Extreme functions with an arbitrary number of slopes Amitabh Basu - - PowerPoint PPT Presentation

Extreme functions with an arbitrary number of slopes

Amitabh Basu ∗ Michele Conforti† Marco Di Summa† Joseph Paat ∗

∗Johns Hopkins University

†Dipartimento di Matematica, Universit‘a degli Studi di Padova, Italy.

Aussois 2016

J.Paat JHU

The Problem

Let b ∈ (0, 1), k ∈ N with k ≥ 2. Find π : [0, 1] → R+ so that (i) π(0) = π(1) = 0, (ii) (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2), for all r1, r2 ∈ [0, 1], (iii) (Symmetry) π(r) + π(b − r) = 1, for all r ∈ [0, 1], (iv) (Extreme) If π1, π2 satisfy (i)-(iii) and π = π1+π2

2

then π = π1 = π2. (v) (k-slopes) Piecewise linear, continuous and has k different slopes.

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

J.Paat JHU

The Problem

Let b ∈ (0, 1), k ∈ N with k ≥ 2. Find π : [0, 1] → R+ so that (i) π(0) = π(1) = 0, (ii) (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2), for all r1, r2 ∈ [0, 1], (iii) (Symmetry) π(r) + π(b − r) = 1, for all r ∈ [0, 1], (iv) (Extreme) If π1, π2 satisfy (i)-(iii) and π = π1+π2

2

then π = π1 = π2. (v) (k-slopes) Piecewise linear, continuous and has k different slopes.

Why do we care?

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

• Ex. k = 2, b = 1/2

J.Paat JHU

The Problem

Let b ∈ (0, 1), k ∈ N with k ≥ 2. Find π : [0, 1] → R+ so that (i) π(0) = π(1) = 0, (ii) (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2), for all r1, r2 ∈ [0, 1], (iii) (Symmetry) π(r) + π(b − r) = 1, for all r ∈ [0, 1], (iv) (Extreme) If π1, π2 satisfy (i)-(iii) and π = π1+π2

2

then π = π1 = π2. (v) (k-slopes) Piecewise linear, continuous and has k different slopes.

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

• Ex. k = 2, b = 1/2

Why do we care?

J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ Rn×m. Consider the feasible region Ax = b x ≥ 0, x ∈ Zm ABxB + ANxN = b x ≥ 0, xB ∈ Zn, xN ∈ Zm−n −A−1

B ANxN = xB − A−1 B b

xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n −A−1

B ANxN = xB − A−1 B b

xB, xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n Let AB be a basis from col(A)

• x ∈ Zm−n

+

: −A−1

B ANx ∈ Zn − A−1 B b

• −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n · The convex hull of Step 4 is Gomory’s corner polyhedron · If A−1

b b ∈ Zn (i.e. xN = 0) then we

want to separate it from the solutions

J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ Rn×m. Consider the feasible region Ax = b x ≥ 0, x ∈ Zm −A−1

B ANxN = xB − A−1 B b

xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n −A−1

B ANxN = xB − A−1 B b

xB, xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n

• x ∈ Zm−n

+

: −A−1

B ANx ∈ Zn − A−1 B b

• Let AB be a

basis from col(A) ABxB + ANxN = b x ≥ 0, xB ∈ Zn, xN ∈ Zm−n −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n · The convex hull of Step 4 is Gomory’s corner polyhedron · If A−1

b b ∈ Zn (i.e. xN = 0) then we

want to separate it from the solutions

J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ Rn×m. Consider the feasible region Ax = b x ≥ 0, x ∈ Zm −A−1

B ANxN = xB − A−1 B b

xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n

• x ∈ Zm−n

+

: −A−1

B ANx ∈ Zn − A−1 B b

• Let AB be a

basis from col(A) ABxB + ANxN = b x ≥ 0, xB ∈ Zn, xN ∈ Zm−n Rearrange −A−1

B ANxN = xB − A−1 B b

xB, xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n · The convex hull of Step 4 is Gomory’s corner polyhedron · If A−1

b b ∈ Zn (i.e. xN = 0) then we

want to separate it from the solutions

J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ Rn×m. Consider the feasible region Ax = b x ≥ 0, x ∈ Zm

