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

Cutting planes for mixed-integer sets 1 • Consider the mixed-integer set: P IP = { x ∈ R n : Ax ≥ b } ∩ ( Z n 1 × R n 2 ) where A and b are rational, and n 1 + n 2 = n . • An inequality ax ≥ b is valid for P IP if P IP ⊆ { x ∈ R n : ax ≥ b } and a cutting plane if P LP = { x ∈ R n : Ax ≥ b } �⊆ { x ∈ R n : ax ≥ b } . • P LP P IP � � strengthened with cutting planes gives a better approximation of conv

Cutting Planes for MILP 2 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.

Split sets and split cuts 3 Consider the split set for M = ( Z n 1 × R n 2 ) : x ∈ R n : γ + 1 > πx > γ } � S ( π, γ ) = where π ∈ Z n , γ ∈ Z , and π j � = 0 only if j ≤ n 1 . Clearly P LP ⊇ conv P LP \ S ( π, γ ) ⊇ P IP = P LP ∩ M. � � S P LP

Split closure 4 • Elementary split closure of P LP is polyhedral: P LP \ S ( π, γ ) � � SC ( P LP ) = � � conv π ∈ Π γ ∈Z where Π = Z n 1 × { 0 } n 2 (integer vectors with last n 2 elements zero.) • There is an alternate definition of this closure P LP ∩ { x : πx ∈ Z} � � SC ( P LP ) = � conv π ∈ Π S ( π, γ ) ∩ { x : πx ∈ Z} = ∅ Note: SC 2 ( P ) = SC ( SC ( P )) , SC 3 ( P ) , SC 4 ( P ) , . . . • Repeat: • There is a set in Z 2 × R + that has facets with unbounded split rank. . [Cook/Kannan/Schrijver ’90]

Some further results on split cuts 5 Let S ∗ be the collection of all split sets for Z n 1 × R n 2 . • For any S ⊂ S ∗ P LP \ S � � SC ( P LP , S ) = � conv S ∈S is also polyhedral. [Andersen/Cornu´ ejols/Li ’05] • Given S ⊆ S ∗ , there exists a finite S F ⊆ S such that for any S ∈ S P LP \ S ′ � P LP \ S � � � ⊆ conv conv for some S ′ ∈ S F . [Averkov ’12]

Two generalizations of split cuts 6 The split closure: P LP \ S P LP ∩ { x : πx ∈ Z} � � � � Cl ( P LP , S ∗ ) � � = = conv conv S ∈S∗ π ∈ Π where S ∗ is the collection of all split sets for ( M = Z n 1 × R n 2 ) . • t-branch split closure: P LP \ T � � Cl ( P LP , T ∗ ) = � conv T ∈T ∗ where T ∗ is the collection of all T i =1 S i where S i ∈ S ∗ . such that T = ∪ t • k-lattice closure: P LP ∩ { x : π i x ∈ Z for i = 1 , . . . , k } � � Cl ( P LP , H ∗ ) = � conv ( π 1 ,...,πk ) ∈H∗ where H ∗ is the collection of all ( π 1 , . . . , π k ) ∈ Π k .

a remark 7 • In general P LP \ ( S 1 ∪ S 2 ) P LP \ S 1 P LP \ S 2 � � � � � � � ⊆ conv conv conv and the inclusion can be strict when P LP ∩ S 1 ∩ S 2 � = ∅ . . x 2 Let S k = { 1 > x k > 0 } for k = 1 , 2 1 Then P LP P LP \ ( S 1 ∪ S 2 ) � � = ∅ conv But P LP \ S 1 P LP \ S 2 � � � � ∩ conv conv = ( 1 / 2 , 1 / 2 ) x 1 1 .

t-branch split closure 8 • Given t ∈ Z , a (rational) mixed integer set P IP = { x ∈ R n 1+ n 2 : Ax ≥ b } ∩ ( Z n 1 × R n 2 ) and a subset T ⊆ T ∗ (a collection of T = ∪ t i =1 S i where S i ∈ S ∗ ), P LP \ T � � Cl ( P LP , T ) = � conv T ∈T is a polyhedron. • Furthermore, there exists a finite subset T F ⊆ T such that for all T ∈ T P LP \ T ′ � P LP \ T � � � ⊆ conv conv for some T ′ ∈ T F is dominated by some T ′ ∈ T F .) (i.e. each T ∈ T . [Dash/Gunluk/Moran ’15] • For all t there is a mixed-integer set with unbounded t -branch split rank. . [Dash/Gunluk ’13]

Today’s talk 9 k-lattice closure: P LP ∩ { x : π i x ∈ Z for i = 1 , . . . , k } � � Cl ( P LP , H ) = � conv ( π 1 ,...,πk ) ∈H where H ⊆ H ∗ and H ∗ is the collection of all ( π 1 , . . . , π k ) ∈ Z n × k . The results: and H ⊆ H ∗ , the k-lattice closure • Given a (rational) polyhedral set P LP Cl ( P LP , H ) is a polyhedron. • There is a finite subset H F ⊆ 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.

