Range Spaces

1. Small-size  -nets for Axis-Parallel Rectangles and BoxesBoris Aronov EstherEzra Micha sharirpolytechnic Duke Tel-AvivInstitute of NYU University University

2. Range Spaces Range space (X, R) : X – Ground set (the “universe”). R – Ranges: Subsets of X . |R|  2|X| Abstract form: Hypergraphs. X – vertices. R – hyperedges.

3. Geometric Range Spaces specification: X d, R = set of simply-shaped regions in d . X – Points on the real line. R – Intervals. X – Points on the plane. R – halfplanes, disks,… For simplicity, assume X is finite: |R| is polynomial in|X|.

4.  -nets for range spaces Given: • A range space (X, R) , assume X is finite, |X| = n . • A parameter 0 <  < 1 , An  -net for (X, R) is a subset N  X that hits every range Q  R, with |Q  X|   n . N is a hitting set for all the ``heavy'' ranges. Example: Points and intervals on the real line: |N| = 1/ . Captures at least an  -fraction of the universe. Bound does not depend on n.  n

5. The hitting-set problem A hitting set for (X, R) is a subset H  X, s.t., for any Q  R , Q  H   . Goal: find smallest hitting set. Useful applications: art-gallery, sensor networking, and more.

6. Hardness of hitting sets Finding a hitting set of smallest size is NP-hard, even for geometric range spaces! Use an approximation algorithm instead. Abstract range spaces [Chvatal 79]:Greedy algorithm. Approximation factor: O(log |X|) Geometric range spaces [Bronimann-Goodrich95], [Clarkson 93]: Achieve improved approximation factor! Approximation factor: O(log OPT), or smaller! This is achieved via -nets: Small-size -nets imply small approximation factors! OPT = size of the smallest hitting set.

7. An upper bound for the  -net size The  -net theorem [Haussler-Welzl 87]: If the ranges are simply-shaped regions, then, for any > 0, a random sample of size O(1/ log (1/ )) is an  -net, with constant probability. Remark: In fact, it is sufficient to assume that the number of ranges is only polynomial in n. Is it optimal? Bound does not depend on n.

8. The lower bound Theorem [Komlos, Pach, Woeginger 92]: The bound is tight! The construction: Artificial on abstract hypergraphs (non-geometric!). No lower bound better than (1/ ) is known in geometry. What is the actual bound? O(1/ ) ? Goal:Obtain smaller bounds for geometric range spaces. Ideally O(1/ ), but anything better than O(1/ log (1/ ))is `exciting‘ ! Achieved by points and intervals on the real line.

9. Previous results Points and halfspaces in 2D, 3D. O(1/ )[Matousek 92], [Pyrga, Ray 08], [Har-Peled et al. 08] Points and disks, or pseudo-disks in 2D: O(1/ ) [Matousek, Seidel, Welzl 90], [Pyrga, Ray 08]. Pseudo-disks

10. Our results Points and axis-parallel rectangles in the plane.  -net size is O(1/ log log (1/ )). Points and axis-parallel boxes in 3-space.  -net size is O(1/ log log (1/ )). Points and -fat triangles in the plane.  -net size is O(1/ log log (1/ )). Points uniformly distributed over the unit-cube, and axis-parallel boxes in d-space.  -net size is O(1/ log log (1/ )). Each of the angles  

11. Improved approximation factorsfor geometric hitting sets Ranges previous bound new bound Axis-parallel rectangles log OPT log log OPT Axis-parallel 3-boxes log OPT log log OPT -fat triangles log OPT log log OPT Axis-parallel d-boxes log OPT log log OPT Uniformly distributed points in [0,1]d .