• x ∈ Zm−n

+

: −A−1

B ANx ∈ Zn − A−1 B b

• Let AB be a

basis from col(A) ABxB + ANxN = b x ≥ 0, xB ∈ Zn, xN ∈ Zm−n Rearrange −A−1

B ANxN = xB − A−1 B b

xB, xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n Drop nonnegativity

• n xB

−A−1

B ANxN = xB − A−1 B b

xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n · The convex hull of Step 4 is Gomory’s corner polyhedron · If A−1

b b ∈ Zn (i.e. xN = 0) then we

want to separate it from the solutions

J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ Rn×m. Consider the feasible region Ax = b x ≥ 0, x ∈ Zm

• x ∈ Zm−n

+

: −A−1

B ANx ∈ Zn − A−1 B b

• Let AB be a

basis from col(A) ABxB + ANxN = b x ≥ 0, xB ∈ Zn, xN ∈ Zm−n Rearrange −A−1

B ANxN = xB − A−1 B b

xB, xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n Drop nonnegativity

• n xB

−A−1

B ANxN = xB − A−1 B b

xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n ∼ = −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n · The convex hull of Step 4 is Gomory’s corner polyhedron · If A−1

b b ∈ Zn (i.e. xN = 0) then we

want to separate it from the solutions

J.Paat JHU

Let’s start by relaxing an integer linear program— Let A ∈ Rn×m. Consider the feasible region Ax = b x ≥ 0, x ∈ Zm Let AB be a basis from col(A) ABxB + ANxN = b x ≥ 0, xB ∈ Zn, xN ∈ Zm−n Rearrange −A−1

B ANxN = xB − A−1 B b

xB, xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n Drop nonnegativity

• n xB

−A−1

B ANxN = xB − A−1 B b

xN ≥ 0, xB ∈ Zn, xN ∈ Zm−n ∼ = −A−1

B ANxN ∈ Zn − A−1 B b

xN ≥ 0, xN ∈ Zm−n · The convex hull of Step 4 is Gomory’s corner polyhedron · If A−1

b b ∈ Zn (i.e. xN = 0) then we

want to separate it from the solutions

J.Paat JHU

So for R ∈ Rn×k and b ∈ Rn we have the feasible region

• x ∈ Zk

+ : Rx ∈ Zn + b

• .

J.Paat JHU

So for R ∈ Rn×k and b ∈ Rn we have the feasible region

• x ∈ Zk

+ : Rx ∈ Zn + b

• .

If R = (r1, r2, . . . , rk) then Rn R= . . . r1 r2 . . . rk . . . x = . . . ← 0 [x(r1) x(r2) . . . x(rk)] 0 →. . .

J.Paat JHU

So for R ∈ Rn×k and b ∈ Rn we have the feasible region

• x ∈ Zk

+ : Rx ∈ Zn + b

• .

If R = (r1, r2, . . . , rk) then Rn R= . . . r1 r2 . . . rk . . . x = . . . ← 0 [x(r1) x(r2) . . . x(rk)] 0 →. . .

J.Paat JHU

So for R ∈ Rn×k and b ∈ Rn we have the feasible region

• x ∈ Zk

+ : Rx ∈ Zn + b

• .

If R = (r1, r2, . . . , rk) then Rn R= . . . r1 r2 . . . rk . . . x = . . . ← 0 [x(r1) x(r2) . . . x(rk)] 0 →. . . Gomory and Johnson introduced the n-row infinite group relaxation Rb(Rn, Zn) :=

• x : Rn → Z+ :
• r∈Rn

rxr ∈ Zn + b, x has finite support

• J.Paat

JHU

Idea of a cut-generating function— Recall: For a fixed R, we want to separate x = 0 from conv

• x ∈ Zk

+ : Rx ∈ Zn + b

• . A valid inequality looks like

k

• i=1

γixi ≥ 1, where γi ≥ 0 for each i ∈ [k].

J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R, we want to separate x = 0 from conv

• x ∈ Zk

+ : Rx ∈ Zn + b

• . A valid inequality looks like

k

• i=1

γixi ≥ 1, where γi ≥ 0 for each i ∈ [k]. We can write γi =: π(ri).

J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R, we want to separate x = 0 from conv

• x ∈ Zk

+ : Rx ∈ Zn + b

• . A valid inequality looks like

k

• i=1

γixi ≥ 1, where γi ≥ 0 for each i ∈ [k]. We can write γi =: π(ri). A cut-generating function π : Rn → R+ satisfies

• r∈Rn

π(ri)xi ≥ 1, for every x ∈ Rb(Rn, Zn).

