On sub-determinants and the diameter of polyhedra Martin Niemeier, - - PowerPoint PPT Presentation

On sub-determinants and the diameter of polyhedra

Martin Niemeier, EPF Lausanne Joint work with: Nicolas Bonifas, Marco Di Summa, Friedrich Eisenbrand, Nicolai H¨ ahnle January 9-13, 2012

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Diameter of Polyhedra

We consider polyhedra defined by m inequalities in dimension n, i.e. P = {x ∈ Rn : Ax ≤ b}, A ∈ Rm×n Vertex: 0-dim face, Edge: 1-dim face Polyhedral graph: Graph defined by vertices and edges of polyhedron in natural way.

Diameter

∆(n, m) is the largest diameter of polyhedral graph with m inequalities, dimension n.

Fundamental Open Problem

Is ∆(n, m) polynomial in n and m? Related: Does the simplex algorithm terminate in polynomial time?

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Hirsch conjecture

Hirsch conjecture (1957)

∆(n, m) ≤ m − n Recently disproved (Santos 2010) Best known bound: ∆(n, m) ≤ m1+log n (Kalai, Kleitmann 1992) Huge gap! Special cases: 0/1-polytopes: Hirsch conjecture holds true (Naddef 1989) flow polytopes: quadratic upper bound (Orlin 1997) transportation polytopes: linear upper bound (Brightwell, v.d. Heuvel and Stougie 2006) P = {x ∈ Rn : Ax ≤ b} with A totally unimodular: upper bound O(m16n3 log(mn)3) (Dyer and Frieze 1994)

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Our result

Our bound

P = {x ∈ Rn : Ax ≤ b} polytope with A totally unimodular. Then diameter at most O(n3.5 log n) (Previous bound: O(m16n3 log(mn)3) (Dyer and Frieze 1994)) More general:

Our bound

P = {x ∈ Rn : Ax ≤ b} polytope, all subdeterminants of A bounded by ∆ (in absolute value). Then diameter at most O(n3.5∆2 log n∆)

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A more geometric view

Consider polytope P P nondegenerate, i.e. each vertex v ∈ P has exactly n tight facets. Normal cone Cv: Cone defined by normal vectors of these facets. Normal fan: Set of all normal cones Normal fan partitions Rn.

Crucial observation

Two vertices in the polyhedral graph are adjacent iff their normal cones share a facet.

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A more geometric view (contd.)

Want to assign volume to each vertex/normal cone Intersect normal fan with euclidean ball Bn Spherical (simplicial) cones: Sv := Cv ∩ Bn. Assign volume: vol(v) := vol(Sv)

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Proof method

Perform BFS from any vertex v ∈ P. Show: After O(n3.5∆2 log n∆) iterations, vertices seen during BFS have combined volume ≥ 1

2 vol(Bn).

Two BFS started from two different discover common node after this number of iterations. Diameter bound follows.

Volume Expansion Lemma

I ⊆ V , with vol(I) ≤ 1

2 vol(Bn)

N(I) neighborhood of I (in polyhedral graph) We have vol(N(I)) ≥

• 2

π 1 ∆2n2.5 · vol(I). Implies iteration bound with standard methods: After k iterations of BFS: vol ≥

• 2

π 1 ∆2n2.5

k vol(Sv).

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How to prove the Volume Expansion Lemma

Volume Expansion Lemma

I ⊆ V , with vol(I) ≤ 1

2 vol(Bn). Then

vol(N(I)) ≥

• 2

π 1 ∆2n2.5 · vol(I). SI :=

v∈I Sv

v ∈ N(I) iff surfaces of Sv and SI touch ((n − 1)-dim intersection). Dockable surface D(SI): Surface of SI disjoint from surface of sphere. D(SN(I)) ≥ D(SI) Sufficient to show: ∆2n3 · vol(Sv) ≥ D(Sv) (1) D(SI) ≥

• 2n

π · vol(SI) (2)

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How to prove the Volume Expansion Lemma

Volume Expansion Lemma

I ⊆ V , with vol(I) ≤ 1

2 vol(Bn). Then

vol(N(I)) ≥

• 2

π 1 ∆2n2.5 · vol(I). Dockable surface D(SI): Surface of SI disjoint from surface of sphere. D(SN(I)) ≥ D(SI) Sufficient to show: ∆2n3 · vol(Sv) ≥ D(Sv) (1) D(SI) ≥

• 2n

π · vol(SI) (2) (1) implies ∆2n3 · vol(SN(I)) ≥ D(SN(I))

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Proving inequality (1)

∆2n3 · vol(Sv) ≥ D(Sv) (1) F facet of Sv y vertex not contained in F Q := conv(F, y) ⊆ Sv is a simplex. hF = d(y, H(F)) vol(Q) = area(F)·hF

n

y F Summing over facets D(Sv) =

• facetF

area(F) =

• facetF

n hF · vol(Q) ≤

• facetF

n hF · vol(Sv)

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Proving inequality (1): Bounding hF

y a1 F F y 1 a1 hF b1 β β A′ = (a1, . . . , an), B = adj(A′) = det(A′) · A′−T hF = cos β =

a1,b1 a1·b1 ≤ 1 n∆2

Recall: D(Sv) ≤

facetF n hF · vol(Sv)

We get: ∆2n3 · vol(Sv) ≥ D(Sv) (1)

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Proving inequality (2)

D(SI) ≥

• 2n

π · vol(SI) (2)

(a) Dock- able surface

• f S.

Area: D(S) (b) Base of S. Area: B(S) (c) Relative boundary of the base of S. Length: L(S)

Basic integration: vol(S) = B(S)/n, D(S) = L(S)/(n − 1) Inequality (2) equivalent to: L(S) B(S) ≥

• 2

π n − 1 √n . (2a)

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Proving inequality (2a)

L(S) B(S) ≥

• 2

π n − 1 √n (2a) Given a volume V , which geometric object of that fixed volume has minimum surface area? Answer (Folklore): The euclidean ball. Here: Given a surface area A, which geometric object on the sphere minimizes length of relative boundary? Answer: A spherical cap (L´ evy’s isoperimetric inequality) Which spherical cap is worst? Answer: Half-sphere (Proof next slide) For the half sphere, bound (2a) holds. (Proof skipped)

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Half-sphere is worst case

Consider a spherical cone S (Cone whose base is a spherical cap). Cap of S separated by hyperplane H. Consider half-sphere over S: Σ ≥ B(S). L(S): Relative boundary of S and half sphere identical Hence: L(S)/B(S) ≥ L(S)/Σ B(S)

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Wrapping things up

Have shown inequalities: ∆2n3 · vol(Sv) ≥ D(Sv) (1) D(SI) ≥

• 2n

π · vol(SI) (2) They imply Volume Expansion Lemma

Volume Expansion Lemma

I ⊆ V , with vol(I) ≤ 1

2 vol(Bn)

N(I) neighborhood of I (in polyhedral graph) Then vol(N(I)) ≥

• 2

π 1 ∆2n2.5 · vol(I). Implies that the number of iterations of the BFS is bounded by O(n3.5∆2 log n∆) Implies diameter bound

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Final remarks

Theorem (B.D.E.H.N., 2012)

P = {x ∈ Rn : Ax ≤ b} polytope, all subdeterminants of A bounded by ∆ (in absolute value). Then diameter at most O(n3.5∆2 log n∆). Can be generalized to polyhedra via Lov´ asz-Simonovits inequality. Diameter bound gets O(n4∆2 log n∆). Open Problems: Constructive version? (Simplex pivoting rule...) Polynomial upper bound for ∆(n, m)?

Thank you for your attention!

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