Convex Set Operators and Polynomial Integer
Minimization in Fixed Dimension
Robert Hildebrand
IFOR - ETH Zürich
Joint work with
Alberto Del Pia, Robert Weismantel and Kevin Zemmer
IBM
ETH
ETH
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
1 / 30
Polynomial IP
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
2 / 30
Introduction
Consider the problem
min{f (x) : x ∈ P ∩ Zn },
(1)
where P is the rational polyhedron P = {x ∈ Rn : Ax ≤ b} with
A ∈ Zm×n , b ∈ Zm , f ∈ Z[x] has degree d.
• Complexity: We want time-complexity when input is binary encoding
of A, b and the coefficients of f .
• Fixed: Dimension n, degree d of f .
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
3 / 30
Flatland: Life in the plane can be Hard!
1
[DHKW, 2005] NP-Hard in 2 variables degree
4, P bounded.
min (x2 − ay − c)2
1 ≤ x ≤ c − 1,
(c−1)2 −a
1−a
,
b ≤y ≤
b
x, y ∈ Z.
Implies AN1 problem - given three positive
integers a, b, c, determine if there exist x ∈ Z
such that x2 ≡ a (mod b) with x < c.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
4 / 30
Flatland: Life in the plane can be Hard!
1
[DHKW, 2005] NP-Hard in 2 variables degree
4, P bounded.
min (x2 − ay − c)2
1 ≤ x ≤ c − 1,
(c−1)2 −a
1−a
,
b ≤y ≤
b
x, y ∈ Z.
Implies AN1 problem - given three positive
integers a, b, c, determine if there exist x ∈ Z
such that x2 ≡ a (mod b) with x < c.
2
[Del-Pia and Weismantel 2014] Polynomial
time for quadratic polynomials in 2 variables.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
4 / 30
Other complexity results
Negative
1
[Matiyasevich 1977, Jones 1982 (Hilbert’s 10th problem, 1900)]
Undecidable in 58 variables with degree 4.
Positive
1
FPTAS for non-negative maximization in fixed dimension (DeLoera,
Hemmeke, Köppe, Weismantel 2011)
2
NP Integer quadratic programming (Del Pia, Dey, Molinaro 2014)
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
5 / 30
Complexity in fixed dimension
f is a polynomial of degree d,
min{f (x) : x ∈ P ∩ Zn },
gen.
d=1
d=2
d=3
d≥4
hom.
d=1
d=2
d=3
d≥4
Robert Hildebrand (ETH)
n=1
P
P
P
P
n=1
P
P
P
P
n=2
P
P
n=3
P
NP
n = 58
P
NP
n fixed
P
NP
NPH
n=2
P
P
NPH
n=3
P
NP
Und
n = 59
P
NP
Und
n fixed
P
NP
NPH
Und
Und
Polynomial IP
MIP Workshop July, 2014
6 / 30
Complexity in fixed dimension
f is a polynomial of degree d,
min{f (x) : x ∈ P ∩ Zn },
gen.
d=1
d=2
d=3
d≥4
hom.
d=1
d=2
d=3
d≥4
Robert Hildebrand (ETH)
n=1
P
P
P
P
n=1
P
P
P
P
n=2
P
P
P
NPH
n=2
P
P
P
P/ I
n=3
P
NP
n = 58
P
NP
n fixed
P
NP
NPH
n=3
P
NP
Und
n = 59
P
NP
Und
n fixed
P
NP
NPH
Und
Und
Polynomial IP
MIP Workshop July, 2014
6 / 30
Main Results
Theorem (Del Pia, H., Weismantel, Zemmer)
The problem min{f (x) : x ∈ P ∩ Z2 } is
1
polynomial time when f is cubic,
2
polynomial time when f is homogeneous, fixed degree, P is bounded,
3
"Intractable" when f is homogeneous, degree 4, P unbounded.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
7 / 30
Main Results
Theorem (Del Pia, H., Weismantel, Zemmer)
The problem min{f (x) : x ∈ P ∩ Z2 } is
1
polynomial time when f is cubic,
2
polynomial time when f is homogeneous, fixed degree, P is bounded,
3
"Intractable" when f is homogeneous, degree 4, P unbounded.
"Intractable" := Requires compact representation.
