Outline
Interval CLP
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[Benhamou et al]
humbly presented by
Johan Lübcke and Dan Lawesson
Preliminaries
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Narrowing
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F-intervals
F = E ∪ {−∞, ∞}, E ⊂ <, E
F-interval : (a, b), {a, b} ⊆ F
F
nite
Choice points
Implementation
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Typeset by FoilTEX Algorithm
Example
Convexity
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F-intervals
Approximation
Newton
Box consistency
Splitting
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Approximations
If ρ is a subset of <n then apx(ρ) is the smallest F-interval
containing ρ.
is the set of all F-intervals
apx(X 2 + Y 2 ≤ 1) = ([−1, 1], [−1, 1])
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Constraint systems
Narrowing
−
u, the narrowing function →
ρ : Fn →
For every F-block −
ρ (u) = apx(u ∩ ρ)
F n of ρ satises →
V
is a set of variables
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A constraint is an expression ρ(x1, . . . , xn), where
ρ ⊆ <n and every xi ∈ V ∪ E
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A system Σ = (i, S), where i : V ∪ E → F and S is
a nite set of constraints
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A solution σ : V ∪ E → < of a system Σ satises
(i)
(ii)
(iii)
(iv)
→
(i)-(iv) : u, ρ, u ∩ ρ, −
ρ
Contractance
Correctness
−
→
ρ (u) ⊆ u
→
u∩ρ=−
ρ (u) ∩ ρ
Monotonicity
→
→
u⊆v⇒−
ρ (u) ⊆ −
ρ (v)
Idempotence
−
→
→
→
ρ (−
ρ (u)) = −
ρ (u)
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Introduce c2 = {(X, Y ) ∈ <2 | X 2 + Y 2 ≤ 1}
c1 = {(X, Y ) ∈ <2 | Y ≥ X}
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c0 = {X ∈ < | X ≥ 0}
Narrowing example
Terminating!
−
As dened earlier : →
ρ (u) = apx(u ∩ ρ)
Intuition: Calculate narrowing x point. Whenever an
interval (denoted X ) becomes narrower, re-evaluate all
constraints containing X .
Σ = (i, S) = ({X 7→ [0, ∞], Y 7→ [0, ∞]}, {c0(X), c1(X, Y )})
Narrowing algorithm
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The resulting stable system is Σ = ({X 7→ [0, 1], Y 7→ [0, 1]}, {c0, c1, c2})
∀x ∈ E, σ(x) = x
∀x ∈ V, σ(x) ∈ i(x)
∀ρ(x1, . . . , xn) ∈ S, (σ(x1), . . . , σ(xn)) ∈ ρ
Interval convexity
Choice points
A constraint ρ is interval convex if for every F-block u and
every i ∈ {1, . . . , n}, πi(ρ ∩ u) is an F-interval.
add = {(x, y, z) ∈ <3 | x + y = z}
Assume ρ = ρ1 ∪ ρ2, i.e. c = c1 ∨ c2
→
−
1. →
ρ (u) = apx(u ∩ ρ2),
ρ (u) = apx(u ∩ ρ1) ∨ −
disjunction implemented by choice point.
→
2. −
ρ (u) = apx(u∩(ρ1 ∪ρ2)), splitting after propagation
(similar to labeling) creates choice point.
is interval convex, but
mult = {(x, y, z) ∈ <3 | xy = z}
Example: mult = mult+ ∪ mult−, where
is not.
mult+ = {(x, y, z) ∈ <3 | x ≥ 0, xy = z}
Example: Assume u = ([−2, 3], [−4, 5], [1, 1]), then
mult ∩ u = ([−2, 3], [−3, − 12 ] ∪ [ 13 , 5], [1, 1]) and
apx(mult ∩ u) = u.
and
mult− = {(x, y, z) ∈ <3 | x < 0, xy = z}
Both are interval convex.
([−0.3, 2.4], [1.5, 2.5])
([0, 2], [2, 2])
([0, 2), [2, 2]) ∪ ∅
([0, 1], [2, 2])
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Newton
Extending Prolog, Newton combines constraints over reals,
integers and booleans.
Newton uses a relaxed implementation of narrowing called
box consistency.
