Opposite-quadrant depth in the plane

Transcription

1 Oosite-quadrant deth in the lane Hervé Brönnimann Jonathan Lenchner János Pach Comuter and Information Science, Polytechnic University, Brooklyn, NY 1101, IBM T. J. Watson Research Center, Yorktown Heights, NY 10598, Courant Institute, NYU, and City College, CUNY 51 Mercer Street, New York, NY 1001, Abstract Given a set S of n oints in the lane, the oosite-quadrant deth of a oint S is defined as the largest number k such that there are two oosite axis-aligned closed quadrants (NW and SE, or SW and NE) with aex, each containing at least k elements of S. We rove that S has a oint with oosite-quadrant deth at least n/8. If the elements of S are in convex osition, then we can guarantee the existence of an element whose oosite-quadrant deth is at least n/4. Both results are asymtotically best ossible. 1 Introduction Let S be a set of n oints in the lane. Given any oint, the horizontal and vertical lines through define four closed quadrants NW(), NE(), SE(), and SW(). We define the oosite-quadrant deth o(, S) of a oint with resect to S as the largest number k such that there are two oosite quadrants NW() and SE() or NE() and SW(), each containing at least k elements of S. In notation: o(, S) = max (min ( NW() S, SE() S ), o(s) = max o(, S). R min ( NE() S, SW() S )), A oint is said to be α-dee, for some α > 0, if its oosite-quadrant deth is at least αn. A (not necessarily unique) oint of maximum oosite-quadrant deth is called a deeest oint, and a oint which is α-dee is also called an α-centeroint. If we do not require the centeroint to be in S, then there always exists a oint that is 1 4-dee at the intersection of the vertical and horizontal Research artially suorted by NSF grant CCF Research artially suorted by NSF grant CCF Research artially suorted by grants from NSF, NSA, OTKA, and the Israeli-US Binational Research Foundation.

2 halving lines. In fact, 1 4 is the best general bound that can be established on the maximum deth for any oint in the lane, as shown by a set of n oints uniformly distributed in a square, a disk, or even on a circle. There are also sets with deeer oints, e.g., any set of n collinear oints has a oint of deth n/. In what follows, we require the centeroint to belong to the set S. In this case, we cannot guarantee that there always exists a 1 4-dee oint in S. However, we can show that S always contains a 1 8 -centeroint. Theorem 1 Any set of n oints in the lane has an element with oosite-quadrant deth at least n/8. Moreover, such a oint can be found in O(n) time. The roof is insired by the catline of Hubert and Rousseuw [6]. Somewhat surrisingly, Theorem 1 is essentially tight. Theorem For infinitely many values of n, there are n-element sets S n with max Sn o(, S n ) = (n + 4)/8. If the oints of S are in convex osition, that is, if they form the vertex set of a convex olygon, we can also guarantee the existence of a 1 4-centeroint in S. In fact, this statement remains true under the weaker condition that S is an exosed set, that is, for any S, at least one of the four quadrants NW(), NE(), SE(), and SW() contains no element of S in its interior. With a little twist of terminology, in this case we say that such a quadrant is emty (although it necessarily contains the oint and may erhas contain some other boundary oints that belong to S). Theorem 3 Any exosed set of n oints in the lane has an element with oosite-quadrant deth at least n/4. Moreover, such a oint can be found in O(n) time. The above notions are closely related to half-sace deth, multivariate regression deth, and other measures of statistical deth [7]. For instance, the half-sace deth of a oint with resect to an n-element set S in d-sace is the minimum number of oints of S covered by a closed halfsace containing. It is well-known that one can always find a oint of half-sace deth at least n d+1 with resect to S []. In shar contrast to the oosite-quadrant deth, however, now we cannot guarantee that S has an element of large half-sace deth. Indeed, if the oints of S are in strictly convex osition, none of them has half-sace deth larger than one. Proofs of Theorems 1 and 3 First, notice that having oints on the same vertical or horizontal line does not create a roblem, since the quadrants were defined to be closed. In fact, we can slightly erturb our original oint set S by assigning to each element i S a vector v i = (i, i ). Then there exists a sufficiently small ε > 0 such that the modified sets S λ = { i + λv i } are in general osition for any λ (0, ε], in the sense that no air of oints have the same x or y coordinate. Consequently, varying λ in this small interval will not change the horizontal or vertical order of the elements of S λ. If our theorems hold for sets in general osition, we can aly them to S ε, and find an α-centeroint i + εv i. Then i + λv i is α-dee with resect to S λ for all λ (0, ε]. Bringing back our oints to their original ositions in S only adds oints on the boundary of the two oosite quadrants of i. Therefore, i is also an α-centeroint for S.

