Stabbing Convex Bodies with Lines and Flats
SStabbing Convex Bodies with Lines and Flats
Sariel Har-Peled ∗ Mitchell Jones † July 22, 2020
Abstract
We study the problem of constructing weak ε -nets where the stabbing elements are linesor k -flats instead of points. We study this problem in the simplest setting where it is stillinteresting—namely, the uniform measure of volume over the hypercube [0 , d . We show that inthis setting one can construct ( k, ε )-nets of size m = (cid:101) O (1 /ε − k/d ), where (cid:101) O hides polylogarithmicfactors. That is, there is a set of m k -flats, such that any convex body in [0 , d of volume largerthan ε is stabbed by one of these k -flats. As a concrete example, in three dimensions all ε -heavybodies in [0 , can be stabbed by (cid:101) O (1 /ε / ) lines. Note, that these bounds are sublinear in 1 /ε ,and are thus somewhat surprising.We also present deterministic constructions of weak ε -nets, under the same volume measureof [0 , d . While our bounds are still weaker than known bounds, the new constructions aresimple and explicit, unlike previous constructions.
1. Introduction
Range spaces and ε -nets. A range space is a pair X = ( U , R ), where U is the ground set (finite or infinite) and R is a (finite or infinite) family of subsets of U . The elements of R are ranges .Suppose that U is a finite set. For a parameter ε ∈ (0 , S ⊆ U is an ε -net for X if every range r ∈ R with | r ∩ U | ≥ ε |U | has r ∩ S (cid:54) = ∅ . The ε -net theorem of Haussler and Welzl[HW87] implies the existence of ε -nets of size O ( δε − log ε − ), where δ is the VC dimension of therange space X . The use of ε -nets is widespread in computational geometry [Mat02, Har11]. Weak ε -nets. Consider the range space ( P, C ), where C is the collection of all compact convexbodies in R d and P ⊂ R d is a point set of size n . This range space has infinite VC dimension—thestandard ε -net constructions do not work for this range space. The notion of weak ε -nets bypassesthis issue by allowing the net S to use points outside of P . Specifically, any convex body Ξ thatcontains at least εn points of P must contain a point of S . Matouˇsek and Wagner [MW04] constructweak ε -nets of size O ( ε − d log f ( d ) ε − ), where f ( d ) = O ( d log d ). Recently, Rubin [Rub18] gave an ∗ Department of Computer Science; University of Illinois; 201 N. Goodwin Avenue; Urbana, IL, 61801, USA; [email protected] ; http://sarielhp.org/. Work on this paper was partially supported by NSF AF awardCCF-1907400. † Department of Computer Science; University of Illinois; 201 N. Goodwin Avenue; Urbana, IL, 61801, USA; [email protected] ; http://mfjones2.web.engr.illinois.edu/. Work on this paper was partially supportedby NSF AF award CCF-1907400. a r X i v : . [ c s . C G ] J u l mproved bound for points in the plane, showing existence of weak ε -nets of size O ( ε − (3 / α ) ) forarbitrarily small α >
0. As for a lower bound, Bukh et al. [BMN09] gave constructions of point setsfor which any weak ε -net must have size Ω( ε − log d − ε − ). Closing this gap remains a major openproblem.( k, ε ) -nets and uniform measure. A natural extension of weak ε -nets is to allow the net S tocontain other geometric objects. Given a collection of n points P ⊂ R d and a parameter 0 ≤ k < d ,we define a (weak) ( k, ε )-net to be a collection of k -flats S such that if Ξ is a convex body containingat least εn points of P , then there exists a k -flat in S intersecting Ξ. Note that (0 , ε )-nets are exactlyweak ε -nets.In general, one would expect that as k increases, the size of the ( k, ε )-net shrinks. For example,a (1 , ε )-net for a collection of points in R can be constructed by projecting the points down ontothe xy -plane and applying Rubin’s construction in the plane to obtain a weak ε -net S of size O ( ε − (3 / α ) ) [Rub18]. Lifting S up back into three dimensions results in a (1 , ε )-net of the same size,which is smaller than the best known weak ε -net size in R [MW04]. However, one might expectthat a (1 , ε )-net of even smaller size is possible in R , as this construction uses a set of parallel lines(i.e., one would expect the lines in an optimal net to be arbitrarily oriented).Here, we study an even simpler version of the problem, where the ground set is the hypercube B = [0 , d . In particular, for ε ∈ (0 ,
1) and 0 ≤ k < d , we are interested in computing the smallestset K of k -flats, such that if Ξ is a convex body with vol (Ξ ∩ B ) ≥ ε , then there is a k -flat in K which intersects Ξ. For sake of exposition, throughout the rest of the paper we refer to this set K as a ( k, ε ) -net . We note that [0 , d can be replaced with any arbitrary compact convex body inthe definition (the size of the ( k, ε )-net increases by a factor depending on d , see Appendix B). Notation.
Throughout, we use O d , Ω d , and Θ d to hide constants depending on the dimension d .In this paper we develop constructions for ( k, ε )-nets.(A) In Section 2 we show that any ( k, ε )-net must have size Ω d (1 /ε − k/d ) (Lemma 2.2), and showthat there exist ( k, ε )-nets of size O d ((1 /ε − k/d ) log(1 /ε )) (Theorem 2.7). The construction israndomized.One key ingredient in the proof of Theorem 2.7 is showing that a ( k, ε )-net can be constructedfrom a family of (0 , ε )-nets. Hence, any improvement in the size of a (0 , ε )-net (or weak ε -netin this specific setting) immediately implies a smaller ( k, ε )-net for our problem.(B) To make the construction of Theorem 2.7 deterministic and explicit, we develop a deterministicconstruction of (0 , ε )-nets. In particular, we show that it suffices to restrict the problem towhen the convex bodies are always ellipsoids. Thus, we show the existence of a collection ofpoints P ⊂ [0 , d , of size O d ((1 /ε ) log d − (1 /ε )), such that any ellipsoid E ⊆ [0 , d with volumeat least ε contains a point of P . See Section 3. This implies a deterministic construction of( k, ε )-nets of size O d ((1 /ε − k/d ) log d − k − (1 /ε )) (Lemma 3.4).While deterministic constructions of ε -nets are known, and can be applied in this case, theyare not explicit—they follow by derandomizing discrepancy constructions for point sets, andbuilding ε -nets for them. As such, while our construction is asymptotically inferior, it is both2imple and explicit. In particular, it can be generated on the fly and by demand, unlike otherdeterministic constructions (see the discussion below).(C) For specific values of k , one would expect that the bound given by Lemma 3.4 is notoptimal. Indeed, in Section 4, we show deterministic constructions of (1 , ε )-nets of size O ((1 /ε − /d ) log γ ( d ) (1 /ε )), where γ ( d ) = d − − (cid:80) d − i =1 /i . Observe that this improves overthe bound given in Lemma 3.4 by a logarithmic factor for d ≥ , ε )-nets for axis parallelboxes in R d , and is briefly mentioned in [BMN09]. In this case, one can construct (0 , ε )-nets ofsize O d (1 /ε ) using Van der Corput sets in two dimensions, and Halton-Hammersely sets in higherdimensions. For completeness, we describe these construction in Appendix A. Deterministic vs. explicit constructions of ε -nets. For the regular concept of ε -nets, thereare known deterministic constructions. They work by repeatedly halving the input point set, usingdeterministic discrepancy constructions, until the set is of the desired size [Mat99, Cha01]. On theone hand, for our setting (i.e., the measure is uniform area on the unit hypercube) it is not clearwhat the generated ε -net is without running this construction algorithm outright. On the otherhand, we develop a construction of weak ε -nets—for uniform area measure over the hypercube forellipsoids—which are much simpler and are explicit (importantly, you can easily compute the i thpoint in this net using polylogarithmic space).
