Covering Spheres with Spheres

Discrete Comput Geom 2007) 38: 665 679 DOI 10.1007/s00454-007-9000-7 Covering Spheres with Spheres Ilya Dumer Received: 1 June 2006 / Revised: 14 July 2007 / Published online: 2 September 2007 Springer Science+Business Media, LLC 2007 Abstract Given a sphere of any radius r in an n-dimensional Euclidean space, we study the coverings of this sphere with solid spheres of radius one. Our goal is to design a covering of the lowest covering density, which defines the average number of solid spheres covering a point in a bigger sphere. For growing dimension n, we design a covering that gives the covering density of order n ln n)/2 for a sphere of any radius r>1 and a complete Euclidean space. This new upper bound reduces two times the order n ln n established in the classic Rogers bound. 1 Introduction Spherical coverings. Let Br n x) be a ball solid sphere) of radius r centered at some point x = x 1,...,x n ) of an n-dimensional Euclidean space R n : { Br n def n x) = z R n z i x i ) 2 r }. 2 i=1 We also use a simpler notation Br n if a ball is centered at the origin x = 0. For any subset A R n, we say that a subset CovA, ε) R n forms an ε-covering an ε-net) of A if A is contained in the union of the balls of radius ε centered at points x CovA, ε). In this case, we use notation CovA, ε) : A Bε n x). x CovA,ε) By changing the scale in R n, we can always consider the rescaled set A/ε and the new covering CovA/ε, 1) with unit balls B1 n x). Without loss of generality, below I. Dumer ) University of California, Riverside, USA e-mail: dumer@ee.ucr.edu

666 Discrete Comput Geom 2007) 38: 665 679 we consider these unit) coverings. One of the classical problems is to obtain tight bounds on the covering size CovBr n, 1) for any ball Bn r of radius r and dimension n. Another related covering problem arises for a sphere S n r { def n+1 = z R n+1 zi }. 2 = r2 Then a unit ball B n+1 1 x) intersects this sphere with a spherical cap i=1 C n r ρ, y) = Sn r Bn+1 1 x), which has some center y Sr n, half-chord ρ 1, and the corresponding half-angle α = arcsin ρ/r. The biggest possible cap Cr n 1, y) is obtained if the center x of the corresponding ball B n+1 1 x) is centered at the distance x = r 2 1 1) from the origin. To obtain a minimal covering, we shall consider the biggest caps Cr n 1, y) assuming that all the centers x satisfy 1). Covering density. Given a set A R n,let A denote n-dimensional volume Lebesque measure) of A. We then consider any unit covering CovA, 1) and define minimum covering density ϑa)= min CovA,1) x CovA,1) B n 1 x) A. A Minimal coverings have been long studied for the spheres S n r and the balls Bn r.the celebrated Coxeter Few Rogers lower bound [1] shows that for a sufficiently large ball B n r, ϑsr n ) c 0n. 2) Here and below c i denote some universal constants. A similar result also holds for any sphere Sr n of radius r n see Example 6.3 in [4]). Various upper bounds on the minimum covering density are obtained for Br n and Sr n by Rogers in the classic papers [2] and [3]. In particular, it follows from these papers that for a sufficiently large radius r, a ball Br n and a sphere Sn r can be covered with density ln ln n ϑ 1 + ln n + 5 ) n ln n. 3) ln n Despite recent improvements obtained in [4] and [5], respectively, for spheres Sr n and balls Br n of a relatively small radius r, the Rogers bound 3) is still the best asymptotic bound known for sufficiently large spheres, balls, and complete spaces R n of growing dimension n.

