A linear algorithm for integer programming in the plane

Transcription

A linear algorithm for integer programming in the
plane
Friedrich Eisenbrand
joint work with Sören Laue
Max Planck Institute for Computer Science
Saarbrücken
Fritz Eisenbrand, MPI für Informatik
A linear algorithm for integer programming in the plane – p.1/30
Integer Programming in the plane
Given m constraints of binary encoding length s
Find integral x∗ ∈
constraints
2
with largest first component satisfying all
constraints
2
P
constraints
2
P
Outline
History
Megiddo’s Prune & Search for linear programming
Integer programming for a triangle
A Prune & Search algorithm with optimal running time
Integer programming in any fixed dimension
Open problems
History I: Euclidean algorithm
Theorem. gcd(a, b) = min{x a + y b | x, y ∈ , x a + y b > 1}
minimize
condition
xa+yb
xa+yb > 1
x, y ∈ .
History I: Euclidean algorithm
Theorem. gcd(a, b) = min{x a + y b | x, y ∈ , x a + y b > 1}
minimize
condition
xa+yb
xa+yb > 1
x, y ∈ .
Integer Programming: Combinatorics & Number Theory
History: II
m: Number of constraints
s: largest binary encoding length of coefficient
Method
Complexity
Kannan 1980, Scharf 1981
Lenstra 1983
Feit 1984
Zamanskij and Cherkasskij 1984
Kanamaru, Nishizeki and Asano 1994
E. and Rote 2000
E. 2003
E. & Laue
polynomial
O(m s + s2 )
O(m log m + m s)
O(m log m + m s)
O(m log m + s)
O(m + (log m) s)
O(m + (log m) s)
O(m + s)
Feasibility test + Euclidean algorithm
O(m + s)
any fixed dimension
Prune & Search: Dealing with the
combinatorics
Megiddo’s Algorithm for LP in the plane
Partition constraints
into “down” and “up”
constraints
acements
lle f t
lright
lmedian
constraints
Pair “up-constraints”
arbitrarily
acements
lle f t
lright
lmedian
constraints
arbitrarily
Compute median of
intersections
acements
lle f t
lright
lmedian
constraints
arbitrarily
Compute median of
intersections
acements
lle f t
lright
lmedian
Decide whether
optimum is left or
right
constraints
arbitrarily
Compute median of
intersections
acements
lle f t
lright
lmedian
Decide whether
optimum is left or
right
Prune 1/4-th of constraints
Each round at least 1/4-th of the constraints pruned
Each round costs linear time
Overall cost is linear
Theorem (Megiddo 1983). A linear program in the plane can be solved
in linear time.
Basis reduction: Dealing with the
numbers
Flatness theorem
Width of P along c, wc (P): max{cT x | x ∈ P} − min{cT x | x ∈ P}
/ then there exists
Theorem (Khinchine’s flatness theorem). If PI = 0,
integral c 6= 0 such that width of P along c is 6 constant f n
P
Flatness theorem
Width of P along c, wc (P): max{cT x | x ∈ P} − min{cT x | x ∈ P}
/ then there exists
Theorem (Khinchine’s flatness theorem). If PI = 0,
integral c 6= 0 such that width of P along c is 6 constant f n
P
Lenstra’s IP algorithm
Lenstra’s Algorithm
Compute width of P and corresponding c
If width is larger than f n , then P is feasible
If width is 6 f n , then
For each α ∈ ∩ [min{cT x | x ∈ P}, max{cT x | x ∈ P}]
Decide feasibility of P ∩ (cT x = α).
Reduction to a constant number of feasibility problems of lower
dimension
Width of a triangle
W.l.o.g. 0 is a vertex
c
tl
Width of a triangle
Width along c is
max{cT u, cT v, 0} − min{cT u, cT v, 0}
c
u
tl
v
0
Width of a triangle
Width along c is
T T
T
u
6 2 max{|c u|, |c v|} = 2 k vT ck∞
c
u
tl
v
0
Width of a triangle
Width along c is
T T
T
u
6 2 max{|c u|, |c v|} = 2 k vT ck∞
T T
T
u
> max{|c u|, |c v|} = k vT ck∞
c
u
tl
v
0
Width of a triangle
Width along c is
T T
T
u
6 2 max{|c u|, |c v|} = 2 k vT ck∞
T T
T
u
> max{|c u|, |c v|} = k vT ck∞
Width of triangle ≈ length of shortT u
est vector of lattice Λ = { vT x | x ∈
2
}.
c
u
tl
v
0
Width of a triangle
Σ = conv{0, v1 , v2 } triangle
Let A be the matrix with rows vT1 , vT2
Width of Σ is about length of shortest vector of Λ(A).
Shortest Vector computation ⇔ IP feasibility
Combining both techniques
Partitioning the Polygon
x(1)
g replacements
Upper right kind
`
x(1)
g replacements
Upper right kind
`
x(1)
Upper right kind
g replacements
`
Upper right kind
Consider P ∩ x(1) > `