2
min{ x2 − N y 2 : (x, y) ∈ Z2≥1 }, N = 52k+1
• Optimal objective value is 1.
• Minimal size solution satisfies Negative Pell Equation x2 − N y 2 = −1.
• Lagarias (1980) - minimal solution has binary encoding size Ω(5k ).
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
7 / 30
Tools #1a: Quasi-convex Integer Minimization
Let C ⊆ Rn be a convex set.
• A function f : C → R is quasi-convex on C if {x ∈ C : f (x) ≤ α} is
convex for all α ∈ R.
Lemma (Convex Lemma (Khachiyan-Porkolab 2000))
1
C is a bounded convex semi-algebraic set,
2
f : Rn → R, polynomial, quasi-convex on C.
In polynomial time (fixed dimension), we can solve the problem
min{f (x) : x ∈ C ∩ Zn }.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
8 / 30
Tools #1b: Quasi-concave Integer Minimization
Let C ⊆ Rn be a convex set.
• A function f : C → R is quasi-concave on C if {x ∈ C : f (x) ≥ α} is
convex for all α ∈ R.
Lemma (Concave Lemma)
1
P a bounded polyhedron,
2
f : P → R, quasi-concave on P .
In polynomial time (fixed dimension), we can solve the problem
min{f (x) : x ∈ P ∩ Zn }.
Proof.
Let x∗ be an optimal vertex. Then
PI = conv(vert(PI )) ⊆ conv({x : f (x) ≥ f (x∗ )}) = {x : f (x) ≥ f (x∗ )}
Enumerate vert(PI ) with Cook-Kannan-Hartman-McDiarmid (1992).
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
9 / 30
Quadratic Minimization Review
Del Pia-Weismantel (2014) - quadratic solved in polynomial time.
f (x, y) ≈ x2 + y 2
Convex: Apply
Khachiyan-Porkolab
2010 test feasibility +
binary search.
Robert Hildebrand (ETH)
f (x, y) ≈ −x2 − y 2
Concave: implies
solution on vertex of
integer hull PI .
Enumerate vertices
using CKHM 1992.
Polynomial IP
f (x, y) ≈ x2 − y 2
Divide problem into
quasi-convex and
quasi-concave regions.
MIP Workshop July, 2014
10 / 30
Homogeneous Case
f (x) homogeneous if
f (λx) = λd f (x) for all λ ∈ R.
f (x) homogeneous polynomial if
every monomial has the same degree,
i.e., f (x, y) = x5 + 3y 2 x3 .
Quasi-convex/Quasi-concave division
• Zeros occur on lines.
• Function no longer
quasi-concave/quasi-convex in
positive/negative regions.
• Domain can still be divided into
quasi-concave/quasi-convex
regions.
• Numerically approximate
regions.
(x + y)(x − y)(x2 + y 2 )4
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
11 / 30
Homogeneous Case
f (x) homogeneous if
f (λx) = λd f (x) for all λ ∈ R.
f (x) homogeneous polynomial if
every monomial has the same degree,
i.e., f (x, y) = x5 + 3y 2 x3 .
Quasi-convex/Quasi-concave division
• Zeros occur on lines.
• Function no longer
quasi-concave/quasi-convex in
positive/negative regions.
• Domain can still be divided into
quasi-concave/quasi-convex
regions.
• Numerically approximate
regions.
(x + y)(x − y)(x2 + y 2 )4
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
11 / 30
Homogeneous Case
f (x) homogeneous if
f (λx) = λd f (x) for all λ ∈ R.
f (x) homogeneous polynomial if
every monomial has the same degree,
i.e., f (x, y) = x5 + 3y 2 x3 .
Quasi-convex/Quasi-concave division
• Zeros occur on lines.
• Function no longer
quasi-concave/quasi-convex in
positive/negative regions.
• Domain can still be divided into
quasi-concave/quasi-convex
regions.
• Numerically approximate
regions.
(x + y)(x − y)(x2 + y 2 )4
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
11 / 30
Homogeneous Case
f (x) homogeneous if
f (λx) = λd f (x) for all λ ∈ R.
f (x) homogeneous polynomial if
every monomial has the same degree,
i.e., f (x, y) = x5 + 3y 2 x3 .
Quasi-convex/Quasi-concave division
• Zeros occur on lines.
• Function no longer
quasi-concave/quasi-convex in
positive/negative regions.
