On sub-determinants and the diameter of polyhedra

On sub-determinants and the diameter of polyhedra ∗ N icolas Bonifas † Mar co Di S umma ‡ F riedr ich Eisenbrand § N icolai Hähnle ¶ Martin N iemeier k A ugust 28, 2018 Abstract W e derive a new upper bound on the diameter of a polyhedron P = { x ∈ R n : A x É b }, where A ∈ Z m × n . The bound is polynomi al in n and the largest absolute value of a sub-deter minan t of A , denoted by ∆ . Mor e precise ly , we sho w that the diameter of P is bounded by O ¡ ∆ 2 n 4 log n ∆ ¢ . If P is bounded, then we show that the diameter of P is at most O ¡ ∆ 2 n 3.5 log n ∆ ¢ . F or the special case in which A is a totally unimodular matrix, the bounds are O ¡ n 4 log n ¢ and O ¡ n 3.5 log n ¢ respect ively . This impro ves over the previou s b est bound of O ( m 16 n 3 (log m n ) 3 ) due to Dyer and Frieze [DF94]. 1 I ntroduction One of the fu ndamental open p robl ems in optimization and discrete geometry is the question whether the diameter of a polyhedron can be bounded by a polynomial in the di mension and t he num- ber of its defining inequalities. The problem is readily explained: A polyhedron is a set of the for m P = { x ∈ R n : A x É b }, where A ∈ R m × n is a matr ix and b ∈ R m is an m -dimensi onal vector . A vertex of P is a point x ∗ ∈ P such tha t there exist n linearly independent ro ws of A whose corresponding inequalities of A x É b are satisfied by x ∗ with equality . Throughout this paper , we assume that th e polyhedron P is point ed , i.e . it has vertices, which is equivalent t o saying that the matrix A has full column-rank. T wo different vert ices x ∗ and y ∗ are neig hbors if they are th e en dpoints of an edge of the polyhedron, i.e. there exist n − 1 linearly independent ro ws of A whose corresponding inequalities of A x É b are satisfied with equality both by x ∗ and y ∗ . I n this way , we obtain the undirected polyhedral graph with edges being pairs of neighboring vertices of P . This graph is connected. The diameter of P is the smallest natural number t hat boun ds the length of a shortest path between any pair of vertices in this graph. The question is n o w as follo ws: Can the diameter of a p olyhedron P = { x ∈ R n : A x É b } be bounded by a polynomial in m and n ? The belief in a positive answer to t his question is called the pol ynomial Hirsch conjecture . Despite a lot of researc h ef fort du ring the last 50 years, the gap bet ween lo wer and upper bounds on t he diameter remains huge. While, when the dimension n is fixed, the diameter can be bounded by a ∗ An extended abstract of this paper was presented at the 28-th annual ACM symposium on Computatio nal Geometry (SOCG 12) † LIX, École P o lytechnique, P alaiseau and IBM, Gentilly (France). bonifas@lix.po lytechnique .fr ‡ Dipart imento di Matematica, U n iversità di Pado v a (Italy). disumma@ma th.unipd.it § Ecole P o lytechnique Fédérale de Lausanne (S w itzerland). friedrich. eisenbrand@ epfl.ch ¶ U niv ersity of Bonn (Germany). haehnle@or.uni-bo nn.de k T echnische U n iversität Berlin (Germany). martin.ni emeier@tu-b erlin.de 1 linear function of m [Lar 70 , Bar74], f or the general case the best upper bound, due to Kalai and Kleit- man [KK92], is m 1 + log n . Th e best lo wer bound is of t he form (1 + ε ) · m for some ε > 0 in fixed and sufficiently large dimension n . This is due to a celebrated result of Santos [San12] who dispro ved