3 Thus, from now on we assume that S is in general osition. To simlify the resentation, we also assume that n = S is divisible by 8, so that we do not have to deal with uer and lower integer arts. The roof of Theorem 1 roceeds by first slitting the set into three arts by two vertical lines, not assing through any element of S, such that in the left and right half-lanes there are recisely n/4 oints and the vertical stri in the middle contains n/ oints. Next, we find a horizontal line, not assing through any oint of S, that divides S into two equal halves. In this way, we obtain a 3 grid, whose six regions will be called cells. We say that a cell is heavy if it contains at least as many oints as the other cell in the same column, that is, the cell below or above it. Note that if every cell in a row is heavy, then each of them contains recisely as many elements as the cell below or above it. Otherwise, there could not be n/ oints in the other row. Thus, in this case, all six cells must be heavy. Therefore, we can always ick one heavy cell er column so that not all of them belong to the same row. We distinguish two cases u to symmetries (reflections about the coordinate axes). (a) Bitonic configuration: The uer cells H 1 and H 3 of the first and third columns, resectively, and the lower cell H of the second column are all heavy. (See Figure 1(a).) (b) Monotone configuration: The uer cells H 1 and H of the first two columns and the lower cell H 3 of the third column are all heavy. (See Figure 1(b).) H1 H11 H1 H H1 H1 H H'1 (a) Bitonic configuration (b) Monotone configuration Figure 1: Proof of Theorem 1. We let H 1 be the lower cell in the left column, H 11 := H 1 NW(), H 1 := H 1 SW(), H 1 := H NW(), and H := H NE(). Case (a): H 1, H 3, and one of H 1 or H have n/8 oints. Case (b): H 3 and one of NW() or SW() have n/8 oints. If NW() does not, both SW() and H have n/8 oints. First we rove Theorem 1 for bitonic configurations. Let be the highest oint of the lower (heavy) cell H of the middle column. The quadrants NW() and NE() contain the heavy cells H 1 and H 3, therefore both of them have at least n/8 oints of S. All the at least n/4 oints in the heavy cell H that belong to S lie in the half-lane SW() SE(). Thus, at least one of the two quadrants, SW() or SE(), must contain at least n/8 oints. This quadrant together with the oosite one forms a good air, concluding the roof for bitonic configurations. Next we establish Theorem 1 for monotone configurations. We have to use a slightly different argument, because we have no information about the relative vertical order of the oints of H 1 and H. Choose to be the lowest oint in S that belongs to H, the uer cell in the middle column. Since SE() contains the heavy cell H 3, it has at least n/8 oints of S. If NW(), the

