Eppstein's bound on intersecting triangles revisited
EEppstein’s bound on intersecting triangles revisited
Gabriel Nivasch ∗ Micha Sharir † School of Computer ScienceTel Aviv UniversityTel Aviv 69978, Israel
July 27, 2008
Abstract
Let S be a set of n points in the plane, and let T be a set of m triangles with verticesin S . Then there exists a point in the plane contained in Ω( m / ( n log n )) triangles of T .Eppstein (1993) gave a proof of this claim, but there is a problem with his proof. Here weprovide a correct proof by slightly modifying Eppstein’s argument. Keywords:
Triangle; Simplex; Selection Lemma; k -Set Let S be a set of n points in the plane in general position (no three points on a line), and let T be a set of m ≤ (cid:0) n (cid:1) triangles with vertices in S . Aronov et al. [2] showed that there alwaysexists a point in the plane contained in the interior ofΩ (cid:18) m n log n (cid:19) (1)triangles of T . Eppstein [5] subsequently claimed to have improved this bound toΩ (cid:18) m n log n (cid:19) . (2)There is a problem in Eppstein’s proof, however. In this note we provide a correct proof of(2), by slightly modifying Eppstein’s argument. ∗ [email protected] . Work was supported by ISF Grant 155/05 and by the HermannMinkowski–MINERVA Center for Geometry at Tel Aviv University. † [email protected] . Work was partially supported by NSF grant CCF-05-14079, by a grant from theU.S.-Israel Binational Science Foundation, by ISF Grant 155/05, and by the Hermann Minkowski–MINERVACenter for Geometry at Tel Aviv University. The very last sentence in the proof of Theorem 4 (Section 4) in [5] reads: “So (cid:15) = 1 / i +1 , and x = m(cid:15)/y = O ( m/ i ), from which it follows that x/(cid:15) = O ( n ).” This is patently false, since what actually follows is that x/(cid:15) = O ( m ), and the entire argument falls through. a r X i v : . [ c s . C G ] J u l .1 The Second Selection Lemma and k -sets The above result is the special case d = 2 of the following lemma (called the Second SelectionLemma in [6]), whose proof was put together by B´ar´any et al. [3], Alon et al. [1], and ˇZivaljevi´cand Vre´cica [8]:
Lemma 1. If S is an n -point set in R d and T is a family of m ≤ (cid:0) nd +1 (cid:1) d -simplices spannedby S , then there exists a point p ∈ R d contained in at least c d (cid:16) mn d +1 (cid:17) s d n d +1 (3) simplices of T , for some constants c d and s d that depend only on d . (Note that m/n d +1 = O (1), so the smaller the constant s d , the stronger the bound.) Thus,for d = 2 the constant s in (3) can be taken arbitrarily close to 3. The general proof ofLemma 1 gives very large bounds for s d ; roughly s d ≈ (4 d + 1) d +1 .The main motivation for the Second Selection Lemma is deriving upper bounds for themaximum number of k -sets of an n -point set in R d ; see [6, ch. 11] for the definition and details. We assume that m = Ω( n log / n ), since otherwise the bound (2) is trivial. The proof,like the proof of the previous bound (1), relies on the following two one-dimensional selectionlemmas [2]: Lemma 2 (Unweighted Selection Lemma) . Let V be a set of n points on the real line, andlet E be a set of m distinct intervals with endpoints in V . Then there exists a point x lyingin the interior of Ω( m /n ) intervals of E . Lemma 3 (Weighted Selection Lemma) . Let V be a set of n points on the real line, and let E be a multiset of m intervals with endpoints in V . Then there exists a multiset E (cid:48) ⊆ E of m (cid:48) intervals, having as endpoints a subset V (cid:48) ⊆ V of n (cid:48) points, such that all the intervals of E (cid:48) contain a common point x in their interior, and such that m (cid:48) n (cid:48) = Ω (cid:18) mn log n (cid:19) . The proof of the desired bound (2) proceeds as follows:Assume without loss of generality that no two points of S have the same x -coordinate. Foreach triangle in T define its base to be the edge with the longest x -projection. For each pairof points a, b ∈ S , let T ab be the set of triangles in T that have ab as base, and let m ab = | T ab | .