Discrete Comput Geom 2007) 38: 665 679 667 For a sphere Sr n of any dimension n 3 and an arbitrary radius r>1, the best universal upper bound known to date is obtained in [4] see Corollary 1.2 and Remark 5.1): ϑsr n ) 1 + 2 ) ) ln ln n e 1 + ln n ln n + n ln n. 4) n ln n Our main result is presented in Theorem 1, which reduces about two times the present upper bounds 3) and 4) forn. Theorem 1 Unit balls can cover a sphere Sr n of any radius r>1and any dimension n 3 with density 1 ϑsr n ) 2 + 2lnlnn + 5 ) n ln n. 5) ln n ln n For n, there exists o1) 0 such that ϑs n r ) 1 2 + 3 2 + o1) ) ln ln n ln n ) n ln n. 6) The following corollary to Theorem 1 see also [8]) shows that the Rogers bound can also be reduced about two times for the coverings of complete Euclidean spaces R n. Corollary 2 For n, unit balls can cover the entire Euclidean space R n with density ) 1 ϑr n ) 2 + o1) n ln n. 7) 2 Preliminaries: Embedded Coverings In this section, we obtain the estimates on ϑsr n ) that are similar to 3) and 4). However, we will introduce here a slightly different technique of embedded coverings that will be substantially extended in Section 3 to improve the former bounds 3) and 4). We will also employ most of our calculations performed in this section. Consider a sphere Sr n of some dimension n 3 and radius r>1. We use notation Cρ,y) for a cap Cr n ρ, y) whenever parameters n and r are fixed; we also use a shorter notation Cρ) when a specific center y is of no importance. In this case, Covρ) will denote any covering of Sr n with spherical caps Cρ). By definition, a covering Covρ) has covering density ϑ ρ = Ω ρ Covρ) where Ω ρ is the fraction of the surface of the sphere S n r covered by a cap Cρ), Ω ρ = Cρ) S n r.

668 Discrete Comput Geom 2007) 38: 665 679 For any τ<ρ 1, we extensively use inequality see Corollary 3.2 ii) in [4]): Ω τ Ω ρ τ ρ ) n 8) and its particular version Ω τ Ω 1 τ n obtained for ρ = 1. We begin with two preliminary lemmas, which will simplify our calculations. Let f 1 x) and f 2 x) be two positive differentiable functions. We say that f 1 x) moderates f 2 x) for x a if for all x a, f 1 x) f 1 x) f 2 x) f 2 x). Lemma 3 Consider m functions f i x) such that f 1 x) moderates each function f i x), i 2, for x a. Then inequality f 1 x) holds for any x a if it is valid for x = a. m f i x) i=2 Proof Note that f i x) = f i a) exp{s i x)}, where x s i x) def f i = t) a f i t) dt. Also, s i x) s 1 x) for all i 2. Therefore, m m m f 1 x) f i a) exp{s 1 x)} f i a) exp{s i x)}= f i x), i=2 i=2 i=2 which completes the proof. Lemma 4 For any n 4, 1 1 ) n < 1 + 1/ln n + 1/ln 2 n. 9) n ln n Proof For n = 4,...,7, the above inequality is verified numerically. Let t = 1 ln n.we take the logarithm of the left part of 9) and use power series. This gives the following inequalities n ln 1 t ) = n i=1 t i n i 1 <t+ t2 i ) 1 i t + t2 2n 15. 10) Here we observe that t< 1 2 for n 8, and replace the left-hand side of the first inequality with geometric series in its right-hand side, which in turn is bounded by i=1

Discrete Comput Geom 2007) 38: 665 679 669 taking n = 8. We also note that for any n 8, t + t2 15 < 1 2. 11) For any z< 1 2 we can also use inequality e z = i=1 z i i! 1 + z + 2z2 3. 12) By combining 10) and 12), we obtain the required estimate 1 1 ) n ) < 1 + t + t2 + 2 ) 2 t + t2 < 1 + t + t 2. n ln n 15 3 15 The latter inequality can be verified for any t<1/2directly, or by applying Lemma 3 for t = 1 z and z 2. An embedded algorithm. We employ the following parameters ε = 1, ρ = 1 ε, 13) n ln n ln ln n λ = 1 + ln n + 2 n. 14) To design a covering Cov1), we also use another covering Covε) : Sr n Cε,u) u Covε) with smaller caps Cε,u). In our first step, we randomly choose N points y Sr n, where λn ln n λn ln n 1 <N. 15) Ω ρ Ω ρ Consider the set {Cρ,y)} of N caps. In the second step, we take all centers u Covε) that are left uncovered by the set {Cρ,y)} and form the extended set {x}={y} {u }. This set {x} covers the entire set Covε) and therefore forms a unit covering, if the caps Cρ,x) are expanded to the caps C1, x), Cov1) : Sr n C1, x). x {x} Lemma 5 For any n 8, covering {x} has density ln ln n ϑ 1 + ln n + 3 ) n ln n. 16) ln n

