Bounds for Covering Symmetric Convex Bodies by a Lattice Congruent to a Given Lattice

In this paper, we focus on lattice covering of centrally symmetric convex body on R2. While there is no constraint on the lattice in many other results about lattice covering, in this study, we only consider lattices congruent to a given lattice to retain more information on the lattice. To obtain some upper bounds on the infimum of the density of such covering, we will say a body is a coverable body with respect to a lattice if such lattice covering is possible, and try to suggest a function of a given lattice such that any centrally symmetric convex body whose area is not less than the function is a coverable body. As an application of this result, we will suggest a theorem which enables one to apply this to a coverable body to suggesting an efficient lattice covering for general sets, which may be non-convex and may have holes.


Introduction
The covering problem of centrally symmetric convex bodies, especially related to the density of covering, is a famous problem in discrete geometry.In this paper, we will deal with lattice coverings, which is fundamental when we deal with centrally symmetric convex bodies.For a body A and a lattice Λ, C = {A + λ|λ ∈ Λ} is called a lattice arrangement.If the members of C cover the whole plane, C is called a lattice covering.The density of a lattice covering can be expressed as S (A) det Λ (Pach & Agarwal, 2011), where S (A) is the area of A and det Λ is the area of the smallest lattice parallelogram of Λ.There are many studies about the upper bounds on the infimum of the density of lattice covering when A is a given body and Λ is any lattice.Because we may choose an appropriate Λ for minimizing the density of covering, the upper bounds are near 1 (Fary, 1950).Especially when A is a centrally symmetric convex body, it is well known that it is 2π √ 27 ≃ 1.2092 (Pach & Agarwal, 2011).In this study, we will consider the same problem when A is a given centrally symmetric convex body and Λ is any lattice congruent with a given lattice Λ 0 .Since the condition of Λ is stronger, this upper bound is a lot bigger than 1.2092.This cannot be a constant, since it can be arbitrarily big depending on the given lattice.Thus, we aim to suggest a function of Λ 0 and S (A) which is always less than or equal to inf A+Λ=R 2 ,Λ≡Λ 0 S (A) det Λ This is equivalent to suggesting a function f of lattice Λ such that for every centrally symmetric convex body A whose area is not less than f (Λ), there exists Λ ′ ≡ Λ such that A + Λ ′ = R 2 .We will call A a coverable body with respect to Λ if A is a centrally symmetric convex body and there exists a lattice To suggest the function f , we will first prove some properties of centrally symmetric convex bodies.Then, we will define several new functions related to Λ and prove some properties of them.Using these, we will prove the main result of this paper, which gives the function f .The condition that the lattice is congruent to a given lattice can be used to suggesting an efficient lattice covering of general sets which need not be convex and may have holes.This will be discussed in the application chapter of this paper.

Geometric Properties of Centrally Symmetric Convex Bodies
In this section, some properties of centrally symmetric convex bodies, which are important lemmas for the main results, are suggested.
The next lemma states a method to determine whether a given set (ii) Given a centrally symmetric convex set A, if there exists a lattice triangle XYZ ⊂ A, A + Λ = R 2 . Proof.
(i) For any set S , denote its boundary by ∂S .For any two distinct points P, Q, ← → PQ denotes the line containing both of them, and PQ may denote either the segment connecting P, Q or the length of such segment.
Let p be an element of the intersection.Then since p ∈ ∂A 1 ⊂ ∂(A + L), any neighborhood of p contains a point p ′ such that p ′ A + L, while (ii) Since A is centrally symmetric, it can be shown that there exists a hexagon The following is a corollary of Lemma 1 (ii).
Corollary 2. If A is a centrally symmetric convex body and there exists a triangle in A which is congruent to a lattice triangle of a lattice Λ, A is a coverable body with respect to Λ.
From now, we will denote Ω as a centrally symmetric convex body.
Lemma 3.There exist polar coordinates such that the origin O is the center of Ω and the four rays θ = πk 2 , k = 0, 1, 2, 3 divide Ω into four parts of the same area.
Proof.First consider polar coordinates whose origin is O.
. Thus there exists t ∈ [0, π 2 ] such that f (t) = 0. Therefore, by rotating the polar coordinates through t, we obtain the polar coordinates satisfying this lemma.
In this section, we will always use the polar coordinates suggested in Lemma 3.
Proof.Since Ω is centrally symmetric, thus there exists ϕ such that r(ϕ) 2 + r(ϕ + π 2 ) 2 = 1.Let W, X, Y, Z the intersections of the boundary of Ω and the rays The following lemma is a key theorem in showing the existence of a certain inscribed parallelogram.
Lemma 5.For any function f : [0, π 8 ] → (0, π) such that its derivative f ′ exists and is continuous on [0, π 8 ], and f (0) = f ( π 8 ) = π 2 , the following holds: 2 also holds, thus we will suppose , thus sin y(x) ≥ sin y 0 (x).For all x, since |y ′ (x Suppose that this y 0 doesn't satisfy this inequality.For any t, let y t be y t (x) = y 0 (x) − tx.Since there exists a > 0 such that Since y 0 is a decreasing function, y a is a strictly decreasing function.
Let z be π 2 − y a and let h be z( π 16 ).Since z is a strictly increasing function and z(0 it is sufficient to prove the following statement for bounded continuous function v whose infimum is at least 1.
Then, it is sufficient to prove the following: . For all n ∈ N, let It can be easily shown that h n P n k=1 b k = τ.If there exist i, j ∈ {1, ..., n} such that then for sufficiently small ϵ > 0, it can be shown that Therefore, the values of

