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New short coverings of Fq3 from pairwise weakly linearly independent sets
Linear Algebra and its Applications 567 (2019) 1–13
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Linear Algebra and its Applications www.elsevier.com/locate/laa
New short coverings of F3q from pairwise weakly linearly independent sets A.N. Martinhão a,∗,1 , E.L. Monte Carmelo b,2 a b
Colegiado de Matemática, Universidade Estadual do Paraná, Paranavaí, Brazil Departamento de Matemática, Universidade Estadual de Maringá, Brazil
a r t i c l e
i n f o
Article history: Received 10 May 2017 Accepted 21 December 2018 Available online 28 December 2018 Submitted by R. Brualdi MSC: primary 94B65 secondary 11T71, 51E20, 94B25
a b s t r a c t Pairwise weakly linearly independent (PWLI) sets were introduced in order to compute sharp cyclic coverings of a finite module over a finite ring. In this work, new relationships between PWLI sets and short coverings of the linear space F3q are explored. As a consequence, we investigate the minimum cardinality of a short covering of F3q when q is odd, in order to improve a previous upper bound. © 2018 Elsevier Inc. All rights reserved.
Keywords: Linear independence Short covering code Quotient Square number
1. Introduction Let C be a subset of the three-dimensional space F3q over the finite field Fq , in which q denotes a prime power. We say that C is a covering of F3q , if any vector v in F3q can * Corresponding author. E-mail addresses: [email protected] (A.N. Martinhão), [email protected] (E.L. Monte Carmelo). 1 The author was partially supported by CAPES. 2 The author is partially supported by CNPq/MCT grants: 311703/2016-0. https://doi.org/10.1016/j.laa.2018.12.024 0024-3795/© 2018 Elsevier Inc. All rights reserved.
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be written as a sum of a vector c ∈ C and a linear combination of at most 1 canonical vector of F3q . In other words, v = c + βej , in which β ∈ Fq and ej denotes a vector of the canonical base {e1 , e2 , e3 }. Kalbfleish and Stanton [6] proved that the minimum cardinality of a covering of F3q is q 2 /2 . A generalization for higher dimensions has been considered a central problem in the combinatorial coding theory since the seminal paper by Tauskky and Todd [13] in 1948. The study of covering codes has been motivated by several applications: data transmission, storage system, data compression and the football pool problem. We refer to [3] for an overview on covering codes. Contributions on similar problems with algebraic constraints appear in literature, as shown in [4] for instance. We deal with the following variant in this work: a subset H of F3q is a short covering of F3q , if any vector v in F3q is written as a liner combination of a vector in H and one in {e1 , e2 , e3 }, that is, v = αu + βej , in which u ∈ H and α, β ∈ Fq . Considering that c(q) denotes the cardinality of a minimum short covering of F3q , the best known bounds are
⎧ ⎪ ⎪ ⎨(q + 3)/2 if q ≡ 3 (mod 4) q+1 ≤ c(q) ≤ (q + 5)/2 if q ≡ 1 (mod 4) ⎪ 2 ⎪ ⎩3(q + 4)/4 if q is even,
(1.1)
according to [8] and their references. Another closely related problem that is worth mentioning is: given a finite commutative ring A with identity, An denotes the A-module in the usual way. A subset M of An is called a cyclic covering of An if An is the union of the cyclic submodules [u] = Au = {au : a ∈ A}, in which u ∈ M. For that purpose, we illustrate the simplest case: the minimum cardinality of a cyclic covering of Fnq is (q n −1)/(q −1), which induces a vector partition of the space into lines. Vector partitions of spaces, in a more general context, are explored in [10]. A lot of tools from linear algebra have been investigated to construct good upper bounds on covering codes and their variants: perfect linear