th e , until then longstanding, or igi nal Hirsch conject ur e for polytopes . The Hirsch conjectur e stated that the diameter of a bounded polyhedron 1 is at most m − n . Inter e stingly , this h uge gap (polynomial versus quasi-polynomial) is also not closed in a ver y simple combinator ial a bstrac tion of polyhe- dral graphs [E HRR10 ]. H ow ever , it was sho wn by V ershynin [V er 09] that ever y polyhedron can be pertu rbed by a small random amount so that the expected diameter of the resulting polyhedron is bounded by a polynomial in n and ln m . See K im an d Santos [KS10] for a recent sur vey . I n light of the importance and apparent difficulty of the open question abo ve, many resear cher s have sho wn that it can be answered in an affirmat ive way in some special cases. N addef [Nad89] pro ved that the Hi rsch conjecture holds tru e for 0/1-polytopes . Orlin [O rl97] pro vided a quadratic upper bound for flo w-polytopes. Brightwell et al. [BvdHS06] sho wed that the diameter of the trans- portation polytope is linear in m and n , an d a similar result holds for th e dual of a transportation polytope [Bal84] and the axial 3-way transpor tation polytope [DLK OS09]. The results on flo w polytopes and classical transportation polytopes conc ern polyhedra defined by totall y u nimodular matrices , i.e., integer matr ices whose sub-deter minants are 0, ± 1. F or such polyhedra Dyer and Frieze [DF94] ha d previousl y sho wn that th e diameter is bounded by a polyno- mial in n and m . Their bound is O ( m 16 n 3 (log m n ) 3 ). Their result is also a lgor ithmic: they sho w that there exists a randomized simplex-algorithm that solves linear progr ams defined by totally unimod- ular matr ices in polynomial time . Our main result is a generalization and considerable impro vement of the diameter bound of D yer and Frieze. W e show that the diameter of a p oly hedron P = { x ∈ R n : A x É b }, with A ∈ Z m × n is bounded by O ¡ ∆ 2 n 4 log n ∆ ¢ . Her e, ∆ denotes the largest absolute value of a sub-determin ant of A . If P is bounded, i.e., a polyt ope , then we can sho w tha t the diameter of P is at most O ¡ ∆ 2 n 3.5 log n ∆ ¢ . T o compare our bound with the one of Dyer and Frieze one has to set ∆ abo ve to one an d obtains O ¡ n 4 log n ¢ and O ¡ n 3.5 log n ¢ respec tively . N otice that our boun d is independent of m , i.e., the num- ber of ro ws of A . The pr oof metho d Let u and v be two ver tices of P . W e estimate the maximum number of iterations of two breadth- first-search explorations of the p oly hedral graph, one initiated at u , the other initiated at v , until a common vertex is disco vered. The diameter of P is at most twice this n umber of iterations . The main idea in the analysis is to reaso n about the normal cones of ver tices of P and to exploit a certa in volume expansion property . W e can assume tha t P = { x ∈ R n : A x É b } is non-degenerate , i.e ., each ver tex ha s exactly n tight inequalities . This can be achieved by slightly pert urbing the right-hand side vector b : in this way the diameter can only gro w . N otice that the polyhedron is then also full-dimensional. W e denote the polyhedral graph of P by G P = ( V , E ). Let v ∈ V no w be a vertex of P . The normal cone C v of v is the set of all vectors c ∈ R n such that v is an optimal solution of the linear program max{ c T x : x ∈ R n , A x É b }. The nor mal cone C v of a ver tex of