670 Discrete Comput Geom 2007) 38: 665 679 Proof Any point u is covered by some cap Cρ,y) with probability Ω ρ. Our goal is to estimate the expected number N of centers u left uncovered after N trials. Here N = 1 Ω ρ ) N Covε). Note that 1 Ω ρ e Ω ρ1+ω ρ /2). Also, according to estimate 15), NΩ ρ >λnln n Ω ρ. Thus, 1 Ω ρ ) N e λn ln n Ω ρ)1+ω ρ /2) e λn ln n. 17) Here the second inequality holds since Ω ρ <Ω 1 < 1/2, whereas λn ln n>16. Next, we estimate the covering size Covε). Let this covering have some density ϑ ε. Then inequality 8) gives the estimates Covε) =ϑ ε /Ω ε n ln n) n ϑ ε /Ω 1, N e λn ln n n ln n) n ϑ ε /Ω 1 ϑ ε /n 2 Ω 1 ). According to our design, there exists a covering {x} with caps C1, x) that has size at most N + N. Next, we combine last inequality with 15), and estimate the covering density of {x}: ϑ 1 = Ω 1 N + N ) λn ln n)ω 1 /Ω ρ + ϑ ε /n 2. 18) Finally, we estimate Ω 1 /Ω ρ using inequalities 8) and 9): Thus, we can rewrite our estimate ϑ 1 as using notation Ω 1 /Ω ρ 1 ε) n 1 + 1/ln n + 1/ln 2 n. 19) ϑ 1 ϑ 1 1/n 2 ) + ϑ ε /n 2 20) ϑ = λn ln n 1 + 1/ln n + 1/ln2 n 1 1/n 2. 21) For any given n, bound 20) only depends on the density ϑ ε. Now let us assume that ϑ 1 meets some known upper bound ϑ 1 valid for all or sufficiently large) radii r. For example, we can use 4) or any weaker estimate. Next, we can change the scale in R n+1 and replace a covering Cov1) of a sphere Sr/ε n with the covering Covε) of the sphere Sr n. This rescaling shows that we can replace ϑ ε in 20) with any known upper bound ϑ 1. In turn, inequality 20) shows that the bound ϑ 1 will be further reduced by this replacement as long as ϑ 1 >ϑ. Thus, this iteration process yields the upper bound ϑ 1 ϑ. Our final step is to verify that the upper bound ϑ of 21) also satisfies inequality 16). Here we use Lemma 3. Then straightforward calculations show that the right-hand side of 16) exceeds the right-hand side of 21)forn = 8 and moderates it for n 8, due to the bigger remaining term 3/ ln n in 16).

Discrete Comput Geom 2007) 38: 665 679 671 Remark More detailed arguments show that 16) holds for n 3, whereas for n 8, ln ln n ϑ 1 + ln n + 2 ) n ln n. ln n 3 New Covering Algorithm for a Sphere S n r Covering design. In this section, we obtain a covering of the sphere Sr n with asymptotic density n ln n)/2. The new design will use both the former covering Covε) with slightly different parameters) and another covering Covμ) with a larger radius μ that has asymptotic order of n 1/2. Namely, we use parameters ε = 1 2n ln n, ρ = 1 ε, β = 1 ln n + 2ln 2 ln n, 22) λ = β + 5 1 2 ln n, μ = n β 2 3, d = 1 2ε μ 2 and proceed as follows. A. Let a sphere Sr n be covered with two different coverings Covμ) and Covε): Covμ) : Sr n Cμ,z), Covε) : S n r z Covμ) u Covε) Cε,u). We assume that both coverings have the former density ϑ of 16) or less. B. Randomly choose N points y Sr n and consider the corresponding spherical caps Cρ,y), where λn ln n Ω d 1 <N λn ln n Ω d. 23) C. Let Cμ, z) be any cap in Covμ) that contains at least one center u Covε) not covered by the ρ-caps. We consider all such centers z and form the joint set {x}={y} { z}. This set covers Covε) with ρ-caps and therefore forms the required covering Cov1) : Sr n C1, x). x {x} We now proceed with preliminary discussion, which outlines the main steps of the proof.

