IIIACSIC Bellaterra Spain pedroiiiacsices 2 Overview Definitions Tree search backtracking Arc consistency Hybrids arc consistency tree search FC MAC ID: 920348
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Slide1
Constraint Programming
Pedro Meseguer
IIIA-CSIC
Bellaterra, Spain
pedro@iiia.csic.es
Slide22
Overview
Definitions
Tree
search
:
backtracking
Arc
consistency
Hybrids
(
arc
consistency
+
tree
search
): FC, MAC
Modelling
Global
constraints
Soft
constraints
Branch
and
bound
Slide33
Some Definitions
Constraint Network (CN): (
X
,
D
,
C
)
X
= {
x
1
,
x
2
,…,
x
n
} variables
D
= {
d
1
,
d
2
,…,
d
n
} domains (finite)
C
= {
c
1
,
c
2
,…,
c
r
} constraints
c
C var
(
c
) = {
x
i
,
x
j
,…,
x
k
}
scope
rel
(
c
)
d
i
x
d
j
x .. x
d
k
permitted tuples
arity
(
c
)
=
|
var
(
c
)|
(
unary, binary, ternary
,…)
Constraint Satisfaction Problem (CSP):
CN solving: assig. satisfying every constraint
NP-complete task
Slide44
Example: n-queens
GOAL: Locate
n
queens in an
n
x
n
chessboard,
such that they do not attack each other
x
1
x
2
x
3x4
1 2 3 4
4-queens
Formulation:
Variables: one queen per row
Domains: available columns
Constraints:
different columns and different diagonals
x
i
xj | xi - xj | | i - j |
x
1
x
2
x
3
x
4
Constraint Graph:
Slide55
Search Tree
State
space
:
explored
as a
tree
root
: empty
one
variable per
level
sucessors of a node: one sucessor per value of the level
variable
meaning: variable
value
x
a
b
c
Tree:
each branch defines an assignment
depth
n
(number of variables)
branching factor
d
(domain size)
Slide66
Search tree for 4-queens
x
1
x
2
x
3
x
4
1
2
3
4
(1,1,1,1)
(4,4,4,4)
(2,1,1,1)
(3,1,1,1)
(4,1,1,1)
Slide77
Backtracking Algorithm
Depth-first tree traversal (DFS)
At each node:
check every completely
assigned
constraint
if consistent, continue DFS
otherwise, prune current branch
continue DFS
Complexity: O(
d
n
)
Slide88
Backtracking on 4-queens
2
3
2
1
1
2
3
4
1
2
3
3
4
4
2
4
1
1
1
2
3
4
1
1
2
3
x
1
x
2
x
3
x
4
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
Q
solution
25 nodes
Slide99Inference
P’ is equivalent to P: Sol(P) = Sol(P’) P’ is presumably easier to solve than Psmaller search spaceconstraints are more explicit Inference can be: complete: produces the solution adaptive consistency incomplete: requires further search arc consistency
legal operations
on variables,
domains,
constraints
Inference:
P P’
Slide1010
Incomplete Inference: Local Consistency
P constraint network of n variables:
Is P solvable
?
Simpler
problem
:
P
1i, P2j, P
3k
,…subnetworks
of P
of 1, 2,3,…variablesAre they
solvable?YES, but there are values that do not appear in any solution
they
can be removed from P
NO
, P
has no solution
#vars in subnet: 1 2 3 …….
node arc
path consistency
consistency consistency
Empty domain: P has no solution !!
