Constraint Satisfaction Problems ECE457 Applied Artificial Intelligence Spring 2008 Lecture #4

Outline Defining constraint satisfaction problems (CSP) CSP search algorithms Russell & Norvig, chapter 5 ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 2

States What do we know about states? A state might be a goal (goal test) A state has a value (cost or payoff) An agent moves from state to state using actions The state space can be discreet or continuous Ties in with the problem definition Initial state, goal test, set of actions and their costs defined in problem ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 3

Definitions A constraint-satisfaction problem has A set of variables V = {X1, X2, …, Xn} Each variable has a domain of values Xi  Di = {di1, di2, …, din} A set of constraints on the values each variable can take C = {C1, C2, …, Cm} A state is a set of assignment of values S1 = {X1 = d12, X4 = d45} ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 4

Definitions Consistent (legal) state Complete state Goal state Does not violate any constraints Complete state All variables have a value Goal state Consistent and complete Might not exist Proof of inconsistency ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 5

Example: 8-queen Variables: 64 squares, number of queens V = {S1,1, S1,2, …, S8,8, Number_of_queens} Values: Queen or no queen Si,j  DS = {queen, empty} Number_of_queens  DN = [0, 64] Constraints: Attacks, queen count {Number_of_queens = 8, Si,j = queen  Si,j+n = empty, Si,j = queen  Si+n,j = empty, Si,j = queen  Si+n,j+n = empty} States: All board configurations 2.8x1014 complete states 1.8x1014 complete states with 8 queens 92 complete and consistent states 12 unique complete and consistent states ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 6

Example: 8-queen ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 7

Example: The Einstein Puzzle There are 5 houses in 5 different colours. In each house lives a person with a different nationality. These 5 owners drink a certain beverage, smoke a certain brand of cigar and keep a certain pet. No owners have the same pet, smoke the same brand of cigar or drink the same drink. Who keeps the fish? The English lives in a red house. The Swede keeps dogs as pets. The Dane drinks tea. The green house is on the left of the white house. The green house owner drinks coffee. The person who smokes Pall Mall rears birds. The owner of the yellow house smokes Dunhill. The man living in the house right in the centre drinks milk. The Norwegian lives in the first house. The man who smokes Blend lives next to the one who keeps cats. The man who keeps horses lives next to the man who smokes Dunhill. The owner who smokes Blue Master drinks beer. The German smokes Prince. The Norwegian lives next to the blue house. The man who smokes Blend has a neighbour who drinks water. ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 8

Example: The Einstein Puzzle 1 2 3 4 5 25 variables V = {N1, …, N5, C1, …, C5, D1, …, D5, S1, …, S5, P1, …, P5} Domains Ni  {English, Swede, Dane, Norwegian, German} Ci  {green, yellow, blue, red, white} Di  {tea, coffee, milk, beer, water} Si  {Pall Mall, Dunhill, Blend, Blue Master, Prince} Pi  {dog, cat, horse, fish, birds} ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 9

Example: The Einstein Puzzle The Norwegian lives in the first house. N1 = Norwegian The English lives in a red house. (Ni = English)  (Ci = Red) The green house is on the left of the white house. (Ci = green)  (Ci+1 = white) (C5 ≠ green) (C1 ≠ white) ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 10

Example: Map Colouring Colour map of the provinces of Australia 3 colours (red, green, blue) No adjacent provinces of the same colour Define CSP ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 11

Example: Map Colouring CSP Variables {WA, NT, SA, Q, NSW, V, T} Domain {R, G, B} Constrains {WA  NT, WA  SA, NT  SA, NT  Q, SA  Q, SA  NSW, SA  V, Q  NSW, NSW  V} ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 12

Solving CSP Iterative improvement methods Tree searching Start with random complete state, improve until consistent Tree searching Start with empty state, make consistent variable assignments until complete ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 13

Iterative Improvement Min-Conflict Algorithm: Assign to each variable a random value While state not consistent Pick a variable (randomly or with a heuristic) Find its values that minimize the number of violated constraints If there is only one such value Assign that value to the variable If there are several values Assign a random value from that set to the variable ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 14

Min-Conflict Algorithm ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 15

Min-Conflict Algorithm Advantage Very efficient Disadvantages Only searches states that are reachable from the initial state Might not search all state space Does not allow worse moves Might get stuck in a local optimum Not complete! Might not find a solution even if one exists ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 16

Tree Search Formulate CSP as tree Root node: no variables are assigned a value Action: assign a value if it does not violate any constraints Solution node at depth n for n-variable problem ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 17

Backtracking Search Start with empty state While not complete Pick a variable (randomly or with heuristic) If it has a value that does not violate any constraints Assign that value Else Go back to previous variable Assign it another value ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 18

Backtracking Search Depth-first search algorithm Algorithm complete Goes down one variable at a time In a dead end, back up to last variable whose value can be changed without violating any constraints, and change it If you backed up to the root and tried all values, then there are no solutions Algorithm complete Will find a solution if one exists Will expand the entire (finite) search space if necessary Depth-limited search with limit = n ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 19

Example: Backtracking Search WA = G WA = R WA = B NT = G NT = B Q = B Q = R NSW = G NSW = G SWA = B NSW = R SA = B SA = ? SA = ? V = R T = G T = B T = R ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 20

Conflict-Directed Backjumping Suppose we colour Australia in this order: WA – R NSW – R T – B NT – B Q – G SA - ? Dead-end at SA No possible solution with WA = NSW Backtracking will try to change T on the way, even though it has nothing to do with the problem, before going to NSW ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 21