• Domain can still be divided into
quasi-concave/quasi-convex
regions.
• Numerically approximate
regions.
(x + y)(x − y)(x2 + y 2 )4
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
11 / 30
Homogeneous Case
f (x) homogeneous if
f (λx) = λd f (x) for all λ ∈ R.
f (x) homogeneous polynomial if
every monomial has the same degree,
i.e., f (x, y) = x5 + 3y 2 x3 .
Quasi-convex/Quasi-concave division
• Zeros occur on lines.
• Function no longer
quasi-concave/quasi-convex in
positive/negative regions.
• Domain can still be divided into
quasi-concave/quasi-convex
regions.
• Numerically approximate
regions.
(x + y)(x − y)(x2 + y 2 )4
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
11 / 30
Homogeneous Case
f (x) homogeneous if
f (λx) = λd f (x) for all λ ∈ R.
f (x) homogeneous polynomial if
every monomial has the same degree,
i.e., f (x, y) = x5 + 3y 2 x3 .
Quasi-convex/Quasi-concave division
• Zeros occur on lines.
• Function no longer
quasi-concave/quasi-convex in
positive/negative regions.
• Domain can still be divided into
quasi-concave/quasi-convex
regions.
• Numerically approximate
regions.
(x + y)(x − y)(x2 + y 2 )4
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
11 / 30
Bordered Hessian, Homogeneous, Volume Argument
1
Bordered Hessian
0 ∇hT
Dh = det
∇h ∇2 h
2
3
4
hom.
=
−d
h(x) · det(∇2 h(x))
d−1
Dh < 0 ⇒ quasi-convex while Dh > 0 ⇒ quasi-concave (R2 only).
Dh ≡ 0 ⇔ h(x) = (cT x)d (Hemmer , 1995)
Dh is a homogeneous polynomial ⇒ zeros occur on lines.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
12 / 30
Bordered Hessian, Homogeneous, Volume Argument
1
Bordered Hessian
0 ∇hT
Dh = det
∇h ∇2 h
2
3
4
hom.
=
−d
h(x) · det(∇2 h(x))
d−1
Dh < 0 ⇒ quasi-convex while Dh > 0 ⇒ quasi-concave (R2 only).
Dh ≡ 0 ⇔ h(x) = (cT x)d (Hemmer , 1995)
Dh is a homogeneous polynomial ⇒ zeros occur on lines.
α1
x2 = R
α2
α3
1
Compute zeros of Dh (x1 , ±R)
to appropriate accuracy
2
Create boxes containing zero
lines
**
0
Lemma (BOWW 2013)
x2 = −R
β1
Robert Hildebrand (ETH)
β2
**
β3
If C ⊆ R2 is convex, vol(C) < 1/2,
then dim(C ∩ Z2 ) ≤ 1.
Polynomial IP
MIP Workshop July, 2014
12 / 30
Homogeneous Minimization
Theorem - Quasiconvex/Quasiconcave division
Let f be a homogeneous translatable polynomial. In polynomial time we
can find a polynomial number of rational polyhedra Pi , Qj and rational
lines Lk such that
• f is quasi-convex on Pi ,
• f is quasi-concave on Qj , and


`1
`2
`3
[
[
[
P ∩ Z2 =  Pi ∪
Qj ∪
Lk  ∩ Z2 .