672 Discrete Comput Geom 2007) 38: 665 679 Outline of the proof. Let us first assume that we keep the design of Section 2 but apply it to the new covering Covμ) instead of Covε). This will require taking ρ = 1 μ to cover the centers of the caps Cμ,z) and then expanding ρ to 1 to cover the whole μ-caps. Contrary to our former choice of ρ = 1 ε in 19), this expansion will exponentially increase the covering density. Namely, straightforward calculations show that Ω 1 /Ω 1 ε 1, Ω 1 /Ω 1 μ = exp{n 1/2 }, n. To circumvent this problem, we keep ρ = 1 ε in 22) but change our design as follows. 1. Given any cap Cμ,z), we say that a cap Cρ,y) is d-close if y falls within distance d<ρto z. In our proof, we refine the selection of the caps Cρ,y) and count only d-close caps, instead of the ρ-close caps considered in Section 2. Itis easy to verify that distance d of 22) is so close to ρ that Ω ρ /Ω d 1, n. 24) For this reason, counting only d-close caps instead of the former ρ-close caps will carry no overhead to the covering size 23). 2. On the other hand, we will show in Lemma 6 that the μ-cap becomes almost completely covered by a cap Cρ,y) when the latter becomes d-close instead of ρ-close. Namely, only a vanishing fraction ω exp 3 2 ln2 n) of a μ-cap is left uncovered in this case. 3. We shall also use the fact that a typical μ-cap is covered by multiple d-close caps. According to 23), the average number Ω d N of these caps has the exact order of λn ln n: λn ln n Ω d <Ω d N λn ln n. 25) In our proof, we first define insufficiently covered μ-caps. Namely, we call a cap Cμ,z ) non-saturated if it has only s or fewer d-close caps, where s = n/q, q = 3lnlnn. 26) This choice of s will achieve two goals. 4. We prove in Lemma 7 that non-saturated caps typically form a very small fraction of order exp[ λn ln n] among all μ-caps. On the other hand, it is easy to see that the quantity Covμ) ϑ /Ω μ exceeds N by the exponential factor Ω d /Ω μ exp[βnln n] or less. Then our choice of λ and β in 22) gives the expected number N = on) of non-saturated caps. Thus, non-saturated caps typically form a vanishing fraction of not only μ-caps but also ρ-caps. 5. Next, we proceed with saturated μ-caps and count all those centers u Covε) that are left uncovered by random ρ-caps. All caps Cμ,z ) that contain uncovered centers u are called porous. For a given s, we show in Lemma 8 that

Discrete Comput Geom 2007) 38: 665 679 673 Fig. 1 Two intersecting caps Cμ,Z) and Cρ,Y) with bases PQRSA and PMRTB the set {u } forms a very small portion of Covε), with expected fraction of ω s+1 exp[ 3 2n ln n]. Note that the quantity Covε) ϑ /Ω ε exceeds N by the exponential factor Ω d /Ω ε exp[n ln n] or less. Therefore, the expected size of {u } is N = on). 6. Finally, the centers of all non-saturated and porous caps are combined into the set z ={z, z }. Then the set {x} ={y, z} completely covers the set Covε) with the caps Cρ,x). Therefore, {x} also covers Sr n with unit caps. Main proofs. To prove Theorem 1, we first observe by numerical comparison) that the existing bound 16) is tighter for n 100 than bound 5) of Theorem 1 or even its refined version 39) considered below. Thus, Theorem 1 holds for n 100. For this reason, we shall only consider dimensions n 100. In the end of the proof, we also address the asymptotic case n, wherein the calculations are more straightforward. The proof is based on three lemmas. Consider two caps Cμ,Z) and Cρ,Y) with centers Y and Z, which are d-close. These caps are represented in Fig. 1. Here the origin O is the center of Sr n. Lemma 6 For any cap Cμ,Z), a randomly chosen d-close cap Cρ,Y) fails to cover any given point x of Cμ,Z) with probability px) ω, where ω = 1 1 3 ) n 1 2 4lnn n ln2 n 1 4lnn exp 3 ) 2 ln2 n. 27) Proof The caps Cμ,Z) and Cρ,Y) have bases PQRSA and PMRTB, which form the balls Bμ na) and Bn ρ B). Below we consider the boundary PQRS of Cμ,Z), which forms the sphere Sμ n 1 A). The bigger cap Cρ,Y) covers this boundary, with the exception of the cap PQR centered at Q. We first consider the case, when x is a boundary point and belongs to PQRS. Then the probability px) is the fraction Ω of the entire boundary contained in uncovered cap PQR. We first estimate the half-angle α = PAQ formed by the cap PQR.

