CS B551: Elements of Artificial Intelligence Instructor: Kris Hauser http://cs.indiana.edu/~hauserk

Constraint Propagation … … is the process of determining how the constraints and the possible values of one variable affect the possible values of other variables It is an important form of “least-commitment” reasoning 2 2

Forward Checking 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 n = number of variables d = size of initial domains s = maximum number of constraints involving a given variable (s  n-1) Forward checking takes O(nsd) time 3

Forward Checking in Map Coloring Empty set: the current assignment {(WA  R), (Q  G), (V  B)} does not lead to a solution WA NT Q NSW V SA T RGB R GB G B RB 4 4

Forward Checking in Map Coloring Contradiction that forward checking did not detect T WA NT SA Q NSW V WA NT Q NSW V SA T RGB R GB G B RB 5 5

Forward Checking in Map Coloring Contradiction that forward checking did not detect T WA NT SA Q NSW V Detecting this contradiction requires a more powerful constraint propagation technique WA NT Q NSW V SA T RGB R GB G B RB 6 6

Constraint Propagation for Binary Constraints REMOVE-VALUES(X,Y) removed  false For every value v in the domain of Y do If there is no value u in the domain of X such that the constraint on (X,Y) is satisfied then Remove v from Y‘s domain removed  true Return removed 7

Constraint Propagation for Binary Constraints AC3 Initialize queue Q with all variables (not yet instantiated) While Q   do X  Remove(Q) For every (not yet instantiated) variable Y related to X by a (binary) constraint do If REMOVE-VALUES(X,Y) then If Y’s domain =  then exit Insert(Y,Q) 8

Edge Labeling We consider an image of a scene composed of polyhedral objects such that each vertex is the endpoint of exactly three edges 9 9

Edge Labeling An “edge extractor” has accurately extracted all the visible edges in the image. The problem is to label each edge as convex (+), concave (-), or occluding () such that the complete labeling is physically possible 10 10

Concave edges Occluding edges Convex edges 11 11

+ + + - - + + + The arrow is oriented such that the object is on the right of the occluding edge + + + - - + + + 12 12

One Possible Edge Labeling + - 13 13

Junction Types L Fork T Y 14 14

- - - - Junction Label Sets + + + + (Waltz, 1975; Mackworth, 1977) 15

Edge Labeling as a CSP A variable is associated with each junction The domain of a variable is the label set associated with the junction type Constraints: The values assigned to two adjacent junctions must give the same label to the joining edge 16 16

AC3 Applied to Edge Labeling Q = (X1, X2, X3, ...) X5 X3 X1 X2 X12 X4 X8 17 17

AC3 Applied to Edge Labeling Q = (X1, ...) + - X5 X1 + - 18 18

Q = (X1, ...) + - X5 X1 + - 19 19

Q = (X5, ...) + - X5 + - 20 20

Q = (X5, ...) + + X5 + - X3 21 21

Q = (X5, ...) + + X5 + - X3 22 22

Q = (X3, ...) + + + - X3 23 23

Q = (X3, ...) + + X3 + - + X8 + 24 24

Q = (X3, ...) + + X3 + - + X8 + 25 25

Q = (X8, ...) + + + - + X8 + 26 26

Q = (X8, ...) + - + + + X12 - + X8 27 27

Complexity Analysis of AC3 n = number of variables d = size of initial domains s = maximum number of constraints involving a given variable (s  n-1) Each variables is inserted in Q up to d times REMOVE-VALUES takes O(d2) time AC3 takes O(ndsd2) = O(nsd3) time Usually more expensive than forward checking AC3 Initialize queue Q with all variables (not yet instantiated) While Q   do X  Remove(Q) For every (not yet instantiated) variable Y related to X by a (binary) constraint do If REMOVE-VALUES(X,Y) then If Y’s domain =  then exit Insert(Y,Q) REMOVE-VALUES(X,Y) removed  false For every value v in the domain of Y do If there is no value u in the domain of X such that the constraint on (X,Y) is satisfied then Remove v from Y‘s domain removed  true Return removed 28

Is AC3 all that we need? No !! AC3 can’t detect all contradictions among binary constraints X Z Y XY XZ YZ {1, 2} 29 29

Is AC3 all that we need? No !! AC3 can’t detect all contradictions among binary constraints X Z Y XY XZ YZ {1, 2} REMOVE-VALUES(X,Y) removed  false For every value v in the domain of Y do If there is no value u in the domain of X such that the constraint on (X,Y) is satisfied then Remove v from Y‘s domain removed  true Return removed 30 30

Is AC3 all that we need? No !! AC3 can’t detect all contradictions among binary constraints X Z Y XY XZ YZ {1, 2} REMOVE-VALUES(X,Y,Z) removed  false For every value w in the domain of Z do If there is no pair (u,v) of values in the domains of X and Y verifying the constraint on (X,Y) such that the constraints on (X,Z) and (Y,Z) are satisfied then Remove w from Z‘s domain removed  true Return removed REMOVE-VALUES(X,Y) removed  false For every value v in the domain of Y do If there is no value u in the domain of X such that the constraint on (X,Y) is satisfied then Remove v from Y‘s domain removed  true Return removed 31 31