4 quadrant oosite to SE(), also has at least n/8 oints, we are done. Otherwise, the number of oints in SW() must exceed n/8, because NW() SW() covers the n/4 oints in the first column. Moreover, if NW() has fewer than n/8 oints, then there must be more than n/8 oints in NE() H. Therefore, in this case both quadrants SW() and NE() have n/8 oints. All of the above stes can be erformed in linear time, first using a linear-time selection algorithm [1] to find the vertical lines with the roerties described in the roof above, and the halving horizontal line, and finally erforming several asses through the oints to determine whether one has a bitonic or monotone configuration and to find the highest and lowest oint in the aroriate cell. Hence one can find a oint of deth at least n/8 from an arbitrary oint set in O(n) time. This comletes the roof of Theorem 1. H1 H1 H H'3 H (a) Bitonic configuration Monotone configuration (b) NE() is emty. H H1 H H1 H'1 H' Monotone configuration (c) NW() is emty. Monotone configuration (d) SW() is emty. Figure : Proof of Theorem 3 when all oints belong to the convex curve shown. For each i = 1,, 3, we let H i be the comlement of H i in column i. Case (a): if SE() is emty, then SW() contains all oints in H and NE() contains all oints in the last column. Case (b): if NE() is emty, then NW() contains all the n/4 oints of H, and SE() H 3 H 3 contains all the n/4 oints of the right column; (c) if NW() is emty, then the situation is symmetric; and (d) if SW() is emty, then NW() contains all the n/4 oints in the left column and SE() contains all the n/ oints in the bottom row. Now we rove Theorem 3. We use the same configurations as for Theorem 1. (See Figure.) We want to use the assumtion that every oint S is exosed, that is, one of the four quadrants determined by is emty. We first rove Theorem 3 for bitonic configurations. As in the roof of Theorem 1, let S

5 be the highest oint in H. Observe that neither NE() nor NW() can be emty since each of them contains a heavy cell (H 1 or H 3 ). Suose without loss of generality that SE() is emty (see Figure (a)). Then is not only the highest, but also the rightmost oint in its cell. Hence, SW() contains all at least n/4 oints in H, while NE() contains all n/4 elements of the last column. It remains to establish Theorem 3 for monotone configurations. Again, let be the lowest element of S in H. Now one of the four quadrants at must be emty. Observe that SE() covers the heavy cell H 3, so it is not emty. If NE() is emty, then NW () contains all the at least n/4 oints in H, and SE() contains all the n/4 oints of S in the third column, so we are done (Figure (b)). By symmetry, if NW () is emty, both NE() and SW() contain at least n/4 oints (Figure (c)). Finally, if SW () is emty, then NW () contains all the n/4 oints of S in the first column, and SE() contains all the n/ oints of S in the second row, so we are also done in this case (Figure (d)). The running time analysis for finding such a oint of deth at least n/4 is the same as for Theorem 1. This comletes the roof of Theorem 3. 3 Proof of Theorem Let ε be a small ositive number, let θ be the counterclockwise rotation by π let S 1 denote the set consisting of the following four oints: about the origin, and T := ( ε, 1), L := ( 1, ε), B := (ε, 1), R := (1, ε). Setting T 1 = {T }, L 1 = {L}, B 1 = {B}, R 1 = {R}, we have S 1 = T 1 L 1 B 1 R 1. Now let k > 1, and suose that T k 1, L k 1, B k 1, R k 1, and S k 1 = T k 1 L k 1 B k 1 R k 1 have already been defined. Let T k := T k 1 (ε k ), L k := L k 1 (ε k θ( )), B k := B k 1 (ε k θ ( )), R k := R k 1 (ε k θ 3 ( )), where = {L, B, R} and stands for the Minkowski sum. Finally, define S k := T k L k B k R k. See Figure 3. Obviously, we have S k = 4 3 k 1, and the set S k is invariant under the rotation θ. Clearly, o(, S 1 ) = 1 holds for every S 1. It is easy to show by induction on k that NW() T k 3k 1 + 1, for every T k. For k = 1 there is nothing to rove. Let k > 1, and suose that we have already verified the statement for k 1. Write T k in the form q (ε k r), for some q T k 1 and r. Using the induction hyothesis, we obtain that NW(q), and hence also NW(), contains at most 3k +1 1 oints of T k 1 in its interior, each giving rise to recisely three oints of T k. In addition, by the construction of, the quadrant NW() contains at most two elements of the trile {q (ε k r) r }. That is, we have ( 3 k + 1 NW() T k 3 ) 1 + = 3k 1 + 1,