v is a full-dimensional simpli cial polyhedral cone . T wo vertices v an d v ′ are adjacent if and only if C v and C v ′ share a facet. N o two distinc t nor ma l cones shar e an interior point. Fur thermore , if P is a polytope, then the union of the nor mal cones of vertices of P is the complete space R n . W e now define the volume of a set U ⊆ V of ver tices as the volume of the union of t h e nor mal 1 A counterexample to the same conjecture for un bounded polyhedra was fo und in 1967 by Klee and W alkup [KW67]. 2 cones of U intersected with the unit ball B n = { x ∈ R n : k x k 2 É 1}, i.e., vol( U ) : = vol µ [ v ∈ U C v ∩ B n ¶ . Consider an iteration of breadth-first-search. Let I ⊆ V be the set of ver tices that have been disco vered so far . Br eadth-first-search will nex t disco ver the neighborh ood of I , which we denote by N ( I ). T ogether with the integrality of A , the bound ∆ on the subdeterminants guarantees that the a ngle between one facet of a n or mal cone C v and t he opposite ray is not too small. W e combine this fact, which we for maliz e in L emma 3, with an isoperimetric inequality to show that t he volume of N ( I ) is large relative to the volume of I . Lemma 1. Let P = { x ∈ R n : A x É b } be a polytope wher e all su b-determinants of A ∈ Z m × n are bounded by ∆ in absolu te value and let I ⊆ V be a s et of vertices of G P with vol( I ) É (1/2) · vol( B n ) . T hen the volume of the neighborhood of I is at least vol( N ( I )) Ê r 2 π 1 ∆ 2 n 2.5 · vol( I ). W e pro vide the proof of this lemma in t h e next section. Our diameter bound for polytopes is a n easy consequence: Theor em 2. Let P = { x ∈ R n : A x É b } be a polytope w her e all subdet erminants of A ∈ Z m × n are bounded by ∆ i n absolute value. The diamete r of P is bounded by O ¡ ∆ 2 n 3.5 log n ∆ ¢ . Proof . W e estimate t he maximum number of iterations of breadth-first-search u n til the total volume of the disco vered ver tices exceeds (1/2) · vol( B n ). This is an upper bound on the aforementioned maximum number of iterations of two breadth-first-search explorations unt il a common ver tex is disco vered. S uppose we star t at ver tex v a nd let I j be the ver tices tha t have been disco vered du ring the first j iterations . W e h ave I o = { v }. I f j Ê 1 and vol( I j − 1 ) É (1/2) · vol( B n ) we have by L emma 1 vol( I j ) Ê Ã 1 + r 2 π 1 ∆ 2 n 2.5 ! vol( I j − 1 ) Ê Ã 1 + r 2 π 1 ∆ 2 n 2.5 ! j vol( I 0 ). The condition vol( I j ) É (1/2) · vol ( B n ) implies    1 + 1 q π 2 ∆ 2 n 2.5    j vol( I 0 ) É 2 n . This is equivalent to j · ln    1 + 1 q π 2 ∆ 2 n 2.5    É ln(2 n /vol( I 0 )). F or 0 É x É 1 one has ln(1 + x ) Ê x /2 a nd thus the inequality a bo ve implies j É p 2 π ∆ 2 n 2.5 · ln(2 n /vol( I 0 )). (1) 3 T o finish t he p roo f we need a lo wer bound on vol( I 0 ), i.e ., the n -dimensi onal volume of t he set C v ∩ B n . The normal cone C v contains the f ull-dimens ional simplex spanned by 0 a nd t he n row -vectors a i 1 , . . . , a i n of A that correspond to the inequalities of A x É b th at are tight at v . S ince A is integral, the volume of this simplex is at least 1/ n !. Fur thermore , if this simplex is scaled by 1/ max{ k a i k k : k = 1, . . . , n }, t hen it is contained in the unit ball. Since each component of A is itself a sub-determinan t , one has max{ k a i k k : k = 1, . . . , n } É p n ∆ and thus vol( I 0 ) Ê 1/( n ! · n n /2 ∆ n ). It