Slide1111
Binary Arc Consistency (AC)
AC: Constraint Cij is directional arc consistent (i j) iff for all aDi there is b
Dj such that (a, b)Rij
Constraint
C
ij
is AC iff it is directional AC in both
directions
A problem is AC iff every constraint is AC
Slide1212AC: Example
a
b
c
a
b
c
X
i
X
j
a
b
c
X
k
Microstructure:
permitted
AC
AC
¬AC
AC
¬AC
¬AC
Slide1313
Filtering by Arc Consistency
If for a Di there not exists b
Dj such that (a
,
b
)
R
ij
,a can be removed from Di (a will not be in any sol)Domain filtering:
Remove arc-inconsistent values
Until no changes
blue,red,
green
blue, green
blue
blue, red
X
2
X
3
X
4
X
1
≠
≠
≠
≠
Example:
Slide1414Example: 3-queens
c
12 is not arc-consistent because value 2 of d1
c
21
is not arc-consistent
because value 2 of
d
2
c
32
is not arc-consistent
because value 2 of d
3
x
1
x
2
x
3
1 2 3
x
1
x
2
x
3
1 2 3
x
1
x
2
x
3
1 2 3
x
1
x
2
x
3
1 2 3
Slide1515Constraint Propagation
AC
(c): procedure to make c arc consistent To make P arc-consistent, process each constraint ? c1 c2
… cr
c
1
c
2
…
cr
c
1
c2 …
cr c1 c2 … cr
AC
But
AC
(
c
) may render other constraints arc-inconsistent
To make
P
arc-consistent, iterate:
Apply AC on {c1,c2,…,cr } Until no changes in domains: fix point
Slide1616Example: 3-queens
value 2 of
d3 was removed(to make c23 arc-consistent)
x
1
x
2
x
3
1 2 3
x
1
x
2
x
3
1 2 3
x
1
x
2
x
3
1 2 3
x
1
x
2
x
3
1 2 3
c
13
is not arc-consistent
because value 1 of
d
1
this makes
c
13
arc-inconsistent
x
1
x
2
x
3
1 2 3
c
13
is not arc-consistent
because value 3 of
d
1
x
1
x
2
x
3
1 2 3
domain
d
1
empty
no solution !!
Slide1717Example: Equations on Integers
y
0
1 2 3 4 5 6 7 8 9
x
0
1 2 3 4 5 6 7 8 9
y
0
1 2 3 4 5 6 7 8 9
x
0
1 2 3 4 5 6 7 8 9
y
0
1 2 3 4 5 6 7 8 9
x
0
1 2 3 4 5 6 7 8 9
fix point
2x
+
4y
=
24
x
+
y
=
9
x
+
y
=
9
2x
+
4y
=
24
x
+
y
=
9
2x
+
4y
=
24
2x
+
4y
=
24
x
+
y
=
9
x
+
y
=
9
2x
+
4y
=
24
x
+
y
=
9
2x
+
4y
=
24
2x
+
4y
=
24
x
+
y
=
9
2x
+
4y
=
24
x
+
y
=
9
y
0
1 2 3 4 5 6 7 8 9
x
0
1 2 3 4 5 6 7 8 9
y
0
1 2 3 4 5 6 7 8 9
x
0
1 2 3 4 5 6 7 8 9
y
0
1 2 3 4 5 6 7 8 9
x
0
1 2 3 4 5 6 7 8 9
2x
+
4y
=
24
x
+
y
=
9
Slide1818 Generalized Arc Consistency
c is arc-consistent iff: every possible value of every variable in var (c) appears in rel(c)
If
P
is arc-consistent
P
has solution
If
c
is
not arc-consistent because a D
x
:
a
will
not be in any solution
a can be removed:
Dx
Dx
– {a}
if Dx
becomes empty, P has no
solution P is arc-consistent iff: every constraint is arc-consistent
inference incompleteinference !!