674 Discrete Comput Geom 2007) 38: 665 679 Let dh, G) denote the distance between any two points H and G. Also, let σh) be the distance from a point H to the line OBY that connects the origin O with the center B of the bigger base Bρ n B) and with the center Y of the cap Cρ,Y). Weuse inequalities σa) σz) dz, Y) d. On the other hand, consider the base PNR of the uncovered cap PQR. Here N denotes the center of this base. Then both lines AN and BN are orthogonal to this base. Also, db, P) = ρ, and dn, P) da, P) = μ. Thus, db, N) = d 2 B, P) d 2 N, P) ρ 2 μ 2 ρ μ 2. 28) The latter inequality follows from the trivial inequality 2ρ 1 μ 2. Finally, the center line OBY is orthogonal to the entire base PMRTB and its line BN. Then db, N) = σn) and da, N) σn) σa) ρ μ 2 d ε. 29) Now we consider the right triangle ANP and deduce that 3 cos α = da, N)/dA, P) ε/μ = ln n. 30) n Note that the latter expression exceeds 1 for n<42, which simply implies that a d-close cap entirely covers μ-cap for small n.) Given α, we can estimate the fraction Ω of the boundary sphere Sμ n 1 A) contained in the uncovered cap PQR. This fraction can be defined as see [4]) { Ω<{2πn 1)} 1/2 sinn 1 α cos α < 6π n 1 } 1/2 sin n 1 α < sinn 1 α n ln n 4lnn. The last inequality follows from the fact that 6π1 n 1 )>16. Now we use 30) and approximation 1 x e x x2 /2. This gives inequalities sin n 1 α 1 3 ) n 1 { 2 n ln2 n exp 3 2 ln2 n 9 exp 3 ) 2 ln2 n, Ω 1 4lnn exp 3 ) 2 ln2 n. ) 4n ln4 n 1 1 n Finally, consider the second case, when a point x does not belong to the boundary PQRS. Therefore, x is taken from a smaller cap Cμ, Z) Cμ,Z) with the same center Z and a half-chord μ <μ. Similarly to the first case, we define the boundary S n 1 μ A ) of Cμ, Z), where A is some center on the line AZ. Again,px) is the fraction Ω of this boundary left uncovered by the bigger cap Cρ,Y). To obtain the )}

Discrete Comput Geom 2007) 38: 665 679 675 upper bound on Ω, we only need to replace μ with a smaller μ in 30). This gives the smaller angle α arccosε/μ ), which is reduced to 0 if ε μ. In particular, the center Z of the cap is always covered by any d-close cap. Thus, we see that any internal layer S n 1 μ A ) of the cap Cμ,Z) has a smaller uncovered fraction Ω Ω. This gives the required condition px) Ω Ω<ω for any point x and proves our lemma. Remark Our choice of d in 22) is central to the above proof, and even a marginal increase in d will completely change our setting. Namely, it can be proven that about half the base of the μ-cap is uncovered if a ρ-cap is d + ε)-close. Our next goal is to estimate the expected number N of non-saturated caps Cμ,z ) left after N trials. Lemma 7 For n 100, the expected number N of non-saturated caps Cμ,z ) is N < 2 n/4 N. 31) Proof Given any center z, a randomly chosen center y is d-close to z with the probability Ω d. Then the probability to obtain s or fewer such caps is Note that for any i s, P = s ) N Ωd i 1 Ω d) N i. i=0 i 1 Ω d ) N i exp{ Ω d + Ωd 2 /2)N i)}. Now we use inequality Ω d 1/2 and the lower bound NΩ d >λnln n Ω d of 25). Then Ω d + Ωd 2 /2)N i) = NΩ d 1 + Ω ) d iω d 1 + Ω ) d 2 2 λn ln n Ω d ) + Ω d 2 [λn ln n Ω d s2 + Ω d )] λn ln n. Here last inequality follows from the fact that s2 + Ω d ) 5n/6lnlnn) and therefore λn ln n Ω d s2 + Ω d ) 2. Then s λn ln n P e i=0 Ω d N) i i! s λn ln n e i=0 λn ln n) i. 32) i!