follo ws that (1) implies j = O ¡ ∆ 2 n 3.5 log n ∆ ¢ . Remar ks. The result of D y er and Frieze [DF94] is also based on analyzing ex p ansion proper ties via isoperimetric inequalities . It is our choice of nor mal cones as the na tural geometr ic repr esentation, and the fact that we only ask f or volume ex pansion instead of expan sio n of the graph itself, that allo ws us to get a bett er bound. Expansion properties of th e graphs of general classes of polytopes have also been studied elsewhere in the literature, e.g. [Kal91, Kai04]. Organiz ation of t he p aper The nex t section is devoted to a proof of the volume-expa n sio n proper ty , i.e., Lemma 1. Th e main tool th a t is used here is a classical isoperi metric inequalit y that states that among measurable subsets of a sphere with fixed volume, spher ical caps have the smallest cir cumference . Section 3 dea ls with unbounded polyhedra. Compar ed to the case of polytopes , the problem t hat a rises here is the fact that the union of the normal cones is not the complete space R n . T o tackle this case , we rely on an isoperimetric inequality of L o vász and Simono vits [LS93]. Finally , we disc uss h o w our bound can be fur t her generalized. I n fact, not all sub-deter minants of A need t o be a t most ∆ but merely th e entr ies of A and the ( n − 1)-dimensional sub-deter minants have to be bounded by ∆ , which yields a slightly stronger result. 2 V olume expansion This section is devoted to a proof of Lemma 1. Throughout this section, we assume tha t P = { x ∈ R n : A x É b } is a polytope. W e begin with some useful notation. A (not necessarily convex) cone is a subset of R n that is closed under the multiplication with non-negative scalars. The intersection of a cone with the un it ball B n is called a spherical cone . Recal l that C v denotes the nor mal cone of the vertex v of P . W e denote the spher ical cone C v ∩ B n by S v and, for a subset U ⊆ V , t he spherical cone S v ∈ U S v by S U . Our goal is to sho w t hat t he follo wing inequality holds for each I ⊆ V with vol( S I ) É 1 2 vol( B n ): vol( S N ( I ) ) Ê r 2 π 1 ∆ 2 n 2.5 · vol( S I ). (2) Rec all that two ver tices are adjacent in G P if and only if th eir nor mal cones have a common fa cet. This mean s that the n eighbors of I ar e those vertices u for which S u has a facet which is p a rt of the surface of the spher ical cone S I . In an iteration of breadth-first-sear ch we thu s augment the set of disco vered ver t ices I by those ver tices u th at can “ dock ” on S I via a common facet. W e call the ( n − 1)- dimensional volume of the surface of a spherical cone S tha t is not on t he spher e, the doc kable surface D ( S ), see Figur e 1. The base of S is the intersection of S with the unit sphere . W e denote th e a r ea of the base b y B ( S ). By ar ea we mean the ( n − 1)-dimensional measure of some sur f ace . Fur thermore , L ( S ) denotes the length of the relative boun dary of the base of S . W e use the ter m len gth to denote the measure of an ( n − 2)-dimensional volume, see Figure 1. 