domain filtering
Slide1919Forward CheckingFC is a combination of:
Search: backtracking
Inference: at each node, AC on constraints with assigned and unassigned variablesWhen a domain becomes empty : No solutions following current branch Prune current branch and backtrack Caution: Values removed by AC at level i, have to be restored when bactracking at level i or above
Slide2020
Example: FC on 4-queens
2
2
3
4
4
1
1
3
x
1
x
2
x
3
x
4
x
1
x
2
x
3
x
4
1 2 3 4
x
1
x
2
x
3
x
4
1 2 3 4
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
Q
Q
solution
8 nodes
Slide2121Maintaining Arc ConsistencyMAC is a combination of:
Search: backtracking
Inference: at each node, AC on all constraints Preprocess: subproblems are ACWhen a domain becomes empty : No solutions following current branch Prune current branch and backtrack Caution: Values removed by AC at level i, have to be restored when bactracking at level i or above
Slide2222MAC vs FC: AC on futures
AC (2,3)
AC (3,4)
AC (2,3)
Q
Q
Q
Q
empty domain
Slide2323Example: MAC on 4-queens
2
3
4
1
1
x
1
x
2
x
3
x
4
x
1
x
2
x
3
x
4
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
Q
x
1
x
2
x
3
x
4
1 2 3 4
Q
Q
Q
Q
Q
solution
5 nodes
Slide2424Constraint ProgrammingCP: provides a platform for solving CSPs
proven useful in many real applications
Platform: set of common structures to reuse best known algorithms for propagation & solvingTwo stages: modelling solving
Slide2525CP: Modelling
Modelling decisions
: select among alternatives the choice of the variables the choice of the domains how we state the constraintssearch space size
space reduction
Example: Map Colouring
variables: are regions or colours?
Any CSP can be modelled in different ways
Efficiency of algorithms can vary dramatically
No strong results are known
Formulating an effective model is not easy,
requires considerable skills in modelling
Slide2626N-queens: Model 1
Variables
:
n
2
, one per cell, matrix B
n
x
n
Domains
: {0,1}, B[a,b]=0, no queen
B[a,b]=1, queen
Constraints
: If B[a,b] = 1 then
same row
B[_,b]=0
same column
B[a,_]=0
same diagonal
B[a+
d
,b+
d
]=0, B[a-d,b-d
]=0 same diagonal B[a-
d,b+d]=0, B[a+d,b-d]=0
Slide2727N-queens: Model 2
Variables
: n, one per row Domains: {0,1,…,n-1}, queen column
Constraints: different columns
x
i
x
j different diagonals | xi -
xj
|
|
i - j |
x
1
x
2x3
x4
1 2 3 4
Different row constraint is included in the formulation!!
Slide2828N-queens: Model 3
Variables
: n, one per row Domains: {0,1,…,n-1}, queen column
Constraints: different columns all-different(
x
1
,
x
2
,…,xn ) different diagonals | xi
-
xj
|
| i -
j |
x
1
x2
x3
x4
1 2 3 4
Different row constraint is included in the formulation!!
Slide2929N-queens Models
Model 1
Model 2
Model 3
Search space size
d
#vars
2
n2
4 65,536
1.27 E30
20 ERROR!!
n
n
256
1.00 E10
1.05 E26
n
n
256
1.00 E10
1.05 E26
Constraints
number
prunning
n rows
n columns
2(n-1) diagonals
n columns
2(n-1) diagonals
Equal model 1
1 all-diff
2(n-1) diagonals
More than model 2
Slide3030Constraint FormulationsBinary (arity
2) : conceptually simple, easy to implement may generate weak formulationsNon-binary (arity > 2) : more complex constraints GAC: stronger (filter more) than AC on equivalent binary decompositionEquivalence: any non-binary CSP can be reformulated as a binary one
Slide3131Global Constraints
Real-life constraints
