Packing and Covering with Centrally Symmetric Convex Disks

Discrete & Computational Geometry, Dec 2013

Dan Ismailescu, Brian Kim

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Packing and Covering with Centrally Symmetric Convex Disks

Discrete Comput Geom Packing and Covering with Centrally Symmetric Convex Disks Dan Ismailescu 0 1 Brian Kim 0 1 0 B. Kim Fu Foundation School of Engineering and Applied Sciences, Columbia University , 2920 Broadway, New York, NY 10027 , USA 1 D. Ismailescu Department of Mathematics, 103 Hofstra University , Hempstead, New York, NY 11549 , USA Given a convex disk K (a convex compact planar set with nonempty interior), let δL(K ) and θL(K ) denote the lattice packing density and the lattice covering density of K , respectively. We prove that for every centrally-symmetric convex disk K we have that The left inequality is tight and it improves a 10-year old result. Arrangements of convex disks; Packing density; Covering density 1 Introduction In this paper, we consider arrangements of convex disks in the Euclidean plane. A convex disk is a compact convex set with nonempty interior; its area will be denoted by A(K ). An arrangement of congruent copies (translates) of a convex disk K is a family A of convex disks, each of which is congruent to (is a translate of) K . The arrangement is a packing if its members’ interiors are mutually disjoint, and it is a covering if the union of its members is the whole plane. For any pair of independent vectors u and v in E, the lattice generated by u and v is the set of vectors L(u, v) = {nu + mv : n, m integers}. An arrangement of translates of K by all vectors of a lattice is a lattice arrangement. Thus, we may have lattice packings and lattice coverings of E with translates of K . The density of an arrangement A is, intuitively speaking, the ratio between the sum of the areas of members of A contained in an arbitrarily large region and the area of the region and is rigorously defined by an appropriate limit (see [5, 15]). Given a convex disk K , the packing density of K , denoted by δ(K), is the maximum density of any packing with congruent replicas of K . If we restrict ourselves to packings with translates of K , or just to lattice packings, then the maximum density defines the translational packing density of K or the lattice packing density of K , denoted by δT (K) and by δL(K), respectively. The analogous notions of the covering density, translational covering density, and lattice covering density denoted by θ (K), θT (K), and θL(K), respectively, are defined as the minimum density of coverings of a corresponding type. Obviously, the following inequalities hold for every convex disk K : δL(K) ≤ δT (K) ≤ δ(K) ≤ 1 ≤ θ (K) ≤ θT (K) ≤ θL(K). For a given convex disk K , let Hmin denote a hexagon of minimum area, which contains K and let hmax denote a hexagon of maximum area which is contained in K . L. Fejes Tóth [7] proved that for every convex disk K we have δ(K) ≤ A(K)/A(Hmin) and θT (K) ≥ A(K)/A(hmax). By a theorem of Dowker [3], if K is centrally symmetric, then hmax and Hmin can be obtained by inscribing in, respectively circumscribing about K a centrally symmetric hexagon. Since translates of any centrally symmetric hexagon can be arranged to form a plane lattice tiling, it immediately follows that for each centrally symmetric convex disk K , we have ( 1 ) ( 2 ) ( 3 ) ( 4 ) δ(K) = δL(K) = A(K)/A(Hmin), θT (K) = θL(K) = A(K)/A(hmax). It is conjectured (see [8], p. 205) that θ (K) = θL(K) holds for all centrally symmetric convex disks K ; this conjecture is supported by the partial result of [7] about crossing-free coverings. Improving previous results of Mahler and Ennola [4, 14], Tammela [18] proved that under the assumption of central symmetry for K we have δL(K) ≥ 0.892656 . . . The conjectured greatest lower bound for δL(K) over all centrally symmetric convex bodies is δL(K0) = (8 − √ √ 32 − ln 2)/( 8 − 1) = 0.90241418 . . . attained by the “smoothed-octagon” K0 constructed by Reinhardt in [16]. The analogous problem for coverings in the case of central symmetry of K has been settled as a corollary of a result proved by Sas [17]. More precisely, the inequality holds for each centrally symmetric convex disk K , and the equality occurs if and only if K is an ellipse, which shows that the above inequality is sharp. Let us mention another related result of L. Fejes Tóth (see [8]), which states that for every centrally symmetric convex disk K √ θL(K) ≤ 2π/ 27 δ(K)/θ (K) ≥ 3/4. Kuperberg [11] proved that this inequality remains true if one drops the condition of central symmetry. The reader interested in a more comprehensive account of the theory of packing and covering is referred to the survey papers of G. Fejes Tóth and W. Kuperberg [5, 6] and to the book of Pach and Agarwal [15]. 2 The Main Problem and Summary of Previous Results Given a family C of convex disks let Ω(C) be the set of all pairs (x, y) with the property that there exists K ∈ C such that θ (K) = x and δ(K) = y; the set ΩL(C) is defined analogously, with the difference that we restrict our attention to lattice packings and coverings. The problem of determining Ω(C) and ΩL(C) is undoubtedly a very difficult one, even if we restrict C to be the family of all centrally symmetric convex disks (see, for instance, the gap between the estimates ( 3 ) and ( 4 )). Questions regarding the sets Ω(C) and ΩL(C), where C is the set of all convex disks, were raised by G. Fejes Tóth and W. Kuperberg in [6, 11]. Since every centrally symmetric convex disk can be approximated arbitrarily close by a centrally symmetric polygon, a natural approach is trying to determine ΩL(P2n) where P2n is the class of all centrally symmetric convex polygons with 2n vertices. Obviously, ΩL(P4) = ΩL(P6) = {( 1, 1 )} since every parallelogram or centrally symmetric hexagon tiles the plane in a lattice manner. An explicit description of ΩL(P8) appeared in [10]. Theorem 2.1 ΩL(P8) = √ (θ , δ) : 1 ≤ θ ≤ 4 − 2 2, 5θ 2 − 12θ + 8 2θ 2 − 5θ + 4 ≤ δ ≤ θ (θ + 4 + √θ 2 − 8θ + 8) 4θ + 2 . The shape of ΩL(P8) suggested the following. Theorem 2.2 [10] For every centrally symmetric convex disk K , 1 − δL(K) ≤ θL(K) − 1 ≤ 1.25 1 − δL(K). ( 6 ) ( 7 ) ( 8 ) The main results of this paper are given in the following. Theorem 2.3 For every centrally symmetric convex disk K Observation It is clear that the left inequality in ( 9 ) improves the left inequality in ( 8 ) as the geometric mean of two different numbers is always smaller than the arithmetic mean. Moreover, our proof has the merit of being very short. Obviously, the left inequality is best possible since if K is a tile then θL(K) = δL(K) = 1. As for the right bound in (9), it is most likely quite far from being optimal. On one hand, an easy upper bound is obtained from (5): 2π δL(K)θL(K) ≤ θL(K) ≤ 3√3 = 1.209199 . . . . From the other direction, we have that δL(K)θL(K) = π 2/9 = 1.096622 . . . if K is a circle. Although it is tempting to conjecture that this is an universal upper bound for the product δL(K)θL(K), it is not hard to check that there exist centrally symmetric octagons which exceed this value. A straightforward application of Theorem 2.1 reveals that there exists a centrally symmetric octagon for which δL(G)θL(G) = 1≤θm≤4a−x2√2 θ 2(θ + 4 + √θ 2 − 8θ + 8) 4θ + 2 = 1.10254 . . . . We expect this value to be quite close to the optimal one; in other words, we feel that the upper estimate δL(K)θL(K) ≤ 1.172559 . . . is rather weak. The rest of the paper is dedicated to proving Theorem 2.3. 