Lattice closure of polyhedra Oktay G unl uk Math Sciences, IBM - - PowerPoint PPT Presentation

Lattice closure of polyhedra

Oktay G¨ unl¨ uk Math Sciences, IBM Research joint work with Sanjeeb Dash and Diego Moran January 2017 – Aussois

1

Cutting planes for mixed-integer sets

• Consider the mixed-integer set:

P IP = {x ∈ Rn : Ax ≥ b} ∩ (Zn1 × Rn2) where A and b are rational, and n1 + n2 = n .

• An inequality ax ≥ b is valid for P IP

if P IP ⊆ {x ∈ Rn : ax ≥ b} and a cutting plane if P LP = {x ∈ Rn : Ax ≥ b} ⊆ {x ∈ Rn : ax ≥ b}.

• P LP

strengthened with cutting planes gives a better approximation of conv

• P IP

2

Cutting Planes for MILP

LP IP Cut-off region

• The region cut-off by the valid inequality is strictly lattice-free (i.e. no integer points).
• Conversely, excluding lattice-free regions from LP gives valid inequalities.
• Proving that a set is lattice-free is hard in general.
• Easier to work with sets known to be lattice-free for cut generation.

3

Split sets and split cuts

Consider the split set for M = (Zn1 × Rn2) : S(π, γ) =

• x ∈ Rn : γ + 1 > πx > γ}

where π ∈ Zn, γ ∈ Z, and πj = 0 only if j ≤ n1. Clearly P LP ⊇ conv

• P LP \ S(π, γ)
• ⊇ P IP = P LP ∩ M.

P LP S

4

Split closure

• Elementary split closure of P LP

is polyhedral: SC(P LP) =

• π∈Π
• γ∈Z

conv

• P LP \ S(π, γ)
• where Π = Zn1 × {0}n2 (integer vectors with last n2 elements zero.)
• There is an alternate definition of this closure

SC(P LP) =

• π∈Π

conv

• P LP ∩ {x : πx ∈ Z}
• Note:

S(π, γ) ∩ {x : πx ∈ Z} = ∅

• Repeat:

SC2(P ) = SC(SC(P )) , SC3(P ), SC4(P ), . . .

• There is a set in Z2 × R+ that has facets with unbounded split rank.

. [Cook/Kannan/Schrijver ’90]

5

Some further results on split cuts

Let S∗ be the collection of all split sets for Zn1 × Rn2 .

• For any S ⊂ S∗

SC(P LP, S) =

• S∈S

conv

• P LP \ S
• is also polyhedral.

[Andersen/Cornu´ ejols/Li ’05]

• Given S ⊆ S∗ , there exists a finite SF ⊆ S such that for any S ∈ S

conv

• P LP \ S′

⊆ conv

• P LP \ S
• for some S′ ∈ SF .

[Averkov ’12]

6

Two generalizations of split cuts

The split closure: Cl(P LP, S∗) =

• S∈S∗

conv

• P LP \ S
• =
• π∈Π

conv

• P LP ∩ {x : πx ∈ Z}
• where S∗ is the collection of all split sets for (M = Zn1 × Rn2) .
• t-branch split closure:

Cl(P LP, T ∗) =

• T ∈T ∗

conv

• P LP \ T
• where T ∗ is the collection of all T

such that T = ∪t

i=1Si where Si ∈ S∗ .

• k-lattice closure:

Cl(P LP, H∗) =

• (π1,...,πk)∈H∗

conv

• P LP ∩ {x : πix ∈ Z for i = 1, . . . , k}
• where H∗ is the collection of all (π1, . . . , πk) ∈ Πk .

7

a remark

• In general

conv

• P LP \ (S1 ∪ S2)
• ⊆ conv
• P LP \ S1

conv

• P LP \ S2
• and the inclusion can be strict when P LP ∩ S1 ∩ S2 = ∅.

x1 x2 P LP 1 1 . . Let Sk = {1 > xk > 0} for k = 1, 2 Then conv

• P LP \ (S1 ∪ S2)
• = ∅

But conv

• P LP \ S1
• ∩conv
• P LP \ S2
• = (1/2, 1/2)

8

t-branch split closure

• Given t ∈ Z , a (rational) mixed integer set

P IP = {x ∈ Rn1+n2 : Ax ≥ b} ∩ (Zn1 × Rn2) and a subset T ⊆ T ∗ (a collection of T = ∪t

i=1Si where Si ∈ S∗ ),

Cl(P LP, T ) =

• T ∈T

conv

• P LP \ T
• is a polyhedron.
• Furthermore, there exists a finite subset TF ⊆ T

such that for all T ∈ T conv

• P LP \ T ′

⊆ conv

• P LP \ T
• for some T ′ ∈ T F

(i.e. each T ∈ T is dominated by some T ′ ∈ T F .) . [Dash/Gunluk/Moran ’15]

• For all t there is a mixed-integer set with unbounded t-branch split rank.

