European Journal of Combinatorics

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1 European Journal of Combinatorics 30 (2009) Contents lists available at ScienceDirect European Journal of Combinatorics ournal homepage: Generalizations of Heilbronn s triangle problem Hanno Lefmann Fakultät für Informatik, TU Chemnitz, D Chemnitz, Germany a r t i c l e i n f o a b s t r a c t Article history: Available online 2 April 2009 For given integers d, 2 and any positive integers n, distributions of n points in the d-dimensional unit cube [0, 1] d are investigated, where the minimum volume of the convex hull determined by of these n points is large. In particular, for fixed integers d, k 2 the existence of a configuration of n points in [0, 1] d is shown, such that, simultaneously for = 2,..., k, the volume of the convex hull of any points among these n points is Ω(1/n ( 1)/(1+ d +1 ) ). Moreover, a deterministic algorithm is given achieving this lower bound, provided that d + 1 k Elsevier Ltd. All rights reserved. 1. Introduction Originally, Heilbronn s triangle problem asks for the supremum value 3 (n), over all distributions of n points in the unit square [0, 1] 2, of the minimum area of a triangle among n points in [0, 1] 2. For prime numbers n, the points (1/n) (i mod n, i 2 mod n), i = 0,..., n 1, show that 3 (n) = Ω(1/n 2 ) as has been observed by Erdős. By using results on the independence number of certain 3- uniform hypergraphs, this lower bound has been improved by Komlós, Pintz and Szemerédi [11] to 3 (n) = Ω(log n/n 2 ). Upper bounds on 3 (n) have been proved by Roth [18 21] and Schmidt [22], and the currently best upper bound 3 (n) = O(2 c log n /n 8/7 ), for some constant c > 0, is due to Komlós, Pintz and Szemerédi [10], compare [6] for an algorithmic version. A variant of this problem, which has been considered by Schmidt [22], asks, for integers 2 n, for the supremum value (n), over all distributions of n points in [0, 1] 2, of the minimum area of the convex hull of among n points. Schmidt obtained 4 (n) = Ω(1/n 3/2 ), and observed that (n) +1 (n) for 2. For fixed 3, the currently best lower bound (n) = Ω((log n) 1/( 2) /n ( 1)/( 2) ) has been obtained in [14], while for fixed 4 only the upper bound (n) = O(1/n) is known. Extensions of Heilbronn s triangle problem to higher dimensions have been investigated in several papers. For integers d,, n 2, let,d (n) be the supremum, over all distributions of n points in address: lefmann@informatik.tu-chemnitz.de /$ see front matter 2009 Elsevier Ltd. All rights reserved. doi: /.ec

2 H. Lefmann / European Journal of Combinatorics 30 (2009) the d-dimensional unit cube [0, 1] d, of the minimum (( 1)-dimensional for d + 1) volume of a -point simplex among the n points. In [2] Barequet has shown that d+1,d = Ω(1/n d ) for fixed integers d 2, and in [4] Barequet and Naor proved the bounds f (k, d)/n k d k+1 k,d (n) (k k/d d k/2 )/(k! n k d ) for arbitrary fixed integers 1 k d, where the function f (k, d) depends on d and k only. For fixed integers d,, 3 d + 1, the currently best lower bounds on,d (n) are,d (n) = Ω((log n) 1/(d +2) /n ( 1)/(d +2) ), compare [12] and [15]. Notice, that the parameter,d (n) is denoted in [4] by H 1,d (n). An on-line version of Heilbronn s triangle problem the points have to be positioned one after the other in [0, 1] d and suddenly this process stops has been considered by Barequet for dimensions d = 3 and d = 4 in [3], where he obtained a configuration of n points in [0, 1] d such that for the minimum volume of any simplex determined by (d + 1) among these n points the lower bounds are Ω(1/n 10/3 ) and Ω(1/n 127/24 ), respectively. These results have been extended in the on-line situation by Barequet and Shaikhet [5,23] to an arbitrary fixed dimension by using a