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  1. Constraint Satisfaction Problems Russell and Norvig: Chapter 5 CMSC 421 – Fall 2005

  2. Intro Example: 8-Queens • Purely generate-and-test • The “search” tree is only used to enumerate all possible 648 combinations

  3. Intro Example: 8-Queens Another form of generate-and-test, with no redundancies  “only” 88 combinations

  4. Intro Example: 8-Queens

  5. What is Needed? • Not just a successor function and goal test • But also a means to propagate the constraints imposed by one queen on the others and an early failure test •  Explicit representation of constraints and constraint manipulation algorithms

  6. Constraint Satisfaction Problem • Set of variables {X1, X2, …, Xn} • Each variable Xi has a domain Di of possible values • Usually Di is discrete and finite • Set of constraints {C1, C2, …, Cp} • Each constraint Ck involves a subset of variables and specifies the allowable combinations of values of these variables

  7. Constraint Satisfaction Problem • Set of variables {X1, X2, …, Xn} • Each variable Xi has a domain Di of possible values • Usually Di is discrete and finite • Set of constraints {C1, C2, …, Cp} • Each constraint Ck involves a subset of variables and specifies the allowable combinations of values of these variables • Assign a value to every variable such that all constraints are satisfied

  8. Example: 8-Queens Problem • 64 variables Xij, i = 1 to 8, j = 1 to 8 • Domain for each variable {yes,no} • Constraints are of the forms: • Xij = yes  Xik = no for all k = 1 to 8, kj • Xij = yes  Xkj = no for all k = 1 to 8, kI • Similar constraints for diagonals

  9. Example: 8-Queens Problem • 8 variables Xi, i = 1 to 8 • Domain for each variable {1,2,…,8} • Constraints are of the forms: • Xi = k  Xj k for all j = 1 to 8, ji • Similar constraints for diagonals

  10. NT Q WA SA NT NSW Q V WA SA T NSW V T Example: Map Coloring • 7 variables {WA,NT,SA,Q,NSW,V,T} • Each variable has the same domain {red, green, blue} • No two adjacent variables have the same value: • WANT, WASA, NTSA, NTQ, SAQ, SANSW, SAV,QNSW, NSWV

  11. 2 3 4 1 5 Example: Street Puzzle Ni = {English, Spaniard, Japanese, Italian, Norwegian} Ci = {Red, Green, White, Yellow, Blue} Di = {Tea, Coffee, Milk, Fruit-juice, Water} Ji = {Painter, Sculptor, Diplomat, Violonist, Doctor} Ai = {Dog, Snails, Fox, Horse, Zebra}

  12. 2 3 4 1 5 Example: Street Puzzle Ni = {English, Spaniard, Japanese, Italian, Norwegian} Ci = {Red, Green, White, Yellow, Blue} Di = {Tea, Coffee, Milk, Fruit-juice, Water} Ji = {Painter, Sculptor, Diplomat, Violonist, Doctor} Ai = {Dog, Snails, Fox, Horse, Zebra} The Englishman lives in the Red house The Spaniard has a Dog The Japanese is a Painter The Italian drinks Tea The Norwegian lives in the first house on the left The owner of the Green house drinks Coffee The Green house is on the right of the White house The Sculptor breeds Snails The Diplomat lives in the Yellow house The owner of the middle house drinks Milk The Norwegian lives next door to the Blue house The Violonist drinks Fruit juice The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s Who owns the Zebra? Who drinks Water?

  13. T1 T2 T4 T3 Example: Task Scheduling • T1 must be done during T3 • T2 must be achieved before T1 starts • T2 must overlap with T3 • T4 must start after T1 is complete • Are the constraints compatible? • Find the temporal relation between every two tasks

  14. Finite vs. Infinite CSP • Finite domains of values  finite CSP • Infinite domains  infinite CSP

  15. Finite vs. Infinite CSP • Finite domains of values  finite CSP • Infinite domains  infinite CSP • We will only consider finite CSP

  16. Binary constraints NT T1 Q WA T2 NSW T4 SA V T3 T Constraint Graph Two variables are adjacent or neighbors if they are connected by an edge or an arc

  17. CSP as a Search Problem • Initial state: empty assignment • Successor function: a value is assigned to any unassigned variable, which does not conflict with the currently assigned variables • Goal test: the assignment is complete • Path cost: irrelevant

  18. CSP as a Search Problem • Initial state: empty assignment • Successor function: a value is assigned to any unassigned variable, which does not conflict with the currently assigned variables • Goal test: the assignment is complete • Path cost: irrelevant n variables of domain size d O(dn) distinct complete assignments