. [Dash/Gunluk ’13]

9

Today’s talk

k-lattice closure: Cl(P LP, H) =

• (π1,...,πk)∈H

conv

• P LP ∩ {x : πix ∈ Z for i = 1, . . . , k}
• where H ⊆ H∗ and H∗ is the collection of all (π1, . . . , πk) ∈ Zn×k .

The results:

• Given a (rational) polyhedral set P LP

and H ⊆ H∗ , the k-lattice closure Cl(P LP, H) is a polyhedron.

• There is a finite subset HF ⊆ H that gives member-wise domination.
• There is a (rational) mixed integer set whose integer hull cannot be obtained by applying

the k-lattice closure repeatedly.

10

Why call it lattice closure?

For k = 2 let π1, π2 ∈ Zn \ {0} , and M(π1, π2) = {x : π1x ∈ Z, π2x ∈ Z}. Then for any q ∈ Z \ {0} M(π1, π2) = M(π1, qπ1 + π2). ⇒ the pair {π1, π2} does not uniquely define the “mixed-lattice” M(π1, π2). The lattice generated by set of rational vectors {π1, . . . , πk} in Rn is L = {x ∈ Rn : x = u1π1 + · · · + ukπk, u ∈ Zk}. and its dual lattice is L∗ = {x ∈ span(L) : yTx ∈ Z for all y ∈ L}, = {x ∈ span(L) : πT

i x ∈ Z for i = 1, . . . , k}.

The mixed lattice in Rn generated by the dual lattice of L is M = L∗ + span(L)⊥

11

Recap of main result with lattice notation

For π ∈ Zn \ {0} , let M(π) = {x ∈ Rn : πTx ∈ Z}.

• Define

M1

n = {M(π) : π ∈ Zn \ {0}}

• and

Mk

n =

• ∩k

j=1 Mj : Mj ∈ M1 n for all j ∈ {1, . . . , k}

• .

As Zn ⊂ M(π) for all π ∈ Zn \ {0} , any M ∈ Mk

n contains Zn .

Given a rational polyhedron P ⊂ Rn and M ⊆ Mk

n , the lattice closure

Cl(P, M) =

• M∈M

conv (P ∩ M) . is a polyhedron and there exists finite MF ⊆ M that dominates M elementwise. ( k = 1 ⇒ split closure; k = 2 ⇒ crooked-cross closure)

12

Polytopes vs polyhedra: Integer hulls and lattice closure

• We know that if P LP is an unbounded rational polyhedron; then

P LP = conv (v1, . . . , vm) + cone (r1, . . . , rt) where all rays ri are integral. (can be done by scaling)

• Letting P LP = QLP + C define

¯ QLP = QLP +

• t
• i=1

λiri : 0 ≤ λi ≤ 1 for i = 1, . . . t

• Meyer’s theorem: If nonempty, P IP = ¯

QIP + C

• Extension: If nonempty, conv
• P LP ∩ M
• = conv

¯ QLP ∩ M

• +C for M ∈ Mk

n

M 1 dominates M 2 on P LP if and only if M 1 dominates M 2 on ¯ QLP

13

Fairly well ordered qosets

Given P ⊂ Rn and M 1, M 2 ∈ Mk

n define the binary relation P

as M 1 P M 2 if and only if conv

• P ∩ M 1

⊆ conv

• P ∩ M 2

. Note that P defines a quasi-order on Mk

n as it is

• 1. reflexive (i.e., M P M for all M ∈ Mk

n ), and

• 2. transitive (i.e., if M 1 P M 2 and M 2 P M 3 , then M 1 P M 3 ).

(Not a partial order as it is not antisymmetric) ⇒ (Mk

n, P) is a qoset.

A qoset (X, ) is called fairly well-ordered if each X′ ⊆ X has a finite subset X′

F ⊆ X′ such that for all x ∈ X′ , there exists y ∈ X′ F

such that y x .

14

Recap of main result in qoset language

A qoset (X∗, ) is called fairly well-ordered if each X ⊆ X∗ has a finite subset XF ⊆ X such that for all x ∈ X , there exists y ∈ XF such that y x . Main result: For any rational polyhedra P ⊂ Rn , the qoset (Mk

n, P) is fairly well-ordered.

proof:

• Given M ⊆ Mk

n , induction on the dimension dim(P) .