Solving the equation we obtain
, where c n is the solution of As n goes to infinity, c n converges to the solution c of Theorem 6.If S (Ω) = π 2 , there exists an inscribed parallelogram PQRS such that S (PQRS ) ≥ 1 and PR, QS divide Ω into four parts of the same area.
Proof.Let f (x) be 1 2 R x 0 r 2 (θ)dθ, let g be its inverse, let ψ(x) be the parallelogram whose vertices are the intersections of the lines whose directions are g(x), g(x + π 8 ) and the boundary of Ω, and let s(x) be S (ψ(x)).Suppose that s(x) < 1 holds for all x.Define functions p, q as p(x) = g(x) + g(x + π 8 ) , q(x) = g(x + π 8 ) − g(x).Then since 64 sin 2 q(x) + q ′ (x) 2 < p ′ (x) always holds.Therefore, Since q( nπ 8 ) = π 2 holds by Lemma 3 for all n ∈ Z, by Lemma 5, the following inequality holds: This is a contradiction, thus there exists x such that s(x) ≥ 1.It can be easily shown that ψ(x) satisfies the theorem.

Without loss of generality, suppose S (P
Let α be the angle such that X(α) = P 1 .For all t, let Y(t) be the point on the boundary of Ω such that X * (t)Y(t) ∥ P 1 * Q 1 and let ψ(t) be the parallelogram , there exists γ between α, β such that S (X(γ)Y(γ)) Thus ψ(γ) satisfies all conditions of this lemma.

Upper Bounds on the Area of Non-coverable Set
In this section, we will suggest a function f such that for any given lattice Λ, any centrally symmetric convex body Ω is a coverable body with respect to Λ if S (Ω) ≥ f (Λ).Also, for more efficient covering, we will suggest a certain lattice Λ * such that det Λ * = 1 and any centrally symmetric convex body Ω is a coverable body with respect to Λ * if S (Ω) ≥ π 2 .The followings are definitions related to the lattice, which are required to construct the function f .Definition 10.An elementary segment is a segment connecting two lattice points X, Y such that no lattice point exists on XY \ {X, Y}.An elementary triangle is a triangle whose vertices are lattice points X, Y, Z such that no lattice point exists on XYZ \ {X, Y, Z}.Theorem 12.For all lattice Λ, µ(D(Λ)) ≤ √ 3.
Proof.Let X, Y be the points such that , the followings can be shown: The following theorem shows how to find µ(D(Λ)) in finite steps.
).Also, it can be easily shown that d 5 = OY 3 , d 6 = OZ, d 7 = OY 4 .Thus  The next lemma shows two inequalities related to the chords of Ω.For any two sets X, Y ⊂ R 2 we will denote d(X, Y) as the distance between X, Y.
Lemma 16.Suppose S (Ω) = π 2 .Let PQRS be an inscribed parallelogram such that S (PQRS ) ≥ 1, S (PQ) Proof.Let l 0 be the line such that l 0 ∥ PQ and O ∈ l 0 .Let UV be a chord between ← → PQ and l 0 such that UV Let u, v, x, y be the tangent lines of • 1 RS Definition 17.Given an elementary segment XY of a lattice Λ, let l be a line such that l ∥ XY and d(l, XY) = 1 XY det Λ.Let T be the union of l ∩ Λ and its reflection with respect to the orthogonal bisector of XY.Let k be the maximum distance between two adjacent points in T .Then the lattice rate of XY is k XY .Remark 18.Let Z be a point on l ∩ Λ such that ∠ZXY, ∠ZY X ≤ π 2 and let H be the point on XY such that ZH ⊥ XY.Let H ′ be the reflection of H with respect to the midpoint of XY.Then since the projection of T onto  Proof.Consider a scaling which transforms the area of Ω to π 2 .It is sufficient to prove that Ω is a coverable body with respect to Λ if S (Ω) = π 2 and max Suppose that Ω is not admissible.Then by Lemma 8, there exists a parallelogram P 0 Q 0 R 0 S 0 ⊂ Ω such that P 0 Q 0 = 1 and S (P 0 Q 0 R 0 S 0 ) ≥ 1.Since d 1 ≤ 1 and d 1 ≥ det Λ, it can be shown that there exists a parallelogram WXYZ ⊂ P 0 Q 0 R 0 S 0 , w form a triangle.Thus, −u + p, −v + p, −w + p form a lattice triangle of −Λ + p ≡ Λ.
which shall be degenerated.Then R 2 = H + Λ ⊂ A + Λ can be shown as the following figure.

For
any lattice Λ, define elementary segments d 1 , d 2 , ... as follows: For all i ∈ N, d i is a shortest segment among all the elementary segments which are not parallel with d 1 , ..., d i−1 .Definition 11.For any lattice Λ, D(Λ) is the set of the lengths of d 2 , d 3 , d 4 , d 5 ....For any set S of positive real numbers, if S = {s 1 , s 2 , ...} and s 1 < s 2 < ..., µ(S ) := sup s i+1 s i .The length of d 1 is excluded from D(Λ) to make µ(D(Λ)) be bounded.The next theorem shows an upper bound of µ(D(Λ)).