J.Paat JHU

Idea of a cut-generating function— Recall: For a fixed R, we want to separate x = 0 from conv

• x ∈ Zk

+ : Rx ∈ Zn + b

• . A valid inequality looks like

k

• i=1

γixi ≥ 1, where γi ≥ 0 for each i ∈ [k]. We can write γi =: π(ri). A cut-generating function π : Rn → R+ satisfies

• r∈Rn

π(ri)xi ≥ 1, for every x ∈ Rb(Rn, Zn).

Andersen, Averkov, Balas, Basu, Borozan, Campelo, Conforti, Cornu´ ejols, Daniilidis, Dash, Dey, Gomory, G¨ unl¨ uk, Hildebrand, Hong, Johnson, K¨

• ppe,

Letchford, Li, Lodi,Miller, Molinaro, Richard, Wolsey, Yıldız, Zambelli, Zhou...

J.Paat JHU

A famous cut-generating function — For b ∈ (0, 1), the Gomory function is GMIb(r) =     

1 br,

0 ≤ r < b

1 1−b(1 − r),

b ≤ r < 1 π(r − j), r ∈ [j, j + 1), j ∈ Z \ {0}.

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

GMI1/2

J.Paat JHU

A famous cut-generating function — For b ∈ (0, 1), the Gomory function is GMIb(r) =     

1 br,

0 ≤ r < b

1 1−b(1 − r),

b ≤ r < 1 π(r − j), r ∈ [j, j + 1), j ∈ Z \ {0}.

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

GMI1/2 Pictures generated using SAGE software by C.Y. Hong, M. K¨

• ppe, and Y. Zhou.

https: //github.com/mkoeppe/infinite-group-relaxation-code

J.Paat JHU

Some cut-generating functions are better than others.

J.Paat JHU

Some cut-generating functions are better than others. A minimal cut-generating function π is one such that no other cut-generating function π′ = π satisfies π′(r) ≤ π(r) for all r ∈ Rn.

J.Paat JHU

Some cut-generating functions are better than others. A minimal cut-generating function π is one such that no other cut-generating function π′ = π satisfies π′(r) ≤ π(r) for all r ∈ Rn. Theorem (RG–EJ 1972) A function π is a minimal cut-generating function if and only if

1 π ≥ 0 and π(0) = 0, 2 (Periodicity) π(r1) = π(r1 + z) for all r1 ∈ Rn and z ∈ Zn, 3 (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2) for all r1, r2 ∈ Rn, 4 (Symmetry) π(r1) + π(b − r1) = 1 for all r1 ∈ Rn

Cornu´ ejols and Yıldız extend this beyond S = Zn + b [GC–SY 2015].

J.Paat JHU

Some cut-generating functions are better than others. A minimal cut-generating function π is one such that no other cut-generating function π′ = π satisfies π′(r) ≤ π(r) for all r ∈ Rn. Theorem (RG–EJ 1972) A function π is a minimal cut-generating function if and only if

1 π ≥ 0 and π(0) = 0, 2 (Periodicity) π(r1) = π(r1 + z) for all r1 ∈ Rn and z ∈ Zn, 3 (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2) for all r1, r2 ∈ Rn, 4 (Symmetry) π(r1) + π(b − r1) = 1 for all r1 ∈ Rn

Cornu´ ejols and Yıldız extend this beyond S = Zn + b [GC–SY 2015]. Note: Periodicity lets us consider functions defined on [0, 1]n, π(z) = 0 for all z ∈ Zn, and b ∈ [0, 1]n \ Zn. [3]

J.Paat JHU

Even stronger than minimal is extreme. An extreme cut-generating function π is one such that if π1, π2 are cut-generating functions and π = π1+π2

2

then π = π1 = π2.

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2

J.Paat JHU

Even stronger than minimal is extreme. An extreme cut-generating function π is one such that if π1, π2 are cut-generating functions and π = π1+π2

2

then π = π1 = π2. Theorem (2-Slope Theorem (RG–EJ 1972)) If π : R → R+ is continuous, piecewise linear, and minimal with 2-slopes then π is extreme for Rb(R, Z). [6]

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2

J.Paat JHU

2-slopes is not necessary...

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

dr projected sequential merge 3 slope gj forward 3 slope

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

chen 4 slope hildebrand 5 slope 22 1 J.Paat JHU

2-slopes is not necessary...