  19. Remark • Finite CSP include 3SAT as a special case (see class on logic) • 3SAT is known to be NP-complete • So, in the worst-case, we cannot expect to solve a finite CSP in less than exponential time

  20. Commutativity of CSP The order in which values are assigned to variables is irrelevant to the final assignment, hence: Generate successors of a node by considering assignments for only one variable Do not store the path to node

  21. empty assignment 1st variable 2nd variable 3rd variable Assignment = {}  Backtracking Search

  22. empty assignment 1st variable 2nd variable 3rd variable  Backtracking Search Assignment = {(var1=v11)}

  23. empty assignment 1st variable 2nd variable 3rd variable  Backtracking Search Assignment = {(var1=v11),(var2=v21)}

  24. empty assignment 1st variable 2nd variable 3rd variable  Backtracking Search Assignment = {(var1=v11),(var2=v21),(var3=v31)}

  25. empty assignment 1st variable 2nd variable 3rd variable  Backtracking Search Assignment = {(var1=v11),(var2=v21),(var3=v32)}

  26. empty assignment 1st variable 2nd variable 3rd variable  Backtracking Search Assignment = {(var1=v11),(var2=v22)}

  27. empty assignment 1st variable 2nd variable 3rd variable  Backtracking Search Assignment = {(var1=v11),(var2=v22),(var3=v31)}

  28. partial assignment of variables Backtracking Algorithm CSP-BACKTRACKING({}) CSP-BACKTRACKING(a) • If a is complete then return a • X select unassigned variable • D  select an ordering for the domain of X • For each value v in D do • If v is consistent with a then • Add (X= v) to a • result CSP-BACKTRACKING(a) • If resultfailure then return result • Return failure

  29. {} NT Q WA WA=red WA=green WA=blue SA NSW V WA=red NT=green WA=red NT=blue T WA=red NT=green Q=red WA=red NT=green Q=blue Map Coloring

  30. Your Turn #1

  31. Questions • Which variable X should be assigned a value next? • In which order should its domain D be sorted?

  32. Questions • Which variable X should be assigned a value next? • In which order should its domain D be sorted? • What are the implications of a partial assignment for yet unassigned variables? ( Constraint Propagation)

  33. NT WA NT Q WA SA SA NSW V T Choice of Variable • Map coloring

  34. Choice of Variable • 8-queen

  35. Choice of Variable #1: Minimum Remaining Values (aka Most-constrained-variable heuristic): Select a variable with the fewest remaining values

  36. NT Q WA SA SA NSW V T Choice of Variable #2: Degree Heuristic (aka Most-constraining-variable heuristic): Select the variable that is involved in the largest number of constraints on other unassigned variables

  37. NT WA NT Q WA SA NSW V {} T Choice of Value

  38. NT WA NT Q WA SA NSW V {blue} T Choice of Value #3: Least-constraining-value heuristic: Prefer the value that leaves the largest subset of legal values for other unassigned variables

  39. Constraint Propagation … … is the process of determining how the possible values of one variable affect the possible values of other variables

  40. Forward Checking After a variable X is assigned a value v, look at each unassigned variable Y that is connected to X by a constraint and deletes from Y’s domain any value that is inconsistent with v

  41. NT Q WA NSW SA T V Map Coloring

  42. NT Q WA NSW SA T V Map Coloring

  43. NT Q WA NSW SA T V Map Coloring

  44. Your Turn #2

  45. NT Q WA NSW SA T V Impossible assignments that forward checking do not detect Map Coloring

  46. Removal of Arc Inconsistencies REMOVE-INCONSISTENT-VALUES(Xi, Xj) • removed  false • For each label x in Domain(Xi) do • If no value y in Xj that satisfies Xi, Xj constraint • Remove x from Domain(Xi) • removed  true • Return removed

  47. Arc-Consistency for Binary CSPs Algorithm AC3 • Q queue of all constraints • while Q is not empty do • (Xi, Xj) RemoveFirst(Q) • If REMOVE-INCONSISTENT-VALUES(Xi,Xj) • For every variable Xk adjacent to Xi do • add (Xk, Xi) to Q

  48. {1, 2} {1, 2} X  Y X Y X  Z Y  Z Z {1, 2} Is AC3 All What is Needed? NO!

  49. Solving a CSP • Interweave constraint propagation, e.g., • forward checking • AC3 • and backtracking • + Take advantage of the CSP structure

  50. X1 {1,2,3,4} X2 {1,2,3,4} 1 2 3 4 1 2 3 4 X3 {1,2,3,4} X4 {1,2,3,4} 4-Queens Problem