packing argument the lower bound Ω(1/n (d+1) ln(d 2) 0.265d ) for fixed d 5, compare [13] for the case of triangles in [0, 1] d. In connection with some range searching problems Chazelle [9] has investigated the function,d (n) for values, which depend on n, and he showed the asymptotically correct order Θ(/n) for log n n and fixed d 2. In view of this result, it might be of interest to know whether there exists for fixed d, k 3, a single configuration of n points in [0, 1] d, which yields good lower bounds on the minimum volume of any points among the n points, simultaneously for = 3,..., k. We investigate this for fixed k. Moreover, for the range d + 1 k, the arguments can be made constructive by a deterministic polynomial in n time algorithm. Theorem 1. Let d, k 2 be fixed integers. (i) Then, for any integer n k, there exists a configuration of n points in the d-dimensional unit cube [0, 1] d, such that, simultaneously for = 2,..., k, the volume of the convex hull of any points among these n points is Ω(1/n ( 1)/(1+ d +1 ) ). (ii) Moreover, for fixed k d+1 and any integer n k, there is a deterministic polynomial time algorithm, which finds a configuration of n points in [0, 1] d, which, simultaneously for = d+1,..., k, achieves the lower bound Ω(1/n ( 1)/(1+ d +1 ) ) on the volume of the convex hull of any among the n points in [0, 1] d. For fixed integers d, 2, Theorem 1 gives,d = Ω(1/n ( 1)/(1+ d +1 ) ). Concerning upper bounds, a partition of the unit cube [0, 1] d into d-dimensional subcubes each of side length Θ(1/n 1/d ), yields,d (n) = O(1/n ( 1)/d ) for d+1 and,d (n) = O(1/n) for d+1. For even integers, the upper bound can be improved to,d (n) = O(1/n ( 1)/d+( 2)/(2d(d 1)) ), see Brass [7] for = d + 1, and [15] for 4 d + 1. Somewhat surprisingly, achieving by a deterministic polynomial time algorithm for the same n points in [0, 1] d the lower bound,d (n) = Ω(1/n ( 1)/(1+ d +1 ) ), simultaneously for = 2,..., k, where d, k 2 are fixed integers, causes so far some difficulties with respect to the simplices of low dimension, for = 4 and d 5 say, in particular, counting in the d-dimensional N N-standard lattice the number of configurations of lattice points with volume at most v. 2. Proof of Theorem 1(i) Let dist (P i, P ) be the Euclidean distance between the points P i, P [0, 1] d. A simplex given by the points P 1,..., P [0, 1] d, 2 d+1, is the set of all points P 1 + λ i=2 i(p i P 1 ) with λ i=2 i 1 and λ 2,..., λ 0. The (( 1)-dimensional) volume of a simplex given by points P 1,..., P [0, 1] d, 2 d + 1, is defined by vol (P 1,..., P ) := 1/( 1)! i=2 dist (P i; P 1,..., P i 1 ), where dist (P i ; P 1,..., P i 1 ) is the Euclidean distance of the point P i from the affine space P 1,..., P i 1 (1)

3 1688 H. Lefmann / European Journal of Combinatorics 30 (2009) over the reals generated by the vectors P 2 P 1,..., P i 1 P 1 and attached to the point P 1. For points P 1,..., P [0, 1] d, d + 1, let vol (P 1,..., P ) be the (d-dimensional) volume of the convex hull of the points P 1,..., P. Next we prove part (i) of Theorem 1. Proof. Let d, k 2 be fixed integers and let n k be any integer. We select uniformly at random and independently of each other N := k n points P 1, P 2,..., P N from the unit cube [0, 1] d. Set v := β /n ( 1)/(1+ d +1 ) for constants β > 0, = 2,..., k, which will be specified later. Let V := {P 1, P 2,..., P N } be the random set of chosen points in [0, 1] d. For = 2,..., k, let E be the set of all -element subsets {P i1,..., P i } [V] of points such that vol (P i1,..., P i ) v. We give upper bounds on the expected numbers E( E ) of -element sets in E, = 2,..., k, and we show that for a suitable choice of v 2,..., v k all numbers E( E ) are not too big, i.e., E( E 2 ) + + E( E k ) (k 1) n. Thus, there exists a choice