2. Constructing ( k, ε ) -nets Definition 2.1.
For parameters ε ∈ (0 , k ∈ { , , . . . , d − } , a set K of k -flats is a ( k, ε ) -net ,if for any convex body Ξ ⊆ R d , with vol (cid:0) Ξ ∩ [0 , d (cid:1) ≥ ε , then there exists a flat ϕ ∈ K such that ϕ ∩ Ξ (cid:54) = ∅ . Lemma 2.2.
For a parameter ε ∈ (0 , , any ( k, ε ) -net must have size Ω d (1 /ε − k/d ) .Proof: Let K be a ( k, ε )-net. For each k -flat ϕ ∈ K , let H ( ϕ, r ) be the locus of points in [0 , d within distance at most r from ϕ (for k = 1 in three dimensions, this is the intersection of [0 , d and the cylinder with radius r centered at the line ϕ ). Note that a ball b with center c and radius ρ intersects a k -flat ϕ if and only if c ∈ H ( ϕ, r ) and ρ ≥ r .Fix r = ( ε/µ ) /d , where µ is a constant to be determined shortly. We claim that by choosing µ appropriately, if K is a ( k, ε )-net, then the collection of objects { H ( ϕ, r ) | ϕ ∈ K } covers [0 , d .Indeed, suppose not. Then there exists a point p ∈ [0 , d not covered by any of the objects H ( ϕ, r ).This implies that a ball b centered at p with radius r does not intersect any k -flat of K , and itsvolume is c d r d = c d ε/µ , where c d is a constant that depends on d . Choose µ = c d so that b hasvolume at least ε , but does intersect any k -flat of K . A contradiction to the required net property.Hence, by the choice of r , any ( k, ε )-net must satisfy the condition that { H ( ϕ, r ) | ϕ ∈ K } covers[0 , d . For any k -flat ϕ , we have β = vol ( H ( ϕ, r )) = O d ( r d − k ) = O d ( ε − k/d ). Thus, to cover [0 , d ,we have that | K | ≥ /β = Ω d (1 /ε − k/d ). 3 .2. Upper bounds Let e , . . . , e d be the d vectors forming the standard basis for R d . Specifically, e i is zero in allcoordinates except the i th coordinate, where it is one. For a compact convex body Ξ and a k -flat ϕ ,let Ξ ↓ ϕ denote the orthogonal projection of Ξ onto ϕ . Before describing the construction, we needthe following facts. Lemma 2.3.
Let E be a d -dimensional ellipsoid centered at the origin with α = vol ( E ) . For i = 1 , . . . , d , let h i be the hyperplane passing through the origin and orthogonal to the i th axis e i .Then there exists an i such that vol ( E ↓ h i ) ≥ α − /d /d. Proof:
For i = 1 , . . . , d , let s i be the segment of maximum length contained in E which is parallelto e i . It is easy to verify that s i must pass through the center of E (the origin) because of thesymmetry of E . Let λ i = | s i | and relabel the axes such that λ ≥ . . . ≥ λ d . Let β = (cid:0) ( d − (cid:1) − /d ≥ (cid:18) e ( d − d − / e − ( d − (cid:19) /d = (cid:18) √ d − e d − ( d − d (cid:19) /d = ed − (cid:18) √ d − e (cid:19) /d ≥ d , since √ πn n +1 / e − n ≤ n ! ≤ n n +1 / e − n (this is a refinement of Stirling’s formula [Rob55]).Suppose that λ d ≤ α /d /β . Define f : R d − → R + , for a point p ∈ R d − , to be the length of thelongest segment contained in E which passes through p and is parallel to e d . We have that α = vol ( E ) = (cid:90) p ∈E ↓ hd f ( p ) d p ≤ λ d (cid:90) p ∈E ↓ hd p = λ d vol ( E ↓ h d ) ⇐⇒ vol ( E ↓ h d ) ≥ αλ d ≥ βα − /d , as claimed.Otherwise, λ ≥ · · · ≥ λ d > α /d /β . In which case, vol ( E ↓ h i ) ≥ α − /d /d , for all i . Indeed, thevolume of E ↓ h is at least the volume of the scaled cross-polytope conv (cid:0) ∪ di =2 s i (cid:1) : vol (cid:16) conv (cid:16) ∪ di =2 s i (cid:17)(cid:17) ≥ | s | · · · | s d | ( d − λ · · · λ d ( d − > α ( d − /d β d − ( d − β d α ( d − /d β d − = βα − /d , since β = (cid:0) ( d − (cid:1) − /d . Lemma 2.4.