Conflict-Directed Backjumping Backtracking goes back one level in the search tree at a time Chronological backtrack Not rational in cases where the previous step is not involved to the conflict Conflict-directed backjumping (CBJ) Should go back to a variable involved in the conflict Skip several levels if needed to get there Non-chronological backtrack ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 22

Conflict-Directed Backjumping Maintain a conflict set for each variable List of previously-assigned variables that are related by constraints conf(WA) = {} conf(NSW) = {} conf(T) = {} conf(NT) = {WA} conf(Q) = {NSW,NT} conf(SA) = {WA,NSW,NT,Q} When we hit a dead-end, backjump to the deepest variable in the conflict set ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 23

Conflict-Directed Backjumping Learn from the conflict by updating the conflict set of the variable we jumped to Conflict at Xj, backjump to Xi conf(Xi)={X1,X2,X3} conf(Xj)={X3,X4,X5,Xi} conf(Xi) = conf(Xi)  conf(Xj) – {Xi} conf(Xi) = {X1,X2,X3,X4,X5} Xi absorbed the conflict set of Xj ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 24

Conflict-Directed Backjumping conf(WA) = {} conf(NSW) = {} conf(T) = {} conf(NT) = {WA} conf(Q) = {NSW,NT} conf(SA) = {WA,NSW,NT,Q} conf(NSW) = {WA} conf(NT) = {WA,NSW} conf(Q) = {WA,NSW,NT} NT backjump to NSW conf(NSW) = {WA} Skips T, which is irrelevant in this conflict Discovers the relationship between NSW and WA, which is not present in our constraints Q backjump to NT conf(NT) = {WA,NSW} Meaning: “There is no consistent solution from NT onwards, given the preceding assignments of WA and NSW together” SA backjump to Q conf(Q) = {WA,NSW,NT} Meaning: “There is no consistent solution from Q onwards, given the preceding assignments of WA, NSW and NT together” ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 25

Heuristics Backtracking and CDJ searches are variations of depth-limited search Uninformed search technique Can we make it an informed search? Add some heuristics Which variable to assign next? In which order should the values be tried? How to detect dead-ends early? ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 26

Variable & Value Heuristics Most constrained variable Choose the variable with the fewest legal values remaining in its domain aka Minimum remaining values Most constraining variable Choose the variable that’s part of the most constraints Useful to pick first variable to assign aka Degree heuristic Least constraining variable Pick the variable that’s part of the fewest constrains Keeps maximum flexibility for future assignments ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 27

Variable & Value Heuristics Values: 2 Constraints: 2 Most constrained variable Values: 2 Constraints: 2 Values: 2 Constraints: 1 Values: 2 Constraints: 2 Most constrained & least constraining variable Most constrained variable Most constraining variable Most constraining variable Most constrained variable Values: 2 Constraints: 1 Least constraining variable ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 28

Forward Checking How to detect dead-ends early? Keep track of the domain of unassigned variables Use constraints to prune domain of unassigned variables Backtrack when a variable has an empty domain Do not waste time exploring further ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 29

Example: Forward Checking B NT G Q R NSW GB V RGB T SA WA B NT G Q RB NSW RGB V T SA R WA RGB NT Q NSW V T SA WA B NT RG Q RGB NSW V T SA ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 30

Problem with Forward Checking NT G Q NSW V T RGB SA R ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 31

Constraint Propagation Propagate the consequences of a variable’s constraints onto other variables Represent CSP as constraint graph Nodes are variables Arcs are constraints Check for consistency NT Q WA NSW SA V T ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 32

Checking for Consistency Node consistency Unary constraint (e.g. NT  G) A node is consistent if and only if all values in its domain satisfy all unary constraints Arc consistency Binary constraint (e.g. NT  Q) An arc Xi  Xj or (Xi, Xj) is consistent if and only if, for each value a in the domain of Xi, there is a value b in the domain of Xj that is permitted by the binary constraints between Xi and Xj. Path consistency Can be reduced to arc consistency ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 33

Checking for Consistency Node consistency Simply scan domain of values of each variable and remove those that are not valid Arc consistency Examine edges and delete values from domains to make arcs consistent ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 34

Checking for Consistency Remove inconsistent values from a variable’s domain Backtrack if empty domain Maintaining node and arc consistency reduces the size of the tree More computationally expensive than Forward Checking, but worth it ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 35

AC-3 Algorithm Keep queue of arcs (Xi, Xj) to be checked for consistency If checking an arc removes a value from the domain of Xi, then all arcs (Xk, Xi) are reinserted in the queue ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 36

AC-3 Algorithm Add all arcs to queue While queue not empty Get next arc (Xi, Xj) from queue For each value di of Xi If there is no consistent value dj in Xj Delete di If a value di was deleted For each neighbour Xk of Xi Add arc (Xk, Xi) to queue ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 37

AC-3 Algorithm WA R NT RGB Q NSW V B T SA Partial Arc Queue (SA,WA) (NT,WA) (SA,V) (NSW,V) (NT,SA) (NSW,SA) (Q,SA) (Q,NT) (Q,NSW) WA R NT RGB Q NSW V B T SA (WA,SA) (V,SA) (SA,NT) (WA,NT) (V,NSW) (SA,NSW) ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 38

AC-3 Algorithm Advantages Disadvantage Prunes tree, reduces amount of searching For n-node CSP that’s n-consistent, solution is guaranteed with no backtracking Disadvantage Computationally expensive If pruning takes longer than searching, it’s not worth it ECE457 Applied Artificial Intelligence R. Khoury (2008) Page 39