1. C ONSTRAINT S ATISFACTION P ROBLEMS Instructor: Kris Hauser http://cs.indiana.edu/~hauserk 1

2. C ONSTRAINT P ROPAGATION Place a queen in a square Remove the attacked squares from future consideration 2

3. C ONSTRAINT P ROPAGATION Count the number of non-attacked squares in every row and column Place a queen in a row or column with minimum number Remove the attacked squares from future consideration 3 66555566655556 5 5 5 5 5 6 7

4. C ONSTRAINT P ROPAGATION Repeat 4 344335344335 4 3 3 3 4 5

5. C ONSTRAINT P ROPAGATION 5 4323443234 3 3 3 4 3

6. C ONSTRAINT P ROPAGATION 6 4221342213 3 3 3 1

7. C ONSTRAINT P ROPAGATION 7 221221 2 2 1 7

8. C ONSTRAINT P ROPAGATION 8 2121 1 21 2

9. 9 1 1

10. 10

11. W HAT DO WE NEED ? More than just a successor function and a goal test We also need: A means to propagate the constraints imposed by one queen’s position on the positions of the other queens An early failure test Explicit representation of constraints Constraint propagation algorithms 11

12. C ONSTRAINT S ATISFACTION P ROBLEM (CSP) Set of variables {X 1, X 2, …, X n } Each variable X i has a domain D i of possible values. Usually, D i is finite Set of constraints {C 1, C 2, …, C p } Each constraint relates a subset of variables by specifying the valid combinations of their values Goal: Assign a value to every variable such that all constraints are satisfied 12

13. M AP C OLORING 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 13 WA NT SA Q NSW V T WA NT SA Q NSW V T 13

14. 8-Q UEEN P ROBLEM 8 variables X i, i = 1 to 8 The domain of each variable is: {1,2,…,8} Constraints are of the forms: X i = k  X j  k for all j = 1 to 8, j  i Similar constraints for diagonals 14 All constraints are binary (in this class) 14

15. S UDOKU 81 variables Domain of each variable: {1,…,9} Constraints: x i  x j for all i  j in same row, same col, same cell x i =v i for fixed cells 15

16. S TREET P UZZLE 16 1 2 34 5 N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {Tea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, Violinist, Doctor} A i = {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 Violinist 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? Who owns the Zebra? Who drinks Water? 16

17. S TREET P UZZLE 17 1 2 34 5 N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {Tea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, Violinist, Doctor} A i = {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 Violinist drinks Fruit juice The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s  i,j  [1,5], i  j, N i  N j  i,j  [1,5], i  j, C i  C j... 17

18. S TREET P UZZLE 18 1 2 34 5 N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {Tea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, Violinist, Doctor} A i = {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 Violinist drinks Fruit juice The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s (N i = English)  (C i = Red) (N i = Japanese)  (J i = Painter) (N 1 = Norwegian) left as an exercise (C i = White)  (C i+1 = Green) (C 5  White) (C 1  Green)

19. S TREET P UZZLE 19 1 2 34 5 N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {Tea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, Violinist, Doctor} A i = {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 Violinist drinks Fruit juice The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s (N i = English)  (C i = Red) (N i = Japanese)  (J i = Painter) (N 1 = Norwegian) (C i = White)  (C i+1 = Green) (C 5  White) (C 1  Green) unary constraints

20. S TREET P UZZLE 20 1 2 34 5 N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {Tea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, Violinist, Doctor} A i = {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  N 1 = Norwegian 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  D 3 = Milk The Norwegian lives next door to the Blue house The Violinist drinks Fruit juice The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s 20

21. S TREET P UZZLE 21 1 2 34 5 N i = {English, Spaniard, Japanese, Italian, Norwegian} C i = {Red, Green, White, Yellow, Blue} D i = {Tea, Coffee, Milk, Fruit-juice, Water} J i = {Painter, Sculptor, Diplomat, Violinist, Doctor} A i = {Dog, Snails, Fox, Horse, Zebra} The Englishman lives in the Red house  C 1  Red The Spaniard has a Dog  A 1  Dog The Japanese is a Painter The Italian drinks Tea The Norwegian lives in the first house on the left  N 1 = Norwegian 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  D 3 = Milk The Norwegian lives next door to the Blue house The Violinist drinks Fruit juice  J 3  Violinist The Fox is in the house next to the Doctor’s The Horse is next to the Diplomat’s