Upper right kind
Width of triangle is
about width of
P ∩ x(1) > `
Upper right kind
about width of
P ∩ x(1) > `
Determine position `,
for which width of
triangle is f 2 + ε
Upper right kind
about width of
P ∩ x(1) > `
Determine position `,
for which width of
triangle is f 2 + ε
Reduce problem to a
constant number of
problems on the line
Prune & Search
Principle: Improve lle f t and
lright
acements
lle f t
lright
lmedian
Prune & Search
lright
Pair constraints arbitrarily
acements
lle f t
lright
lmedian
Prune & Search
lright
Compute median of
intersections
acements
lle f t lmedian
lright
Prune & Search
lright
Compute median of
intersections
Compute width of triangle
defined by median
acements
lle f t lmedian
lright
Prune & Search
lright
Compute median of
intersections
defined by median
acements
Update bounds
lle f t
lright
lmedian
Prune & Search
lright
Compute median of
intersections
defined by median
acements
Update bounds
lle f t
lright
Prune 1/4-th of constraints
lmedian
Analysis: Part 1
Each round 1/4-th of constraints pruned
Analysis: Part 1
Computing median is linear
Analysis: Part 1
Running time without width checking: O(m)
Analysis: Part 1
Running time without width checking: O(m)
Number of checked triangles: O(log m)
The checked triangles
acements
α
α
acements
α
acements
Query: Given α ∈ , β ∈
uT
Λ = {α βvT x | x ∈ 2 }
compute shortest vector of lattice
Batching the width checks
Theorem. Shortest vector of Λ = {
x/y convergent of b/a.
ab
0 c
x|x∈
2
−x a+y b
yc
} is
Query: Compute shortest vector of Λ = {α
a b
0 βc
, where
x|x∈
2
}
Preprocessing: Compute list of convergents
x(1)/y(1), . . . , x(k)/y(k) of b/a
Complexity: O(s)
Λ = {α
a b
0 βc
x|x∈
2
} incoming query:
Search convergent x( j)/y( j) with minimal
max{| − x( j) a + y( j) b|, |β y( j) c|}
Λ = {α
a b
0 βc
x|x∈
2
} incoming query:
max{| − x( j) a + y( j) b|, |β y( j) c|}
Sequence | − x( j) a + y( j) b| is decreasing
Λ = {α
a b
0 βc
x|x∈
2
} incoming query:
max{| − x( j) a + y( j) b|, |β y( j) c|}
Sequence |β y( j) c| is increasing
Λ = {α
a b
0 βc
x|x∈
2
} incoming query:
max{| − x( j) a + y( j) b|, |β y( j) c|}
Binary search: One query costs O(log(s))
Λ = {α
a b
0 βc
x|x∈
2
} incoming query:
max{| − x( j) a + y( j) b|, |β y( j) c|}
Preprocessing and O(log m) queries: O(s + log m · log s)
Λ = {α
a b
0 βc
x|x∈
2
} incoming query:
max{| − x( j) a + y( j) b|, |β y( j) c|}
Preprocessing and O(log m) queries: O(s + log m · log s)
With prune & search O(m + s)
Total complexity
Theorem (E. & Laue ). IP in the plane can be solved in O(m + s).
Integer programming in any fixed
dimension
Sliding objective
Sliding objective
Sliding objective
Sliding objective
Sliding objective
Sliding objective
Sliding objective
Sliding objective
Two-layer simplices
Σ is two-layer simplex if vertices partition into two sets V and
W such that
v(1) = 0 and w(1) = w0 (1) for all v ∈ V, w, w0 ∈ W.
Example in 3D
cements
µ w1
µ w2
)v1 + µw1
)v1 + µw2
0
v1
w1
w2
Example in 3D
cements
v1
0
(1 − µ)v1 + µw1
(1 − µ)v1 + µw2
µ w1
µ w2
w1
w2
Example in 3D
cements
v1
0
(1 − µ)v1 + µw1
(1 − µ)v1 + µw2
µ w1
Width of truncation is
approximately width of
simplex spanned by V
and µW .
µ w2
w1
w2
Width and shortest vector
Width of ΣV,µW is “about” the length of shortest vector in Λ(Aµ,k ),
where
A is matrix with rows wT1 , . . . , wTk , vT1 , . . . , vTn−k
Aµ,k results from A by scaling first k rows with µ
Parametric shortest vector
The following problem has to be solved:
PARAMETRIC SHORTEST VECTOR :
Given matrix A ∈ n×n and parameter U ∈ , find parameter
p ∈ such that U 6 SV(Λ(A p,k )) 6 2n+1/2 ·U
Theorem (E. 2003). The parametric shortest vector problem can be
solved in linear time in fixed dimension.
Complexity of IP
Using Clarkson’s algorithm for LP-type problems:
Theorem (E. 2003). An integer program with m constraints, each
involving coefficients of size at most s can be solved in expected time
O(m + (log m)s).
Open problem: Can IP be solved in time O(m + s)?

A linear algorithm for integer programming in the plane

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