of N points P 1, P 2,..., P N [0, 1] d such that E E k (k 1) n. Then, for = 2,..., k, we delete one point from each -element set of points in E. The set of the remaining points has cardinality at least k n (k 1) n = n, and the volume of the convex hull of any points of these at least n points is bigger than v. Lemma 1. Let d, k 2 be fixed integers. For = 2,..., k, there exist constants c,d > 0 such that for every number v > 0 it is (2) E( E ) c,d N v 1+ d +1. (3) Proof. For given v > 0 and random points P 1,..., P [0, 1] d we give an upper bound on the probability Prob (vol (P 1,..., P ) v ). Let the points P 1,..., P be numbered such that for 2 g h and g d + 1 it is dist (P g ; P 1,..., P g 1 ) dist (P h ; P 1,..., P g 1 ). (4) The point P 1 can be anywhere in [0, 1] d. Given P 1, the probability, that the point P 2 [0, 1] d has from P 1 a Euclidean distance within the infinitesimal range [r 1, r 1 + dr 1 ], is at most the difference of the volumes of the d-dimensional balls with center P 1 and with radii (r 1 + dr 1 ) and r 1, respectively, hence Prob (r 1 dist (P 1, P 2 ) r 1 + dr 1 ) d C d r d 1 1 dr 1, where C d is the volume of the d-dimensional unit ball in R d. Given the points P 1 and P 2 with dist (P 1, P 2 ) = r 1, the probability that the Euclidean distance of the point P 3 [0, 1] d from the affine line P 1, P 2 is within the infinitesimal range [r 2, r 2 + dr 2 ] is at most the difference of the volumes of two cylinders centered at the affine line P 1, P 2 with radii r 2 + dr 2 and r 2, respectively, and, by assumption (4), with height 2 r 1 = 2 dist (P 1, P 2 ), thus Prob (r 2 dist (P 3 ; P 1, P 2 ) r 2 + dr 2 ) 2 r 1 (d 1) C d 1 r d 2 2 dr 2. In continuing in this manner, given the points P 1,..., P g, g < and g < d + 1, with dist (P x ; P 1,..., P x 1 ) = r x 1, x = 2,..., g, by (4) for g 2 and g d 1 the proection of the point P g+1 onto the affine space P 1,..., P g is contained in a (g 1)-dimensional box with volume 2 g 1 r 1 r g 1, hence Prob (r g dist (P g+1 ; P 1,..., P g ) r g + dr g ) 2 g 1 r 1 r g 1 (d g + 1) C d g+1 r d g g dr g. (5) For g = 1 < d, in order to satisfy vol (P 1,..., P ) v, we must have 1 ( 1)! dist (P i ; P 1,..., P i 1 ) v. (6) i=2

4 H. Lefmann / European Journal of Combinatorics 30 (2009) By (4) the proection of the point P onto the affine space P 1,..., P 1 is contained in a ( 2)- dimensional box with volume 2 2 r 1 r 2, and by (6) the point P has Euclidean distance at most (( 1)!v )/(r 1 r 2 ) from the affine space P 1,..., P 1, which happens with probability at most ( 1)! d v r 1 r 2 C d +2. (7) r 1 r 2 For d g 1, again by (4), the proection of the point P g+1 onto the affine space P 1,..., P d is contained in a (d 1)-dimensional box with volume at most 2 d 1 r 1 r d 1, and by (6), the point P g+1 has Euclidean distance at most (d! v )/(r 1 r d 1 ) from the affine space P 1,..., P d, which happens with probability at most 2 d 1 r 1 r d 1 2 d! v r 1 r d 1 = d! 2 d v. (8) Thus, for d with (5) and (7), and for some constants c,,d c,d > 0 we obtain Prob (vol (P 1,..., P ) v ) g=1 r 2 =0 r 1 =0 2 2 Cd +2 (( 1)!) d +2 v d +2 (r 1 r 2 ) d g 1 r 1 r g 1 (d g + 1) C d g+1 r d g g dr 2... dr 1 c,d vd +2 r 2 =0 2 r 1 =0 g=1 r 2 2g 3 g dr 2 dr 1 c,d vd +2 as 2 2 g 3 1. (9) Similarly, for = d + 1,..., k, by (5) and (8) for constants c,,d c,d > 0 we infer Prob (vol (P 1,..., P ) v ) g=1 r d 1 =0 r 1 =0 (d! 2 d v ) d d 1 2 g 1 r 1 r g 1 (d g + 1) C d g+1 r d g g drd 1... dr 1 c,d v d r d 1 =0 d 1 r 1 =0 g=1 r 2d 2g 1 g drd 1 dr 1 c,d v d as 2 d 2 g 1 1. (10) By (9) and (10), we have Prob (vol (P 1,..., P ) v ) c,d v1+ d +1 for constants c >,d 0, = 2,..., k. Since there are choices for out of the N random points P 1,..., P N [0, 1] d, inequality (3) follows. N By (3) and Markov s inequality there exist N = k n points P 1,..., P N in [0, 1] d such that for = 2,..., k the corresponding sets E fulfill E k c,d N v 1+ d +1. (11) Fix constants β := 1/(c,d k +1 ) 1/(1+ d +1 ), = 2,..., k. We claim that it is E N/k. (12)