676 Discrete Comput Geom 2007) 38: 665 679 Note that consecutive summands in 32) differ at least λn ln n)/s λq ln n times. Therefore s λn ln n) i i=0 i! λn ln n)s s! λq ln n) i i=0 λn ln n)s λn ln n)s 2 s! s/e) s eλq ln n) n/q. Here the sum of the geometric series λq ln n) i was first bounded from above by c n <c 100 < 2. Then we used the Sterling formula in the form s! >2πs) 1/2 s/e) s and removed for simplicity) the vanishing term 22πs) 1/2. Finally, the last inequality follows from the fact that its left-hand side increases in s for any s<λnln n. Summarizing, these substitutions give where h n = P exp{nh n λn ln n}, lneλq ln n) q = 1 3 + lneλq). q Next, we recalculate ϑ of 16) using parameter λ of 22) and the lower bound of 25). It is easy to verify that for any n, ϑ 2λn ln n 1/2) 2Ω d N. Now we estimate the size of Covμ) using 8) and 22) as follows Covμ) ϑ /Ω μ 2NΩ d /Ω μ 2Nμ n 2N exp{n[β ln n + ln12)/2]}. 33) Thus, the expected number of non-saturated caps is N Covμ) P 2N exp{n[h n λ β)ln n + ln12)/2]} 2N exp{n[h n 5 ln 12)/2]}. 34) Now we see that the quantity Ψ n in the brackets of 34) consists of the declining positive function h n and the negative constant. Thus, Ψ n is a declining function of n. Direct calculation shows that Ψ 100 < 0.257. Therefore estimate 31) holds. Consider now the saturated caps Cμ,z) and the centers u Covε) inside them. Lemma 8 For any n 100, the number of centers u Covε) left uncovered in all saturated caps Cμ,z) has expectation N < 2 n/2 N. Proof We first estimate the total number Covε) of centers u. Similarly to 33), Covε) ϑ /Ω ε 2NΩ d /Ω ε 2N2n ln n) n.

Discrete Comput Geom 2007) 38: 665 679 677 Any cap Cμ,z) intersects with at least s + 1 randomly chosen caps Cρ,y). According to Lemma 6, any single ρ-cap fails to cover any given point x Cμ,z) with probability ω or less. Therefore any point u Cμ,z) is not covered with probability ω s+1 or less. Note that ω s+1 <ω n/q where q = 3lnlnn. Then we use the upper bound 27)forω and deduce that N Covε) ω n/q 2N exp{ nφ n }, 35) where ln 2 n Φ n = 2lnlnn + ln 4 ) ln n ln ln n ln 2 + 1 3lnlnn 3. 36) Direct calculation shows that Φ 100 > 0.71. It is also easy to see that the first term of Φ n in parentheses) moderates both terms ln n and ln ln n. Thus, Φ n > 0.71 for all n 100, and the lemma is proved. Proof of Theorem 1 Consider any cap Cμ, z) that contains at least one uncovered center u Covε). Such a cap is either non-saturated or porous and therefore { z}={z } {z }. Equivalently, we can directly consider the set {z } {u }.) Then, according to Lemmas 7 and 8, { z} has expected size N N + N < 2 1 n/4 N. Thus, there exist N randomly chosen centers y that leave at most 2 1 n/4 N centers z. The extended set {x} ={y} { z} forms a unit covering of Sr n. This covering has density ϑ Ω 1 N + N) Ω 1 N1 + 2 1 n/4 ) λn ln n1 + 2 1 n/4 )Ω 1 /Ω d. Similarly to inequality 9), we now verify that for n 100, Ω 1 /Ω d 1 1 ) n n ln n n 2β < 1 + 1 ln n + 1 ln 2 n. 37) Indeed, replace parameter t = ln 1 n used in the proof of Lemma 4 with the new value t = 1 ln n + n1 2β = 1 ln n + 1 ln 4 n. This new t also satisfies inequality 11) forn 100, which in turn proves inequality 37). As a secondary remark, note that 37) also proves asymptotic equality 24). Finally, we take λ of 9) and combine the last inequalities for ϑ and Ω 1 /Ω d as follows ϑ 1 n ln n 2 + 2lnlnn + 5 ) 1 + 1 ln n 2lnn ln n + 1 ) ln 2 1 + 2 1 n/4 ) n < 1 2 + 2lnlnn ln n + 5 ln n.