12. Main idea :Use two-level sampling Primary sampling step: Obtain an initial sample S of ~1/ points of X. On average, each heavy rectangle Q must satisfy Q S  . Second sampling step (repair step): In each heavy rectangle Q  R , with Q S =  , sample additional points to guarantee that Q is stabbed by the net. S contains at least  n points Q

13. The  -net construction Input:X - a set of n points. Parameters:r := 1/ . Primary sample: Produce a random sample S  X of size r . Make S part of the output.|S| = r. Apply the second sampling step in each empty rectangle… Instead of processing all input rectangles, we consider a smaller set of representative rectangles.

14. The set of maximal S-empty rectangles A maximal S-empty rectangle M satisfies int(M)  S =  , and for each rectangle M’  M, int(M’)  S   . M is defined by  4 points of S. M - set of all maximal S-empty rectangles. Apply repair-step on M instead on the input rectangles. M S

15. Why is it sufficient to consider M? For each input heavy rectangle Q, with Q  S =  , expand Q until each of its sides touches a point of S or continues to  . Since Q is heavy, a sufficiently large sample in M will hit Q, with high probability. Otherwise, done! M Q

16. The repair step [CF-90, CV-07] Consider a heavy rectangle M, with |M  X | = t n/r, 1  t  log r . Second sampling step: Construct (1/ t)-net NM inside M , by sampling O(t log t) points in M. According to the  -net theorem, each input (empty) rectangle Q  R, Q  M, with |Q|  n/r , must be stabbed by NM ! r=1/ According to the  -net theorem The excess of M M Q The “universe” size is now t n/r

17. The final  -net Output: The union of S andM  M NM . What is the expected size of the  -net ? r +E{t1 t log t|Mt | } Exponential Decay Lemma: [Chazelle, Friedman 90], [Agarwal Matousek, Schwarzkopf 98] E{ |Mt | } = O( 2-t E{ |M| }) , The number of heavy rectangles decreases exponentially! Mt= set of rectangles inMwith excesst Expected number of maximal empty rectangles

18. An improved  -net Theorem: E{ |M| } = O(r log r) E{ |Mt| } = O(r log r). The expected  -net size is O(1/ log(1/ )) . Key observation: Use oversampling. Choose a slightly larger primary sample, and repair only rectangles M with excess t c log log r . t  1 No improvement yet… c > 1 |S| = r |S| = c r log log r |M  X | n/r

19. What have we gained? “On average”, an S-empty rectangle contains now at most O(n/(r loglog r)) << n/r points. So M cannot be an “average” S-empty rectangle. It is much heavier. Exponential Decay Lemma:E{ |Mt | } = O( 2-t E{ |M| }) . # maximal heavy S-empty rectangles is much smaller! E{ |Mt| } = O(s log s / polylog r) = o(s) = o(r). The number of heavy (empty) rectangles is only sublinear in r ! The expected  -net size is O(rlog log r) . s = |S| = c r log log rt = clog log r .

20. Oversampling: A trick or a technique? By oversampling at the preliminary step, we significantly decrease the size of the secondary sample. Note: The number of maximal S-empty rectangles is O(s log s) , however, we do not traverse all of them, but only the heavy ones! New Concept: The sample points and the maximal S-empty rectangles are two different entities.

21. merci beaucoup!

22. Bounding the number of maximal S-empty rectangles Upper bound:O(s2) . Each rectangle is determined by its two opposite corners. problem: The bound O(s2) is bad for the analysis, and yields an  -net of size O(1/ 2) ! Recall s=c 1/ log log 1/

23. Quadratic Lower bound construction A staircase construction: Each point in the upper staircase is matched with each point in the lower staircase. (s2) empty rectangles. We can prune away most of these rectangles and remain only with O(s log s) rectangles .