3 A Short Proof of δL(K)θL(K) ≥ 1 The main ingredient in our proof is a particular case of the Brunn–Minkowski inequality. Some background is presented below. Given two convex regions K and L, recall that A(K, L), the mixed area of K and L, is defined by the equality A sK + (1 − s)L = s2A(K) + 2s(1 − s)A(K, L) + (1 − s)2A(L). The Brunn–Minkowski inequality states that A(K, L) ≥ A(K) · A(L). ( 10 ) Let h and H be two convex hexagons whose corresponding sides are parallel—see Fig. 1. Then, following [13], A(h, H ) can be easily described as follows. Let O be a fixed point inside Q. If li is the length of a side of h, denote by di the distance from O to the corresponding side of H . Then A2(D) ≥ A(h) · A(H ). ( 11 ) ( 12 ) On the other hand, it is easy to see that the area of any dodecagon D formed by the vertices of h and six other vertices—one chosen along each side of H —is also equal to 12 in=1 li di (see Fig. 1). Applying Brunn–Minkowski inequality in this particular situation, we obtain that Let now K be a centrally symmetric convex disk and let hmax := P1P2P3P4P5P6 be a hexagon of maximum area inscribed in K . We then consider the supporting lines of K , which are parallel to the edges of hmax; these lines determine a hexagon H := Q1Q2Q3Q4Q5Q6 circumscribed about K —see Fig. 2. Denote the contact points between H and K by R1, R2, R3, R4, R5, R6 as shown. We obtain a dodecagon D := P1R1P2R2P3R3P4R4P5R5P6R6, which is contained in K . Applying now inequality ( 12 ) to the polygons hmax, H and D it follows that A2(K) ≥ A2(D) ≥ A(hmax) · A(H ) from which by ( 1 ) and ( 2 ) A(K) A(K) A2(D) δL(K)θL(K) = A(Hmin) · A(hmax) ≥ A(H )A(hmax) ≥ 1, as desired. This ends the first part of the paper, the one dedicated to proving the lower bound. We still have to prove the right inequality stated in Theorem 2.3. We will present the proof throughout the next three sections. Let K be a centrally symmetric convex disk having O as its center. As already mentioned, a result of Dowker [3] implies the existence of a hexagon of minimum area containing K , which is also centrally symmetric with respect to O. We will also need the following theorem proved by Chakerian and Lange in [1] (see also Day [2]). Theorem 4.1 Let K be a convex disk and n > 3 a given integer. Let P be a convex n-gon of minimum area containing K . Then the midpoints of the sides of P lie on the boundary of K . Combining the two results above it follows that if K has a symmetry center at O, there exists Hmin := A1A2A3A4A5A6, a hexagon of minimum area circumscribed about K such that Hmin is also symmetric about O and Hmin has the additional property that B1, B2, B3, B4, B5, and B6—the midpoints of its sides—lie on the boundary of K . Please refer to Fig. 3. Following an idea of Ledermann and Mahler [12], it is convenient for our purposes to specify Hmin by the vectors 2r1, 2r2, 2r3, −2r1, −2r2, −2r3, which form the sides of the circumscribed hexagon of minimum area as shown in Fig. 3. We will use the outer product operation of two vectors to express the area of various polygons. The outer product (u, v) := u1v2 − u2v1 represents the signed area of the parallelogram formed by the vectors u = (u1, u2) and v = (v1, v2). Let us recall the basic properties of outer product. (a) Skew-symmetry: (u, v) = −(v, u). (b) Linearity: (au, v) = a · (u, v). (u, v + w) = (u, v) + (u, w). (c) Nondegeneracy: (u, v) = 0 if and only if u and v are dependent. For an elementary presentation on the geometric applications of the outer product, the reader is referred to [9]. From the convexity of the hexagon Hmin, it follows that the quantities a := (r1, r2), b := (r2, r3), c := (r1, r3), ( 13 ) are all positive. With the notations above, the area of Hmin can be easily expressed as A(Hmin) = 2A(A1A2A3) + 2A(A1A3A4) = (2r1, 2r2) + (2r1 + 2r2, 2r3) = 4a + 4b + 4c. ( 14 ) Also, notice that from the convexity of K it follows that the hexagon h1 := B1B2B3B4B5B6 is contained in K . Moreover, it is easy to compute the area of h1 by removing from Hmin the triangles B1A2B2, B2A3B3, . . . , B6A1B1. It follows that A(h1) = A(B1B2B3B4B5B6) = 4(a + b + c) − a − b − c = 3(a + b + c). ( 15 ) Further, denote by M2, M3, . . . , M1 the midpoints of the sides B1B2, B2B3, . . . , B6B1 