4 (a) Dockable sur face of S . (b) Base of S . (c) Relative boundar y o f the base of S . Figur e 1: Ill ustration of D ( S ), B ( S ) and L ( S ). Given any spherical cone S in the unit ball, the follo wing well-kno wn relations f ollo w f rom basic integration: vol( S ) = B ( S ) n , D ( S ) = L ( S ) n − 1 . (3) T o obtain t he volume expansion relation (2) we need to bound the dockable sur face of a spherical cone from belo w by its volume a nd, for a sim plicial spherical cone, we need a n u pper bound on the dockable surface b y its volume. Mor e prec isely , we sho w th a t for ever y simplicial spher ical cone S v one has D ( S v ) vol( S v ) É ∆ 2 n 3 (4) and for any spherical cone one has D ( S ) vol( S ) Ê r 2 n π . (5) Once inequalities (4) and (5 ) are derived, the bound (2 ) can be obtained as follo ws . All of the dockable surfa ce of S I must be “ consumed ” by the n eighbors of I . Usi ng (5) one has thu s X v ∈ N ( I ) D ( S v ) Ê D ( S I ) Ê r 2 n π · vol( S I ). (6) On th e oth er han d, (4) implies X v ∈ N ( I ) D ( S v ) É ∆ 2 n 3 · X v ∈ N ( I ) vol( S v ) = ∆ 2 n 3 · vol( S N ( I ) ). (7) These last two inequalities imply inequa lity (2 ). The remainder of this section is devoted t o pro ving (4 ) and (5). 2.1 Ar ea to volume ratio of a spherical simplicial cone W e will first der ive inequality (4). Lemma 3. Let v be a ver tex of P . One has D ( S v ) vol( S v ) É ∆ 2 n 3 . 5 y a 1 F r Figur e 2: Pro of of Lemma 3. Proof . Let F be a f acet of a spherical cone S v . Let y be th e ver tex of S v not contained in F . L et Q denote the convex hu ll of F and y (see Figure 2). W e have Q ⊆ S v because S v is convex. M oreo ver , if h F is the Euclidean distance of y from t he hyperplan e containing F , then vol( S v ) Ê vol ( Q ) = area( F ) · h F n . S umming o ver the fa cets of S v , we find D ( S v ) vol( S v ) = X facet F area( F ) vol( S v ) É n · X facet F 1 h F . (8) I t remains t o pro vide a lo wer bound on h F . Let a 1 , . . . , a n be the ro w-vectors of A defining the extreme rays of the normal cone of v , an d let A v be the non-singular matr ix whose ro ws are a 1 , . . . , a n . Further- more , suppose that t he vertex y lies on the ray generated b y a 1 . Let H be the hyperplan e generated by a 2 , . . . , a n . The distance d ( y , H ) of y to H is equal to d ( a 1 , H )/ k a 1 k . Let b 1 , . . . , b n be the columns of the adjugate of A v . The column-vector b 1 is integral and each component of b 1 is bounded b y ∆ . Further mor e b 1 is orthogonal to each of a 2 , . . . , a n . Thus d ( a 1 , H ) is the length of the projection of a 1 to b 1 , which is |〈 a 1 , b 1 〉| / k b 1 k Ê 1/( p n · ∆ ), since a 1 and b 1 are int egral. Thus h F = d ( y , H ) Ê 1 n ∆ 2 . Plugg ing this into (8) completes the proof. 2.2 An isoperimetric inequality for spherical cones W e no w der ive t he lo wer boun d (5) on th e ar ea to volume ratio for a general spherical cone. T o do that, we assume that the spher ical cone has the least fa vorable shape for the a r ea to volume ratio and der ive the inequality for cones of this shape. H ere one uses classical isoperimetric inequalities . The basic isoperimetric inequality states t hat the measurable subset of R n with a prescribed volume and minimal area is the ball of this volume . I n this paper , we need L évy’ s isoper imetric inequality , see e.g. [ F LM77, Theorem 2.1], which can be seen as an analogous result for spheres : it states that a measurable subset of the sphere of prescribed area and minimal bounda ry is a spherical cap. A spherical cone S is a cone of r evolution if there ex ist a vector v and a n angle 0 < θ É π /2 such that S is th e set of vectors in th e u nit ball th at for m an angle of at most θ with v : S = ½ x ∈ B n : v T x k v kk x k Ê cos θ ¾ . 