CS B551: E LEMENTS OF A RTIFICIAL I NTELLIGENCE Instructor: Kris Hauser 1

C ONSTRAINT P ROPAGATION … … 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

F ORWARD C HECKING 3 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

F ORWARD C HECKING IN M AP C OLORING 4 WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RGBGBGRGBRGBRGBRGBGBGBRGBRGB RBGRBRBBBRGBRGB Empty set: the current assignment {(WA  R), (Q  G), (V  B)} does not lead to a solution 4

F ORWARD C HECKING IN M AP C OLORING 5 T WA NT SA Q NSW V Contradiction that forward checking did not detect WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RGBGBGRGBRGBRGBRGBGBGBRGBRGB RBGRBRBBBRGBRGB 5

F ORWARD C HECKING IN M AP C OLORING 6 T WA NT SA Q NSW V Contradiction that forward checking did not detect WANTQNSWVSAT RGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGBRGB RGBGBGRGBRGBRGBRGBGBGBRGBRGB RBGRBRBBBRGBRGB Detecting this contradiction requires a more powerful constraint propagation technique 6

C ONSTRAINT P ROPAGATION FOR B INARY C ONSTRAINTS REMOVE-VALUES(X,Y) 1. removed  false 2. 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 a. Remove v from Y‘s domain b. removed  true 3. Return removed 7

C ONSTRAINT P ROPAGATION FOR B INARY C ONSTRAINTS 8 AC3 1. Initialize queue Q with all variables (not yet instantiated) 2. While Q   do a. X  Remove(Q) b. For every (not yet instantiated) variable Y related to X by a (binary) constraint do – If REMOVE-VALUES(X,Y) then i. If Y’s domain =  then exit ii. Insert(Y,Q)

E DGE L ABELING 9 We consider an image of a scene composed of polyhedral objects such that each vertex is the endpoint of exactly three edges 9

E DGE L ABELING 10 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

11 Convex edges Concave edges Occluding edges 11

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

O NE P OSSIBLE E DGE L ABELING

J UNCTION T YPES 14 Fork L T Y 14

J UNCTION L ABEL S ETS (Waltz, 1975; Mackworth, 1977) 15

E DGE L ABELING 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

AC3 A PPLIED TO E DGE L ABELING 17 Q = (X 1, X 2, X 3,...) X1X1 X5X5 X3X3 X8X8 X 12 X2X2 X4X4 17

X1X1 X5X5 Q = (X 1,...) 18 AC3 A PPLIED TO E DGE L ABELING

X1X1 X5X5 Q = (X 1,...) 19

X5X5 Q = (X 5,...) 20

Q = (X 5,...) X5X5 X3X3 21

Q = (X 5,...) X5X5 X3X3 22

Q = (X 3,...) X3X3 23

Q = (X 3,...) X3X3 X8X8 + 24

Q = (X 3,...) X3X3 X8X8 + 25

Q = (X 8,...) X8X8 + 26

X 12 X8X8 Q = (X 8,...) 27

C OMPLEXITY A NALYSIS 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(d 2 ) time AC3 takes O(n  d  s  d 2 ) = O(n  s  d 3 ) time Usually more expensive than forward checking 28 AC3 1.Initialize queue Q with all variables (not yet instantiated) 2.While Q   do a.X  Remove(Q) b.For every (not yet instantiated) variable Y related to X by a (binary) constraint do – If REMOVE-VALUES(X,Y) then i.If Y’s domain =  then exit ii.Insert(Y,Q) REMOVE-VALUES(X,Y) 1. removed  false 2. 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 a. Remove v from Y‘s domain b. removed  true 3. Return removed

I S AC3 ALL THAT WE NEED ? No !! AC3 can’t detect all contradictions among binary constraints 29 X Z Y XYXY XZXZ YZYZ {1, 2} 29

I S AC3 ALL THAT WE NEED ? No !! AC3 can’t detect all contradictions among binary constraints 30 X Z Y XYXY XZXZ YZYZ {1, 2} 30 REMOVE-VALUES(X,Y) 1. removed  false 2. 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 a. Remove v from Y‘s domain b. removed  true 3. Return removed