of the hexagon h1. It is easy to see that 1 B−−1−M→2 = −M−2−B→2 = 2 (r1 + r2), 1 B−−3−M→4 = −M−4−B→4 = 2 (r3 − r1), 1 B−−2−M→3 = −M−3−B→3 = 2 (r2 + r3), 1 B−−4−M→5 = −M−5−B→5 = 2 (−r1 − r2), Finally, for every i = 1,...,6 denote by Ni the intersection point between the segment AiMi and the boundary of K. Denote by x the ratio between the lengths of A2N2 and A2M2. Similarly, let y be the ratio between the lengths of A3N3 and A3M3 and let z be the ratio between the lengths of A4N4 and A4M4. Keeping in mind that the entire figure is symmetric about O we have that: A−−2−N→2 = xA−−2−M→2, A−−3−N→3 = yA−−3−M→3, A−−4−N→4 = zA−−4−M→4, A−−5−N→5 = xA−−5−M→5, A−−6−N→6 = yA−−6−M→6, A−−1−N→1 = zA−−1−M→1. Obviously, x,y,z ∈ [0,1]. A new hexagon contained in K is formed: h2 := N1N2N3N4N5N6—see Fig. 3. Notice that the sides of h2 are not necessarily parallel to the sides of Hmin. Our next goal is to express the area of h2 in terms of a, b, c, x, y, and z. On one hand, it is easy to see that −O−N→2 = −O−A→2 + A−−2−N→2 = −r2 − r3 + r1 + x2(−r1 + r2), y −O−N→3 = −O−A→3 + A−−3−N→3 = −r3 + r1 + r2 + 2(−r2 + r3). From these, by using (13) and the properties of the outer product we can find the area of triangle ON2N3. 2A(ON2N3) = (−O−N→2,−O−N→3) = 2(a + b) − x2(2a + b − c) − 2(a + 2b − c) + x4y (a + b − c). y Similarly, we can find expressions for the areas of triangles ON3N4 and ON4N5. 2A(ON3N4) = (−O−N→3,−O−N→4) = 2(b + c) − 2(2b + c − a) − 2(b + 2c − a) + y4z(b + c − a), y z 2A(ON4N5) = (−O−N→4,−O−N→5) = 2(c + a) − 2z(2c + a − b) − 2(c + 2a − b) + x4z(c + a − b). x Summing the last three equalities and using the central symmetry, it follows that A(h2) = 4(a + b + c) − 2(ax + by + cz) 1 + 4 xy(a + b − c) + yz(−a + b + c) + xz(a − b + c) . ( 16 ) At this point, we have expressions for the areas of Hmin, the hexagon of minimum area containing K , as well as for the areas of h1 and h2, two centrally symmetric hexagons inscribed in K . We still need to obtain an upper bound for the area of K . 5 An Upper Bound for A(K): Cutting Corners For every i = 1, . . . , 6 construct the tangent line to K at Ni . Notice that this line is not necessarily parallel to the segment Bi−1Bi . Denote by Pi and Qi the points where this supporting line intersects the boundary of Hmin as shown in Fig. 4. By removing the triangles Pi Ai Qi from the circumscribed hexagon Hmin we obtain a dodecagon D which contains K . Our next objective is to find an upper estimate for A(D); this is going to be achieved by finding first a lower bound for the areas of the corner triangles Pi Ai Qi . Let us focus on one particular corner triangle, say P2A2Q2. The reasoning for the other ones is absolutely similar. Denote P2A2 = λr1, A2Q2 = μr2. Clearly, by convexity, we have that λ, μ ∈ [0, 1]—see Fig. 5. Recall that A−−2−N→2 = x A−−2−M→2. Since the vectors P−−2−N→2 and −Q−2−N→2 are collinear, their outer product is 0. We obtain: 0 = (P−−2−N→2, −Q−2−N→2) = x x λr1 + 2 (−r1 + r2), −μr2 + 2 (−r1 + r2) , from which after expanding gives −2λμa + λxa + μxa = 0, that is, μ = λx/(2λ − x). A(Hmin) = 4U 2; A(h1) = 3U 2; A(h2) = 4U 2 − 2W + R/4; A(K) ≤ 4U 2 − V 2. We distinguish two cases. Case 1. 2V ≥ U In this case, we have A(K) A(K) A2(K) δL(K)θL(K) = A(Hmin) · A(hmax) ≤ A(Hmin) · A(h1) It follows that 2A(P2A2Q2) = (λr1, μr2) = λμa = λ2ax/(2λ − x). But it is easy to see that the last quantity is minimized when λ = x so in the end we get that 2A(P2A2Q2) ≥ ax2. Similar reasoning for the other triangles of type Pi Ai Qi turns out that 2A(P3A3Q3) ≥ by2 and 2A(P4A4Q4) ≥ cz2. We thus obtained the desired upper bound on A(D): A(K) ≤ A(D) ≤ A(Hmin) − A(Pi Ai Qi ) ≤ 4(a + b + c) − ax2 + by2 + cz2 . 