6 B ( S ) K H θ Σ Figur e 3: Pro of of Lemma 5. N ote that a spherical cone is a cone of revol ution if and only if its base is a spher ical cap. W e also observe that two spherical cones of r evolution, defined by two different vectors but b y the same an gle , are always congruent. Th er ef or e, in the follo wing we will only specify th e angle of a cone of revol ution. Lemma 4. The spheric al cone of given volume with minimum lateral surface is a cone of r evolution. Proof . By the fir st equation of (3), ever y spher ical cone of volume V intersects the unit sphere in a surface of area n V . Further mor e, by the second equation of (3), the length of the boundar y of th is surface is proportional to the area of t he lateral surface of t he cone. Then the problem of finding the spherical cone of volume V with the minimum lateral sur face can be rephrased a s follo ws: Find a sur- face of area n V on the unit sphere having the boundar y of minimum length. By L évy’ s isoperimetric inequality for spheres, the optimal shape for such a surface is a spher ical cap . As obser ved above , this corres ponds to a cone of revo lution. Lemma 5. Let S be a s pherical cone of r evolut ion of angle 0 < θ É π /2 . The n D ( S ) vol( S ) Ê r 2 n π . Proof . Us ing (3), we h a ve to show that L ( S ) B ( S ) Ê r 2 π n − 1 p n . (9) This is done in two steps . W e first pro ve th at this ratio is minimal for S being the half-ball, i.e., θ = π /2. Then we sho w that L ( S ) B ( S ) Ê q 2 π n − 1 p n holds for th e h alf-ball. Let H be the hyperplan e containing the boundar y of the base of S . Then H divides S into two part s: a tr uncated cone K and the convex hull of a spher ical cap. The radius r of the base of K is bounded by one. Consider now the h a lf-ball that contains B ( S ) and tha t has H ∩ B n as its flat-surfa ce , see Figure 3, and let Σ denote the area of the corresponding half-sphere . One ha s B ( S ) É Σ a nd thu s L ( S ) B ( S ) Ê L ( S ) Σ . N ow Σ and L ( S ) a r e the surf ace of an ( n − 1)-dimensional half-sphere of radius r and the length of its boun dary r espectively . If w e scale this h a lf-spher e by a factor of 1/ r , we obtain the un it half-ball and its length respectively . Since scaling by a factor of 1/ r increases areas by a factor of 1/ r n − 1 and 7 lengths by a f actor of 1/ r n − 2 , we have tha t L ( S ) Σ is at least t h e length of the u nit-half-ball divided by the area of the base of the half-ball. S uppose now that S is the h a lf-unit-ball. W e sho w that th e inequality L ( S )/ B ( S ) Ê q 2 π n − 1 p n holds . The base of S is a half unit sphere and L ( S ) is the length of the boundar y of a unit ball of dimension n − 1. Thus B ( S ) = n 2 π n /2 Γ ¡ n 2 + 1 ¢ , L ( S ) = ( n − 1) π ( n − 1)/2 Γ ¡ n − 1 2 + 1 ¢ , where Γ is th e well-kno wn Gamma function. U sing the fact that Γ ( x + 1/2)/ Γ ( x ) Ê q x − 1 4 for all x > 1 4 (see , e.g., [M er96]), one easily verifies t h at Γ ³ n 2 + 1 ´ Ê r n 2 · Γ µ n − 1 2 + 1 ¶ . I t follo ws that L ( S ) B ( S ) = 2 p π n − 1 n Γ ¡ n 2 + 1 ¢ Γ ¡ n − 1 2 + 1 ¢ Ê r 2 π · n − 1 p n . Finally we are now ready to consider the case of an arbitrar y spher ical cone. Lemma 6. Let S be a (not nec essarily conve x) s pherical cone with vol( S ) É 1 2 vol( B n ) . Then D ( S ) vol( S ) Ê r 2 n π . Proof . Let S ∗ be a spherical cone of rev olution with the same volume as S . By L emma 4, D ( S ) Ê D ( S ∗ ). N ow , using Lemma 5 one has D ( S ) vol( S ) Ê D ( S ∗ ) vol( S ∗ ) Ê r 2 n π . This was the final step in t h e proof of Lemma 1 and thus we ha ve also pro ved Theorem 2, our main result on polytopes. The next section is devoted to u nbounded polyhedra. 