(Thus, (cid:80) ab m ab = m .)Discard all sets T ab for which m ab < m/n . We discarded at most (cid:0) n (cid:1) m/n < m/ T (cid:48) of at least m/ m ab = 0or m ab ≥ m/n for each base ab . This critical discarding step is missing in [5], and that is why the proof there does not work. E , E , . . . , E k for k = log ( n /m ),so that each E j contains all the bases ab for which4 j − mn ≤ m ab < j mn . (4)Let T j = (cid:83) ab ∈ E j T ab denote the set of triangles with bases in E j , and m j = | T j | denote theirnumber. There must exist an index j for which m j ≥ − ( j +1) m, since otherwise the total number of triangles in T (cid:48) would be less than m/
2. From now on wefix this j , and work only with the bases in E j and the triangles in T j .For each pair of triangles abc , abd having the same base ab ∈ E j , project the segment cd into the x -axis, obtaining segment c (cid:48) d (cid:48) . We thus obtain a multiset M of horizontal segments,with | M | ≥ m j (cid:18) j − mn − (cid:19) = Ω (cid:18) j m n (cid:19) . (Each of the m j triangles in T j is paired with all other triangles sharing the same base, andeach such pair is counted twice.)We now apply the Weighted Selection Lemma (Lemma 3) to M , obtaining a multiset M of segments delimited by n distinct endpoints, all segments containing some point z in theirinterior, with | M | n = Ω (cid:18) | M | n log n (cid:19) = Ω (cid:18) j m n log n (cid:19) . Let (cid:96) be the vertical line passing through z . For each horizontal segment c (cid:48) d (cid:48) ∈ M , eachof its (possibly multiple) instances in M originates from a pair of triangles abc , abd , wherepoints a and c lie to the left of (cid:96) , and points b and d lie to the right of (cid:96) . Let p be theintersection of (cid:96) with ad , and let q be the intersection of (cid:96) with bc . Then, pq is a verticalsegment along (cid:96) , contained in the union of the triangles abc , abd (see Figure 1). Let M bethe set of all these segments pq for all c (cid:48) d (cid:48) ∈ M .Note that the vertical segments in M are all distinct, since each such segment pq uniquelydetermines the originating points a , b , c , d (assuming z was chosen in general position).3et n be the number of endpoints of the segments in M . We have n ≤ nn , since eachendpoint (such as p ) is uniquely determined by one of n “inner” vertices (such as d ) and oneof at most n “outer” vertices (such as a ).Next, apply the Unweighted Selection Lemma (Lemma 2) to M , obtaining a point x ∈ (cid:96) that is contained in Ω (cid:18) | M | n (cid:19) = Ω (cid:32) n (cid:18) | M | n (cid:19) (cid:33) = Ω (cid:18) j m n log n (cid:19) segments in M . Thus, x is contained in at least these many unions of pairs of triangles of T j . But by (4), each triangle in T j participates in at most 4 j m/n pairs. Therefore, x iscontained in Ω (cid:18) m n log n (cid:19) triangles of T j . Eppstein [5] also showed that there always exists a point in R contained in Ω( m/n ) trianglesof T . This latter bound is stronger than (2) for small m , namely for m = O ( n / log n ).On the other hand, as Eppstein also showed [5], for every n -point set S in general positionand every m = Ω( n ), m ≤ (cid:0) n (cid:1) , there exists a set T of m triangles with vertices in S , suchthat no point in the plane is contained in more than O ( m /n ) triangles of T . Thus, withthe current lack of any better lower bound, the bound (2) appears to be far from tight. Evenachieving a lower bound of Ω( m /n ), without any logarithmic factors, is a major challengestill unresolved.It is known, however, that if S is a set of n points in R in general position (no four pointson a plane), and T is a set of m triangles spanned by S , then there exists a line (in fact, aline spanned by two points of S ) that intersects the interior of Ω( m /n ) triangles of T ; see[4] and [7] for two different proofs of this. References [1] N. Alon, I. B´ar´any, Z. F¨uredi, and D. Kleitman, Point selections and weak (cid:15) -nets for convexhulls,
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