1. 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 ?

2. 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

3. 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}

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}

5. 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

6. 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

7. 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 !!

8. 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)

9. 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

10. 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 …

11. 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

12. Lookahead checking at workB 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