I S AC3 ALL THAT WE NEED ? No !! AC3 can’t detect all contradictions among binary constraints 31 X Z Y XYXY XZXZ YZYZ {1, 2} 31 REMOVE-VALUES(X,Y) 1. removed  false 2. 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 a. Remove v from Y‘s domain b. removed  true 3. Return removed REMOVE-VALUES(X,Y,Z) 1. removed  false 2. 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 a. Remove w from Z‘s domain b. removed  true 3. Return removed

I S AC3 ALL THAT WE NEED ? No !! AC3 can’t detect all contradictions among binary constraints Not all constraints are binary 32 X Z Y XYXY XZXZ YZYZ {1, 2} 32

T RADEOFF 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

M ODIFIED B ACKTRACKING A LGORITHM WITH AC3 34 CSP-BACKTRACKING(A, var-domains) 1.If assignment A is complete then return A 2.Run AC3 and update var-domains accordingly 3.If a variable has an empty domain then return failure 4.X  select a variable not in A 5.D  select an ordering on the domain of X 6.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 7.Return failure

A C OMPLETE E XAMPLE : 4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 1) The modified backtracking algorithm starts by calling AC3, which removes no value 35

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 2)The backtracking algorithm then selects a variable and a value for this variable. No heuristic helps in this selection. X 1 and the value 1 are arbitrarily selected 36

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 3)The algorithm performs forward checking, which eliminates 2 values in each other variable’s domain

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 4)The algorithm calls AC3 38

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 4)The algorithm calls AC3, which eliminates 3 from the domain of X 2 X 2 = 3 is incompatible with any of the remaining values of X 3 39 REMOVE-VALUES(X,Y) 1. removed  false 2. 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 a. Remove v from Y‘s domain b. removed  true 3. Return removed

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 4)The algorithm calls AC3, which eliminates 3 from the domain of X 2, and 2 from the domain of X 3 40

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X3{1,2,3,4}X3{1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 4)The algorithm calls AC3, which eliminates 3 from the domain of X 2, and 2 from the domain of X 3, and 4 from the domain of X 3 41

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X3{1,2,3,4}X3{1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 5)The domain of X 3 is empty  backtracking 42

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 6)The algorithm removes 1 from X 1 ’s domain and assign 2 to X 1 43

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 7)The algorithm performs forward checking 44

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 8)The algorithm calls AC3 45

4-Q UEENS P ROBLEM X 1 {1,2,3,4} X 3 {1,2,3,4} X 4 {1,2,3,4} X 2 {1,2,3,4} 8)The algorithm calls AC3, which reduces the domains of X 3 and X 4 to a single value 46

E XPLOITING THE S TRUCTURE OF CSP If the constraint graph contains several components, then solve one independent CSP per component 47 T WA NT SA Q NSW V 47

E XPLOITING THE S TRUCTURE OF CSP If the constraint graph is a tree, then : 48 1.Order the variables from the root to the leaves  (X 1, X 2, …, X n ) 2.For j = n, n-1, …, 2 call REMOVE-VALUES(X j, X i ) where X i is the parent of X j 3.Assign any valid value to X 1 4.For j = 2, …, n do Assign any value to X j consistent with the value assigned to its parent X i X YZ U V W  (X, Y, Z, U, V, W) 48

E XPLOITING THE S TRUCTURE OF CSP Whenever a variable is assigned a value by the backtracking algorithm, propagate this value and remove the variable from the constraint graph 49 WA NT SA Q NSW V 49

E XPLOITING THE S TRUCTURE OF CSP Whenever a variable is assigned a value by the backtracking algorithm, propagate this value and remove the variable from the constraint graph 50 WA NT Q NSW V If the graph becomes a tree, then proceed as shown in previous slide 50

Representation Variables X ij for i, j in {1,..,9} Domains {1,…,9} Constraints: X ij  X ik, for j  k X ij  X kj, for i  k X ij  X mn, for (i,j), (m,n) in same cell

Representation Variables X ij for i, j in {1,..,9} Domains {1,…,9} Constraints: X ij  X ik, for j  k X ij  X kj, for i  k X ij  X mn, for (i,j), (m,n) in same cell

Representation Variables X ij for i, j in {1,..,9} Domains {1,…,9} Constraints: X ij  X ik, for j  k X ij  X kj, for i  k X ij  X mn, for (i,j), (m,n) in same cell Can we detect this using constraint propagation?

