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 1 / 16
Diameter of Polyhedra We consider polyhedra defined by m inequalities in dimension n , i.e. P = { x ∈ R n : Ax ≤ b } , A ∈ R m × 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? 2 / 16
Hirsch conjecture Hirsch conjecture (1957) ∆( n , m ) ≤ m − n Recently disproved (Santos 2010) Best known bound: ∆( n , m ) ≤ m 1+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 ∈ R n : Ax ≤ b } with A totally unimodular: upper bound O ( m 16 n 3 log( mn ) 3 ) (Dyer and Frieze 1994) 3 / 16
Our result Our bound P = { x ∈ R n : Ax ≤ b } polytope with A totally unimodular . Then diameter at most O ( n 3 . 5 log n ) (Previous bound: O ( m 16 n 3 log( mn ) 3 ) (Dyer and Frieze 1994)) More general: Our bound P = { x ∈ R n : Ax ≤ b } polytope, all subdeterminants of A bounded by ∆ (in absolute value). Then diameter at most O ( n 3 . 5 ∆ 2 log n ∆) 4 / 16
A more geometric view Consider polytope P P nondegenerate, i.e. each vertex v ∈ P has exactly n tight facets. Normal cone C v : Cone defined by normal vectors of these facets. Normal fan: Set of all normal cones Normal fan partitions R n . Crucial observation Two vertices in the polyhedral graph are adjacent iff their normal cones share a facet. 5 / 16
A more geometric view (contd.) Want to assign volume to each vertex/normal cone Intersect normal fan with euclidean ball B n Spherical (simplicial) cones : S v := C v ∩ B n . Assign volume: vol( v ) := vol( S v ) 6 / 16
Proof method Perform BFS from any vertex v ∈ P . Show: After O ( n 3 . 5 ∆ 2 log n ∆) iterations, vertices seen during BFS have combined volume ≥ 1 2 vol( B n ). 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( B n ) N ( I ) neighborhood of I (in polyhedral graph) We have � 2 1 vol( N ( I )) ≥ ∆ 2 n 2 . 5 · vol( I ) . π Implies iteration bound with standard methods: � k �� 2 1 After k iterations of BFS: vol ≥ vol( S v ). ∆ 2 n 2 . 5 π 7 / 16
How to prove the Volume Expansion Lemma Volume Expansion Lemma I ⊆ V , with vol( I ) ≤ 1 2 vol( B n ). Then � 2 1 vol( N ( I )) ≥ ∆ 2 n 2 . 5 · vol( I ) . π S I := � v ∈ I S v v ∈ N ( I ) iff surfaces of S v and S I touch (( n − 1)-dim intersection). Dockable surface D ( S I ) : Surface of S I disjoint from surface of sphere. D ( S N ( I ) ) ≥ D ( S I ) Sufficient to show: � 2 n ∆ 2 n 3 · vol( S v ) ≥ D ( S v ) D ( S I ) ≥ π · vol( S I ) (1) (2) 8 / 16
How to prove the Volume Expansion Lemma Volume Expansion Lemma I ⊆ V , with vol( I ) ≤ 1 2 vol( B n ). Then � 2 1 vol( N ( I )) ≥ ∆ 2 n 2 . 5 · vol( I ) . π Dockable surface D ( S I ) : Surface of S I disjoint from surface of sphere. D ( S N ( I ) ) ≥ D ( S I ) Sufficient to show: � 2 n ∆ 2 n 3 · vol( S v ) ≥ D ( S v ) D ( S I ) ≥ π · vol( S I ) (1) (2) (1) implies ∆ 2 n 3 · vol( S N ( I ) ) ≥ D ( S N ( I ) ) 9 / 16
Proving inequality (1) ∆ 2 n 3 · vol( S v ) ≥ D ( S v ) (1) F facet of S v y y vertex not contained in F Q := conv( F , y ) ⊆ S v F is a simplex. h F = d ( y , H ( F )) vol( Q ) = area( F ) · h F n Summing over facets n n � � � · vol( Q ) ≤ · vol( S v ) D ( S v ) = area( F ) = h F h F facet F facet F facet F 10 / 16
Proving inequality (1): Bounding h F a 1 y a 1 b 1 y β F h F 1 β F A ′ = ( a 1 , . . . , a n ), B = adj( A ′ ) = det( A ′ ) · A ′− T � a 1 , b 1 � 1 h F = cos β = � a 1 �·� b 1 � ≤ n ∆ 2 n Recall: D ( S v ) ≤ � h F · vol( S v ) facet F We get: ∆ 2 n 3 · vol( S v ) ≥ D ( S v ) (1) 11 / 16
Proving inequality (2) � 2 n D ( S I ) ≥ π · vol( S I ) (2) (a) Dock- (b) Base of S . (c) Relative able surface Area: B ( S ) boundary of of S . Area: the base of D ( S ) S . Length: L ( S ) D ( S ) = L ( S ) / ( n − 1) Basic integration: vol ( S ) = B ( S ) / n , Inequality (2) equivalent to: � n − 1 L ( S ) 2 √ n . B ( S ) ≥ (2 a ) π 12 / 16
Proving inequality (2a) � L ( S ) 2 n − 1 B ( S ) ≥ √ n (2 a ) π 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) 13 / 16
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 ) 14 / 16
Wrapping things up Have shown inequalities: � 2 n ∆ 2 n 3 · vol( S v ) ≥ D ( S v ) D ( S I ) ≥ π · vol( S I ) (2) (1) They imply Volume Expansion Lemma Volume Expansion Lemma I ⊆ V , with vol( I ) ≤ 1 2 vol( B n ) N ( I ) neighborhood of I (in polyhedral graph) � 2 1 Then vol( N ( I )) ≥ ∆ 2 n 2 . 5 · vol( I ) . π Implies that the number of iterations of the BFS is bounded by O ( n 3 . 5 ∆ 2 log n ∆) Implies diameter bound 15 / 16
Final remarks Theorem (B.D.E.H.N., 2012) P = { x ∈ R n : Ax ≤ b } polytope, all subdeterminants of A bounded by ∆ (in absolute value). Then diameter at most O ( n 3 . 5 ∆ 2 log n ∆) . Can be generalized to polyhedra via Lov´ asz-Simonovits inequality. Diameter bound gets O ( n 4 ∆ 2 log n ∆). Open Problems: Constructive version? (Simplex pivoting rule...) Polynomial upper bound for ∆( n , m )? Thank you for your attention! 16 / 16
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