22. T ASK S CHEDULING Four tasks T1, T2, T3, and T4 are related by time constraints: 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? What are the possible time relations between two tasks? What if the tasks use resources in limited supply? How to formulate this problem as a CSP? 22 T1T1 T2T2 T3T3 T4T4

23. 3-SAT n Boolean variables u 1,..., u n p constraints of the form u i *  u j *  u k *= 1 where u* stands for either u or  u Known to be NP-complete 23

24. F INITE VS. I NFINITE CSP 24

25. CSP AS A S EARCH P ROBLEM n variables X 1,..., X n Valid assignment: {X i 1  v i 1,..., X ik  v ik }, 0  k  n, such that the values v i 1,..., v ik satisfy all constraints relating the variables X i 1,..., X ik Complete assignment: one where k = n [if all variable domains have size d, there are O(d n ) complete assignments] States : valid assignments Initial state : empty assignment {}, i.e. k = 0 Successor of a state: {X i 1  v i 1,..., X ik  v ik }  {X i 1  v i 1,..., X ik  v ik, X ik+1  v ik+1 } Goal test : k = n 25

26. 26 {X i 1  v i 1,..., X ik  v ik } r = n - k variables with s values  r  s branching factor {X i 1  v i 1,..., X ik  v ik, X ik+1  v ik+1 } 26

27. A K EY PROPERTY OF CSP: C OMMUTATIVITY The order in which variables are assigned values has no impact on the reachable complete valid assignments 27 Hence: 1)One can expand a node N by first selecting one variable X not in the assignment A associated with N and then assigning every value v in the domain of X [  big reduction in branching factor] 27

28. 28 The depth of the solutions in the search tree is un-changed ( n ) {X i 1  v i 1,..., X ik  v ik } {X i 1  v i 1,..., X ik  v ik, X ik+1  v ik+1 } r = n - k variables with s values  r  s branching factor r = n - k variables with s values  s branching factor

29. 29  4 variables X 1,..., X 4  Let the valid assignment of N be: A = {X 1  v 1, X 3  v 3 }  For example pick variable X 4  Let the domain of X 4 be {v 4,1, v 4,2, v 4,3 }  The successors of A are all the valid assignments among: {X 1  v 1, X 3  v 3, X 4  v 4,1 } {X 1  v 1, X 3  v 3, X 4  v 4,2 } {X 1  v 1, X 3  v 3, X 4  v 4,3 } 29

30. A K EY PROPERTY OF CSP: C OMMUTATIVITY The order in which variables are assigned values has no impact on the reachable complete valid assignments 30 Hence: 1)One can expand a node N by first selecting one variable X not in the assignment A associated with N and then assigning every value v in the domain of X [  big reduction in branching factor] 2)One need not store the path to a node  Backtracking search algorithm

31. B ACKTRACKING S EARCH Essentially a simplified depth-first algorithm using recursion 31

32. B ACKTRACKING S EARCH (3 VARIABLES ) 32 Assignment = {} 32

33. B ACKTRACKING S EARCH (3 VARIABLES ) 33 Assignment = {(X 1,v 11 )} X1X1 v 11 33

34. B ACKTRACKING S EARCH (3 VARIABLES ) 34 Assignment = {(X 1,v 11 ), (X 3,v 31 )} X1X1 v 11 v 31 X3X3 34

35. B ACKTRACKING S EARCH (3 VARIABLES ) 35 Assignment = {(X 1,v 11 ), (X 3,v 31 )} X1X1 v 11 v 31 X3X3 X2X2 Assume that no value of X 2 leads to a valid assignment Then, the search algorithm backtracks to the previous variable (X 3 ) and tries another value 35

