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 Xi yang mempunyai values dari domain Di 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).
CSP Example: Map-Coloring
Variables: WA, NT, Q, NSW, V, SA, T Domains: Di = {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),...}
CSP Example: Map-Coloring
Solutions are complete and consistent assignments, e.g. {WA=red, NT=green, Q=red, NSW=green, V=red, SA=blue, T=green}
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
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 3. Karena path irrelevan kita dapat gunakan algoritma local search
Backtracking Example: Map Coloring
Backtracking Example: Map Coloring
Backtracking Example: Map Coloring
Backtracking Example: Map Coloring
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 ?
Forward Checking Idenya : • Simpan nilai valid untuk variable yang belum diassign • Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid maka pencarian dihentikan
Forward Checking Idenya : • Simpan nilai valid untuk variable yang belum di-assign • Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid maka search dihentikan
Forward Checking Idenya : • Simpan nilai valid untuk variable yang belum diassign • Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid maka search dihentikan
Forward Checking Idenya : • Simpan nilai valid untuk variable yang belum diassign • Bila salah satu variable tidak mempunyai kemungkinan nilai yang valid search dihentikan
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)
R7 R6 R2
R3
R4
R5
R1
Isikan bidang (R1..R7) di atas dengan warna: merah, kuning, hijau, biru. Bidang bertetangga tidak boleh memiliki warna yang sama. 1. 2. 3.
Apakah variabel yang Anda gunakan? Apakah domain yang tersedia? Bagaimana Anda mengevaluasi constraints-nya?
Variabel yang harus diisi: R1, .. R7 2. Domain yang tersedia: warna (merah, 1.
kuning, hijau, biru) 3. Constraints: 1. 2. 3. 4. 5. 6.
R1 <> R2, …, R7, R2 <> R3, R3 <> R4, R4 <> R5, R5 <> R6, R6 <> R7
Backtracking
Backtracking
Backtracking
Backtracking
Backtracking
Backtracking
Backtracking
Backtracking
Most Constrained Variable Heuristic Pilih variabel dengan kemungkinan nilai legal paling sedikit, constraint terbesar. (Cari variabel yang paling susah untuk diisi)
Dikenal juga dengan heuristik Minimum Remaining Values (MRV)
Least Constraining Value Diberikan sebuah variabel, pilihlah yang memiliki nilai constraint paling sedikit (legalitas terbesar, variabel yang paling mudah diisi)
Degree Heuristic Pilih variabel dengan constraint paling besar diantara variabel yang belum terisi (kumpulkan variabel-variabel yang paling sulit diisi)
Tie-breaker diantara MRV variabel
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
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 Deep Fritz.
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 :
???
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.
› e.g. backgammon, monopoly
Deterministic
Stochastic
admissible, perfect info
Checkers, Chess, Go, Othello
Backgammon, Monopoly
not admissible, imperfect info
???
Bridge, Poker, Scrabble
We can model game playing as a search in a state space as we did with other problems before. In order to model a game into a search problem we need to decide
› the states, › operator,
› initial state, › goal, and › utility function
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 game.
Assume the opponent’s moves can be predicted given the computer's moves
How complex would search be in this case? – Worst case: O(bm) branching factor, max depth – Tic-Tac-Toe: ~5 legal moves, max of 9 moves
–
Chess: ~35 legal moves, ~100 moves per game
59 = 1,953,125 states bd ~ 35100 ~10154 states, “only” ~1040 legal states
Common games produce enormous search trees
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 How do we deal with this ?
Search Tree ? How do we implement the search tree ??
• 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)
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.
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.
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(bm) 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 But do we need to explore every path?
• 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 search.
• 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: n
v(s) = w1f1(s) + w2f2(s) + … + wnfn(s) =
w i f i (s) i 1
(where s is board state)
• More important features get more weight • This function can be given by the expert or learned from experience.
The evaluation function
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
The evaluation function : precision? • 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.
Diberikan sebuah pen jadwalan kelas sebagai berikut: ada 4 kelas (C1,… ,C4), dan 3 ruangan (R1, .., R3). Terdapat penjadwalan sebagai berikut:
Terdapat pembatasan sebagai berikut: › Setiap kelas harus menggunakan salah satu dari ketiga ruangan yang tersedia › R3 terlalu kecil untuk C3 › R2 dan R3 terlalu kecil untuk C4
Variabel dan domain apa saja yang dapat diberikan untuk problem penjadwalan tersebut? 2. Tunjukkan kemungkinan isi nilai untuk setiap variabel sesuai dengan constraints di atas. 3. Ekspresikan constraints problem secara formal. 1.
1. 2.
Variabel: C1, C2, C3, C4; Domain: R1, R2, R3 Kemungkinan alokasi variabel dari domain › C1: { R1, R2, R3 } › C2: { R1, R2, R3 } › C3: { R1, R2 } › C4: { R1 }
3.
Constraints yang ada:
› Kelas tidak boleh ada yang bentrok C1 != C2, C1 != C3, C2 != C3, C2 != C4, C3 != C4 › Pembatasan kapasitas ruangan C3 != R3, C4 != R2, C4 != R3
Berikan sekarang solusinya (manfaatkan constraints graph)
Constraints graph C1
C2
C3
C4
time r1
r2
r3
Diberikan sebuah situasi permainan seperti di bawah ini:
X (max player) sedang dalam giliran untuk melanjutkan permainan. Berikan semua situasi berikutnya yang mungkin untuk X
Pilihlah jalur yang tepat sesuai dengan algoritma minmax, jika diketahui fungsi utilitas untuk situasi menang untuk X = +10, kalah = -10, dan draw = 0.