24. An O(s log s) bound for |M| Key observation: Consider a vertical line l, and all points to its left. Claim: The number of maximal S-empty rectangles, anchored at l is only linear. Next step:Use a tree decomposition built on top of X in order to obtain the O(s log s) bound. l Q’ Q v l3 l2 l‘3 l1 l3 l‘2 l‘‘3

25. Dual (Geometric) Range Spaces Flip roles of X and R, and obtain (R, X*) . R = set of regions in d , X* = {Rp | p X}, Rp = {r | r  R , r containsp} . R – Intervals. X* – Subsets of intervals containing a common point in 1 . R – Disks. X* – Subsets of Disks containing a common point in 2 p p

26.  -nets for dual range spaces  -net for (R, X*) is a subset N  R that covers all points at depth  |R| . An  -net is a set cover for all the “deep” points. Upper bound [Haussler-Welzl, 87]: O(/ log (/ )) ,  is the VC-dimension of (R, X*) . depth(p) = #ranges that cover p  X.

27. Previous results Range space (R, X*), s.t., for each T  R, |T| = m, the union T has (a small) complexity o(m log m) : o(1/ log (1/ )) . [Clarkson, Varadarajan 07] Theorem: [Clarkson, Varadarajan 07] The complexity of the union is O(m (m))  -net size is O(1/  (1/ )). In fact, this should be the complexity of the vertical decomposition of the complement of the union.  () is a slowly growing function.

28. More about the Clarkson-Varadarajantechnique Example: disks (or pseudo-disks) and points Input: A set T of m (pseudo) disks. Union complexity: O(m) . [kedem et al. 86]  -net size is O(1/ ). Example: fat triangles and points Input: A set T of m-fat triangles. Union complexity: O(m loglog m) . [Matousek et al. 1994]  -net size is O(1/ log log (1/ )). Each of the angles  

29. Our results: Dual Theorem: [Clarkson, Varadarajan 07] The complexity of the union is O(m (m))  -net size is O(1/  (1/ )). Using the oversampling concept: Theorem (improvement!): The complexity of the union is O(m (m))  -net size is O(1/ log (1/ )) .

30. Proof sketch Draw a random sample S of s = c/log (1/ ) regions. Construct the union of S: Decompose its complement into O(s (s)) “trapezoidal cells”. Each cell  is defined by  4 regions. Claim: With high probability,  meets  (n/s) log sregions of the input. 

31. Proof sketch Apply a repair step on the heavy cells: Sample O(t log t)regions in each cell  that meets t n/s regions, for t  c log  (1/ ) Each point at depth  n is covered by at least one region. Use the Exponential Decay Lemma to show: # regions sampled at the repair step = o(1/ ) . Overall  -net size: O(1/log (1/ )) .

32. New  -net bounds Fat triangles: Union complexity: O(m loglog m)  -net size is O(1/ log log log(1/ )). Locally -fat objects: Union complexity: O(m polylogm)  -net size is O(1/ log log (1/ )). And several other improved bounds. area(D O)    area(D) 0 <   1 O D

33. Open problems • Improve our upper bound O(1/ loglog (1/ ))for points and axis-parallel rectangles.Conservative goal: Obtain a weak -net of size o(1/ loglog (1/ )) . • Extend our bound to points and axis-parallel boxes in d  4.Best known upper bound: O(1/ log (1/ )) . • Dual range spaces for rectangles and points.Best known upper bound: O(1/ log (1/ )) .Can improve to O(1/ loglog (1/ )) ? The points of the  -net are not necessarily chosen from X . p

34. Motivation: Approximation for geometric hitting sets The Bronimann-Goodrich technique / LP-relaxation If (X, R) admits an  -net of size f(1/ ) , then there exists a polynomial-time approximation algorithm that reports a hitting set of size O(f(OPT)) . Idea: Assign weights on X s.t each range Q  R becomes heavy . Construct an -net for the weighed range space. Each range is hit by the -net. Small-size -nets imply small approximation factors!