Let E be a d -dimensional ellipsoid centered at the origin with α = vol ( E ) . For any k > , there exists a set of d − k coordinates such that the ( d − k ) -dimensional linear subspace ϕ that spans these coordinates has vol ( E ↓ ϕ ) ≥ ( d − k )! d ! α ( d − k ) /d . Proof:
We use Lemma 2.3, k times, to find favorable projections. Formally, let E i be the d − i dimensional ellipsoid computed by the end of the i th iteration, that lives in a d − i flat ϕ d − i whichis spanned by d − i of the axes. Initially, E = E .In the i th iteration, by Lemma 2.3 there is a projection E i − down one dimension, onto one ofthe d − i coordinates to obtain E i , such that vol ( E i ) ≥ vol ( E i − ) γ ( i ) / ( d − i + 1) , γ ( i ) = ( d − i ) / ( d − i + 1) <
1. Applying this formula repeatedly, we have that vol ( E i ) ≥ d − i + 1 (cid:32) vol ( E i − ) γ ( i − d − i + 2 (cid:33) γ ( i ) ≥ vol ( E i − ) γ ( i − γ ( i ) ( d − i + 2)( d − i + 1) ≥ ( d − i )! d ! vol ( E ) (cid:81) ij =1 γ ( j ) . Observe that (cid:81) ij =1 γ ( j ) = d − d · d − d − · · · d − id − i +1 = ( d − i ) /d . We conclude that vol ( E i ) ≥ ( d − i )! d ! vol ( E ) ( d − i ) /d ,which establishes the claim. The key step is to show that a ( k, ε )-net can be constructed from a family of (0 , ε )-nets. This followsfrom the facts proved above. As mentioned in the introduction, any improvement in the size of a(0 , ε )-net immediately implies a smaller ( k, ε )-net.
Lemma 2.5.
Suppose there exists an ε -net (i.e., (0 , ε ) -net) for the volume measure over [0 , d for ellipsoids of size T ( ε, τ ) , for τ = 1 , . . . , d . Then one can construct a ( k, ε ) -net for the volumemeasure over [0 , d for ellipsoids of size O (cid:0) ( d/k ) k T ( cε − k/d , d − k ) (cid:1) , where c = ( d − k )! d ! d d − k .Proof: Let δ = cε ( d − k ) /d . For every subset S ∈ (cid:0) (cid:74) d (cid:75) d − k (cid:1) of d − k coordinates of R d , consider the linearsubspace R S they span, and construct a δ -net N S for the unit hypercube of R S . Next, orthogonallylift each point of p ∈ N S into the k -flat { p } × R (cid:74) d (cid:75) \ S . Let N be the union of all these k -flats. Thesize of the resulting net N is (cid:0) dk (cid:1) T ( δ, d − k ) . Consider any convex body Ξ, such that vol (cid:0) Ξ ∩ [0 , d (cid:1) ≥ ε . Let E be the ellipsoid of largestvolume contained inside Ξ ∩ [0 , d . By John’s ellipsoid theorem, we have that E ⊆ Ξ ⊆ d E . Inparticular, vol ( E ) = vol ( d E ) /d d ≥ vol (Ξ) d d ≥ εd d . For E , Lemma 2.4 implies the existence of a set S ∈ (cid:0) (cid:74) d (cid:75) d − k (cid:1) such that the orthogonal projectiononto the ( d − k )-dimensional subspace of S results in an ellipsoid F with vol ( F ) ≥ ( d − k )! d ! vol ( E ) − k/d ≥ ( d − k )! d ! (cid:16) εd d (cid:17) − k/d = ( d − k )! d ! d d − k ε − k/d = δ. In particular, the constructed net N S stabs F . That is, there is a point p ∈ N S ∩ F . But then the k -flat { p } × R (cid:74) d (cid:75) \ S that is in N intersects E , and thus the original body Ξ. Lemma 2.6.
For ε ∈ (0 , , a random sample of size T ( ε, d ) = O (( d /ε ) log ε − ) from [0 , d , is an ε -net for the volume measure over [0 , d for ellipsoids with probability ≥ − ε O ( d ) .Proof: Akama and Irie [AI11] showed that the VC dimension of ellipsoids in d dimensions is D = ( d + 3 d ) /
2. By the ε -net theorem a sample of size Dε log ε is an ε -net with probability ≥ − ε/ D , which establishes the claim. 5igure 3.1: The net constructed. Theorem 2.7.
There is a randomized construction of ( k, ε ) -nets for [0 , d of size O (cid:18) d d + k +3 ε − k/d log dε (cid:19) . The construction succeeds with probability ≥ − ε O ( d ) .Proof: Follows by plugging in the bound for T ( ε, d ) in Lemma 2.6 into Lemma 2.5. Specifically, let c = ( d − k )! d ! d d − k ≥ /d d , and the desired net has size O (cid:0) ( d/k ) k T ( cε − k/d , d − k ) (cid:1) = O (cid:18) ( d/k ) k ( d − k ) cε − k/d log 1 cε (cid:19) = O (cid:18) d d + k +3 ε − k/d log dε (cid:19) . An easy calculation shows that the probability that any of these samples fails is smaller than ε O ( d ) ,which implies that the constructed set is the desired net with probability close to one.
3. Stabbing ellipsoids with points
By Lemma 2.5, the problem of computing ( k, ε )-nets reduces to constructing (0 , ε )-nets when theconvex bodies are restricted to be ellipsoids. Here, we give a deterministic explicit construction ofsuch a (0 , ε )-net. This in turn will imply a deterministic construction of a ( k, ε )-net.
Let E be an ellipse contained in the unit square [0 , with area ( E ) ≥ ε . The following constructionis inspired by a construction of Pach and Tardos [PT13]. Construction.
Let M = 3 + (cid:6) lg ε − (cid:7) . For j = 1 , . . . , M −
1, consider the rectangle R j = [0 , / M − j ] × [0 , / j ] . Consider the natural tiling of [0 , by the rectangle R i , and let P i be the set of vertices of theresulting grid G i in the interior of the unit square. Let N = ∪ i P i . See Figure 3.1.6 (cid:104) E Yy + y / y / y − (cid:104) / (cid:104) / Z (cid:96) k τ k (cid:104) E Figure 3.2: The setup for proof of correctness.
Correctness.
We need the following easy observation, whose proof is included for the sake ofcompleteness.
Claim 3.1.
Let c be the center of an ellipse E , and let (cid:104) be the longest horizontal segment containedin E . The segment (cid:104) passes through c .Proof: By the central symmetry of E , if (cid:104) does not pass through c , then it has a symmetric reflection (cid:104) (cid:48) through c , which is a horizontal segment of the same length. Let (cid:96) be the horizontal line through c , and observe that | (cid:96) ∩ E| ≥ | (cid:104) | by convexity. By the smoothness of E , it follows that | (cid:96) ∩ E| > | (cid:104) | ,which is a contradiction. Lemma 3.2.