5 1690 H. Lefmann / European Journal of Combinatorics 30 (2009) Namely, for = 2,..., k, by (2) and (11) with N = k n, we infer E N/k k c,d N v 1+ d +1 N/k k +1 c,d n 1 v 1+ d +1 1 k +1 c,d β 1+ d +1 1, which holds by the choice of β > 0. By (12), we have E E k ((k 1)/k) N. For = 2,..., k, we discard one point from each -element set in E. Then, the set of remaining points has cardinality at least N/k = n. These at least n points in [0, 1] d satisfy, simultaneously for = 2,..., k, that the volume of the convex hull of each points is bigger than v = β /n ( 1)/(1+ d +1 ), which finishes the proof of Theorem 1(i). 3. Proof of Theorem 1(ii) Throughout this section, let d, k 2 be fixed integers with k d + 1. Let B d (T) denote the d-dimensional ball with radius T around the origin. To provide a deterministic polynomial time algorithm which, for any integer n > 0, finds a configuration of n points in [0, 1] d, such that the volume of the convex hull of sets with few points is large, we discretize the d-dimensional unit cube [0, 1] d by considering all points in B d (T) Z d for T large enough. With this discretization we have to take care of degenerate sets of points. A set {P 1,..., P } B d (T) Z d with d + 1 is said to be degenerate, if all points P 1,..., P are contained in a (d 1)-dimensional affine subspace of R d, otherwise {P 1,..., P } is said to be non-degenerate. In our arguments for proving Theorem 1(ii) we use lattices. A lattice L in Z d is a subset of Z d, which is generated by all integral linear combinations of some linearly independent vectors b 1,..., b m Z d, hence L = Zb Zb m. The parameter m = rank(l) is called the rank of the lattice L, and the set B = {b 1,..., b m } is a basis of L. The set F B := { m α i=1 i b i 0 α i 1, i = 1,..., m} R d is the fundamental parallelepiped of B and its volume is vol(f B ) := (det(g(b) G(B))) 1/2, where G(B) := (b 1,..., b m ) d m is a d m generator matrix of B. If B and B are two bases of a lattice L in Z d, then vol(f B ) = vol(f B ), see [8]. For vectors a = (a 1,..., a d ) R d and b = (b 1,..., b d ) R d let a, b := d i=1 a i b i be the standard scalar product. The length a of a vector a R d is defined by a := a, a. For a lattice L in Z d let span(l) be the linear space over the reals, which is generated by the vectors in L. For a subset S = {P 1,..., P k } R d of points the rank of S is the dimension of the linear space over the reals, which is generated by the vectors P 2 P,..., 1 P k P 1. A vector a = (a 1,..., a d ) Z d \ {0 d } is said to be primitive, if the greatest common divisor of a 1,..., a d is equal to 1 and a > 0 with = min{i a i 0}. A lattice L in Z d is said to be (d 1)- maximal, if rank(l) = d 1 and no other lattice L L in Z d with rank(l ) = d 1 contains L as a proper subset. The next result shows that the (d 1)-maximal lattices in Z d and the primitive vectors a = (a 1,..., a d ) Z d \ {0 d } are in a one-to-one correspondence. Proposition 1. (i) For each lattice L in Z d with rank(l) = d 1 1 there is exactly one primitive vector a L = (a 1,..., a d ) Z d \ {0 d } with a L, x = 0 for every x L. The vector a L is the primitive normal vector of L. (ii) For each lattice L in Z d with rank(l ) = d 1 there is exactly one (d 1)-maximal lattice L in Z d with L L. (iii) There is a biection between the set of all (d 1)-maximal lattices L in Z d and the set of all primitive vectors a L in Z d. For a lattice L in Z d, a residue class of L is a set L of the form L = x + L = {x + v v L} with x Z d. In our arguments, we use the following Lemmas 2 5; their proofs can be found in [17].