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=
=
=
=
u
−→
int(u)
→
−
→−
neq(
int(u))
−→ −→ −→
int(neq(int(u)))
Example:
(using int for two variables)
is a special case of integers
Booleans
neq = {(x, y) ∈ <2 | x < y} ∪ {(x, y) ∈ <2 | x > y}
→
If u = [a, b] then −
int(u) = [dae , bbc],
Integers
Restricted intervals
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A constraint c is arc-consistent wrt (D1, . . . , Dn) and an
integer i, 1 ≤ i ≤ n i
Di ⊆ {ri | ∃r1 ∈ D1, . . . , ri−1 ∈ Di−1,
ri+1 ∈ Dr+1, . . . , rn ∈ Dn : c(r)}
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Another example would be f 0(X, Y ) = [−∞, ∞]...
Example: f 0([x1, x2], [y1, y2]) = [bx1 + y1c , dx2 + y2e] is an interval extension
of f (x, y) = x + y.
∀F1, . . . , Fn ∈ F : x1 ∈ F1, . . . , xn ∈ Fn ⇒ f (x1, . . . , xn) ∈ f 0(F1, . . . , Fn)
is an interval extension of f : <n → < i
f0 : Fn → F
Interval extension
Arc consistency revisited
Arc consistency for systems is dened in the natural way.
Generalized to intervals, a constraint ρ is arc consistent wrt
(F1, . . . , Fn) and an integer i , 1 ≤ i ≤ n i
Fi = apx(Fi ∩ {ri | ∃r1 ∈ F1, . . . , ri−1 ∈ Fi−1,
ri+1 ∈ Fr+1, . . . , rn ∈ Fn : r ∈ ρ})
This is the x point criteria for narrowing, i.e.
→
u=−
ρ (u) = apx(u ∩ ρ)
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Box consistency
Box consistency is a relaxed version of arc consistency
that is less expensive to calculate. An interval constraint
ρ0 = C(F1, . . . , Fn) is box consistent wrt (F1, . . . , Fn)
and an integer i i
Box consistency example
x1 − x1 + x2 = 0 is not arc consistent wrt ([−1, 1], [0, 2])
because assigning x2 6= 0 makes the constraint
unsatisable. I.e. the intervals should be narrowed to
([−1, 1], [0, 0]).
X1 − X1 + X2 = 0 is box consistent wrt ([−1, 1], [0, 2]),
C(F1, . . . , Fi−1, [l, l+], Fi+1, . . . , Fn)
∧
because
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[−1, −1] − [−1, −1] + [0, 2] = [0, 0]
C(F1, . . . , Fi−1, [u−, u], Fi+1, . . . , Fn)
where l = lef t(Fi) and u = right(Fi).
This is equivalent to
Fi = apx({ri ∈ Fi | C(F1, . . . , Fi−1,
apx({ri}), Fi+1, . . . , Fn)})
.
[1, 1] − [1, 1] + [0, 2] = [0, 0]
.
[−1, 1] − [−1, 1] + [0, 0] = [0, 0]
Box consistency diers from arc consistency when ρ0
contains multiple instances of the same variable.
.
[−1, 1] − [−1, 1] + [2, 2] = [0, 0]
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Implementation
The choice of interval extension determines the eciency
of the propagation.
Natural Interval Extension Expensive to calculate,
the natural extension provides maximum propagation.
I.e. the natural interval extension of x1(x2 + x3) is
X1(X2 + X3).
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is silently ignored in this
Distributed Interval Extension Rewriting the constraint
as a sum of terms allows for more ecient calculation
of box consistency, but weaker pruning. The distributed
interval extension of x1(x2 + x3) is X1X2 + X1X3.
presentation.
Result: ∃b ∈ Benchmarks(∃S ∈ OtherSolvers(betterb(N ewton, S)))
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Taylor Interval Extension
Summary
Interval constraints allows for a coherent handling of multiple simultaneous variable
domain types.
Splitting
Sometimes propagation just ain't enough...
Splitting corresponds to labeling with domain splitting. To
achieve smaller boxes, the box given after propagation can
be divided into subranges. The solver is restarted for each
sub range, giving smaller new propagation opportunities.
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