36. B ACKTRACKING S EARCH (3 VARIABLES ) 36 Assignment = {(X 1,v 11 ), (X 3,v 32 )} X1X1 v 11 X3X3 v 32 v 31 X2X2 36

37. B ACKTRACKING S EARCH (3 VARIABLES ) 37 Assignment = {(X 1,v 11 ), (X 3,v 32 )} X1X1 v 11 X3X3 v 32 X2X2 Assume again that no value of X 2 leads to a valid assignment The search algorithm backtracks to the previous variable (X 3 ) and tries another value. But assume that X 3 has only two possible values. The algorithm backtracks to X 1 v 31 X2X2 37

38. B ACKTRACKING S EARCH (3 VARIABLES ) 38 Assignment = {(X 1,v 12 )} X1X1 v 11 X3X3 v 32 X2X2 v 31 X2X2 v 12 38

39. B ACKTRACKING S EARCH (3 VARIABLES ) 39 Assignment = {(X 1,v 12 ), (X 2,v 21 )} X1X1 v 11 X3X3 v 32 X2X2 v 31 X2X2 v 12 v21v21 X2X2 39

40. B ACKTRACKING S EARCH (3 VARIABLES ) 40 Assignment = {(X 1,v 12 ), (X 2,v 21 )} X1X1 v 11 X3X3 v 32 X2X2 v 31 X2X2 v 12 v21v21 X2X2 The algorithm need not consider the variables in the same order in this sub-tree as in the other 40

41. B ACKTRACKING S EARCH (3 VARIABLES ) 41 Assignment = {(X 1,v 12 ), (X 2,v 21 ), (X 3,v 32 )} X1X1 v 11 X3X3 v 32 X2X2 v 31 X2X2 v 12 v21v21 X2X2 v 32 X3X3 41

42. B ACKTRACKING S EARCH (3 VARIABLES ) 42 Assignment = {(X 1,v 12 ), (X 2,v 21 ), (X 3,v 32 )} X1X1 v 11 X3X3 v 32 X2X2 v 31 X2X2 v 12 v21v21 X2X2 v 32 X3X3 The algorithm need not consider the values of X 3 in the same order in this sub-tree 42

43. B ACKTRACKING S EARCH (3 VARIABLES ) 43 Assignment = {(X 1,v 12 ), (X 2,v 21 ), (X 3,v 32 )} X1X1 v 11 X3X3 v 32 X2X2 v 31 X2X2 v 12 v21v21 X2X2 v 32 X3X3 Since there are only three variables, the assignment is complete 43

44. B ACKTRACKING A LGORITHM 44 CSP-BACKTRACKING(A) 1. If assignment A is complete then return A 2. X  select a variable not in A 3. D  select an ordering on the domain of X 4. For each value v in D do a. Add (X  v) to A b. If A is valid then i. result  CSP-BACKTRACKING(A) ii. If result  failure then return result c. Remove (X  v) from A 5. Return failure Call CSP-BACKTRACKING({}) [This recursive algorithm keeps too much data in memory. An iterative version could save memory (left as an exercise)] 44

45. C RITICAL Q UESTIONS FOR THE E FFICIENCY OF CSP-B ACKTRACKING 45 CSP-BACKTRACKING(A) 1.If assignment A is complete then return A 2.X  select a variable not in A 3.D  select an ordering on the domain of X 4.For each value v in D do a.Add (X  v) to A b.If a is valid then i.result  CSP-BACKTRACKING(A) ii.If result  failure then return result c.Remove (X  v) from A 5.Return failure 45

46. C RITICAL Q UESTIONS FOR THE E FFICIENCY OF CSP-B ACKTRACKING 1. Which variable X should be assigned a value next? 2. In which order should X’s values be assigned? 46

47. C RITICAL Q UESTIONS FOR THE E FFICIENCY OF CSP-B ACKTRACKING 1. Which variable X should be assigned a value next? The current assignment may not lead to any solution, but the algorithm does not know it yet. Selecting the right variable X may help discover the contradiction more quickly 2. In which order should X’s values be assigned? 47

