1. Constraint Satisfaction Problems
ECE457 Applied Artificial Intelligence
Spring 2007
Lecture #4

2. Outline Defining constraint satisfaction problems (CSP)
CSP search algorithms
Russell & Norvig, chapter 5
ECE457 Applied Artificial Intelligence R. Khoury (2007) Page 2

3. States What do we know about states?
A state might be a goal (goal test)
A state has a value (cost or payoff)
An agent moves from state to state using actions
The state space can be discreet or continuous
Ties in with the problem definition
Initial state, goal test, set of actions and their costs defined in problem
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4. Definitions A constraint-satisfaction problem has
A set of variables
V = {X1, X2, …, Xn}
Each variable has a domain of values
Xi  Di = {di1, di2, …, din}
A set of constraints on the values each variable can take
C = {C1, C2, …, Cm}
A state is a set of assignment of values
S1 = {X1 = d12, X4 = d45}
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5. Definitions Consistent (legal) state Complete state Goal state
Does not violate any constraints
Complete state
All variables have a value
Goal state
Consistent and complete
Might not exist
Proof of inconsistency
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6. Example: 8-queen Variables: 64 squares Values: Queen or no queen
V = {X1,1, X1,2, …, X64,64}
Values: Queen or no queen
Xi,j  D = {queen, empty}
States: All board configurations
1.8x1014 complete states
Constraints: Attacks
{Xi,j = queen  Xi,j+n = empty, Xi,j = queen  Xi+n,j = empty, Xi,j = queen  Xi+n,j+n = empty}
92 complete and consistent states
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7. Example: 8-queen
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8. Example: The Einstein Puzzle
There are 5 houses in 5 different colours. In each house lives a person with a different nationality. These 5 owners drink a certain beverage, smoke a certain brand of cigar and keep a certain pet. No owners have the same pet, smoke the same brand of cigar or drink the same drink. Who keeps the fish?
The English lives in a red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the left of the white house.
The green house owner drinks coffee.
The person who smokes Pall Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the house right in the centre drinks milk.
The Norwegian lives in the first house.
The man who smokes Blend lives next to the one who keeps cats.
The man who keeps horses lives next to the man who smokes Dunhill.
The owner who smokes Blue Master drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blend has a neighbour who drinks water.
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9. Example: The Einstein Puzzle
25 variables
V = {N1, …, N5, C1, …, C5, D1, …, D5, S1, …, S5, P1, …, P5}
Domains
Ni  {English, Swede, Dane, Norwegian, German}
Ci  {green, yellow, blue, red, white}
Di  {tea, coffee, milk, beer, water}
Si  {Pall Mall, Dunhill, Blend, Blue Master, Prince}
Pi  {dog, cat, horse, fish, birds}
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10. Example: The Einstein Puzzle
The Norwegian lives in the first house.
N1 = Norwegian
The English lives in a red house.
(Ni = English)  (Ci = Red)
The green house is on the left of the white house.
(Ci = green)  (Ci+1 = white)
(C5 ≠ green)
(C1 ≠ white)
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11. Example: Map Colouring
Colour map of the provinces of Australia
3 colours (red, green, blue)
No adjacent provinces of the same colour
Define CSP
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12. Example: Map Colouring CSP
Variables
{WA, NT, SA, Q, NSW, V, T}
Domain
{R, G, B}
Constrains
{WA  NT, WA  SA, NT  SA, NT  Q, SA  Q, SA  NSW, SA  V, Q  NSW, NSW  V}
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13. Solving CSP Iterative improvement methods Tree searching
Start with random complete state, improve until consistent
Tree searching
Start with empty state, make consistent variable assignments until complete
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14. Iterative Improvement
Algorithm:
Start with random complete state
While not consistent
Pick a variable (randomly or with a heuristic)
Change its assignment to minimize number of violated constraints
Not complete
Might not search all state space
Might not find a solution even if one exists
We won’t do that in this course
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15. Tree Search Formulate CSP as tree
Root node: no variables are assigned a value
Action: assign a value if it does not violate any constraints
Solution node at depth n for n-variable problem
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16. Backtracking Search Start with empty state While not complete
Pick a variable (randomly or with heuristic)
If it has a value that does not violate any constraints
Assign that value
Else
Go back to previous variable
Assign it another value
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17. Backtracking Search Depth-first search algorithm Algorithm complete
Goes down one variable at a time
In a dead end, back up to last variable whose value can be changed without violating any constraints, and change it
If you backed up to the root and tried all values, then there are no solutions
Algorithm complete
Will find a solution if one exists
Will expand the entire (finite) search space if necessary
Depth-limited search with limit = n
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18. Example: Backtracking Search
WA = G
WA = R
WA = B
NT = G
NT = B
Q = B
Q = R
NSW = G
NSW = G
SWA = B
NSW = R
SA = B
SA = ?
SA = ?
V = R
T = G
T = B
T = R
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19. Conflict-Directed Backjumping
Suppose we colour Australia in this order:
WA – R
NSW – R
T – B
NT – B
Q – G
SA - ?