The set N constructed above is an ε -net for the volume measure over [0 , forellipses. Furthermore, | N | = O ( ε − log ε − ) .Proof: Observe that for any i , we have area ( R i ) = 2 − ( M − j ) − j = 2 − M ≥ ε/
8. As such, | P i | = O (1 /ε ),and | N | = O ( M/ε ) = O ( ε − log ε − ).Let E ⊆ [0 , be any ellipse with area ( E ) ≥ ε . Let Y denote the projection of E onto the y -axis. Observe that | Y | ≥ ε . Let (cid:104) be the longest horizontal segment contained in E (which passesthrough the center of E by Claim 3.1). The two extreme y -axis points in E , and the segment (cid:104) forms a quadrilateral in E of area | (cid:104) | | Y | /
2, see Figure 3.2. Let Y = [ y − , y + ], and for α ∈ Y , let g ( α ) = | ( y = α ) ∩ E| . We have that | (cid:104) | | Y | / ≤ area ( E ) = (cid:90) y + α = y − g ( α )d α ≤ | (cid:104) | | Y | . Since area ( E ) ≥ ε , we conclude that | (cid:104) | ≥ ε/ | Y | .We set y / = (3 / y − + (1 / y + and y / = (1 / y − + (3 / y + . Consider the two horizontalsegments (cid:104) / = (cid:8) y = y / (cid:9) ∩ E and (cid:104) / = (cid:8) y = y / (cid:9) ∩ E . These two segments are of the samelength and are parallel. Furthermore, γ = (cid:12)(cid:12) (cid:104) / (cid:12)(cid:12) = (cid:12)(cid:12) (cid:104) / (cid:12)(cid:12) ≥ | (cid:104) | /
2, see Figure 3.2. Consider theparallelogram Z formed by the convex hull of (cid:104) / and (cid:104) / . Observe, that for any α ∈ [ y / , y / ],we have that |{ y = α } ∩ Z | = γ . As such, area ( Z ) = | Y | / · | (cid:104) | ≥ ε/
2. Let k be the minimuminteger such that 1 / k +1 ≤ | Y | /
2. Since | Y | ≥ ε , it follows that k < M − G k +1 has a horizontal line (cid:96) k that intersects Z . Furthermore, we have | (cid:96) k ∩ E| ≥ | (cid:96) k ∩ Z | = | (cid:104) | ≥ ε | Y | ≥ ε k − ≥ · k − M = 12 M − k +1 − > M − ( k +1) = β, since M = 3 + (cid:6) lg ε − (cid:7) . Namely, the spacing of the points of G k +1 on the line (cid:96) k (i.e., β ) is shorterthen the interval (cid:96) k ∩ E . It follows that a point of P k +1 ⊆ N lies in E , and thus establishing theclaim. We now extend the previous construction to higher dimensions. The construction is recursive.Namely, we assume that for all d (cid:48) < d , we can construct an ε -net for the volume measure over [0 , d (cid:48) for ellipsoids, of size ( β ( d (cid:48) ) /ε ) lg d (cid:48) − (1 /ε ), where β ( d (cid:48) ) is a constant depending on the dimension d (cid:48) (to be determined shortly). Lemma 3.2 proves the claim when d = 2. Construction.
Label the d axes x , . . . , x d . Let τ = (cid:100) (1 /d ) lg(1 /ε ) (cid:101) (where lg = log ) and definethe function ∆( i ) = 2 i ε /d . We repeat the following construction for each axis x (cid:96) , where (cid:96) = 1 , . . . , d .For each i = 0 , . . . , τ , let M i = (cid:100) lg(1 / ∆( i )) (cid:101) . For each i , and for each j = 0 , . . . , M i , form 2 j + 1evenly spaced hyperplanes which are orthogonal to the axis x (cid:96) (thus consecutive hyperplanes areseparated by distance 2 − j ). For each hyperplane h , we recursively construct a (0 , ε/ ∆( i + 2))-net P (cid:96),i,j for [0 , d − on h ∩ [0 , d . Let P (cid:96) = ∪ τi =1 ∪ M i j =1 P (cid:96),i,j . Finally, we claim the point set P = ∪ d(cid:96) =1 P (cid:96) is the desired (0 , ε )-net. Theorem 3.3.
For ε ∈ (0 , − d ] , there exists a ε -net (i.e., (0 , ε ) -net) for the volume measure over [0 , d for ellipsoids, of size O ( d ) ε − lg d − ε − .Proof: We first bound the size of the resulting net. Since ε ≤ − d , By a direct calculation, | P | ≤ d (cid:88) (cid:96) =1 | P (cid:96) | ≤ d τ (cid:88) i =0 M i (cid:88) j =0 (2 j + 1) · β ( d − · (cid:18) ∆( i + 2) ε lg d − (cid:18) ∆( i + 2) ε (cid:19)(cid:19) ≤ d · β ( d − ε τ (cid:88) i =0 M i +1 · ∆( i ) lg d − (cid:18) ∆( i + 2) ε (cid:19) ≤ d · β ( d − ε τ (cid:88) i =0 lg d − (cid:18) i +2 ε − /d (cid:19) ≤ d · β ( d − ε τ (cid:88) i =0 (cid:18) ( i + 2) + lg (cid:18) ε − /d (cid:19)(cid:19) d − ≤ d · β ( d − ε (cid:20) ( τ + 1) · d − lg d − (cid:18) ε (cid:19)(cid:21) (since i + 2 ≤ τ + 2 ≤ lg(1 /ε ) for ε ≤ − d ) ≤ d · β ( d − ε (cid:20) d lg (cid:18) ε (cid:19) · d − lg d − (cid:18) ε (cid:19)(cid:21) = 2 d +5 · β ( d − ε lg d − (cid:18) ε (cid:19) . In particular, we obtain the recurrence β ( d ) = 2 d +5 β ( d − β ( d ) = 2 O ( d ) . Hence, | P | = 2 O ( d ) ε − lg d − ε − . 8e now argue correctness. Let E be an ellipsoid of volume at least ε . Let B be the smallestenclosing axis-aligned box for E . Suppose that the longest edge of B is along the (cid:96) th axis. Inparticular, along this (cid:96) th axis B has side length s ≥ ε /d , for otherwise vol ( E ) ≤ vol ( B ) ≤ s d < ε .We claim that E intersects a point in the set P (cid:96) .Let L = [ (cid:96) − , (cid:96) + ] be the projection of E onto the (cid:96) th axis, with s = | L | . For x ∈ L , define H ( x )to be the hyperplane orthogonal to the (cid:96) th axis which intersects the (cid:96) th axis at x . Finally, let K bethe hyperplane through the center of E which is orthogonal to the (cid:96) th axis and set F = E ∩ K . Weclaim that vol ( F ) ≥ ε/s . To prove the claim, suppose towards contradiction that vol ( E ∩ K ) < ε/s .Then, vol ( E ) = (cid:90) (cid:96) + (cid:96) − vol ( E ∩ H ( x )) d x < εs (cid:90) (cid:96) + (cid:96) − x = εs | L | = ε, a contradiction.Choose an integer i ≥ s ∈ [∆( i ) , ∆( i + 1)). Let z / = (3 / (cid:96) − + (1 / (cid:96) + and z / = (1 / (cid:96) − + (3 / (cid:96) + . Observe that for all x ∈ [ z / , z / ], vol ( E ∩ H ( x )) ≥ ε/ (2 s ) ≥ ε/ ∆( i + 2).Next, let j be the minimum integer such that 1 / j +1 ≤ s/
2. Note that such an integer exists, as wecan choose j = (cid:100) lg(1 /s ) (cid:101) . Since s ≥ ∆( i ), j ≤ (cid:100) lg(1 / ∆( i )) (cid:101) ≤ M i . Thus, for our choices of i and j ,we have found a hyperplane h which intersects E with vol ( E ∩ h ) ≥ ε/ ∆( i + 2). By our recursiveconstruction, there is a point in the net P (cid:96),i,j which intersects E ∩ h and thus E . Lemma 3.4.