6 H. Lefmann / European Journal of Combinatorics 30 (2009) Lemma 2. Let L be a (d 1)-maximal lattice in Z d with primitive normal vector a L Z d, basis B, and fundamental parallelepiped F B. (i) There exists a point v Z k \ L such that Z d can be partitioned into the residue classes s v + L, s Z, and, for each point x L, it is dist(s v + x, span(l)) = s / a L. (ii) vol(f B ) = a L. Lemma 3. Let d N be fixed. Let S B d (T) Z d be a set of points with rank(s) d 1. Then there exists a (d 1)-maximal lattice L in Z d such that S is contained in some residue class L = v + L of L for some point v Z d, and L has a basis b 1,..., b d 1 Z d with b i = O(T), i = 1,..., d 1. Lemma 4. Let d N be fixed. Let L be a (d 1)-maximal lattice of Z d with primitive normal vector a L Z d, and let B = {b 1,..., b d 1 } be a basis of L with b i = O(T), i = 1,... d 1. Let L be a residue class of L. Then the following hold: a L = O(T d 1 ) and L B d (T) = O ( T d 1 / a L ). For integers g, l N, let r g (l) be the number of g-tuples (x 1,..., x g ) Z g such that x x2 g = l. Lemma 5. Let g, r N be fixed integers. Then, for all integers m N: m { r g (l) ( O m = g/2 r) if g/2 r > 0 l r O (log m) if g/2 r 0. l=1 Now we have all the tools to prove Theorem 1(ii). Proof. Set v := β T d /n ( 1)/( d) for suitable constants β > 0, = d + 1,..., k, which will be fixed later, where T/ log T = ω(n). We construct for = d + 1,..., k two types of -element sets. For points P i1,..., P i B d (T) Z d, let {P i1,..., P i } E if vol (P i1,..., P i ) v and {P i1,..., P i } is not contained in a (d 1)-dimensional affine subspace of R d, i.e., the set {P i1,..., P i } is non-degenerate. Moreover, let {P i1,..., P i } E 0 if {P i1,..., P i } is degenerate. Next we give upper bounds on the sizes E and E 0, = d + 1,..., k. Lemma 6. Let d, k 2 be fixed integers with k d + 1. For = d + 1,..., k, there exist constants c,0 > 0, such that the numbers E 0 of degenerate -element sets of points in Bd (T) Z d satisfy E 0 c,0 T (d 1)+1 log T. (13) Proof. By Lemma 3, each degenerate -element subset of points in B d (T) Z d is contained in a residue class L of some (d 1)-maximal lattice L in Z d, and L has a basis b 1,..., b d 1 Z d with b i = O(T), i = 1,..., d 1. Fix a (d 1)-maximal lattice L in Z d, which is determined by its primitive normal vector a L Z d where a L = O(T d 1 ) by Lemma 4. By Lemma 2(i), there are O(T a L ) residue classes L of L with ( L B d (T). By Lemma 4, from each set L B d (T) we can select points in obtain a degenerate -element set, hence we infer E 0 = O T d 1 / a T a a Z d, a =O(T d 1 ) = O T (d 1)+1 1 a 1 a Z d, a =O(T d 1 ) O(T d 1 / a L ) ) ways to

7 1692 H. Lefmann / European Journal of Combinatorics 30 (2009) = O T (d 1)+1 O(T 2d 2 ) l=1 r d (l) l ( 1)/2 = O ( T (d 1)+1 log T ), as, by Lemma 5, we have m l=1 r d(l)/l ( 1)/2 = O(log m) for d + 1. Lemma 7. Let d, k 2 be fixed integers with k d + 1. For = d + 1,..., k, there exist constants c > 0, such that the numbers E of non-degenerate -element sets of points in B d (T) Z d with volumes of their convex hulls at most v, fulfill E c T d2 v d. (14) Proof. For = d + 1,..., k, consider points P 1,..., P B d (T) Z d with vol (P 1,..., P ) v, where {P 1,..., P } is non-degenerate. Let these points be numbered such that for 2 g h and g d + 1 it is dist (P g ; P 1,..., P g 1 ) dist (P h ; P 1,..., P g 1 ). (15) By Lemma 3, the points P 1,..., P d B d (T) Z d are contained in a residue class L of some (d 1)- maximal lattice L in Z d with primitive normal vector a L Z d, where L has a basis b 1,..., b d 1 Z d with b i = O(T), i = 1,..., d 1. By Lemma 