Running example The 4-houses puzzle: 4 families A, B, C and D live next to each other in houses numbered 1, 2, 3 and 4. D lives in a house with lower number than B, B lives next to A in a house with higher number, There is at least one house between B and C, D does not live in the house with number 2, C does not live in the house with number 4. Which family lives in which house ?

Representation: The variables: A, B, C and D The domains: dA = dB = dC = dD = { 1, 2, 3, 4} Constraints: unary: c(C) = C  4 c(D) = D  2 binary: c(A,B) = B = A + 1 c(A,C) = A  C c(B,D) = D  B c(A,D) = A  D c(B,C) = |B - C|  1 c(C,D) = C  D

Node-consistency: Or: 1-consistency (only 1 variable is involved) Unary constraints are eliminated through domain- reductions: c(C) = C  4 dC = { 1, 2, 3} c(D) = D  2 dD = { 1, 3, 4}

The constraint network: B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 1, 2, 3, 4} { 1, 2, 3} { 1, 3, 4}

The example – AC3 (1): First pass (== Look ahead check): B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 1, 2, 3, 4} { 1, 2, 3} { 1, 3, 4} A B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 3, 4} { 1, 2} { 1, 3} { 1, 2, 3} Deletion_occurred := true

The example AC3 (2): Second pass: Deletion_occurred := true A B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 3, 4} { 1, 2} { 1, 3} { 1, 2, 3} A B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 3, 4} { 1, 2} { 1, 3} { 2, 3} Deletion_occurred := true

The example AC3 (3): Third pass: B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 3, 4} { 1, 2} { 1, 3} { 2, 3} Deletion_occurred := false Result: A (2 or 3) , B (3 or 4), C (1 or 2), D (1 or 3) Consistent, but NOT REALLY A SOLUTION !!

K-consistency: 1-consistency (node-consistency): unary constraints (on 1 variable) are consistent 2-consistency (arc-consistency): binary constraints (on 2 variables) are consistent 3-consistency: all constraints involving 3 variables are consistent Eample: A B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 1, 2, 3, 4} { 1, 2, 3} { 1, 3, 4} A value remains in the domain if there are consistent values in the domains of the 2 other variables (for all connecting constraints)

Path consistency Starting with an arc-consistent graph: B D C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 3, 4} { 1, 2} { 1, 3} { 2, 3} Consistent, but NOT REALLY A SOLUTION !! Path-consistency considers all possible paths (triples of variables) AB -> D, AB->C, AC -> D, AC->B, BD->C, BD->A, CD->A, CD->B, etc etc

Path consistency <A,B>::[<2,3>, <3,4>] vs D, vs C B = A + 1 A  C D  B A  D |B - C|  1 C  D { 3, 4} { 1, 2} { 1, 3} { 2, 3} <A,B>::[<2,3>, <3,4>] vs D, vs C <A,C>::[<2,1>, <3,1>, <3,2>] vs B, vs D <B,C>::[<3,1>, <4,1>, <4,2>] vs A, vs D <B,D>::[<3,1>, <4,1>, <4,3> ] vs A, vs C …

Search: Forward checking at work (starting with 1-consistent network) 3 2 A B C D {1} {1,2,3,4} {1,2,3} {1,3,4} B=A+1 AD AC A B C D {2} {1,2,3,4} {1,2,3} {1,3,4} B=A+1 AD AC A B C D {3} {1,2,3,4} {1,2,3} {1,3,4} B=A+1 AD AC B 3 B 4 B 2 A B C D {2} {3} {1,3} {1,3,4} |B-C|1 D B A B C D {3} {4} {1,2} {1,4} |B-C|1 D B A B C D {1} {2} {2,3} {3,4} |B-C|1 D B C 1 1 C 2 A B C D {3} {4} {1} CD A B C D {3} {4} {2} {1} CD A B C D {2} {3} {1} CD fail fail fail success

Lookahead checking at work B C D {1,2,3,4} {1,2,3} {1,3,4} B=A+1 AD AC D B |B-C|1 CD A B C D {3} {3,4} {1,2} {1,3} B=A+1 AD AC D B |B-C|1 CD 1 A 3 2 A B C D {1} {3,4} {1,2} {1,3} B=A+1 B 4 A B C D {3} {4} {2} {1} D B |B-C|1 CD fail C 2 A B C D {2} {3,4} {1,2} {1,3} B=A+1 AD AC D B |B-C|1 CD A B C D {3} {4} {2} {1} CD fail success