i=1
j=1
(2)
k=1
Theorem - Homogeneous, Bounded
1
In polynomial time, we can minimize a homogeneous translatable
polynomial of fixed degree in 2 variables over the integer points of a
bounded polyhedron.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
13 / 30
Tools #2: Feasibility Using Convex Set Operator
Theorem: Convex Set Operator
Let C be a convex semi-algebraic set. In fixed dimension, we can determine
in polynomial time if the following sets are non-empty:
1
2
(P ∩ C) ∩ Zn
(P
\ C) ∩ Zn
Robert Hildebrand (ETH)
[KP, 2000]
Enumerate vertices of PI [CKHM, 1993]
Polynomial IP
MIP Workshop July, 2014
14 / 30
Tools #2: Feasibility Using Convex Set Operator
Theorem: Convex Set Operator
Let C be a convex semi-algebraic set. In fixed dimension, we can determine
in polynomial time if the following sets are non-empty:
1
2
(P ∩ C) ∩ Zn
(P
\ C) ∩ Zn
Robert Hildebrand (ETH)
[KP, 2000] “Quasiconvex"
Enumerate vertices of PI [CKHM, 1993] “Quasiconcave"
Polynomial IP
MIP Workshop July, 2014
14 / 30
Tools #2: Feasibility Using Convex Set Operator
Theorem: Convex Set Operator
Let C be a convex semi-algebraic set. In fixed dimension, we can determine
in polynomial time if the following sets are non-empty:
1
2
(P ∩ C) ∩ Zn
(P
\ C) ∩ Zn
[KP, 2000] “Quasiconvex"
Enumerate vertices of PI [CKHM, 1993] “Quasiconcave"
Definition
f
on P is a list of
A division description of S≤ω
polyhedra Pi , Qj and lines Lk that cover P ∩ Z2 such that
1
f
C := Pi ∩ S≤ω
is convex
2
f
C := Qj ∩ S>ω
is convex
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MIP Workshop July, 2014
14 / 30
Tools #2: Division description
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MIP Workshop July, 2014
15 / 30
Tools #2: Division description
P1
Q1
P2
Q2
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MIP Workshop July, 2014
15 / 30
Tools #2: Division description
L1
P1
Q1
P2
Q2
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MIP Workshop July, 2014
15 / 30
Tools #2: Division description
P1 , P2 , Q1 , Q2 , L1 is a division
f
description of S≤ω
.
L1
P1
Q1
P2
1
f
f
P1 ∩ S≤ω
, P2 ∩ S≤ω
are convex.
2
f
f
Q1 ∩ S>ω
, Q2 ∩ S>ω
are convex.
3
P ∩ Z2 ⊆ P1 ∪ P2 ∪ Q1 ∪ Q2 ∪ L1
Q2
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
15 / 30
Tools #2: Division description
P1 , P2 , Q1 , Q2 , L1 is a division
f
description of S≤ω
.
L1
P1
Q1
Q2
P2
1
f
f
P1 ∩ S≤ω
, P2 ∩ S≤ω
are convex.
2
f
f
Q1 ∩ S>ω
, Q2 ∩ S>ω
are convex.
3
P ∩ Z2 ⊆ P1 ∪ P2 ∪ Q1 ∪ Q2 ∪ L1
Theorem: Division ⇒ Optimization
Suppose P is bounded and that for
every ω ∈ Z + 12 we can compute a
f
division description of S≤ω
in
polynomial time.
Then, we can solve
min{f (x) : x ∈ P ∩ Zn } in
polynomial time.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
15 / 30
Cubic Minimization
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
16 / 30
Critically Affine
Definition
A function is said to be critically affine if its gradient vanishes exactly on a
finite union of affine spaces.
Theorem
NP-hard example
Suppose f (x, y) is one of the
following:
1
Cubic polynomial,
2
Homogeneous polynomial,
3
Separable polynomial.
(x2 − ay + c)2
Then f (x, y) is critically affine.
Gradient vanishes on
the parabola
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
17 / 30
Division Description - Parametrize by y
f (x, y) = f0 (x) + f1 (x)y + f2 (x)y 2 + f3 (x)y 3
• Constant-Cubic
f (x, y) = f0 (x)
Then the set {(x, y) : f (x, y) ≤ ω} is union of cylinders.
⇒ Division description is easy.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
18 / 30
Division Description - Parametrize by y
f (x, y) = f0 (x) + f1 (x)y + f2 (x)y 2 + f3 (x)y 3
• Constant-Cubic
f (x, y) = f0 (x)
Then the set {(x, y) : f (x, y) ≤ ω} is union of cylinders.
⇒ Division description is easy.