6 T 1 T L 1 R 1 S 1 S L R B 1 B Figure 3: The construction for the roof of Theorem. The set S 1 of four oints (right), and the set S of 1 oints has oosite-quadrant deth, shown with a air of oosite quadrants with and 5 oints, resectively. as required. Analogously, we obtain NE() T k 3k 1 + 1, for every T k, because NE() also contains at most two elements of the trile {q (ε k r) r }. These two bounds imly that o(, S k ) 3k = S k + 4, 8 for every k 1 and T k. The easy verification of the fact that equality can be attained here is left to the reader. Since the construction is symmetric under the rotation theta, we obtain that the same bound is true for every S k. This comletes the roof of Theorem. 4 Concluding remarks The notion of oosite-quadrant deth arose in connection with conflict-free coloring roblems [4,5]. Take n oints in the lane in general osition and connect two of them by an edge if the smallest axis-arallel rectangle containing them covers no other oint of the set. Is it true that this grah contains a large indeendent set? Let f(n) denote the size of the largest indeendent set whose existence can always be guaranteed and suose for a moment that there exists a oint whose oosite-quadrant deth is n/4. That is, NW() and SE(), say, both contain at least n/4 oints. The oint guarantees that no air of oints lying in oosite quadrants are connected by an edge. Thus, after recursively finding indeendents sets in the subgrahs induced by NW() and SE(), their union is always an indeendent set in the whole grah. Thus, we would have f(n) f(n/4) 1, imlying that f(n) = Ω( n). Unfortunately, Theorem 1 yields only f(n) = Ω(n 1/3 ), and, by Theorem, this method cannot lead to a better bound. Nevertheless, by an easy alication of a lemma of Erdős and Szekeres [3], we obtain f(n) = Ω( n), and this bound can be slightly imroved [9]. It is not known whether f(n) = o(n).

7 The natural extension of oosite-quadrant deth to oosite-orthant deth in three dimensions seems to lead to interesting questions. Curiously, the oosite-orthant deth of the intersection oint of three halving lanes can be zero. For instance, lace n/4 oints in small neighborhoods of the oints (1, 1, 1), (1, 1, 1), ( 1, 1, 1) and ( 1, 1, 1). However, it is not hard to see that any set of n oints in three-dimensional sace has an element of oosite-orthant deth at least cn, where c > 0 is an aroriate constant. Indeed, it follows by the reeated alication of the lemma of Erdős and Szekeres [3] mentioned earlier that any 9-tule of oints in three-dimensional sace contains a trile (, q, r) such that and r belong to oosite orthants with resect to q. Clearly, the number of distinct triles that can be obtained in this way is at least ( n ( n 9) 3) ( n 3 ) = 6 Therefore, there exists a oint q which lays the role of the middle element in at least ( n 1 3) n ( 9 3) such triles. At least one quarter of these triles have the same tye, in the sense that the other two elements belong to the same air of oosite orthants. That is, the oosite-orthant deth of q is at least ( n 1 3) 4n ( 9 = Ω(n), 3) as required. The same argument works in higher dimensions. For similar results, consult [8]. An alternate extension of o(, S) to three dimensions is max σ,τ min( S σ, S τ ), over all airs σ, τ of oosite quadrants determined by a air of axis-aligned lanes. There are now six airs of oosite quadrants. With this definition, all lanar lower bounds remain valid in three dimensions, but they are not necessarily tight. Acknowledgements Thanks to Boris Aronov, Joshua Cooer, Raghavan Dhandaani, Jeff Erickson, John Iacono, and Rado s Radoi cić for discussions and comments. References [1] T. Cormen, C. Leiserson, R. Rivest, and C. Stein. Introduction to Algorithms (nd ed.). MIT Press, Cambridge, MA, 001. [] H. Edelsbrunner. Algorithms in Combinatorial Geometry. Sringer, Berlin, [3] P. Erdős and G. Szekeres. A combinatorial roblem in geometry. Comositio Math. : , [4] G. Even, Z. Lotker, D. Ron, and S. Smorodinsky. Conflict-free colorings of simle geometric regions with alications to frequency assignments in cellular networks. SIAM J. Comut. 33(1):94-136, 003. [5] S. Har-Peled and S. Smorodinsky. Conflict-free colorings of oints and simle regions in the lane. Discrete and Comut. Geom. 34:47-70, 005. ( 9 3).

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