6 i=1 6 The Proof of the Upper Bound: δL(K)θL(K) ≤ 8(13√8113−35) At this point, we have the necessary expressions for the relevant quantities A(Hmin), A(h1), A(h2) as well as the upper bound ( 17 ) for A(K). Let us introduce some helpful notations √ U : = a + b + c, V := ax2 + by2 + cz2, W := ax + by + cz, ( 18 ) R : = xy(a + b − c) + yz(−a + b + c) + xz(a − b + c). Recall that a, b, and c are all positive and x, y, and z are in [0, 1]. With these new notations, relations ( 14 ), ( 15 ), ( 16 ), and ( 17 ) become ( 17 ) (19) (20) ≤ as f (t ) = (2−t)8(2+t)2 is strictly increasing in the interval [0, 1/2] and f (1/2) = 75/64. Case 2b. 2V ≤ U and R < 0 This is the trickiest part of the proof. Recall that R = xy(a + b − c) + yz(−a + b + c) + xz(a − b + c). Since R < 0 it follows that one of the quantities a + b − c, a − b + c and −a + b + c is negative. Due to symmetry, we lose no generality if we assume that a > b + c. The following technical lemma is crucial in settling this remaining subcase. Lemma 6.1 (Bounding R from below) Suppose that a > b + c, b > 0, c > 0, R = xy(a + b − c) + yz(−a + b + c) + xz(a − b + c) < 0, and x, y, z ∈ [0, 1]. Then x(y + z) < yz. In particular, y > 0, z > 0. x(2 − y − z) ≤ 1 − yz. R ≥ 2(ax + by + cz) − (a + b + c). (21) (22) (23) Proof Since a > b + c, denote a − b − c = 2d , with d > 0. Then the condition R < 0 can be written in equivalent form as xy(b + d) + xz(c + d) < yzd −→ (xy + xz)d < yzd −→ x(y + z) < yz, and (21) is proved. For proving (22), recall that since y, z ∈ [0, 1] it follows that yz ≤ y and yz ≤ z, from which 2yz ≤ y + z. We have 2yz ≤ y + z → 2yz − yz(y + z) ≤ y + z − yz(y + z) → yz(2 − y − z) ≤ (y + z)(1 − yz) → xyz(2 − y − z) ≤ x(y + z)(1 − yz) (21) −→ xyz(2 − y − z) ≤ yz(1 − yz) → x(2 − y − z) ≤ 1 − yz. This proves (22). Finally, after moving both terms of (23) on the left side, the inequality to prove is equivalent to a (1 − yz) − x(2 − y − z) + b (1 − xz) − y(2 − x − z) + c (1 − xy) − z(2 − x − y) ≥ 0. Notice that by (22) the coefficient of a in the above inequality is nonnegative. Since a > b + c, it would suffice to prove that (b + c) (1 − yz) − x(2 − y − z) + b (1 − xz) − y(2 − x − z) + c (1 − xy) − z(2 − x − y) ≥ 0, which after some straightforward algebraic manipulations becomes 2b(1 − x)(1 − y) + 2c(1 − x)(1 − z) ≥ 0, and that is obviously true. This completes the proof of the lemma. Notice that using notations (18) and (19), inequality (23) can be written as R ≥ 2W − U 2. Combining this with the expression of A(h2) from (20), we obtain that But this implies that A(h2) = 4U 2 − 2W + R/4 ≥ 4U 2 − 2W + 41 2W − U 2 = 145 U 2 − 23 W ≥ 145 U 2 − 23 U V . as g(t ) = (145−−t26)t2 reaches a maximum of 8(13√8113−35) ≈ 1.1725596 when t = 5−√313 . This completes the proof of δL(K)θL(K) ≤ 1.1725596 . . . 7 Conclusions and Open Problems Problems involving packing and covering with convex disks have an undisputed intuitive appeal; however, the problems are difficult and the progress has been slow. Building on the work of many others, we make some progress by proving certain bounds for the product between the packing and covering densities of a centrally symmetric convex disk. We would like to conclude with two questions. Open Problem 1 Find a better upper bound for the product δL(K )θL(K ) where K is a centrally symmetric convex disk. Open Problem 2 Is it true that 1 1 δL(K ) + θL(K ) ≤ 2 for every centrally symmetric convex disk K ? If true, this would represent an improvement over the inequality δL(K ) · θL(K ) ≥ 1 proved earlier in this paper. 1. Chakerian , G.D. , Lange , L.H. : Geometric extremum problems . Math. Mag. 44 , 57 - 69 ( 1971 ) 2. Day , M.M.: Polygons circumscribed about closed convex curves . Trans. Am. Math. Soc . 62 , 315 - 319 ( 1947 ) 3. Dowker , C.H. : On minimum circumscribed polygons . Bull. Am. Math. Soc . 50 , 120 - 122 ( 1944 ) 4. Ennola , V.: On the lattice constant of a symmetric convex domain . J. Lond. Math. Soc . 36 , 135 - 138 ( 1961 ) 5. 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Dan Ismailescu, Brian Kim. Packing and Covering with Centrally Symmetric Convex Disks, Discrete & Computational Geometry, 2013, 495-508, DOI: 10.1007/s00454-013-9562-5