• Linear-Cubic
f (x, y) = f0 (x) + f1 (x)y
Curve f (x, y) = ω, described by y = g(x) := (ω − f0 (x))/f1 (x)
Find inflection points and vertical asymptes of g(x) and divide into
vertical cylinders about these points.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
18 / 30
Quadratic-Cubic: Division Description
• Quadratic-Cubic
f (x, y) = f0 (x) + f1 (x)y + f2 (x)y 2
Then f (x, y) = ω on curves
y± =
Robert Hildebrand (ETH)
−f1 ±
p
f12 − 4(f0 − ω)f2
2(f0 − ω)
1
Compute vertical asymptotes
2
Compute intersections of y+ , y−
3
Compute inflection points
(changes in concavity)
of y+ , y−
Polynomial IP
MIP Workshop July, 2014
19 / 30
Quadratic-Cubic: Division Description
• Quadratic-Cubic
f (x, y) = f0 (x) + f1 (x)y + f2 (x)y 2
Then f (x, y) = ω on curves
y± =
Robert Hildebrand (ETH)
−f1 ±
p
f12 − 4(f0 − ω)f2
2(f0 − ω)
1
Compute vertical asymptotes
2
Compute intersections of y+ , y−
3
Compute inflection points
(changes in concavity)
of y+ , y−
4
Numerical approximations
Polynomial IP
MIP Workshop July, 2014
19 / 30
Quadratic-Cubic: Division Description
• Quadratic-Cubic
f (x, y) = f0 (x) + f1 (x)y + f2 (x)y 2
Then f (x, y) = ω on curves
y± =
Robert Hildebrand (ETH)
−f1 ±
p
f12 − 4(f0 − ω)f2
2(f0 − ω)
1
Compute vertical asymptotes
2
Compute intersections of y+ , y−
3
Compute inflection points
(changes in concavity)
of y+ , y−
4
Numerical approximations
Polynomial IP
MIP Workshop July, 2014
19 / 30
Quadratic-Cubic: Separating y+ and y− on intervals
Lemma
Any line can intersect a cubic level set at most 3 times.
NP-hard Quartic
(x2 − ay + c)2
Cubic Polynomials
Concave/Concave Concave/Convex Convex/Concave
(Convex/Convex)
.
.
.
.
Robert Hildebrand (ETH)
y+
y+
y+
y−
y−
..
y−
Polynomial IP
MIP Workshop July, 2014
20 / 30
Cubic-Cubic: Approximate Rotation to Quadratic-cubic
=⇒
x3 + y 3 + 9
3u2 v − 3uv 2 + v 3 + 9
⇐=
x3 + δx2 y + y 3 + 8.96
0.05u3 + 2.89u2 v − 2.95uv 2 + v 3 + 8.97
Lemma
{(x, y) ∈ P ∩ Z2 : f (x, y) ≤ ω + 12 } = {(x, y) ∈ P ∩ Z2 : f (x, y) ≤ ω + 21 }
Robert Hildebrand (ETH)
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MIP Workshop July, 2014
21 / 30
Cubic-Cubic: Approximate Rotation to Quadratic-cubic
=⇒
x3 + y 3 + 9
3u2 v − 3uv 2 + v 3 + 9
⇐=
x3 + δx2 y + y 3 + 8.96
3 + 2.89u2 v − 2.95uv 2 + v 3 + 8.97
0.05u
Lemma
{(x, y) ∈ P ∩ Z2 : f (x, y) ≤ ω + 12 } = {(x, y) ∈ P ∩ Z2 : f (x, y) ≤ ω + 21 }
Robert Hildebrand (ETH)
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MIP Workshop July, 2014
21 / 30
Cubic Minimization (Bounded)
Theorem - Cubic Division Description
Let f be a cubic polynomial in 2 variables with integer coefficients and let
ω ∈ Z + 21 . In polynomial time, we can determine a division description of
f
S≤ω
on P .
Theorem - Cubic, Bounded
In polynomial time, we can minimize a cubic polynomial in 2 variables over
the integer points of a bounded polyhedron.
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Polynomial IP
MIP Workshop July, 2014
22 / 30
Cubic Minimization (Unbounded)
Robert Hildebrand (ETH)
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MIP Workshop July, 2014
23 / 30
Cubic Unbounded
Theorem: Cubic minimization unbounded
In polynomial time, we can either decide that the problem is
unbounded, or find a polynomial size bound on the feasible region that
must contain the optimal solution.
Case analysis:
f (x, y) = (x3 + xy 2 ) + (x2 + xy − 3x + 2)
| {z } |
{z
}
h(x,y)
g(x,y)
Assume P ∩ Z2 6= ∅ and rec(P ) 6= ∅ is pointed.