678 Discrete Comput Geom 2007) 38: 665 679 Here we again used Lemma 3. Namely, we verify numerically that the last expression exceeds the previous one at n = 100 and moderates it for larger n, due to its bigger remaining term 5/ ln n. This completes the non-asymptotic case n 3. To complete the proof of Theorem 1, we now present similar estimates for n. We take any constant b>3/2 and redefine the parameters in 22) and 26) as follows: β = 1 ln ln n + b 2 ln n, λ = β + 3 1 4 ln n, μ = 1 2 n ln b n, q = ln 2 ln n. First, bounds 30) and 27) can be replaced with cos α ε/μ 1 n ln b 1 n, ω 1 1 ) n 1 2 n ln2b 2 n exp 1 ) 2 ln2b 2 n. 38) Second, bound 34) can be rewritten as { lneλq ln n) N 2N exp n + ln 2 3 )}. q 4 Note that the first term lneλq ln n)/q vanishes for n, and N declines exponentially in n.thus,lemma7 holds. Finally, bound 35) is replaced with { [ N 2N exp n ln2n ln n) 1 2 ln 2b 2 n ln 2 ln n which vanishes faster than exponent in n) for any given b>3/2. Thus, Lemma 8 also holds. Then we proceed similarly to 37). In this case, for sufficiently large n, we obtain the density ϑ n ln n 1 ln ln n + b 2 ln n + 3 4 ]}, ) 1 1 + 1 ln n ln n + 1 ) ln 2 1 ln ln n + b n 2 which proves 6). This completes the proof of Theorem 1. ln n + 3 1 2 ln n, For finite n, we can slightly refine bounds 28 29), and also use exact expression for ω in 27). However, these refinements only marginally improve the main bound 5), which is replaced with ϑ n ln n 1 2 + 2lnlnn + 4lnlnn ln n ln n. 39)

Discrete Comput Geom 2007) 38: 665 679 679 Finally, note that Theorem 1 directly leads to Corollary 2. Indeed, here we can use the well known fact ϑr n 1 ) = lim r ϑsn r ) see a sketch of the proof in [6] or Theorem.1 in[7], where a similar proof is detailed for packings of R n ). Concluding remarks In summary, we prove in this paper that the classic Rogers bound 3) on covering density of a sphere Sr n or the Euclidean space R n can be reduced about two times for large dimensions n. The main open problem is to further reduce the gap between this bound and its lower counterpart 2), which is linear in n. In this regard, note that our design holds if we employ any constant parameter β>1/2 introduced in 22). However, it can be verified that choosing a smaller constant β<1/2will first increase parameter μ = n β and then reduce parameter d of 22) to the order of 1 μ 2 /2. The latter reduction will exponentially increase the covering size, by a factor of Ω ρ /Ω d exp{1 2β}. Therefore, our conjecture is that a completely new design is needed for further asymptotic reductions. Another important problem is to extend the above results to the balls Br n of an arbitrary radius r. Our conjecture is that ϑbr n) 1 2 + o1))n ln n for any r and n. Acknowledgements The author is grateful to an anonymous referee for many helpful comments and remarks. This research was supported in part by NSF grants CCF-0622242 and CCF-0635339. References 1. Coxeter, H.S.M., Few, L., Rogers, C.A.: Covering space with equal spheres. Mathematika 6, 147 157 1959) 2. Rogers, C.A.: A note on coverings. Mathematika 4, 1 6 1957) 3. Rogers, C.A.: Covering a sphere with spheres. Mathematika 10, 157 164 1963) 4. Böröczky, K. Jr., Wintsche, G.: Covering the sphere with equal spherical balls. In: Aronov, B., Bazú, S., Sharir, M., Pach, J. eds.) Discrete Computational Geometry The Goldman Pollack Festschrift, pp. 237 253. Springer, Berlin 2003) 5. Verger-Gaugry, J.-L.: Covering a ball with smaller equal balls in R n. Discrete Comput. Geom. 33, 143 155 2005) 6. Conway, J.H., Sloane, N.J.A.: Sphere Packings, Lattices and Groups. Springer, New York 1988) 7. Levenstein, V.I.: Bounds on packings of metric spaces and some of their applications. Vopr. Kibern. 40, 43 109 1983) in Russian) 8. Dumer, I.: Covering spheres and balls with smaller balls. In: 2006 IEEE International Symposium Information Theory, Seattle, WA, July 9 15 2006, pp. 992 995 2006)