There is a deterministic, explicit construction of ( k, ε ) -nets for [0 , d of size O d (cid:18) ε − k/d log d − k − ε (cid:19) . Proof:
Follows by plugging in the bound for Theorem 3.3 into Lemma 2.5.
Remark 3.5.
Since an ε -net for intervals on the real line has size O (1 /ε ), it follows that one canconstruct a ( d − , ε )-net of size O ( d/ε /d ), which matches the same bound as the simple construction(see Lemma 4.1, naturally extended to higher dimensions). On the other hand, we should expectthat for specific values of k and d , there are better explicit constructions with size smaller thanthat of Lemma 3.4. For example, in Section 4 we prove the existence of (1 , ε )-nets with size at most O d (log γ ( d ) (1 /ε ) /ε − /d ), where γ ( d ) = d − − (cid:80) d − i =1 /i < d − d ≥
4. Improved bounds for lines: (1 , ε ) -nets Lemma 4.1.
For a parameter ε ∈ (0 , , there is a (1 , ε ) -net of size O (1 / √ ε ) .Proof: On [0 , , overlay a grid of sidelength √ ε/
2, and let L be the lines of this grid, so that | L | = O (1 / √ ε ). Let Ξ be any convex body. Note that for Ξ to avoid the lines of L , it must becontained entirely in a grid cell (cid:50) . But then vol (Ξ) ≤ vol ( (cid:50) ) = ε/ < ε .9 − − . . x (vol( C ∩ ( x = α ))) / vol( C ∩ ( x = α )) Figure 4.1: The slice volume, and its 1 / We first give a construction which produces a (1 , ε )-net of size O (1 /ε ). Next, we improve theconstruction to obtain size O ( (cid:112) log(1 /ε ) /ε / ), which is only a O ( (cid:112) log(1 /ε )) factor away from thelower bound, by Lemma 2.2. Additionally, it improves the logarithmic factor given by the generalbound of Theorem 2.7. The Brunn-Minkowski inequality and unimodal functions.
The Ξ be a convex body in R d . For a parameter α ∈ R , let f ( α ) denote the ( d − x = α . The Brunn-Minkowski inequality [Mat02, Har11] implies that the function g ( α ) = f ( α ) / ( d − is concave. In particular, g is unimodal . Namely, there exists a β ∈ R suchthat g is non-decreasing on ( −∞ , β ] and non-increasing on [ β, ∞ ). As such, the function f itself isunimodal. See Figure 4.1. Construction.
On [0 , , we overlay a grid of sidelength t = √ ε/
4. Let L be the lines of thisgrid, so that | L | = O (1 /t ) = O (1 /ε ). Lemma 4.2. L is a (1 , ε ) -net.Proof: We assume without loss of generality that Ξ is contained in [0 , . We want to show that if vol (Ξ) ≥ ε , then it is stabbed by one of the lines of L . Assume towards contradiction that Ξ doesnot intersect any lines of L . We claim that there must be a cell (cid:50) in the grid (which has volume t = ε / /
64) such that vol (Ξ ∩ (cid:50) ) ≥ ε/
4. This is a contradiction, as ε/ ≤ vol (Ξ ∩ (cid:50) ) ≤ vol ( (cid:50) ) = ε / / ⇐⇒ ε ≥ , which does not hold since ε ∈ (0 , α ∈ R , let f ( α ) denote the two-dimensional volume of Ξ intersected with theplane x = α (note f ( α ) = 0 for α (cid:54)∈ [0 , f is unimodal.We now prove the desired claim. First, if f ( α ) ≥ ε/ α = 0 , t, t, . . . , (cid:100) t (cid:101) t , then asimilar argument to Lemma 4.1 implies that the two-dimensional body Ξ (cid:48) = Ξ ∩ { x = α } contains avertex of the grid of side length t on the plane x = α , and as such Ξ is stabbed by one of the lines of L orthogonal to the plane x = α , contradicting our assumption. Now, define β = arg max α ∈ [0 , f ( α ),10igure 4.2: The multi-level grid, and its associated lines.and observe that ε ≤ vol (Ξ) = (cid:90) vol (Ξ ∩ { x = α } ) d α = (cid:90) f ( α ) d α, therefore max α ∈ [0 , f ( α ) = f ( β ) ≥ ε . Choose i ∈ { , , . . . , (cid:100) t (cid:101)} such that β ∈ [ it, ( i + 1) t ]. Let S x = (cid:8) ( x, y, z ) ∈ [0 , | it ≤ x ≤ ( i + 1) t (cid:9) be a vertical slab. Since f is unimodal and f ( β ) ≥ ε , wehave f ( α ) < ε/ α (cid:54)∈ ( it, ( i + 1) t ). As such vol (Ξ \ S x ) < ε/ vol (Ξ ∩ S x ) ≥ ε/ S x , S y , and S z such that vol (Ξ \ S j ) ≤ ε/ j ∈ { x, y, z } .Note that S x ∩ S y ∩ S z corresponds to a cell (cid:50) of the grid in [0 , . On the other hand, vol (Ξ ∩ (cid:50) ) = vol (Ξ) − vol (Ξ \ (cid:50) ) ≥ ε − vol (Ξ \ S x ) − vol (Ξ \ S y ) − vol (Ξ \ S z ) ≥ ε/ , which leads to the contradiction stated above. (1 , ε ) -nets of size O ( (cid:112) log(1 /ε ) /ε / ) in 3D Construction.