Representation Variables X ij for i, j in {1,..,9} Domains {1,…,9} Constraints: X ij  X ik, for j  k X ij  X kj, for i  k X ij  X mn, for (i,j), (m,n) in same cell Must detect 9- way interactions

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: ??? R 1 8 =3 R 2 2 =8 R 3 2 =6 R 3 7 =9 … C 1 5 =4 C 1 2 =9 C 2 8 =4C 3 8 =1 C 3 3 =8 C 3 9 =5 …

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: ??? R 1 8 =3 R 2 2 =8 R 3 2 =6 R 3 7 =9 … C 1 5 =4 C 1 2 =9 C 2 8 =4C 3 8 =1 C 3 3 =8 C 3 9 =5 … X 1 8 = (1,3)

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: ??? R 1 8 =3 R 2 2 =8 R 3 2 =6 R 3 7 =9 … C 1 5 =4 C 1 2 =9 C 2 8 =4C 3 8 =1 C 3 3 =8 C 3 9 =5 … X 1 8 = (1,3)X 2 2 = (3,3)

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: ??? R 1 8 =3 R 2 2 =8 R 3 2 =6 R 3 7 =9 … C 1 5 =4 C 1 2 =9 C 2 8 =4C 3 8 =1 C 3 3 =8 C 3 9 =5 … X 1 8 = (1,3)X 2 2 = (3,3) X 3 2 = (2,3) X 3 7 = (3,3)

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: R i j =k  C k j =I Similar constraints between X’s and R’s, X’s and C’s R i j  R i k for j  k C i j  C i k for j  k X i j  X i k for j  k R 1 8 =3 R 2 2 =8 R 3 2 =6 R 3 7 =9 … C 1 5 =4 C 1 2 =9 C 2 8 =4C 3 8 =1 C 3 3 =8 C 3 9 =5 … X 1 8 = (1,3)X 2 2 = (3,3) X 3 2 = (2,3) X 3 7 = (3,3)

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: R i j =k  C k j =I Similar constraints between X’s and R’s, X’s and C’s R i j  R i k for j  k C i j  C i k for j  k X i j  X i k for j  k C 1 2 = 9 R 1 2 = {1-9}

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: R i j =k  C k j =i Similar constraints between X’s and R’s, X’s and C’s R i j  R i k for j  k C i j  C i k for j  k X i j  X i k for j  k C 2 2 = {1-9}C 1 2 = 9 R 1 2 = {2-9}

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: R i j =k  C k j =I Similar constraints between X’s and R’s, X’s and C’s R i j  R i k for j  k C i j  C i k for j  k X i j  X i k for j  k C 1 2 = 9 R 1 2 = {2-9} X 2 2 = (3,3) X 3 2 = (2,2)

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: R i j =k  C k j =I Similar constraints between X’s and R’s, X’s and C’s R i j  R i k for j  k C i j  C i k for j  k X i j  X i k for j  k C 1 2 = 9 R 1 2 =2 X 2 2 = (3,3) X 3 2 = (2,2)

Representation 2 Variables R i j in {1,…,9} C i j in {1,…,9} X i j in {(1,1),…,(3,3)} Constraints: R i j =k  C k j =I Similar constraints between X’s and R’s, X’s and C’s R i j  R i k for j  k C i j  C i k for j  k X i j  X i k for j  k C 1 2 = 9 R 1 2 =2 X 2 2 = (3,3) X 3 2 = (2,2)

L OCAL S EARCH FOR CSP S Init: Make an arbitrary assignment Repeat: Modify some variable to reduce # of violated constraints 65

B OOLEAN S ATISFIABILITY P ROBLEMS Highly successful local search algorithms WalkSAT See R&N p constraints of form u i *  u j *  u k *= 1 where u i * is either u i or  u i n variables u i, …, u n

O BSERVATIONS … 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 If a CSP has many constraints, local search solves it quickly Constraints “guide” solver to a solution (if one exists) 67

H ARD S UDOKU ’ S |... | | 4.. | |... | | 9. 3 | |. 7. | | 8 5. | |... | |.. 9 | |... |.. 1 Human solvers: Lot of logic (deep constraint propagation) Computer solvers: Lot of backtracking

H ARD & E ASY 3-SAT P ROBLEMS Let R = # of constraints / # of variables As n , the fraction of hard problems reduces to 0 69

R ECAP Constraint propagation, AC3 Taking advantage of CSP structure Local search for CSPs 70

N EXT C LASS Intro to uncertainty R&N 4.3-4,