Is AC3 all that we need? No !! AC3 can’t detect all contradictions among binary constraints Not all constraints are binary X Z Y XY XZ YZ {1, 2} 32 32

Tradeoff Generalizing the constraint propagation algorithm increases its time complexity Tradeoff between time spent in backtracking search and time spent in constraint propagation A good tradeoff when all or most constraints are binary is often to combine backtracking with forward checking and/or AC3 (with REMOVE- VALUES for two variables) 33 33

Modified Backtracking Algorithm with AC3 CSP-BACKTRACKING(A, var-domains) If assignment A is complete then return A Run AC3 and update var-domains accordingly If a variable has an empty domain then return failure X  select a variable not in A D  select an ordering on the domain of X For each value v in D do Add (Xv) to A var-domains  forward checking(var-domains, X, v, A) If no variable has an empty domain then (i) result  CSP-BACKTRACKING(A, var-domains) (ii) If result  failure then return result Remove (Xv) from A Return failure 34

A Complete Example: 4-Queens Problem 1 3 2 4 X1 {1,2,3,4} X3 X4 X2 1) The modified backtracking algorithm starts by calling AC3, which removes no value 35 35

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 2) The backtracking algorithm then selects a variable and a value for this variable. No heuristic helps in this selection. X1 and the value 1 are arbitrarily selected 36 36

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 3) The algorithm performs forward checking, which eliminates 2 values in each other variable’s domain 37

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 4) The algorithm calls AC3 1 2 38 38

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 REMOVE-VALUES(X,Y) removed  false For every value v in the domain of Y do If there is no value u in the domain of X such that the constraint on (X,Y) is satisfied then Remove v from Y‘s domain removed  true Return removed 4-Queens Problem X1 {1,2,3,4} X3 X4 X2 1 3 2 4 X2 = 3 is incompatible with any of the remaining values of X3 4) The algorithm calls AC3, which eliminates 3 from the domain of X2 39 39

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 4) The algorithm calls AC3, which eliminates 3 from the domain of X2, and 2 from the domain of X3 40 40

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 The algorithm calls AC3, which eliminates 3 from the domain of X2, and 2 from the domain of X3, and 4 from the domain of X3 41 41

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 The domain of X3 is empty  backtracking 42 42

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 The algorithm removes 1 from X1’s domain and assign 2 to X1 43 43

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 The algorithm performs forward checking 44 44

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 The algorithm calls AC3 1 2 3 4 45 45

4-Queens Problem X1 {1,2,3,4} X3 X4 X2 The algorithm calls AC3, which reduces the domains of X3 and X4 to a single value 46 46

Exploiting the Structure of CSP If the constraint graph contains several components, then solve one independent CSP per component T WA NT SA Q NSW V 47 47

Exploiting the Structure of CSP If the constraint graph is a tree, then : Order the variables from the root to the leaves  (X1, X2, …, Xn) For j = n, n-1, …, 2 call REMOVE-VALUES(Xj, Xi) where Xi is the parent of Xj Assign any valid value to X1 For j = 2, …, n do Assign any value to Xj consistent with the value assigned to its parent Xi X Y Z U V W  (X, Y, Z, U, V, W) 48 48

Exploiting the Structure of CSP Whenever a variable is assigned a value by the backtracking algorithm, propagate this value and remove the variable from the constraint graph WA NT SA Q NSW V 49 49

Exploiting the Structure of CSP Whenever a variable is assigned a value by the backtracking algorithm, propagate this value and remove the variable from the constraint graph If the graph becomes a tree, then proceed as shown in previous slide WA NT Q NSW V 50 50

8 Representation Variables Xij for i, j in {1,..,9} Domains {1,…,9} Constraints: XijXik, for jk Xij  Xkj, for ik XijXmn, for (i,j), (m,n) in same cell 2 2 7 5 8 9 9 5 7 4 3 6 2 9

8 Representation Variables Xij for i, j in {1,..,9} Domains {1,…,9} Constraints: XijXik, for jk Xij  Xkj, for ik XijXmn, for (i,j), (m,n) in same cell 2 2 7 5 8 9 9 5 7 4 3 6 2 9

Can we detect this using constraint propagation? 8 Representation Variables Xij for i, j in {1,..,9} Domains {1,…,9} Constraints: XijXik, for jk Xij  Xkj, for ik XijXmn, for (i,j), (m,n) in same cell 2 2 7 5 8 9 9 5 7 4 Can we detect this using constraint propagation? 3 6 2 9

Must detect 9-way interactions 8 Representation Variables Xij for i, j in {1,..,9} Domains {1,…,9} Constraints: XijXik, for jk Xij  Xkj, for ik XijXmn, for (i,j), (m,n) in same cell 2 2 7 5 8 9 9 5 7 4 Must detect 9-way interactions 3 6 2 9