The idea behind the construction is to use octtrees. Starting with the entire cube[0 , , we construct three orthogonal planes which split the cube into eight cubes of side length 1 / splitting planes . This process is continued recursively, for i = 0 , . . . , τ ,where τ is to be defined shortly, so that cubes at the i th level of the construction has side length1 / i . The number of such cubes at the i th level is 8 i . Naturally, these cubes together form a gridwith side length 1 / i . See Figure 4.2 for an illustration of the construction in two dimensions.For each splitting plane h at level i ≥
1, which splits cells of side length 1 / i − into cells of sidelength 1 / i , we apply the 2D construction of Lemma 4.1, so that the lines all lie on the plane h ,with the parameter ε i = 2 i − (1 − / lg(1 /ε )) i − ε /ε ) , where lg = log . We collect all lines which lie on all splitting planes at each of the τ levels into our(1 , ε )-net L . Lemma 4.3.
For ε ∈ (0 , / , the set of lines L is a (1 , ε ) -net for [0 , of size O ( (cid:112) log(1 /ε ) /ε / ) .Proof: We first bound the size of L . Let c be the constant hiding in the bound O (1 / √ ε ) of Lemma 4.1.Using the construction described above, the total number of lines in our ( k, ε )-net is at most, for11 ≤ / τ (cid:88) i =0 i c √ ε i +1 = 3 c (cid:114) /ε ) ε τ (cid:88) i =0 (cid:18) − / lg(1 /ε ) (cid:19) i/ = O (cid:32)(cid:114) log(1 /ε ) ε (cid:18) − / lg(1 /ε ) (cid:19) τ/ (cid:33) = O (cid:32)(cid:114) log(1 /ε ) ε (cid:16) / log(1 /ε ) (cid:17) τ/ (cid:33) , where the last inequality follows since 1 − x ≥ − x for x ≤ /
2. Now, setting τ = (cid:100) (1 /
3) lg(1 /ε ) (cid:101) + 1,the number of lines is bounded by O ( (cid:112) log(1 /ε ) /ε / ).Next, we show that L is indeed a (1 , ε )-net. Let Ξ be a convex body contained in [0 , withvolume at least ε . If Ξ is not stabbed by any of the lines of L then we show that there is a cell, ofthe finest grid, which contains a large fraction of the volume of Ξ. This is impossible, as the volumeall such cells are too “small”.The argument is inductive. Specifically, we claim that at the i th level of the octtree, there iscell (cid:50) i of side length 1 / i such that vol (Ξ ∩ (cid:50) i ) ≥ (1 − / lg(1 /ε )) i ε . The base case is i = 0, when (cid:50) = [0 , , which holds by assumption. Now assume the claim holds for all i (cid:48) < i .The proof of the inductive step for i is similar to the analysis of the previous construction.Specifically, we argue that one of the 8 cubes of side length 1 / i contains a fraction of the volume ofΞ (cid:48) = Ξ ∩ (cid:50) i − . For sake of exposition, assume (cid:50) i − = [0 , / i − ] .Let h x , h y , and h z be the three splitting planes used to split (cid:50) i − . For each plane we willdetermine the associated halfspace that contains a large portion of the volume of Ξ (cid:48) . Define thefunction f ( α ) = vol (Ξ (cid:48) ∩ { x = α } ), for α ∈ R . First, note that if f (1 / i ) ≥ ε i , then by constructionthere will be a line (lying on the plane h x ) which stabs Ξ, which contradicts our assumption that Ξis not stabbed by L , so we have that f (1 / i ) < ε i . Let β = arg max α ∈ [0 , f ( α ). If β > / i , thenwe define the halfspace S x as (cid:8) x ≥ / i (cid:9) . Else if β < / i , we define S x to be (cid:8) x ≤ / i (cid:9) . Notethat since f is unimodal, vol (cid:0) Ξ (cid:48) \ S x (cid:1) ≤ ε i / i − = (1 − / lg(1 /ε )) i − ε/ (3 lg(1 /ε )) ≤ vol (Ξ ∩ (cid:50) i − ) / (3 lg(1 /ε )) . Applying the above process to each dimension independently, we obtain a cell (cid:50) i = S x ∩ S y ∩ S z .The cube (cid:50) i has side length 1 / i , with vol (Ξ ∩ (cid:50) i ) ≥ (1 − / lg(1 /ε )) vol (Ξ ∩ (cid:50) i − ) ≥ (1 − / lg(1 /ε )) i ε, finishing the inductive step.In particular, for τ = (cid:100) (1 /
3) lg(1 /ε ) (cid:101) + 1, there is a grid cell (cid:50) τ such that vol (Ξ ∩ (cid:50) τ ) ≥ (1 − / lg(1 /ε )) τ ε ≥ − τ/ lg(1 /ε ) ε ≥ − /
3) lg(1 /ε )+2) / lg(1 /ε ) ε = 2 − / − / lg(1 /ε ) ε ≥ − ε = ε/ , where the second inequality follows as 1 − x ≥ − x when x ≤ /
2, the third inequality follows as τ ≤ (1 /
3) lg(1 /ε ) + 2, and the last since ε ≤ /
16. But this is a contradiction, as ε/ ≤ vol (Ξ ∩ (cid:50) τ ) ≤ vol ( (cid:50) τ ) = 2 − τ ≤ − /
3) lg(1 /ε )+1) = 2 − ε = ε/ , since τ ≥ (1 /
3) lg(1 /ε ) + 1. 12 .4. (1 , ε ) -nets in higher dimensions Naturally, one can extend the previous construction to higher dimensions. Namely, we assume byinduction that there exists a (1 , ε )-net for all d (cid:48) < d . Lemma 4.3 proves the result when d = 3. Construction.