Why call it lattice closure? 10 For k = 2 let π 1 , π 2 ∈ Z n \ { 0 } , and M ( π 1 , π 2 ) = { x : π 1 x ∈ Z , π 2 x ∈ Z} . Then for any q ∈ Z \ { 0 } M ( π 1 , π 2 ) = M ( π 1 , qπ 1 + π 2 ) . ⇒ the pair { π 1 , π 2 } does not uniquely define the “mixed-lattice” M ( π 1 , π 2 ) . The lattice generated by set of rational vectors { π 1 , . . . , π k } in R n is { x ∈ R n : x = u 1 π 1 + · · · + u k π k , u ∈ Z k } . L = and its dual lattice is L ∗ { x ∈ span ( L ) : y T x ∈ Z for all y ∈ L } , = { x ∈ span ( L ) : π T = i x ∈ Z for i = 1 , . . . , k } . The mixed lattice in R n generated by the dual lattice of L is M = L ∗ + span ( L ) ⊥

Recap of main result with lattice notation 11 For π ∈ Z n \ { 0 } , let M ( π ) = { x ∈ R n : π T x ∈ Z} . • Define n = { M ( π ) : π ∈ Z n \ { 0 }} M 1 • and � � M k ∩ k j =1 M j : M j ∈ M 1 n = n for all j ∈ { 1 , . . . , k } . As Z n ⊂ M ( π ) for all π ∈ Z n \ { 0 } , any M ∈ M k n contains Z n . Given a rational polyhedron P ⊂ R n and M ⊆ M k n , the lattice closure � Cl ( P, M ) = conv ( P ∩ M ) . M ∈M is a polyhedron and there exists finite M F ⊆ M that dominates M elementwise. ( k = 1 ⇒ split closure; k = 2 ⇒ crooked-cross closure)

Polytopes vs polyhedra: Integer hulls and lattice closure 12 • We know that if P LP is an unbounded rational polyhedron; then P LP = conv ( v 1 , . . . , v m ) + cone ( r 1 , . . . , r t ) where all rays r i are integral. (can be done by scaling) • Letting P LP = Q LP + C define t � � Q LP = Q LP + ¯ � λ i r i : 0 ≤ λ i ≤ 1 for i = 1 , . . . t i =1 • Meyer’s theorem: If nonempty, P IP = ¯ Q IP + C � ¯ P LP ∩ M Q LP ∩ M + C for M ∈ M k � � � • Extension: If nonempty, conv = conv n M 1 dominates M 2 on P LP M 1 dominates M 2 on ¯ Q LP if and only if

Fairly well ordered qosets 13 Given P ⊂ R n and M 1 , M 2 ∈ M k n define the binary relation � P as M 1 � P M 2 � P ∩ M 1 � � P ∩ M 2 � ⊆ conv if and only if conv . defines a quasi-order on M k Note that � P n as it is 1. reflexive (i.e., M � P M for all M ∈ M k 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) ⇒ ( M k n , � P ) is a qoset. X ′ ⊆ X ( X, � ) A qoset is called fairly well-ordered if each has a finite subset F ⊆ X ′ such that for all x ∈ X ′ , there exists y ∈ X ′ X ′ such that y � x . F

Recap of main result in qoset language 14 ( X ∗ , � ) X ⊆ X ∗ A qoset is called fairly well-ordered if each has a finite subset X F ⊆ X such that for all x ∈ X , there exists y ∈ X F such that y � x . Main result: For any rational polyhedra P ⊂ R n , the qoset ( M k n , � P ) is fairly well-ordered. proof: • Given M ⊆ M k 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 { F i } of P. • Using [Higmans’52], ( M , � Q ) where Q = ∪ i F i is fairly well-ordered. i.e. for some finite M F ⊆ M , Cl ( Q, M ) = Cl ( Q, M F ) • Remaining M ∈ M \ M F have an effect on P only if they exclude a ball B ( δ ) ...

a technical result 15 M ′ ∈ M k P ⊆ R n Lemma : Let be a polytope and be a mixed-lattice. Let n is dominated by M ′ on all facets of M ∈ M k n be such that P ∩ M � = ∅ , and M and M ′ , P but not on P . Then there is a constant κ , that depends only on P M ∈ M k ˜ aff ( P ) ∩ M = aff ( P ) ∩ ˜ such that there is an that satisfies (i) M , (ii) n M = M ( π ) ∩ M 2 where � π � ≤ κ and M 2 ∈ M k − 1 ˜ , and (iii) P �⊂ M ( π ) . n � 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.

Rank result 16 Let P ⊂ R n be a rational polyhedron and P I = { x ∈ R n : x ∈ P, x i ∈ Z for i = 1 , . . . , k } Consider n : M ⊇ Z k × R n − k } . M = { M ∈ M k Clearly � conv ( P ∩ M ) = conv ( P I ) Cl ( P, M ) = M ∈M as ( Z k × R n − k ) ∈ M . Now consider M ∩ M k − 1 , mixed-lattices with lattice dimension at most k − 1 . There exists a rational polyhedron P ⊂ R n for which the repeated closure Cl q ( P, M ∩ M k − 1 ) � = conv ( P I ) , for any finite q > 0 .

The set P ( h ) 17 Consider the n-dimensional simplex n S = { x ∈ R n : � x i ≤ n, x i ≥ 0 for i = 1 , . . . , n } . i =1 ( S is integral and maximal lattice-free.) For x ∈ S , let d ( x ) denote its distance from closest facet. n x i ) / √ n } � d ( x ) = min { x 1 , . . . , x n , ( n − i =1 Let p = (1 / 2 , . . . , 1 / 2) ∈ S and for any h > 0 P ( h ) = conv ( S × { 0 } , { ( p, h ) } ) ⊂ R n +1 Note: P ( h ) ∩ ( Z n × R ) = S × { 0 }