1 h(r) < 0 for some r ∈ rec(P )
Then
f (x + λr) = λ3 h(r) + O(λ2 )ḡ(x, r) ≈ λ3 h(r) → −∞
Problem is unbounded.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
24 / 30
Case analysis: f (x, y) = (x3 + xy 2 ) + (x2 + xy − 3x + 2)
| {z } |
{z
}
h(x,y)
g(x,y)
Assume P ∩ Z2 6= ∅ and rec(P ) 6= ∅ is pointed.
2
h(r) > 0 for all r ∈ rec(P )
I
I
Find lower bound m ≤ h(r) for all r ∈ rec(P ) with krk = 1.
⇒ Use Rouche’s Theorem
Find decomposition P = Q + rec(P ), Q bounded, q ∈ Q
I
f (q + λr) = λ3 h(r) + O(λ2 )ḡ(q, r) ≥ λ3 m + O(λ2 )ḡ(q, r) ≥ f (x)
I
for all λ ≥ M .
Problem is bounded by R := RQ + M
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Polynomial IP
MIP Workshop July, 2014
25 / 30
Case analysis: f (x, y) = (x3 + xy 2 ) + (x2 + xy − 3x + 2)
| {z } |
{z
}
h(x,y)
g(x,y)
Assume P ∩ Z2 6= ∅ and rec(P ) 6= ∅ is pointed.
2
h(r) > 0 for all r ∈ rec(P )
I
I
Find lower bound m ≤ h(r) for all r ∈ rec(P ) with krk = 1.
⇒ Use Rouche’s Theorem
Find decomposition P = Q + rec(P ), Q bounded, q ∈ Q
I
f (q + λr) = λ3 h(r) + O(λ2 )ḡ(q, r) ≥ λ3 m + O(λ2 )ḡ(q, r) ≥ f (x)
I
3
for all λ ≥ M .
Problem is bounded by R := RQ + M
h(r) ≥ 0 for all r ∈ rec(P )
.... Need to analyze r where h(r) = 0.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
25 / 30
Homogeneous cubic polynomials
10
10
5
5
0
0
-5
-5
-10
-10
-5
h(x, y) =
0
5
Q3
10
i=1 (ai x
-10
-10
10
0
0
-5
-5
0
5
-10
-10
10
2
h(x, y) = (a1 x + b1 y) (a2 x + b2 y)
Robert Hildebrand (ETH)
5
10
10
5
-5
0
h(x, y) = (a1 x + b1 y)3
+ bi y)
5
-10
-10
-5
Polynomial IP
-5
0
5
10
h(x, y) = (a1 x + b1 y) q(x, y)
MIP Workshop July, 2014
26 / 30
Homogeneous cubic polynomials
+
−
+
+
−
−
+
h(x, y) =
−
Q3
i=1 (ai x
h(x, y) = (a1 x + b1 y)3
+ bi y)
+
+
−
+
−
−
h(x, y) = (a1 x + b1 y)2 (a2 x + b2 y)
Robert Hildebrand (ETH)
Polynomial IP
h(x, y) = (a1 x + b1 y) q(x, y)
MIP Workshop July, 2014
26 / 30
Homogeneous cubic polynomials
−
+
rec(P )
rec(P )
+
+
−
−
+
h(x, y) =
−
Q3
rec(P )
i=1 (ai x
h(x, y) = (a1 x + b1 y)3
rec(P )
+ bi y)
+
+
−
+
−
−
h(x, y) = (a1 x + b1 y)2 (a2 x + b2 y)
Robert Hildebrand (ETH)
Polynomial IP
h(x, y) = (a1 x + b1 y) q(x, y)
MIP Workshop July, 2014
26 / 30
Rational Zero
Lemma: Rational Zeros
If h(r̄1 , r̄2 ) = 0,and h ≥ 0 on all of rec(P ), then r̄1 /r̄2 is rational and it
can be computed in polynomial time.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
27 / 30
Rational Zero
Lemma: Rational Zeros
If h(r̄1 , r̄2 ) = 0,and h ≥ 0 on all of rec(P ), then r̄1 /r̄2 is rational and it
can be computed in polynomial time.
Fix r̄ ∈ rec(P ) with h(r̄) = 0.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
27 / 30
Rational Zero
Lemma: Rational Zeros
If h(r̄1 , r̄2 ) = 0,and h ≥ 0 on all of rec(P ), then r̄1 /r̄2 is rational and it
can be computed in polynomial time.