3 The case of an unbounded polyhedron If the polyhedron P is u n bounded, then the union of the nor mal cones of all vertices of P for ms a proper subset K ′ of R n : namely , K ′ is the set of objective functions c for which the linear program max{ c T x : x ∈ P } has finite optimum. S imilarly , the set K ′ ∩ B n is a proper subset of B n . Th en, given the un ion of the spherical cones that have already been disco vered by the breadth-first-search (we denote this set by S ), we should redefine the dockable surf ace of S as that par t of t he lateral sur face of S th a t is shared b y some neighboring cones. In oth er words, we should exclude t h e pa rt lying on the boundar y of K ′ ∩ B n . H o wever , this implies that L emma 6 cannot be immediately a pplied. T o o verc ome this difficulty , we make use of the Lovás z-Si mono vits inequali ty , which we no w recall. Belo w we use notation d ( X , Y ) to indicate the Euclidean distance between two subsets X , Y ⊆ R n , i.e., d ( X , Y ) = inf{ k x − y k : x ∈ X , y ∈ Y }. Also , [ x , y ] denotes the segment connecting two points x , y ∈ R n (see Figur e 4). 8 x y K 1 K 2 K 3 Figur e 4: Illustration of the L o vász-Si mono vits inequality . Theor em 7. [LS93] Let K ⊆ R n be a convex compact set, 0 < ε < 1 and ( K 1 , K 2 , K 3 ) be a parti tion of K into three m easurable sets such that ∀ x , y ∈ K , d ([ x , y ] ∩ K 1 , [ x , y ] ∩ K 2 ) Ê ε · k x − y k . (10) Then vol( K 3 ) Ê 2 ε 1 − ε min ( vol( K 1 ), vol( K 2 ) ) . W e now illustrate ho w the abo ve result can be used in our context. Let K = K ′ ∩ B n and observe that K is a convex and compact set. Let S ⊆ K be the union of the spherical cones that have already been disco vered b y t he breadth-first-search. W e define the dockable surface of S as that par t of the lateral sur face of S that is disjoint from the boundar y of K . W e denote by D ′ ( S ) the area of the dockable surface of S . W e can pro ve the follo wing ana log ue of Le mma 6: Lemma 8. If vol( S ) É 1 2 vol( K ) , then D ′ ( S ) Ê vol( S ) . Proof . Let F denote the dockable surf ace of S (thus D ′ ( S ) is t he area of F ). For ever y ε > 0 we defin e K 3, ε = ( F + ε B n ) ∩ K , K 1, ε = S \ K 3, ε , K 2, ε = K \ ( K 1, ε ∪ K 3, ε ), where X + Y den otes t he Minko wski sum of two subsets X , Y ∈ R n , i.e ., X + Y = { x + y : x ∈ X , y ∈ Y }. Clearly ( K 1, ε , K 2, ε , K 3, ε ) is a part ition of K into thr ee measurable sets . Fur thermore , condition (10) is satisfied. Thus Theorem 7 implies that vol( K 3, ε ) 2 ε Ê 1 1 − ε min ¡ vol( K 1, ε ), vol( K 2, ε ) ¢ . W e observe th at vol( K 2, ε ) Ê vol( K \ S ) − vol( K 3, ε ) Ê vol( S ) − vol( K 3, ε ) Ê vol( K 1, ε ) − vol( K 3, ε ). Combini ng those two inequalities, we find vol( F + ε B n ) 2 ε Ê vol( K 3, ε ) 2 ε Ê 1 1 − ε (vol( K 1, ε ) − vol( K 3, ε )). (11) 9 By a well-kno wn r esult in geometr y (see, e.g., [F ed69],) as ε tends to 0 the left-ha nd side of (11) tends to the area of F , which is precisely the dockable sur face D ′ ( S ). Mo reo ver , as ε t en ds to 0, vol( K 3, ε ) tends to 0 an d vol( K 1, ε ) tends to vol( S ). W e conclude that D ′ ( S ) Ê vol( S ). F ollo wing the same app roach as that used for the case of a polytope, one can sho w the follo wing result for p olyhedra. Theor em 9. Let P = { x ∈ R n : A x É b } be a polyhedron, where a ll sub-det erminants of A ∈ Z m × n are bounded by ∆ i n abs olute value. Th en the diameter of P is bounded by O ¡ ∆ 2 n 4 log n ∆ ¢ . In part icular , if A is totally unimodular , then the diameter of P is bounded by O ( n 4 log n ) . 