The construction is similar to Lemma 4.3, using d -dimensional quadtrees. Startingwith the cube [0 , d , we construct d orthogonal hyperplanes which split [0 , d into 2 d cells of sidelength 1 /
2. As before, we refer to these hyperplanes as splitting hyperplanes . The process iscontinued recursively inside each cell, for i = 1 , . . . , τ , where τ d = (cid:100) (1 /d ) lg(1 /ε ) (cid:101) + 1. For eachsplitting hyperplane h , we apply the ( d − h , with the parameter ε i,d = 2 i − (1 − / lg(1 /ε )) i − εd lg(1 /ε ) . We collect all lines which lie on all hyperplanes at each of the τ levels into our (1 , ε )-net L . Theorem 4.4.
Let γ ( d ) = (cid:80) d − i =2 (1 − /i ) = d − − (cid:80) d − i =1 /i . Then for ε ∈ (0 , /d ) and d ≥ there exists a (1 , ε ) -net of size O d (log γ ( d ) (1 /ε ) /ε − /d ) .Proof: By our inductive hypothesis, we assume that one can construct a (1 , ε )-net for [0 , d − ofsize β ( d − · log γ ( d − (1 /ε ) /ε − / ( d − , where β ( d −
1) is a constant depending on the dimension.The correctness of the construction is virtually identical to the proof of Lemma 4.3, and isonly sketched here. Let Ξ be a convex body with vol (cid:0) Ξ ∩ [0 , d (cid:1) ≥ ε . The proof is once again bycontradiction. If Ξ is not stabbed by any of the lines of L , then we show that there is a cell of thefinest grid, containing a large fraction of the volume of Ξ. This will lead to a contradiction, as thevolume of such a cell will be too “small”. In particular, if Ξ is not stabbed by any of the lines of L ,then one can argue that at each level i of the quadtree, there exists a cell (cid:50) i of side length 1 / i suchthat vol (Ξ ∩ (cid:50) i ) ≥ (1 − / lg(1 /ε )) i ε . By our choice of ε i,d this claim can easily be established. Onthe one hand, for our choice of τ , this implies that there is a cell (cid:50) τ such that vol (Ξ ∩ (cid:50) τ ) ≥ ε/ ε/ ≤ vol (Ξ ∩ (cid:50) τ ) ≤ vol ( (cid:50) τ ) = 2 − τd ≤ ε/ , a contradiction.We now bound the size of L : | L | ≤ d τ (cid:88) i =0 i β ( d − ε − / ( d − i +1 ,d log γ ( d − ε i +1 ,d ≤ d · β ( d − · lg − / ( d − (1 /ε ) ε − / ( d − τ (cid:88) i =0 (cid:18) − / lg(1 /ε )) d − (cid:19) i/ ( d − log γ ( d − (cid:18) d log(1 /ε )2 i (1 − / lg(1 /ε )) i ε (cid:19) ≤ d +2 d · β ( d − · lg − / ( d − (1 /ε ) ε − / ( d − (cid:104) d − / lg(1 /ε ) (cid:105) τ/ ( d − lg γ ( d − ε ≤ d +2 d · β ( d − · lg γ ( d ) (1 /ε ) ε − / ( d − lg(1 /ε ) / ( d ( d − = 2 d +2 d · β ( d − · lg γ ( d ) (1 /ε ) ε − / ( d − / ( d ( d − = 2 d +2 d · β ( d − · lg γ ( d ) (1 /ε ) ε − /d .
13n the third inequality, we use the inequalities 1 − x ≥ − x for x ≤ / ε ≤ /d . The fourthinequality follows since γ ( d −
1) + 1 − / ( d −
1) = γ ( d ) and τ ≤ (1 /d ) lg(1 /ε ) + 2. In particular, weobtain the recurrence β ( d ) = 2 d +2 d β ( d −
1) which implies β ( d ) = 2 O ( d ) . As such, L is a (1 , ε )-netof size O d (log γ ( d ) (1 /ε ) /ε − /d ).
5. Conclusion
The main open problem left by our work is bounding the size of ( k, ε )-nets in the general case. Thatis, the input is a set P of n points in R d , and we would like to compute a minimum set of k -flatswhich stab all convex bodies containing at least εn points of P . As noted earlier, there is a ( k, ε )-netof asymptotically the same size as of a weak ε -net in R d − k . This follows by projecting the point setto a subspace of dimension d − k , constructing a regular weak ε -net, and lifting the net back to theoriginal space. Can one do better than this somewhat na¨ıve construction?Note that it is easy to show a lower bound of size Ω(1 /ε ) for (1 , ε )-net s in the general case.Take a point set that consists of (cid:100) /ε (cid:101) equally sized clusters of tightly packed points, such that noline passes through three clusters. Namely, our sublinear results in 1 /ε are special for the uniformmeasure on the hypercube.Finally, many of our constructions presented have size which depends exponentially on thedimension (e.g., 2 O ( d ) ). It may be interesting to see if these hidden constants can be improved. References [AI11] Y. Akama and K. Irie.
VC dimension of ellipsoids . CoRR , abs/1109.4347, 2011. arXiv: .[BMN09] B. Bukh, J. Matouˇsek, and G. Nivasch.
Lower bounds for weak epsilon-nets and stair-convexity . Proc. 25th Annu. Sympos. Comput. Geom. (SoCG) , The discrepancy method: randomness and complexity . New York: CambridgeUniversity Press, 2001.[Har11] S. Har-Peled.
Geometric approximation algorithms . Vol. 173. Math. Surveys & Mono-graphs. Boston, MA, USA: Amer. Math. Soc., 2011.[HW87] D. Haussler and E. Welzl. ε -nets and simplex range queries . Discrete Comput. Geom. , 2:127–151, 1987.[Mat02] J. Matouˇsek.
Lectures on discrete geometry . Vol. 212. Grad. Text in Math. Springer,2002.[Mat99] J. Matouˇsek.
Geometric discrepancy: an illustrated guide . Vol. 18. Algorithms andCombinatorics. Springer, 1999.[MW04] J. Matouˇsek and U. Wagner.
New constructions of weak epsilon-nets . Discrete Comput.Geom. , 32(2): 195–206, 2004.[PT13] J. Pach and G. Tardos.