48. C RITICAL Q UESTIONS FOR THE E FFICIENCY OF CSP-B ACKTRACKING 1. Which variable X should be assigned a value next? The current assignment may not lead to any solution, but the algorithm does not know it yet. Selecting the right variable X may help discover the contradiction more quickly 2. In which order should X’s values be assigned? The current assignment may be part of a solution. Selecting the right value to assign to X may help discover this solution more quickly 48

49. C RITICAL Q UESTIONS FOR THE E FFICIENCY OF CSP-B ACKTRACKING 1. Which variable X should be assigned a value next? The current assignment may not lead to any solution, but the algorithm does not know it yet. Selecting the right variable X may help discover the contradiction more quickly 2. In which order should X’s values be assigned? The current assignment may be part of a solution. Selecting the right value to assign to X may help discover this solution more quickly More on these questions very soon... 49

50. F ORWARD C HECKING 50 A simple constraint-propagation technique: Assigning the value 5 to X 1 leads to removing values from the domains of X 2, X 3,..., X 8 1234567812345678 X1 X2 X3 X4 X5 X6 X7 X8X1 X2 X3 X4 X5 X6 X7 X8 50

51. F ORWARD C HECKING IN M AP C OLORING 51 WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB T WA NT SA Q NSW V Constraint graph 51

52. F ORWARD C HECKING IN M AP C OLORING 52 WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB T WA NT SA Q NSW V Forward checking removes the value Red of NT and of SA 52

53. F ORWARD C HECKING IN M AP C OLORING 53 WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RGBGBRGBRGBRGBRGBRGBRGBGBGBRGBRGB RGBGBGRGBRGBRGBRGBGBGBRGBRGB T WA NT SA Q NSW V 53

54. F ORWARD C HECKING IN M AP C OLORING 54 WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RGBGBRGBRGBRGBRGBRGBRGBGBGBRGBRGB RBGRBRBRGBRGBBRGBRGB RBGRBRBBBRGBRGB T WA NT SA Q NSW V 54

55. F ORWARD C HECKING IN M AP C OLORING 55 WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RGBGBRGBRGBRGBRGBRGBRGBGBGBRGBRGB RBGRBRBRGBRGBBRGBRGB RBGRBRBBBRGBRGB Empty set: the current assignment {(WA  R), (Q  G), (V  B)} does not lead to a solution 55

56. F ORWARD C HECKING (G ENERAL F ORM ) 56 Whenever a pair (X  v) is added to assignment A do: For each variable Y not in A do: For every constraint C relating Y to the variables in A do: Remove all values from Y’s domain that do not satisfy C 56

57. M ODIFIED B ACKTRACKING A LGORITHM 57 CSP-BACKTRACKING(A, var-domains) 1. If assignment A is complete then return A 2. X  select a variable not in A 3. D  select an ordering on the domain of X 4. For each value v in D do a. Add (X  v) to A b. var-domains  forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result  CSP-BACKTRACKING(A, var-domains) (ii) If result  failure then return result d. Remove (X  v) from A 5. Return failure 57

58. 58 No need any more to verify that A is valid M ODIFIED B ACKTRACKING A LGORITHM CSP-BACKTRACKING(A, var-domains) 1. If assignment A is complete then return A 2. X  select a variable not in A 3. D  select an ordering on the domain of X 4. For each value v in D do a. Add (X  v) to A b. var-domains  forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result  CSP-BACKTRACKING(A, var-domains) (ii) If result  failure then return result d. Remove (X  v) from A 5. Return failure

59. Need to pass down the updated variable domains 59 M ODIFIED B ACKTRACKING A LGORITHM CSP-BACKTRACKING(A, var-domains) 1. If assignment A is complete then return A 2. X  select a variable not in A 3. D  select an ordering on the domain of X 4. For each value v in D do a. Add (X  v) to A b. var-domains  forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result  CSP-BACKTRACKING(A, var-domains) (ii) If result  failure then return result d. Remove (X  v) from A 5. Return failure