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1

Figure: 28 slopes — kzh 28 slope 1

Computational methods for testing extremality

Algorithm for rational breakpoints [AB–RH–MK 2013] SAGE software [CYH–MK–YZ 2014] https: //github.com/mkoeppe/infinite-group-relaxation-code

[8]

J.Paat JHU

The Problem

Let b ∈ (0, 1), k ∈ N with k ≥ 2. Find π : [0, 1] → R+ so that (i) π(0) = π(1) = 0, (ii) (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2), for all r1, r2 ∈ [0, 1], (iii) (Symmetry) π(r) + π(b − r) = 1, for all r ∈ [0, 1], (iv) (Extreme) If π1, π2 satisfy (i)-(iii) and π = π1+π2

2

then π = π1 = π2. (v) (k-slopes) Piecewise linear, continuous and has k different slopes.

J.Paat JHU

The Problem

Let b ∈ (0, 1), k ∈ N with k ≥ 2. Find π : [0, 1] → R+ so that (i) π(0) = π(1) = 0, (ii) (Subadditivity) π(r1 + r2) ≤ π(r1) + π(r2), for all r1, r2 ∈ [0, 1], (iii) (Symmetry) π(r) + π(b − r) = 1, for all r ∈ [0, 1], (iv) (Extreme) If π1, π2 satisfy (i)-(iii) and π = π1+π2

2

then π = π1 = π2. (v) (k-slopes) Piecewise linear, continuous and has k different slopes.

This is equivalent to

The Problem Does there exist a continuous, piecewise linear, extreme cut-generating function for Rb(R, Z) with k slopes?

J.Paat JHU

Answer: Yes.

J.Paat JHU

Answer: Yes. Theorem (AB-MC-MDS-JP 2015) Let k ∈ N and b ∈ (0, 1). There exists a continuous, piecewise linear, extreme function for Rb(R, Z) with k slopes.

J.Paat JHU

Answer: Yes. Theorem (AB-MC-MDS-JP 2015) Let k ∈ N and b ∈ (0, 1). There exists a continuous, piecewise linear, extreme function for Rb(R, Z) with k slopes.

The proof uses an iterative constructive.

J.Paat JHU

The construction: b = 1/2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2

π2 = GMI1/2 Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2

π2 = GMI1/2 Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2 Slope 3

π3 = GJ Forward-3 Slope Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2 Slope 3

π3 = GJ Forward-3 Slope Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2 Slope 3

π3 = GJ Forward-3 Slope Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2 Slope ≈ 2×Slope

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2 Slope 3 Slope 4

π4 Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2 Slope ≈ 2×Slope

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2 Slope 3 Slope 4

π4 Slope ≈ 2×Slope Slope ≈ 2×Slope

. . .

b b/4 b/8

J.Paat JHU

The construction: b = 1/2 Slope ≈ 2×Slope Slope ≈ 2×Slope

0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Slope 1 Slope 2 Slope 3 Slope 4 Slope 5

π5

. . .

b b/4 b/8

J.Paat JHU

Theorem (AB-MC-MDS-JP 2015) Let k ∈ N and b ∈ (0, 1). There exists a continuous, piecewise linear, extreme function for Rb(R, Z) with k slopes. Proof: For minimality, use induction on k. For extremality, use induction and the Interval Lemma.

J.Paat JHU

Theorem (AB-MC-MDS-JP 2015) Let k ∈ N and b ∈ (0, 1). There exists a continuous, piecewise linear, extreme function for Rb(R, Z) with k slopes. Proof: For minimality, use induction on k. For extremality, use induction and the Interval Lemma. Corollary There is a continuous extreme function with infinitely many slopes.

J.Paat JHU

What about for Rb(Rn, Zn)?

J.Paat JHU

What about for Rb(Rn, Zn)?

The Problem Does there exist a continuous, piecewise linear, extreme cut-generating function for Rb(Rn, Zn) with k gradients?

J.Paat JHU

So what’s been done for Rb(Rn, Zn)? Some results for Rb(R, Z) can be generalized Theorem ((n+1)-Slope Theorem (AB–RH–MK–MM 2013)) If π : Rn → R+ is continuous, piecewise linear, genuinely n-dimensional, minimal with at most (n + 1)-slopes then π is extreme. [2]

J.Paat JHU

So what’s been done for Rb(Rn, Zn)? Some results for Rb(R, Z) can be generalized Theorem ((n+1)-Slope Theorem (AB–RH–MK–MM 2013)) If π : Rn → R+ is continuous, piecewise linear, genuinely n-dimensional, minimal with at most (n + 1)-slopes then π is extreme. [2] Some new tools have been developed Dey and Richard developed the sequential merge

Let π1 and π2 be minimal (extreme) for Rb1(R, Z) and Rb2(Rn, Zn), respectively. Then the sequential merge π1 ⋄ π2 of π1 and π2 is minimal (extreme) for R(b1,b2)(Rn+1, Zn+1).

J.Paat JHU

The Problem Does there exist a continuous, piecewise linear, extreme cut-generating function for Rb(Rn, Zn) with k gradients?

J.Paat JHU

The Problem Does there exist a continuous, piecewise linear, extreme cut-generating function for Rb(Rn, Zn) with k gradients? Ans: Yes. Extending πk in a trivial way gives an extreme function.

J.Paat JHU

This function isn’t genuinely n-dimensional. π : Rn → R is genuinely n-dimensional if there does not exist a linear T : Rn → Rn−1 and φ : Rn−1 → R such that π = φ ◦ T.

J.Paat JHU

This function isn’t genuinely n-dimensional. π : Rn → R is genuinely n-dimensional if there does not exist a linear T : Rn → Rn−1 and φ : Rn−1 → R such that π = φ ◦ T. The Problem Does there exist a continuous, piecewise linear, extreme cut-generating, genuinely n-dimensional function for Rb(Rn, Zn) with k gradients?

J.Paat JHU

Theorem (AB-MC-MDS-JP 2015) Let n, k ∈ N with k ≥ 2. There exists an extreme function Πk : Rn → R+ for Rb(Rn, Zn) so that Πk is genuinely n-dimensional and has at least k gradients.

J.Paat JHU

Theorem (AB-MC-MDS-JP 2015) Let n, k ∈ N with k ≥ 2. There exists an extreme function Πk : Rn → R+ for Rb(Rn, Zn) so that Πk is genuinely n-dimensional and has at least k gradients. The main tools for this are the previous construction and the sequential merge procedure. [5] The proof follows from results in [SD – JPR 2010]. The only new idea is to show that Πk has at least k slopes.

J.Paat JHU

π2 ⋄ GMI1/2

J.Paat JHU

π3 ⋄ GMI1/2

J.Paat JHU

π4 ⋄ GMI1/2

J.Paat JHU

Some open questions

Question 1: Implementation Question 2: Is there a value of k = k(n) such that k slopes generates the most useful cut-generating functions? Not all extreme functions are piecewise linear and continuous

[SD–OG 2006], [ANL–AL 2002], [RH 2014].

Question 3: From a theoretical point of view, is it enough to consider continuous, piecewise linear functions? [4, 9, 1]

J.Paat JHU

Thank you

J.Paat JHU

Amitabh Basu, Robert Hildebrand, and Matthias K¨

• ppe.

Light on the infinite group problem. http://arxiv.org/abs/1410.8584, 2014. Amitabh Basu, Robert Hildebrand, Matthias K¨

• ppe, and

Marco Molinaro. A (k + 1)-slope theorem for the k-dimensional infinite group relaxation. SIAM Journal on Optimization, 23(2):1021–1040, 2013. Gerard Cornu´ ejols and Sercan Yıldız. Cut-generating functions for integer variables. Sanjeeb Dash and Oktay G¨ unl¨ uk. Valid inequalities based on simple mixed-integer sets. Mathematical Programming, 105:29–53, 2006. Santanu S. Dey and Jean-Philippe P. Richard.

J.Paat JHU

Relations between facets of low- and high-dimensional group problems. Mathematical Programming, 123(2):285–313, June 2010. Ralph E. Gomory and Ellis L. Johnson. Some continuous functions related to corner polyhedra, I. Mathematical Programming, 3:23–85, 1972. Ralph E. Gomory and Ellis L. Johnson. T-space and cutting planes. Mathematical Programming, 96:341–375, 2003. C.Y. Hong, Matthias K¨

• ppe, and Y. Zhou.

Sage program for computation and experimentation with the 1-dimensional gomory–johnson infinite group problem. available from https://github.com/mkoeppe/ infinite-group-relaxation-code., 2012. A.N. Letchford and A. Lodi.

J.Paat JHU

Strengthening chvatal-gomory cuts and gomory fractional cuts. Operations Research Letters, 30(2):74–82, 2002.

J.Paat JHU