8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 C39=5 C15=4 C33=8 … Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: ??? R18=3 2 R22=8 R32=6 2 7 R37=9 … 5 8 9 9 5 7 4 3 6 2 9

X18 = (1,3) 8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 C39=5 C15=4 … C12=9 C28=4 C38=1 X18 = (1,3) 8 Representation 2 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: ??? R18=3 2 R22=8 R32=6 2 7 R37=9 … 5 8 9 9 5 7 4 3 6 2 9

X18 = (1,3) 8 X22 = (3,3) Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 C39=5 C15=4 C33=8 … C12=9 C28=4 C38=1 X18 = (1,3) 8 X22 = (3,3) Representation 2 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: ??? R18=3 2 R22=8 R32=6 2 7 R37=9 … 5 8 9 9 5 7 4 3 6 2 9

X18 = (1,3) 8 X22 = (3,3) X32 = (2,3) X37 = (3,3) Representation 2 2 2 C39=5 C15=4 C33=8 … C12=9 C28=4 C38=1 X18 = (1,3) 8 X22 = (3,3) X32 = (2,3) X37 = (3,3) Representation 2 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: ??? R18=3 2 R22=8 R32=6 2 7 R37=9 … 5 8 9 9 5 7 4 3 6 2 9

X18 = (1,3) 8 X22 = (3,3) X32 = (2,3) X37 = (3,3) Representation 2 2 2 C39=5 C15=4 C33=8 … C12=9 C28=4 C38=1 X18 = (1,3) 8 X22 = (3,3) X32 = (2,3) X37 = (3,3) Representation 2 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: Rij=k  Ckj=I Similar constraints between X’s and R’s, X’s and C’s Rij  Rik for j  k Cij  Cik for j  k Xij  Xik for j  k R18=3 2 R22=8 R32=6 2 7 R37=9 … 5 8 9 9 5 7 4 3 6 2 9

8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 C12 = 9 R12= {1-9} Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: Rij=k  Ckj=I Similar constraints between X’s and R’s, X’s and C’s Rij  Rik for j  k Cij  Cik for j  k Xij  Xik for j  k 2 2 7 5 8 9 9 5 7 4 3 6 2 9

8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 C12 = 9 C22 = {1-9} {2-9} 8 Representation 2 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: Rij=k  Ckj=i Similar constraints between X’s and R’s, X’s and C’s Rij  Rik for j  k Cij  Cik for j  k Xij  Xik for j  k 2 2 7 5 8 9 9 5 7 4 3 6 2 9

8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 X32 = (2,2) C12 = 9 {2-9} Representation 2 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: Rij=k  Ckj=I Similar constraints between X’s and R’s, X’s and C’s Rij  Rik for j  k Cij  Cik for j  k Xij  Xik for j  k 2 2 7 5 8 9 9 5 7 4 3 6 2 9

8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 X32 = (2,2) C12 = 9 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: Rij=k  Ckj=I Similar constraints between X’s and R’s, X’s and C’s Rij  Rik for j  k Cij  Cik for j  k Xij  Xik for j  k 2 2 7 5 8 9 9 5 7 4 3 6 2 9

2 8 Representation 2 2 2 7 5 8 9 9 5 7 4 3 6 2 9 X32 = (2,2) C12 = 9 Variables Rij in {1,…,9} Cij in {1,…,9} Xij in {(1,1),…,(3,3)} Constraints: Rij=k  Ckj=I Similar constraints between X’s and R’s, X’s and C’s Rij  Rik for j  k Cij  Cik for j  k Xij  Xik for j  k 2 2 7 5 8 9 9 5 7 4 3 6 2 9

Local Search for CSPs Init: Make an arbitrary assignment Repeat: Modify some variable to reduce # of violated constraints

Boolean Satisfiability Problems Highly successful local search algorithms WalkSAT See R&N 7.3 p constraints of form ui*  uj*  uk*= 1 where ui* is either ui or ui n variables ui, …, un

Observations… If a CSP has few constraints, local search solves it quickly Random starting assignment not too far from a solution Million-queens puzzles solved in < 1min, c. 1990 If a CSP has many constraints, local search solves it quickly Constraints “guide” solver to a solution (if one exists)

Hard Sudoku’s Human solvers: Lot of logic (deep constraint propagation) 1 . . | . . . | . . 2 . 9 . | 4 . . | . 5 . . . 6 | . . . | 7 . . ------+-------+------ . 5 . | 9 . 3 | . . . . . . | . 7 . | . . . . . . | 8 5 . | . 4 . 7 . . | . . . | 6 . . . 3 . | . . 9 | . 8 . . . 2 | . . . | . . 1 Computer solvers: Lot of backtracking

Hard & Easy 3-SAT Problems Let R = # of constraints / # of variables As n , the fraction of hard problems reduces to 0

Recap Constraint propagation, AC3 Taking advantage of CSP structure Local search for CSPs

Next Class Intro to uncertainty R&N 4.3-4, 13.1-2