60. 60 M ODIFIED B ACKTRACKING A LGORITHM CSP-BACKTRACKING(A, var-domains) 1. If assignment A is complete then return A 2. X  select a variable not in A 3. D  select an ordering on the domain of X 4. For each value v in D do a. Add (X  v) to A b. var-domains  forward checking(var-domains, X, v, A) c. If no variable has an empty domain then (i) result  CSP-BACKTRACKING(A, var-domains) (ii) If result  failure then return result d. Remove (X  v) from A 5. Return failure

61. 61 1) Which variable X i should be assigned a value next?  Most-constrained-variable heuristic  Most-constraining-variable heuristic 2) In which order should its values be assigned?  Least-constraining-value heuristic

62. M OST -C ONSTRAINED -V ARIABLE H EURISTIC 62 1) Which variable X i should be assigned a value next? Select the variable with the smallest remaining domain [Rationale: Minimize the branching factor] 62

63. 8-Q UEENS 63 4 3 2 3 4Numbers of values for each un-assigned variable New assignment Forward checking 63

64. 8-Q UEENS 64 3 2 1 3New numbers of values for each un-assigned variable New assignment Forward checking 64

65. M AP C OLORING 65  SA’s remaining domain has size 1 (value B remaining)  Q’s remaining domain has size 2  NSW’s, V’s, and T’s remaining domains have size 3  Select SA WA NT SA Q NSW V T WA NTSA

66. M OST -C ONSTRAINING -V ARIABLE H EURISTIC 66 1) Which variable X i should be assigned a value next? Among the variables with the smallest remaining domains (ties with respect to the most-constrained- variable heuristic), select the one that appears in the largest number of constraints on variables not in the current assignment [Rationale: Increase future elimination of values, to reduce future branching factors] 66

67. M AP C OLORING 67  Before any value has been assigned, all variables have a domain of size 3, but SA is involved in more constraints (5) than any other variable  Select SA and assign a value to it (e.g., Blue) WA NT SA Q NSW V T SA 67

68. L EAST -C ONSTRAINING -V ALUE H EURISTIC 68 2) In which order should X’s values be assigned? Select the value of X that removes the smallest number of values from the domains of those variables which are not in the current assignment [Rationale: Since only one value will eventually be assigned to X, pick the least- constraining value first, since it is the most likely not to lead to an invalid assignment] [Note: Using this heuristic requires performing a forward-checking step for every value, not just for the selected value]

69. M AP C OLORING 69  Q’s domain has two remaining values: Blue and Red  Assigning Blue to Q would leave 0 value for SA, while assigning Red would leave 1 value {} WA NT SA Q NSW V T WA NT 69

70. M AP C OLORING 70  Q’s domain has two remaining values: Blue and Red  Assigning Blue to Q would leave 0 value for SA, while assigning Red would leave 1 value  So, assign Red to Q {Blue} WA NT SA Q NSW V T WA NT 70

71. M ODIFIED B ACKTRACKING A LGORITHM 71 1) Most-constrained-variable heuristic 2) Most-constraining-variable heuristic 3) Least-constraining-value heuristic CSP-BACKTRACKING(A, var-domains) 1.If assignment A is complete then return A 2.X  select a variable not in A 3.D  select an ordering on the domain of X 4.For each value v in D do a.Add (X  v) to A b.var-domains  forward checking(var-domains, X, v, A) c.If no variable has an empty domain then (i) result  CSP-BACKTRACKING(A, var-domains) (ii) If result  failure then return result d.Remove (X  v) from A 5.Return failure 71

72. A PPLICATIONS OF CSP CSP techniques are widely used Applications include: Crew assignments to flights Management of transportation fleet Flight/rail schedules Job shop scheduling Task scheduling in port operations Design, including spatial layout design 72

73. R ECAP CSPs Backtracking search Constraint propagation 73

74. H OMEWORK R&N 6.3-5 74