Fix r̄ ∈ rec(P ) with h(r̄) = 0.
f (x + λr̄) = λ2 gr̄,2 (x) + λgr̄,1 (x) + gr̄,0 (x).
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
27 / 30
Rational Zero
Lemma: Rational Zeros
If h(r̄1 , r̄2 ) = 0,and h ≥ 0 on all of rec(P ), then r̄1 /r̄2 is rational and it
can be computed in polynomial time.
Fix r̄ ∈ rec(P ) with h(r̄) = 0.
f (x + λr̄) = λ2 gr̄,2 (x) + λgr̄,1 (x) + gr̄,0 (x). Note that
• f (x + µr̄ + λr̄) = λ2 gr̄,2 (x + µr̄) + L.O.T.
• f (x + (µ + λ)r̄) = (λ + µ)2 gr̄,2 (x) + L.O.T.
Matching terms on λ2 , we see that gr̄,2 (x + µr̄) = gr̄,2 (x).
⇒ gr̄,2 (x) = ḡ(x · r̄ ⊥ ).
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
27 / 30
Rational Zero
Lemma: Rational Zeros
If h(r̄1 , r̄2 ) = 0,and h ≥ 0 on all of rec(P ), then r̄1 /r̄2 is rational and it
can be computed in polynomial time.
Fix r̄ ∈ rec(P ) with h(r̄) = 0.
f (x + λr̄) = λ2 gr̄,2 (x) + λgr̄,1 (x) + gr̄,0 (x). Note that
• f (x + µr̄ + λr̄) = λ2 gr̄,2 (x + µr̄) + L.O.T.
• f (x + (µ + λ)r̄) = (λ + µ)2 gr̄,2 (x) + L.O.T.
Matching terms on λ2 , we see that gr̄,2 (x + µr̄) = gr̄,2 (x).
⇒ gr̄,2 (x) = ḡ(x · r̄ ⊥ ).
1
2
3
Handle separately gr̄,2 (x) = 0 on these lower dimensional slices.
If ∃x with gr̄,2 (x) < 0, then problem is unbounded.
If all x satisfy gr̄,2 (x) > 0, then problem is bounded.
• If gr̄,2 (x) ≡ 0, then proceed similarly with gr̄,1 (x).
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
27 / 30
Recap/Comments
Theorem (Del Pia, H., Weismantel, Zemmer)
The problem min{f (x) : x ∈ P ∩ Z2 } is
1
polynomial time when f is cubic,
2
polynomial time when f is homogeneous, fixed degree, P is bounded,
3
"Intractable" when f is homogeneous, degree 4, P unbounded.
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
28 / 30
Recap/Comments
Theorem (Del Pia, H., Weismantel, Zemmer)
The problem min{f (x) : x ∈ P ∩ Z2 } is
1
polynomial time when f is cubic,
2
polynomial time when f is homogeneous, fixed degree, P is bounded,
3
"Intractable" when f is homogeneous, degree 4, P unbounded.
• Techniques apply to low rank minimization problem
min{f (x, y) : (x, y, z) ∈ P ∩ Z × Z × Zn }
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
28 / 30
Open Questions
gen.
d=1
d=2
d=3
d≥4
hom.
d=1
d=2
d=3
d≥4
n=1
P
P
P
P
n=1
P
P
P
P
n=2
P
P
P
NPH
n=2
P
P
P
P/ I
Robert Hildebrand (ETH)
n=3
P
?
?
NPH
n=3
P
?
?
NPH
n = 58
P
?
?
Und
n = 59
P
?
?
Und
Polynomial IP
n fixed
P
?
?
n fixed
P
?
?
Und
MIP Workshop July, 2014
29 / 30
Open Questions
gen.
d=1
d=2
d=3
d≥4
hom.
d=1
d=2
d=3
d≥4
n=1
P
P
P
P
n=1
P
P
P
P
n=2
P
P
P
NPH
n=2
P
P
P
P/ I
n=3
P
?
?
NPH
n=3
P
?
?
NPH
n = 58
P
?
?
Und
n = 59
P
?
?
Und
n fixed
P
?
?
n fixed
P
?
?
Und
How to do get into 3-dimensions and higher?
Robert Hildebrand (ETH)
Polynomial IP
MIP Workshop July, 2014
29 / 30
Thank you for listening!