4 R emarks 4.1 Which sub-determinants enter the bound? F or simplicity , we have a ssumed that a bound ∆ was given for the absolute value of all sub-det er minants of A . Ho wever , our proof only uses the fact the t he sub-deter minants of size 1 (i.e., the entr ies of the matrix) and n − 1 are bounded. Call ing ∆ 1 (res p . ∆ n − 1 ) the bound on the absolute value of the ent ries of A (resp . on the sub-deter minants of A of size n − 1), one easily verifies that all t he results discussed abo ve remain essentially unchanged, except that the statement of Lemma 3 becomes D ( S v ) vol( S v ) É ∆ 1 ∆ n − 1 n 3 and the lo wer bound on vol( I 0 ) becomes vol( I 0 ) Ê 1 n ! n n /2 ∆ n 1 . This implies the follo wing strengthened result: Theor em 10. Let P = { x ∈ R n : A x É b } be a polyhedron, w her e the entri es of A (respectively t he su b- determinant s of A of size n − 1 ) are boun ded in absol ute value by ∆ 1 (respectively ∆ n − 1 ). Then th e diameter of P is bounded by O ¡ ∆ 1 ∆ n − 1 n 4 log n ∆ 1 ¢ . M oreo ver , if P i s a p olytope, its diameter is bounded by O ¡ ∆ 1 ∆ n − 1 n 3.5 log n ∆ 1 ¢ . 4.2 A mor e ge neral geometric setti ng S ince our result was first announced in [BDSE + 12], Brunsch and Röglin [BR13] have f ound an algo- rithm to compute a shor t path between two given vertices of a non-degenerate polyhedron P = { x ∈ R n : A x É b } that ru ns in expected polynomial time in n , m an d 1/ δ , where δ is a lo wer bound on th e sine of the angle of a ro w of A to t he subspace of n − 1 other ro ws of A . The expected length of the path is O ( m n 2 / δ 2 ). If A ∈ Z m × n , then δ Ê 1/( ∆ 1 ∆ n − 1 n ), where ∆ 1 and ∆ n − 1 are , a s bef or e, bounds on the absolute values of 1 × 1 and ( n − 1) × ( n − 1) sub-det er minants. Our proof technique ap p lies in this setting as well. W e have volume expansion since the normal cones cannot be too flat. The parameter δ is a measur e for this flatness. In this setting, Lemma 1 reads as follo ws. Lemma 11. Let P = { x ∈ R n : A x É b } be a polytope and l et I ⊆ V be a set of verti ces w ith vol( I ) É (1/2) · vol( B n ) . Then the volume of the neighborhood of I is at least vol( N ( I )) Ê r 2 π ¡ δ / n 1.5 ¢ · vol( I ). 10 The proof is a long the lines of the proof of Lemma 1 and by ada p ting L emma 3 . Her e one has no w D ( S v )/vol( S v ) É n 2 / δ . Theorem 2 is in the geometric setting now becomes the follo wing. Theor em 12. Let P = { x ∈ R n : A x É b } be a polytope where the si ne of the angle of any ro w of A to t he subspace generated by n − 1 o ther ro ws of A is at least δ . The di ameter of P i s bounded by O ¡ n 2.5 / δ · ln ( n / δ ) ¢ . Again, t h e proof is along the lines of the proof of Theorem 2 where the volume of S v is no w lo wer bounded by δ n − 1 / n !. I n fact, the diameter bound O ¡ n 2.5 / δ · ln ( n / δ ) ¢ holds already for non-degenerate polytopes where each S v contains a ball of radius δ . 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