4, it suffices to consider primitive normal vectors a L Z d with a L = O(T d 1 ). Fix a (d 1)-maximal lattice L in Z d with primitive normal vector a L Z d. By Lemma 2(i), there are O(T a L ) residue classes L of L with L B d (T). By Lemma 4, from each set L B d (T) we ( can select d points P 1,..., P d in O(T d 1 / a L ) d ) ways. With (15) we infer for the (d 1)-dimensional volume vol (P 1,..., P d ) > 0, as otherwise {P 1,..., P } is degenerate. Again by (15) the proection of each point P d+1,..., P B d (T) Z d onto the residue class L is contained in a (d 1)-dimensional box of volume 2 d 1 (d 1)! vol (P 1,..., P d ), which, by Lemma 2(ii), contains at most (d 1)! 2 2d 2 vol (P 1,..., P d )/ a L (16) points of L, as each fundamental parallelepiped contains 2 d 1 lattice points. Now vol (P 1,..., P ) v implies dist (P i, P 1,..., P d ) d v /vol (P 1,..., P d ), and by Lemma 2(i), each point P i B d (T) Z d, i = d + 1,...,, is contained in one of at most a L d v /vol (P 1,..., P d ) 3 a L d v /vol (P 1,..., P d ) (17) residue classes L of L. By (16) in each such residue class L we can choose at most (d 1)! 2 2d 2 vol (P 1,..., P d )/ a L points P i B d (T) Z d, hence with (17), each point P i, i = d + 1,...,, can be chosen in at most d! 3 2 2d 2 v ways, and we infer the upper bound E = O T d 1 / a T a v d d a Z d, a =O(T d 1 ) = O = O T d 2 d+1 v d T d 2 d+1 v d 1 a d 1 a Z d, a =O(T d 1 ) O(T 2d 2 ) l=1 r d (l) l (d 1)/2 as m l=1 r d(l)/l (d 1)/2 = O(m 1/2 ) by Lemma 5. = O(T d2 v d ),

8 H. Lefmann / European Journal of Combinatorics 30 (2009) For fixed integers d,, k 2 the sets E and E 0, can easily be constructed by brute force in time polynomial in T. Let B d (T) Z d = C d T d, where C > d 0 is a constant. We enumerate the lattice points in Bd (T) Z d by P 1,..., P C dt d, and to each point P i we associate a variable p i [0, 1], i = 1,..., C d T d, and define a function F: [0, 1] C d T d R as follows: F(p 1,..., p C dt d) := 2pC d T d /2 + k =d+1 C d T d ( 1 p i 2 i=1 {i 1,...,i } E 0 ) + k =d+1 p i1 p i 2 k p c,0 T (d 1)+1 log T. {i 1,...i } E p i1 p i 2 k p c T d2 v d Fix p 1 := := p C dt d := p = (2 k n)/(c d T d ) in the beginning. Then it is p 1 as T/ log T = ω(n). We infer, using 1 + x e x for every x R and Lemmas 6 and 7, that F(p,..., p) < (2/e) pc d T d /2 + (2k 2d)/(2k). This is less than 1 for p C d T d 7 ln k, which holds for n k 2. By using the linearity of F(p 1,..., p C dt d) in each p i (with fixed p, i), we minimize F(p 1,..., p C dt d) step by step by fixing one after the other p i := 0 or p i := 1 for i = 1,..., C d T d, and finally we obtain F(p 1,..., p C dt d) < 1. The set V = {P i B d (T) Z d p i = 1} of points yields subsets E 0 := [V ] E 0 and E := [V ] E of -element sets, = d + 1,..., k, such that V p C d T d /2 (18) E 2 k p c T d2 v d E 0 2 k p c,0 T (d 1)+1 log T. (20) Namely, otherwise, if one of the inequalities (18), (19), or (20) would not hold, then F(p 1,..., p C d) > dt 1, which would be a contradiction. By the choice of the numbers v, = d + 1,..., k, the running time of this procedure is O(T d + k ( E =d+1 + E 0 )) = O(T dk ), i.e., polynomial in T for fixed integers d, k 2. Lemma 8. For = d + 1,..., k, and 0 < β (C d /(2+2 k +1 c )) 1/( d), it is E V /(2 k). (19) Proof. By (18) and (19) with v := β T d /n ( 1)/( d), and p = (2 k n)/(c d T d ), and with β > 0 it is E V /(2 k) 2 k p c T d2 v d p C d T d /(4 k) ( 2 k 1 n 8 k 2 C d T c d T d2 d β T d 2 +2 k +1 c β d C d, n 1 d ) d C d which holds for β d C d /(2+2 k +1 c ), = d + 1,..., k. Lemma 9. For = d + 1,..., k, and T/(log T) 1/( 1) = ω(n), it is E 0 V /(2 k).

