not use Minimax to solve Chess ?

For chess, b 35, m 100 for “reasonable” games

exact solution completely infeasible

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• Suppose we have 100 seconds to make a move, and we can search 104_{nodes per second.}

• So we can only search 106_{nodes per move}

(Or even fewer, if we spend time deciding which nodes to search.) • Standard approach:

1. Use a cutoff test instead of terminal test (e.g. based on depth limit)

2. Use an evaluation function instead of utility function for the nodes where we cutoff the

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• An evaluation function

v(s)

represents the “goodness” of a

board state (e.g. chance of winning from that position).

• If the features of the board can be evaluated independently,

use a

weighted linear function

:

v(s) = w

₁

f

₁

(s) + w

₂

f

₂

(s) + … + w

_n

f

_n

(s) =

(where s is board state)

• More important features get more weight

• This function can be given by the expert or learned from

experience.

(s) f w n 1 i i i

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w₁ = 9; f₁(s) = (number of white queens) – (number of black queens) w₂ = 3; f₂(s) = (number of white knights) – (number of black knights)

w₃ = 1; f₃(s) = (number of white pawns) - (number of black pawns)

The quality of play depends directly on the quality of the evaluation function

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• Evaluation function is only approximate, and is usually better if we are close to the end of the game.

• Move chosen is the same if we apply a monotonic transformation to the evaluation function.

• Only the order of the numbers matter: payoffs in deterministic games act as an ordinal utility function.

=

The evaluation function :

precision?

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 Diberikan sebuah pen jadwalan kelas sebagai berikut: ada 4

kelas (C1,… ,C4), dan 3 ruangan (R1, .., R3). Terdapat penjadwalan sebagai berikut: