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Some 0 / 1 polytopes need exponential size extended formulations - - PowerPoint PPT Presentation

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Some 0 / 1 polytopes need exponential size extended formulations Thomas Rothvo Department of Mathematics, M.I.T. b b b 0 / 1 polytopes conv( X ) with X { 0 , 1 } n P = (1 , 1) P (0 , 0) (1 , 0) b b b 0 / 1 polytopes conv( X ) with

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Some 0/1 polytopes need exponential size extended formulations

Thomas Rothvoß

Department of Mathematics, M.I.T.

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0/1 polytopes

P = conv(X) with X ⊆ {0, 1}n P (0, 0) (1, 0) (1, 1)

b b b

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0/1 polytopes

P = conv(X) with X ⊆ {0, 1}n → exponential! P (0, 0) (1, 0) (1, 1)

b b b

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0/1 polytopes

P = conv(X) with X ⊆ {0, 1}n → exponential! = {x ∈ Rn | Ax ≤ b} P (0, 0) (1, 0) (1, 1)

b b b

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0/1 polytopes

P = conv(X) with X ⊆ {0, 1}n → exponential! = {x ∈ Rn | Ax ≤ b} → exponential! P (0, 0) (1, 0) (1, 1)

b b b

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0/1 polytopes

P = conv(X) with X ⊆ {0, 1}n → exponential! = {x ∈ Rn | Ax ≤ b} → exponential! P (0, 0) (1, 0) (1, 1)

b b b

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0/1 polytopes

P = conv(X) with X ⊆ {0, 1}n → exponential! = {x ∈ Rn | Ax ≤ b} → exponential! = p(Q) → can be polynomial

◮ p : Rn+k → Rn linear projection ◮ Q = {(x, y) | Bx + Cy ≤ c}

P

b b b b b b b b b b b b b b

Q (0, 0) (1, 0) (1, 1)

b b b

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Extension complexity

Definition

Extension complexity: xc(P) := min   #facets of Q | Q polyhedron p linear projection p(Q) = P    P

b b b b b b b b b b b b b b

Q

p

b b b

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Extension complexity

Definition

Extension complexity: xc(P) := min   #facets of Q | Q polyhedron p linear projection p(Q) = P    If xc(P) ≤ poly(n) ⇒ (Q, p) compact formulation of P. P

b b b b b b b b b b b b b b

Q

p

b b b

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Example: Parity Polytope

P = conv{x ∈ {0, 1}n | # ones in x is odd}

b b b b

(0, 0, 0) (1, 0, 0) (0, 1, 0) (1, 1, 0) (0, 0, 1) (1, 0, 1) (0, 1, 1) (1, 1, 1)

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Example: Parity Polytope

P = conv{x ∈ {0, 1}n | # ones in x is odd} = {0 ≤ x ≤ 1 | x(A) − x([n]\A) ≤ |A| − 1 ∀A ⊆ [n] : |A| even}

b b b b

(0, 0, 0) (1, 0, 0) (0, 1, 0) (1, 1, 0) (0, 0, 1) (1, 0, 1) (0, 1, 1) (1, 1, 1)

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Example: Parity Polytope

P = conv{x ∈ {0, 1}n | # ones in x is odd} = {0 ≤ x ≤ 1 | x(A) − x([n]\A) ≤ |A| − 1 ∀A ⊆ [n] : |A| even} 1T zk = k (k odd) 0 ≤ zk ≤ 1 k = 1 k = 3

b b b b

(0, 0, 0) (1, 0, 0) (0, 1, 0) (1, 1, 0) (0, 0, 1) (1, 0, 1) (0, 1, 1) (1, 1, 1)

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Example: Parity Polytope

P = conv{x ∈ {0, 1}n | # ones in x is odd} = {0 ≤ x ≤ 1 | x(A) − x([n]\A) ≤ |A| − 1 ∀A ⊆ [n] : |A| even} x =

  • k odd

zk 1T zk = k · λk (k odd) 0 ≤ zk ≤ 1 · λk 1T λ = 1 λ ≥ k = 1 k = 3

b b b b

(0, 0, 0) (1, 0, 0) (0, 1, 0) (1, 1, 0) (0, 0, 1) (1, 0, 1) (0, 1, 1) (1, 1, 1)

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What’s known?

Compact formulations:

◮ Perfect Matching in planar graphs [Barahona ’93] ◮ Perfect Matching in bounded genus graphs [Gerards

’91]

◮ O(n log n)-size for Permutahedron [Goemans ’10]

(→ tight)

◮ nO(1/ε)-size ε-apx for Knapsack Polytope [Bienstock ’08] ◮ . . .

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What’s known?

Compact formulations:

◮ Perfect Matching in planar graphs [Barahona ’93] ◮ Perfect Matching in bounded genus graphs [Gerards

’91]

◮ O(n log n)-size for Permutahedron [Goemans ’10]

(→ tight)

◮ nO(1/ε)-size ε-apx for Knapsack Polytope [Bienstock ’08] ◮ . . .

Theorem (Yannakakis)

No symmetric compact formulation for TSP Polytope and Perfect Matching Polytope.

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What’s known?

Compact formulations:

◮ Perfect Matching in planar graphs [Barahona ’93] ◮ Perfect Matching in bounded genus graphs [Gerards

’91]

◮ O(n log n)-size for Permutahedron [Goemans ’10]

(→ tight)

◮ nO(1/ε)-size ε-apx for Knapsack Polytope [Bienstock ’08] ◮ . . .

Theorem (Yannakakis)

No symmetric compact formulation for TSP Polytope and Perfect Matching Polytope.

Theorem (Kaibel, Pashkovich & Theis ’10)

Compact formulation for log n size matchings, but no symmetric one.

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1st Carg` ese Workshop on Combinatorial Optim.

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1st Carg` ese Workshop on Combinatorial Optim.

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1st Carg` ese Workshop on Combinatorial Optim.

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1st Carg` ese Workshop on Combinatorial Optim.

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1st Carg` ese Workshop on Combinatorial Optim.

Open problem I

Does the matching polytope have a compact formulation?

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1st Carg` ese Workshop on Combinatorial Optim.

Open problem I

Does the matching polytope have a compact formulation?

N

  • i

d e a !

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1st Carg` ese Workshop on Combinatorial Optim.

Open problem I

Does the matching polytope have a compact formulation?

N

  • i

d e a !

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1st Carg` ese Workshop on Combinatorial Optim.

Open problem I

Does the matching polytope have a compact formulation?

N

  • i

d e a !

Open problem II (V. Kaibel)

Is there any 0/1 polytope without a compact formulation?

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1st Carg` ese Workshop on Combinatorial Optim.

Open problem I

Does the matching polytope have a compact formulation?

N

  • i

d e a !

Open problem II (V. Kaibel)

Is there any 0/1 polytope without a compact formulation?

Y e s !

Theorem

For every n there exists X ⊆ {0, 1}n s.t. xc(conv(X)) ≥ 2n/2·(1−o(1)).

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Proof strategy X ⊆ {0, 1}n

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Proof strategy X ⊆ {0, 1}n

conv(X)

 x | ∃y : A x + B y ≤ b   

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Proof strategy X ⊆ {0, 1}n

injective map conv(X)

 x | ∃y : A x + B y ≤ b   

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Proof strategy X ⊆ {0, 1}n

injective map conv(X)

 x | ∃y : A x + B y ≤ b   

#0/1 polytopes ≤ # extended form.

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Proof strategy X ⊆ {0, 1}n

injective map conv(X)

 x | ∃y : A x + B y ≤ b   

#0/1 polytopes ≤ # extended form. = 22n

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Proof strategy X ⊆ {0, 1}n

injective map conv(X)

 x | ∃y : A x + B y ≤ b   

#0/1 polytopes ≤ # extended form. = = 22n ∞

√ 2 irrational entries

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Proof strategy X ⊆ {0, 1}n

injective map conv(X)

 x | ∃y : A x + B y ≤ b   

#0/1 polytopes ≤ # extended form. = = 22n ∞

√ 2 irrational entries

D

  • e

s n ’ t w

  • r

k !

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Slack-matrix

Write: P = conv({x1, . . . , xv}) = {x ∈ Rn | Ax ≤ b}

  • non-redundant

S

# facets # vertices Sij Sij = bi − AT

i xj

Slack-matrix

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Slack-matrix

Write: P = conv({x1, . . . , xv}) = {x ∈ Rn | Ax ≤ b}

  • non-redundant

S

# facets # vertices facet i vertex j Sij Sij = bi − AT

i xj

Slack-matrix

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Slack-matrix

Write: P = conv({x1, . . . , xv}) = {x ∈ Rn | Ax ≤ b}

  • non-redundant

S

# facets # vertices

U ≥ V ≥ 0

r r Sij Sij = bi − AT

i xj

Slack-matrix Non-negative rank: rk+(S) = min{r | ∃U ∈ Rf×r

≥0 , V ∈ Rr×v ≥0 : S = UV }

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Example for slack-matrix

P (0, 0) (1, 0) (1, 1) 2x1 ≤ 2 x2 ≥ 0 x1 − x2 ≥ 0

S

1 2 1

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Yannakakis’ Theorem

Theorem (Yannakakis ’91)

Let S be slackmatrix for P:

◮ xc(P) = rk+(S). ◮ For any non-neg. factorization S = UV :

P = {x ∈ Rn | ∃y ≥ 0 : Ax + Uy = b}

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Yannakakis’ Theorem

Theorem (Yannakakis ’91)

Let S be slackmatrix for P:

◮ xc(P) = rk+(S). ◮ For any non-neg. factorization S = UV :

P = {x ∈ Rn | ∃y ≥ 0 : Ax + Uy = b}

◮ For vertex xj: Aixj + UiV j = bi.

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Yannakakis’ Theorem

Theorem (Yannakakis ’91)

Let S be slackmatrix for P:

◮ xc(P) = rk+(S). ◮ For any non-neg. factorization S = UV :

P = {x ∈ Rn | ∃y ≥ 0 : Ax + Uy = b}

◮ For vertex xj: Aixj + UiV j = bi. ◮ Aix > bi =

⇒ Aix + Uiy

  • ≥0

> bi.

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Yannakakis’ Theorem

Theorem (Yannakakis ’91)

Let S be slackmatrix for P:

◮ xc(P) = rk+(S). ◮ For any non-neg. factorization S = UV :

P = {x ∈ Rn | ∃y ≥ 0 : Ax + Uy = b}

◮ For vertex xj: Aixj + UiV j = bi. ◮ Aix > bi =

⇒ Aix + Uiy

  • ≥0

> bi.

◮ “≥” follows from an application of duality.

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Controlling the coefficients

◮ Fix X ⊆ {0, 1}n, P := conv(X) = {x ∈ Rn | Ax ≤ b} and

Slack-matrix S = UV Valid assumption:

◮ A, b integral with A∞, b∞ ≤ 2n log(2n)

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Controlling the coefficients

◮ Fix X ⊆ {0, 1}n, P := conv(X) = {x ∈ Rn | Ax ≤ b} and

Slack-matrix S = UV Valid assumption:

◮ A, b integral with A∞, b∞ ≤ 2n log(2n) ◮ xc(P) ≤ 2n.

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Controlling the coefficients

◮ Fix X ⊆ {0, 1}n, P := conv(X) = {x ∈ Rn | Ax ≤ b} and

Slack-matrix S = UV Valid assumption:

◮ A, b integral with A∞, b∞ ≤ 2n log(2n) ◮ xc(P) ≤ 2n. ◮ |Sij| = |bi − Aixj| ≤ (n + 1) · 2n log(2n) ≤ 2n2

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Controlling the coefficients

◮ Fix X ⊆ {0, 1}n, P := conv(X) = {x ∈ Rn | Ax ≤ b} and

Slack-matrix S = UV Valid assumption:

◮ A, b integral with A∞, b∞ ≤ 2n log(2n) ◮ xc(P) ≤ 2n. ◮ |Sij| = |bi − Aixj| ≤ (n + 1) · 2n log(2n) ≤ 2n2 ◮ U∞, V ∞ ≤ 2n2

S

U V

Slack-matrix

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Controlling the coefficients

◮ Fix X ⊆ {0, 1}n, P := conv(X) = {x ∈ Rn | Ax ≤ b} and

Slack-matrix S = UV Valid assumption:

◮ A, b integral with A∞, b∞ ≤ 2n log(2n) ◮ xc(P) ≤ 2n. ◮ |Sij| = |bi − Aixj| ≤ (n + 1) · 2n log(2n) ≤ 2n2 ◮ U∞, V ∞ ≤ 2n2

◮ Scale s.t. U ℓ∞ = Vℓ∞.

S

U V

Slack-matrix col ℓ row ℓ

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Controlling the coefficients

◮ Fix X ⊆ {0, 1}n, P := conv(X) = {x ∈ Rn | Ax ≤ b} and

Slack-matrix S = UV Valid assumption:

◮ A, b integral with A∞, b∞ ≤ 2n log(2n) ◮ xc(P) ≤ 2n. ◮ |Sij| = |bi − Aixj| ≤ (n + 1) · 2n log(2n) ≤ 2n2 ◮ U∞, V ∞ ≤ 2n2

◮ Scale s.t. U ℓ∞ = Vℓ∞. ◮ But S∞ ≥ U ℓ∞ · Vℓ∞

S

U V

Slack-matrix col ℓ row ℓ ≤ 2n2 largest entry in row/col

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Rounding entries

b b b b b b b b b b b b b b

P Q

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ ∞ : A ·x+ U ·y =b
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Rounding entries

b b b b b b b b b b b b b b

P Q

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b
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Rounding entries

◮ Pick rows I max. | det([AI, UI])|

b b b b b b b b b b b b b b

P Q

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b
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Rounding entries

◮ Pick rows I max. | det([AI, UI])|

b b b b b b b b b b b b b b

P Q

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b
  • √

2

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Rounding entries

◮ Pick rows I max. | det([AI, UI])| ◮ Round Uij → U ′ ij ∈ Z · (1 2)n3

b b b b b b b b b b b b b b

P Q

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b
  • 1.41
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Rounding entries

◮ Pick rows I max. | det([AI, UI])| ◮ Round Uij → U ′ ij ∈ Z · (1 2)n3

b b b b b b b b b b b b b b

P Q

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

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Rounding entries

◮ Pick rows I max. | det([AI, UI])| ◮ Round Uij → U ′ ij ∈ Z · (1 2)n3 ◮ Consider vertex x ∈ X: rounding error

≤ U − U ′∞ · y1 ≤ . . . ≤ (1

2)n2

b b b b b b b b b b b b b b

P Q

b

x

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

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Rounding entries

◮ Consider x /

∈ X.

b b b b b b b b b b b b b b

P Q

b

x

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

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Rounding entries

◮ Consider x /

∈ X.

◮ ∃ violated constraint: Aℓx ≥ bℓ + 1

b b b b b b b b b b b b b b

P Q

b

x

Aℓx ≤ bℓ

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

  • ℓ
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Rounding entries

◮ Consider x /

∈ X.

◮ ∃ violated constraint: Aℓx ≥ bℓ + 1 ◮ (Aℓ, Uℓ) = i∈I det

  • det
  • ·

(Ai, Ui)

b b b b b b b b b b b b b b

P Q

b

x

Aℓx ≤ bℓ

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

  • ℓ
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Rounding entries

◮ Consider x /

∈ X.

◮ ∃ violated constraint: Aℓx ≥ bℓ + 1 ◮ (Aℓ, Uℓ) = i∈I [−1, 1]· (Ai, Ui)

b b b b b b b b b b b b b b

P Q

b

x

Aℓx ≤ bℓ

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

  • ℓ
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Rounding entries

◮ Consider x /

∈ X.

◮ ∃ violated constraint: Aℓx ≥ bℓ + 1 ◮ (Aℓ, Uℓ) = i∈I [−1, 1]· (Ai, Ui) ◮ violation in ℓ ≤ i∈I violation in i :

1 ≤ |Aℓx + Uℓy − b| ≤

  • i∈I

|Aix + Uiy − bi|

b b b b b b b b b b b b b b

P Q

b

x

Aℓx ≤ bℓ

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

  • ℓ
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Rounding entries

◮ Consider x /

∈ X.

◮ ∃ violated constraint: Aℓx ≥ bℓ + 1 ◮ (Aℓ, Uℓ) = i∈I [−1, 1]· (Ai, Ui) ◮ violation in ℓ ≤ i∈I violation in i :

1 ≤ |Aℓx + Uℓy − b| ≤

  • i∈I

|Aix + Uiy − bi|

◮ One i ∈ I violated by 1 |I| − U − U ′∞ · y1 ≫ (1 2)n2

⇒ x not feasible!

b b b b b b b b b b b b b b

P Q

b

x

Aℓx ≤ bℓ

b b b

X =

  • x ∈ {0, 1}n | ∃0 ≤ y ≤ 2n2 : A ·x+ U ·y =b ±

1 2n2

  • ℓ
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Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

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Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

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SLIDE 62

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective

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SLIDE 63

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective 22n ≤

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SLIDE 64

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective

n R n + R

22n ≤

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SLIDE 65

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective

n R n + R

22n ≤

poly(n) bits

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SLIDE 66

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective

n R n + R

22n ≤ 2poly(n)·R2

poly(n) bits

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SLIDE 67

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective

n R n + R

22n ≤ 2poly(n)·R2 ⇒ R ≥ 2n/2·(1−o(1))

poly(n) bits

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SLIDE 68

Conclusion

◮ Let R :=

max

X⊆{0,1}n{xc(conv(X))}.

b b b b b b b b b b b b b b

P Rn RR Q

b b b

X ⊆ {0, 1}n

A U ′b

injective

n R n + R

22n ≤ 2poly(n)·R2 ⇒ R ≥ 2n/2·(1−o(1)) ⇒ ∃X ⊆ {0, 1}n : xc(conv(X)) ≥ 2n/2·(1−o(1))

poly(n) bits

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SLIDE 69

Consequences for matroids

◮ Fact: There are 22n/poly(n) many matroids on n elements

[Duke ’03].

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SLIDE 70

Consequences for matroids

◮ Fact: There are 22n/poly(n) many matroids on n elements

[Duke ’03].

Theorem

For every n there exists a matroid M = ([n], I) such that xc(conv(χ(I)) ≥ 2n/2·(1−o(1)).

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SLIDE 71

Consequences for TSP

Theorem (Folkore)

NP ⊆ P/poly ⇒ no compact formulation for PTSP = conv{χ(C) | C is Hamiltonian in complete graph} with polynomially encodable coefficients.

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Consequences for TSP

Theorem

NP ⊆ P/poly ⇒ no compact formulation for PTSP = conv{χ(C) | C is Hamiltonian in complete graph} with polynomially encodable coefficients. P Rn Rxc(P )

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SLIDE 73

Consequences for TSP

Theorem

NP ⊆ P/poly ⇒ no compact formulation for PTSP = conv{χ(C) | C is Hamiltonian in complete graph} with polynomially encodable coefficients. Given instance G = (V, E) for Hamilto- nian Cycle. Optimize ce :=

  • 1

e ∈ E

  • therwise
  • ver rounded polytope.

b b b b b b b b b b b b b b

P Rn Rxc(P ) Q projx(Q)

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SLIDE 74

Consequences for TSP

Theorem

NP ⊆ P/poly ⇒ no compact formulation for PTSP = conv{χ(C) | C is Hamiltonian in complete graph} with polynomially encodable coefficients. Given instance G = (V, E) for Hamilto- nian Cycle. Optimize ce :=

  • 1

e ∈ E

  • therwise
  • ver rounded polytope.

P + ε

b b b b b b b b b b b b b b

P Rn Rxc(P ) Q projx(Q)

◮ ∃HC ⇒ OPT ≥ n ◮ NO HC ⇒ OPT ≤ n − 1 + εn ≤ n − 1 2

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SLIDE 75

Open problems

Open Problem

Can one prove a super-polynomial lower bound for any explicit 0/1 polytope???

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SLIDE 76

Open problems

Open Problem

Can one prove a super-polynomial lower bound for any explicit 0/1 polytope???

Thanks for your attention