: often complex, non-binary c is global iff: arity(c) > 2 c is logically equivalent to {c1,c2,…,ck } binary AC(c) prunes more than AC(c1,c2,…,
ck ) Propagation: specialized algorithms
exploit constraint semantics
decrease AC complexity
Slide3232Example: all-different
3 binary constraints,
they are AC,no pruning
1 ternary
constraint
,
not
GAC
,GAC pruning ®
empty
domain
no solution
!!logicallyequivalent
Var:
F, N, S
; Val: { }; Ctrs:
N
≠
S ≠ F ≠
N
F
{ }
S
{ }
N { }
≠
≠≠
all-different
F
{ }
S
{ }
N
{ }
Slide3333Example: all-differentEnforcing arc-consistency:
n
variables, d values n(n-1)/2 binary constraints : O(n2 d2) 1 n-ary constraint: general purpose algorithm O(dn)specialized algorithm O(n2 d2)
Slide3434CP Solving: Some Guidelines
Easy/hard problems:
hybrid search dynamic variable ordering: min domain / degree easy: FC / hard: MACOne solution/All solutions: one solution: hybrid search all solutions: hybrid search or complete inferenceFor specific problems (scheduling, routing…) check: formulation, global constraints heuristics, experiences
Slide3535CP: Declarative Programming
Declarative Programming:
you declare Variables Domains Constraints and ask the SOLVER to find a solution!! SOLVER offers:Implementation for variables / domains / constraintsHybrid algorithm: backtracking + incomplete inferenceGlobal constraints + optimized AC propagationEmpty domain detectionEmbedded heuristics
Slide3636
Hard vs Soft Constraints
[90’s]
Hard constraint:
it must be
always
satisfied:
physical constraints
if violated, the solution is invalid
Soft constraint:
it will be satisfied
if possible
:
preferences if violated, the solution is still valid
aggregation of violated constraintsClassical CSP + soft constraints: optimization problems narrow the set of solutions: user intention
Slide3737
Soft CSP (Weighted Model = COP)
Soft CSP: (
X
,
D
,
C
)
X
set of n
variables
D variable domains
C
set of r soft constraints
c
Î
C c :
∏ di ® R
+ i var(c)
0 t satisfies completely c
c (
t) = (0, +¥) t satisfies / violates partially c
+¥ t
violates completely c c (
t) : cost associated with violation of c by t Goal: min ∑
c (t ) (NP-hard)
c C
Slide3838
Example: 3-queens
GOAL:
Locate
3
queens
in
an
3
x
3
chessboard
,
such that the number of attacks is
minimum
x
1
x
2
x3
1 2 3
3-queens
Formulation:
Variables: one queen per row
Domains: available columns
Constraints:
different columns and different diagonals xi xj | xi - xj | | i - j |
Slide39Constraints -> Cost Functions
39
x
1
x
2
1 3
3 1
x
1
x
3
1 2
2 1
2 3
3 2
x
2
x
3
1 3
3 1
3-queens:
CSP:
x
1
x2 1 1 1 1 2 1 1 3 0 2 1 1 2 2 1
2 3 1
3 1 0 3 2 1
3 3 1
COP:
x
1
x
3
1 1 1 1 2 0 1 3
1 2 1 0 2 2 1 2 3 0 3 1 1 3 2 0 3 3 1
x2 x3 1 1 1 1 2 1 1 3 0 2 1 1
2 2 1 2 3 1 3 1 0 3 2 1 3 3 1
Slide4040
COP Search: Branch and Bound
Depth-first
tree
search
:
internal
node:
partial
assignment
leaf
: total assignment
At each node:Upper bound (UB): minimum
cost of
visited
leaves:
cost
of the current
best solution
Lower bound (LB)
: underestimation
of minimum cost
among
leaves below
current node Pruning: UB ≤ LBSimplest LB: LBsearch =
cost (t) = ∑ c (t) c
CP
2
n
current
node
1
Slide4141
Search tree for 3-queens
x
1
x
2
x
3
1
2
3
(1,1,1)
(3,3,3)
(2,1,1)
(3,1,1)
Slide4242
BnB for 3-queens
x
1
x
2
x
3
0
0
0
1 1 0
3
2
2 3 2 3
1
1 2 2 1 1
1 1 1
0 1 1
24 NODES
LB
UB
Slide4343
BnB + soft AC* for 3-queens
x
1
x
2
x
3
C
= 1
3
2
1
9 NODES
C
= 1
C
= 1
C
= 2
C
= 2
C
= 1