35. The repair step repair step: On average, each heavy rectangle Q must satisfy Q S  . The number of “bad” rectangles is small. It is sufficient to consider a set M of maximal S-empty rectangles, instead of R. Mis defined over the points of S. |M| = f(1/)(does not depend on n). S and so does #points sampled at the repair step. M Q

36. An O(s log s) bound for |M| Key observation: Consider a vertical line l, and all points to its left. Claim: The number of maximal S-empty rectangles, anchored at l is only linear. Handling a query rectangle Q: One of the halves Q’of Q contains at least n/(2r) points. Q’ is anchored at l. Expand Q’on “heavier” side of l . l l Q Q’

37. Tree decomposition Each node is a vertical strip • Build balanced binary tree T on X, sorted by x-coordinate • Stop expansion of T when nodes have n/r points. T has O(log r) = O(log s) levels. At each level: #maximal S-empty anchored rectangles: O(s) Overall (over all levels): O(s log s) . v l3 l2 l‘3 l1 l3 l‘2 l‘‘3

38. Query rectangle Q For an input rectangle Q with n/r points: Find the first (highest) node of T whose bounding line lmeets Q. Expand Q within the“heavier” strip v bounded by l . The maximal S-empty anchored rectangles comprise the representative set for R. Q’ Q v l3 l2 l‘3 l1 l3 l‘2 l‘‘3

39. Is the bound optimal? Theorem [Komlos, Pach, Woeginger 92]: The bound is tight! The construction: Artificial on abstract hypergraphs (non-geometric!). No lower bound better than (1/ ) is known in geometry. What is the actual bound? O(1/ ) ? Goal:Obtain smaller bounds for geometric range spaces. Ideally O(1/ ), but anything better than O(1/ log (1/ ))is `exciting‘ ! Achieved by points and intervals on the real line.

40. Bounding the  -net size Exponential Decay Lemma: [Chazelle, Friedman 90], [Agarwal Matousek, Schwarzkopf. 98] E{ |Mt| } = O( 2-t E{ | M'| }) , where: • S' is a smaller random sample, each point chosen with probability s/(t n) . • Mt - all maximal S-empty rectangles M with tMt . • M' - all maximal S'-empty rectangles.

41. Bounding the final  -net size A very useful tool: Exponential Decay Lemma: [Chazelle, Friedman 90], [Agarwal Matousek, Schwarzkopf. 98] E{ |Mt| } = O( 2-t E{ |M| }) , where Mt is all maximal S-empty rectangles M with tMt . The number of heavy rectangles decreases exponentially!

42. A nearly-linear bound for |M| Fix a node v of T and its strip v : Xv = S v , Sv = S v Lemma: The number of maximal Sv-empty anchored rectangles in v is O(Sv) . At a fixed level i of T , overall number is O(s) . Overall: O(s log r) . v Entry side

43. The set-cover problem Primal: A hitting set for (X, R) is a subset H  X, s.t., for any Q  R , Q  H   . Dual: A set cover for (X, R) is a subset S  R, s.t., any x  X is covered by S . A set cover for (X, R) is a hitting set for (R, X*) Finding a set cover of smallest size is NP-hard! (even for geometric range spaces). Achieve improved approximation factors via -nets (using the Bronimann-Goodrich technique / LP-relaxation). f(1/ ) O(f(OPT))

44.  -nets for dual range spaces depth(p) = #ranges that cover p  X.  -net for (R, X*) is a subset N  R that covers all points at depth  |R| . An  -net is a set cover for all the deep points. Example: Intervals and points on the real line: |N| = 1/ .  n

45. Extensions to axis-parallel boxes in 3-space Use similar machinery, with s = c r log log r, and a 3-level range tree decomposition. At each fixed triple-level of the tree, we have a subdivision of space into (clipped) orthants. y-order z-order x-order

46. Axis-parallel boxes in 3-space Fix a orthant . Consider the points in , and the set M of all maximal S-empty boxes anchored at the apex of . Claim:M = O(s). E{|M| } = E{ |M| } = O(s log3s) The expected size of the -net is O(1/ log log (1/ )).  All these boxes grow from a common point. They behave as maximal S-empty orthants!