Tight lower bounds for the size of epsilon-nets . J. Amer. Math.Soc. , 26: 645–658, 2013.[Rob55] H. Robbins.
A remark on Stirling’s formula . Amer. Math. Monthly , 62(1): 26–29, 1955.14igure A.1: The Van der Corput set with n = 16 (left) and n = 128 (right).[Rub18] N. Rubin. An improved bound for weak epsilon-nets in the plane . Proc. 59th Annu. IEEESympos. Found. Comput. Sci. (FOCS) , A. (0 , ε ) -nets when the bodies are axis-aligned boxes Here we show the existence of a (0 , ε )-net of size O (1 /ε ) that intersects any axis-aligned box B with vol (cid:0) B ∩ [0 , (cid:1) ≥ ε . The following constructions are essentially described in [Mat99] (in thecontext of low-discrepancy point sets), however the proofs use similar tools. We give the proofs forcompleteness. Definition A.1 (the Van der Corput set).
For an integer α , let bin ( α ) ∈ { , } (cid:63) denote the binaryrepresentation of α , and rev ( bin ( α )) be the reversal of the string of digits in bin ( α ). We define br ( α ) ∈ [0 ,
1] to be the bit-reversal of α , which is defined as the number obtained by concatenating“0 . ” with the string rev ( bin ( α )). For example, br (13) = 0 . α = (cid:80) ∞ i =0 i b i with b i ∈ { , } , then br ( α ) = (cid:80) ∞ i =0 b i / i +1 .For an integer n , the Van der Corput set is the collection of points p , . . . , p n − , where p i = ( i/n, br ( i )). See Figure A.1. Lemma A.2.
For a parameter ε ∈ (0 , ,there is a collection of O (1 /ε ) points P ⊂ [0 , such thatany axis-aligned box B with vol (cid:0) B ∩ [0 , (cid:1) ≥ ε contains a point of P .Proof: Let n = (cid:100) /ε (cid:101) . We claim that the Van der Corput set of size n is the desired point set P .Let B be a box contained in [0 , of width w and height h , with wh ≥ ε . Let q ≥ / q < h/ ≤ / q − . By the choice of q , the projection of B onto the y -axis contains an interval of the form I = [ k/ q , ( k + 1) / q ) for some integer k . Let B I = B ∩ (cid:8) ( x, y ) ∈ [0 , | y ∈ I (cid:9) be the box restricted to I along the y -axis. Observe that vol ( B I ) = w/ q = w/ (4 · q − ) ≥ wh/ ≥ ε/ ⇐⇒ w ≥ q ε/ . Let S = [0 , × I , so that each p j ∈ P ∩ S has br ( j ) ∈ I . In particular, the first q binary digitsof br ( j ) are fixed. This implies that the q least significant binary digits of j are fixed. In otherwords, P ∩ S contains all points p j such that j ≡ (cid:96) (mod 2 q ) for some integer (cid:96) —the x -coordinates15f the points in P are regularly spaced in the strip S with distance 2 q /n . If the width of B I is atleast 2 q /n , then this implies that B contains a point of P in the strip S . Indeed, by the choice of n ,2 q /n ≤ q ε/ ≤ w .By extending the definition of the Van der Corput set to higher dimensions, the above proofalso generalizes. Definition A.3 (the Halton-Hammersely set).
For a prime number ρ and an integer α = (cid:80) ∞ i =0 ρ i b i , b i ∈ { , . . . , ρ − } , written in base ρ , define br ρ ( α ) = (cid:80) ∞ i =0 b i /ρ i +1 . Note that br = br fromDefinition A.1.For integers n and d , the Halton-Hammersely set is the collection of points p , . . . , p n − ,where p i = ( br ρ ( i ) , br ρ ( i ) , . . . , br ρ d − ( i ) , i/n ), and ρ , . . . , ρ d − are the first d − Lemma A.4.
For a parameter ε ∈ (0 , , there is a collection of O ( d log d ) /ε points P ⊂ [0 , d suchthat any axis-aligned box B with vol (cid:0) B ∩ [0 , d (cid:1) ≥ ε contains a point of P .Proof: The proof is similar to Lemma A.2, with the Chinese remainder theorem as the additionaltool.Let n = (cid:6) (2 d − /ε ) · ( d − (cid:93) (cid:7) , where k(cid:93) is the primorial function, defined as the product of thefirst k prime numbers. It is known that k(cid:93) ≤ exp((1 + o (1)) k log k ), which implies n = 2 O ( d log d ) /ε .We claim that the Halton-Hammersely set of size n is the desired point set P .Denote the side lengths of the box B by s , . . . , s d , with (cid:81) di =1 s i ≥ ε . For each i = 1 , . . . , d − q i be the smallest integer such that 1 /ρ q i i < s i / ≤ /ρ q i − i , where ρ i is the i th prime number.By the choice of q i , the projection of B onto the i th axis contains an interval of the form I i =[ k i /ρ q i i , ( k i + 1) /ρ q i i ] for some integer k i . Let S denote the box I × . . . × I d − × [0 ,
1] and B S = B ∩ S .Observe that vol ( B S ) = s d d − (cid:89) i =1 ρ q i i ≥ s d d − (cid:89) i =1 s i ρ i ≥ ε d − d − (cid:89) i =1 ρ i ⇐⇒ s d ≥ ε d − d − (cid:89) i =1 ρ q i − i . Similar to Lemma A.2, we observe that the point p j ∈ P falls into S when j ≡ (cid:96) i (mod ρ q i i ) forsome integers (cid:96) , . . . , (cid:96) d − . By the Chinese remainder theorem, there is exactly one number in the set (cid:110) , , . . . , (cid:81) d − i =1 ρ q i i − (cid:111) (the d th coordinate of p j ) which satisfies these d − P ∩ S are spaced regularly along the d th axis with distance δ = (1 /n ) (cid:81) d − i =1 ρ q i i . Onceagain, we argue that the length of B along the d th axis is at least δ , which implies the result. Indeed,by our choice of n we have that, δ = 1 n d − (cid:89) i =1 ρ q i i ≤ ε d − d − (cid:89) i =1 ρ q i − i ≤ s d . Making i/n the d th coordinate instead of the 1st coordinate simplifies future notation. . Extension: Replacing [0 , d with other convex bodies We now prove a generalized version of Theorem 2.7.
Lemma B.1.