Dead-end at SA
No possible solution with WA = NSW
Backtracking will try to change T on the way, even though it has nothing to do with the problem, before going to NSW
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20. Conflict-Directed Backjumping
Backtracking goes back one level in the search tree at a time
Chronological backtrack
Not rational in cases where the previous step is not involved to the conflict
Conflict-directed backjumping (CBJ)
Should go back to a variable involved in the conflict
Skip several levels if needed to get there
Non-chronological backtrack
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21. Conflict-Directed Backjumping
Maintain a conflict set for each variable
List of previously-assigned variables that are related by constraints
conf(WA) = {}
conf(NSW) = {}
conf(T) = {}
conf(NT) = {WA}
conf(Q) = {NSW,NT}
conf(SA) = {WA,NSW,NT,Q}
When we hit a dead-end, backjump to the deepest variable in the conflict set
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22. Conflict-Directed Backjumping
Learn from the conflict by updating the conflict set of the variable we jumped to
Conflict at Xj, backjump to Xi
conf(Xi)={X1,X2,X3} conf(Xj)={X3,X4,X5,Xi}
conf(Xi) = conf(Xi)  conf(Xj) – {Xi}
conf(Xi) = {X1,X2,X3,X4,X5}
Xi absorbed the conflict set of Xj
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23. Conflict-Directed Backjumping
SA backjump to Q
conf(Q) = {WA,NSW,NT}
Meaning: “There is no consistent solution from Q onwards, given the preceding assignments of WA, NSW and NT together”
Q backjump to NT
conf(NT) = {WA,NSW}
NT backjump over T to NSW
conf(NSW) = {WA}
conf(WA) = {}
conf(NSW) = {}
conf(T) = {}
conf(NT) = {WA}
conf(Q) = {NSW,NT}
conf(SA) = {WA,NSW,NT,Q}
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24. Heuristics
Backtracking and CDJ searches are variations of depth-limited search
Uninformed search technique
Can we make it an informed search?
Add some heuristics
Which variable to assign next?
In which order should the values be tried?
How to detect dead-ends early?
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25. Variable & Value Heuristics
Most constrained variable
Choose the variable with the fewest legal values remaining in its domain
aka Minimum remaining values
Most constraining variable
Choose the variable that’s part of the most constraints
Useful to pick first variable to assign
aka Degree heuristic
Least constraining variable
Pick the variable that’s part of the fewest constrains
Keeps maximum flexibility for future assignments
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26. Variable & Value Heuristics
Values: 2 Constraints: 2
Most constrained variable
Values: 2 Constraints: 2
Values: 2 Constraints: 1
Values: 2 Constraints: 2
Most constrained & least constraining variable
Most constrained variable
Most constraining variable
Most constraining variable
Most constrained variable
Values: 2 Constraints: 1
Least constraining variable
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27. Forward Checking How to detect dead-ends early?
Keep track of the domain of unassigned variables
Use constraints to prune domain of unassigned variables
Backtrack when a variable has an empty domain
Do not waste time exploring further
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28. Example: Forward CheckingB NT G Q R
NSW GB V
RGB T SA WA B NT G Q RB
NSW
RGB V T SA R WA
RGB NT Q
NSW V T SA WA B NT RG Q
RGB
NSW V T SA
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29. Problem with Forward CheckingNT G Q
NSW V T
RGB SA R
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30. Constraint Propagation
Propagate the consequences of a variable’s constraints onto other variables
Represent CSP as constraint graph
Nodes are variables
Arcs are constraints
Check for consistency NT Q WA
NSW SA V T
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31. Checking for Consistency
Node consistency
Unary constraint (e.g. NT  G)
A node is consistent if and only if all values in its domain satisfy all unary constraints
Arc consistency
Binary constraint (e.g. NT  Q)
An arc Xi  Xj or (Xi, Xj) is consistent if and only if, for each value a in the domain of Xi, there is a value b in the domain of Xj that is permitted by the binary constraints between Xi and Xj.
Path consistency
Can be reduced to arc consistency
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32. Checking for Consistency
Node consistency
Simply scan domain of values of each variable and remove those that are not valid
Arc consistency
Examine edges and delete values from domains to make arcs consistent
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33. Checking for Consistency
Remove inconsistent values from a variable’s domain
Backtrack if empty domain
Maintaining node and arc consistency reduces the size of the tree
More computationally expensive than Forward Checking, but worth it
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34. AC-3 Algorithm
Keep queue of arcs (Xi, Xj) to be checked for consistency
If checking an arc removes a value from the domain of Xi, then all arcs (Xk, Xi) are reinserted in the queue
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35. AC-3 Algorithm Add all arcs to queue While queue not empty
Get next arc from queue
For each value di of Xi
If there is no consistent value dj in Xj
Delete di
If a value di was deleted
For each neighbour Xk of Xi
Add arc (Xk, Xi) to queue
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36. AC-3 Algorithm Advantages Disadvantage
Prunes tree, reduces amount of searching
For n-node CSP that’s n-consistent, solution is guaranteed with no backtracking
Disadvantage
Computationally expensive
If pruning takes longer than searching, it’s not worth it
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