• If the result holds for lower dimensional polyhedra, then it holds for each facet of P.
• Then (M, Fi) is fairly well-ordered for the facets {Fi} of P.
• Using [Higmans’52], (M, Q) where Q = ∪iFi is fairly well-ordered.

i.e. for some finite MF ⊆ M , Cl(Q, M) = Cl(Q, MF)

• Remaining M ∈ M \ MF have an effect on P only if they exclude a ball B(δ)...

15

a technical result

Lemma : Let P ⊆ Rn be a polytope and M ′ ∈ Mk

n

be a mixed-lattice. Let M ∈ Mk

n be such that P ∩ M = ∅ , and M

is dominated by M ′ on all facets of P but not on P . Then there is a constant κ , that depends only on P and M ′, such that there is an ˜ M ∈ Mk

n

that satisfies (i) aff(P ) ∩ M = aff(P ) ∩ ˜ M , (ii) ˜ M = M(π) ∩ M 2 where π ≤ κ and M 2 ∈ Mk−1

n

, and (iii) P ⊂ M(π). π ≤ κ implies:

• All such π form a finite set
• Each such M(π) ∩ P

is the union of a finite number of slices

• Each slice is a lower dimensional polytope.
• Finite dominance as the result holds for lower dimensional polytopes.

16

Rank result

Let P ⊂ Rn be a rational polyhedron and P I = {x ∈ Rn : x ∈ P, xi ∈ Z for i = 1, . . . , k} Consider M = {M ∈ Mk

n : M ⊇ Zk × Rn−k}.

Clearly Cl(P, M) =

• M∈M

conv (P ∩ M) = conv(P I) as (Zk × Rn−k) ∈ M. Now consider M ∩ Mk−1 , mixed-lattices with lattice dimension at most k − 1. There exists a rational polyhedron P ⊂ Rn for which the repeated closure Clq(P, M ∩ Mk−1) = conv(P I), for any finite q > 0.

17

The set P(h)

Consider the n-dimensional simplex S = {x ∈ Rn :

n

• i=1

xi ≤ n, xi ≥ 0 for i = 1, . . . , n}. ( S is integral and maximal lattice-free.) For x ∈ S , let d(x) denote its distance from closest facet. d(x) = min{x1, . . . , xn, (n −

n

• i=1

xi)/√n} Let p = (1/2, . . . , 1/2) ∈ S and for any h > 0 P (h) = conv (S × {0} , {(p, h)}) ⊂ Rn+1 Note: P (h) ∩ (Zn × R) = S × {0}

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Proof of the rank result

Thm: Clq(P (1), M) = P I where M = {M ∈ Mn−1

n+1 : M ⊇ Zn × R}, q ∈ Z.

• proof. Consider a mixed-lattice M = ∩n−1

i=1 M(πi).

• All πi =

π′

i

• where π′

i ∈ Zn for i = 1, . . . , n − 1.

• span(π′

1, . . . , π′ n−1) has dimension strictly less than n.

• There exists a rational v ∈ Rn is orthogonal to all πi
• For all y ∈ Zn and α ∈ R , the point x = y + αv satisfies π′

ix ∈ Z for all i

• ∃x ∈ S ∩ (Zn + span(v)) with d(x) ≥ 1/2n .
• For x ∈ S if d(x) ≥ γ, then (x, 2γh/n) ∈ P (h).
• Therefore (x, 2γh/n) ∈ P (h) ∩ M.
• [By the Height Lemma] For some ǫ > 0 , (p, ǫ) ∈ P (h)∩M for all M ∈ Mn−1

n+1.

P (ǫ) ⊆ Cl(P (1), M) for some ǫ > 0 ⇒ P (ǫq) ⊆ Clq(P (1), M)

19

Cuts from lattice-free sets

Let P I = {x ∈ Rn : x ∈ P LP, xi ∈ Z for i = 1, . . . , k} Consider a lattice-free set F ⊂ Rk i.e. F ∩ Zk = ∅ . Clearly P I ⊆ conv

• P LP \ (F × Rn−k)
• ⊆ P LP.

If the lineality space of F contains a non-zero rational vector, then we can show that P ∩ M = ∅ for some mixed lattice M ∈ Mk−1

k

such that M ⊃ Zk. Therefore, cuts generated by F × Rn−k can also be generated by M ′ = M × Rn−k . Iterated closure of P (1) using such lattice-free sets does not give the integer hull. (Example: t -branch split sets for t < k and unbounded (max) convex lattice-free sets.)

20

thank you