9 1694 H. Lefmann / European Journal of Combinatorics 30 (2009) Proof. By (18) and (20), with p = (2 k n)/(c d T d ), = d + 1,..., k, we infer E 0 V /(2 k) 2 k p c,0 T (d 1)+1 log T p C d T d /(4 k) 2 k 1 n 8 k 2 C d T c d,0 T (d 1) d+1 log T C d 2 +2 k +1 c,0 n 1 log T C T 1 which holds for T/(log T) 1/( 1) = ω(n). d, With say T := n (log n) 2, and β := C d /(2+2 k +1 c ) 1/( d), = d + 1,..., k, all assumptions in Lemmas 8 and 9 are fulfilled. By discarding in time O( V + k ( E 0 =d+1 + E )) = O(T kd ) one point from each -element set in E and E 0, = d + 1,..., k, the remaining points yield a subset V V B d (T) Z d of size at least V /k p C d T d /(2 k) = n. These at least n points in V satisfy that the volume of the convex hull of any of these points, = d + 1,..., k, is at least v, i.e. Ω(T d /n ( 1)/( d) ). After rescaling by the factor (2 T) d, we have n points in the unit cube [0, 1] d such that the volume of the convex hull of any of these points is Ω(1/n ( 1)/( d) ), = d + 1,..., k. Altogether the time of this algorithm is O((n (log n) 2 ) dk ) for fixed d, k 2, hence polynomial in n, which finishes the proof of Theorem 1(ii). 4. Remarks By more involved techniques using results on the independence number of certain hypergraphs for some range of k the lower bound (1) in Theorem 1 can be improved by a polylogarithmic factor, as is indicated below, and complements the work from [15]. For a hypergraph G the notation G = (V, E 2 E k ) means that V is its vertex set and E i is the set of all i-element edges in G, i = 2,..., k. The independence number α(g) of a hypergraph G = (V, E) is the largest size of a subset I V which contains no edges from E. A hypergraph G is said to be linear if distinct edges intersect in at most one vertex. The following result from [16] extends an earlier one by Atai, Komlós, Pintz, Spencer and Szemerédi [1] from uniform to arbitrary linear hypergraphs. Theorem 2. Let k 3 be fixed. Let G = (V, E 3 E k ) be a linear hypergraph on V = N vertices, such that for some number T 1 for i = 3,..., k the average degrees i E i /N for the i- element edges are at most T i 1 (log T) (k i)/(k 1). Then, the independence number α(g) fulfills α(g) = Ω((N/T) (log T) 1/(k 1) ). With Theorem 2 one can show the following existence result by more involved arguments than those used here (the details are given elsewhere): Let d 2 and k 3 with k d + 1 be fixed. For every n k there exist n points in the unit cube [0, 1] d, such that, simultaneously for = 3,..., k, the ( 1)- dimensional volume of any -point simplex among these n points is Ω((log n) 1/(d +2) /n ( 1)/(d +2) ). References [1] M. Atai, J. Komlós, J. Pintz, J. Spencer, E. Szemerédi, Extremal uncrowded hypergraphs, Journal of Combinatorial Theory Series A 32 (1982) [2] G. Barequet, A lower bound for Heilbronn s triangle problem in d dimensions, SIAM Journal on Discrete Mathematics 14 (2001) [3] G. Barequet, The on-line Heilbronn s triangle problem, Discrete Mathematics 283 (2004) [4] G. Barequet, J. Naor, Large k D simplices in the D-dimensional unit cube, Far East Journal of Applied Mathematics 24 (2006) [5] G. Barequet, A. Shaikhet, The on-line Heilbronn s triangle problem in d dimensions, Discrete & Computational Geometry 38 (2007)

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