code, MDS code and the matrix method. See [1–3,5,7] for instance. In particular, a set of vectors is pairwise weakly linearly independent (PWLI) if none of the vectors is a scalar multiple of another one. PWLI sets were introduced in [11] in order to compute sharp cyclic covering of a finite module over a finite ring. Since PWLI sets produce cyclic coverings, we can ask how PWLI sets and short coverings are related. The contributions of this work focus on this question for the case in which A = Fq and n = 3. As the first goal, new connections between PWLI sets and short coverings are explored. Then, we present in Theorem 3.8 a criterion for checking if a subset of F3q is in fact a short covering. As a numerical application, this criterion
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enables us to compute the following refinement of Eq. (1.1), in order to improve the previous upper bound when q ≡ 1 (mod 4). Theorem 1.1. If q is an odd prime power, then c(q) ≤ (q + 3)/2. The paper is organized as the following: in Section 2 we explore new relationships between PWLI sets and certain quotients. The problem of finding minimum short coverings of F3q is discussed in Section 3. Finally, Theorem 1.1 is proved in Section 4 through a construction from suitable PWLI sets. 2. Tools: pairwise weakly linearly independent sets and quotients In this section we briefly describe the main tools adopted in this work. / P. We say that P is pairwise weakly Definition 2.1. Let P be a subset of Fnq such that 0 ∈ linearly independent (PWLI) if it satisfies the property: for every pair of distinct vectors u and v in P, the inequality u = λv holds for any λ ∈ Fq . As usual, a PWLI set P is maximal in a subset W of Fnq if P ⊆ W and P ∪{v} do not satisfy the property above for every v ∈ W \ P. For instance, the subset {(1, x) : x ∈ Fq } ∪ {(0, 1)} is a maximal PWLI in F2q . We refer to [11] for more details on PWLI sets. Notation 2.2. Consider u = (u1 , u2 , u3 ) as a vector in F3q . For indices 1 ≤ k < l ≤ 3, the quotient is denoted by
Γkl (u) =
ul u−1 k
if uk = 0
∞
if uk = 0,
in which ∞ denotes a symbol. More generally, the quotients generated by a set L are represented by Γkl (L) = {Γkl (u) : u ∈ L}. This way of defining a quotient has been successfully employed in constructions of remarkable discrete structures: projective geometries, combinatorial designs and Mathieu groups. See the book [12] for instance. Notation 2.3. The three projections from F3q to F2q (x, y, z) → (x, y), (x, y, z) → (x, z), and (x, y, z) → (y, z) are denoted by π12 , π13 , and π23 , respectively. The weight of a vector v, denoted by wt(v), is the number of non-zero coordinates of v.
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In this section, we discuss relationships between PWLI sets and quotients. Proposition 2.4. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors of F3q such that wt(u) ≥ 2 and wt(v) ≥ 2. Given indices 1 ≤ k < l ≤ 3, the set {πkl (u), πkl (v)} is linearly dependent if and only if Γkl (u) = Γkl (v). Proof. Note that πkl (u) = (0, 0) and πkl (v) = (0, 0) because wt(u) ≥ 2 and wt(v) ≥ 2. Suppose that {πkl (u), πkl (v)} is linearly dependent. There are scalars λ, μ in Fq , with λ = 0 or μ = 0, such that λπkl (u) + μπkl (v) = (0, 0).
(2.1)
Claim 1: We claim that λ = 0 and μ = 0. Indeed, suppose the contradiction that λ = 0 and μ = 0. Then, Eq. (2.1) implies the absurd μπkl (v) = (0, 0) (because πkl (v) = (0, 0)). The case in which λ = 0 and μ = 0 follows analogously. Eq. (2.1) and Claim 1 yield (vk , vl ) = πkl (v) = (−μ−1 λ)πkl (u) = (−μ−1 λ)(uk , ul ).
(2.2)
We analyze two cases: if vk = 0, then Eq. (2.2) yields uk = 0, and consequently, Γkl (u) = ∞ = Γkl (v), otherwise, vk = 0. Hence, Eq. (2.2) assures that uk = 0. The following equalities follow from Eq. (2.2), Γkl (v) = vl vk−1 = (−μ−1 λ)ul [(−μ−1 λ)uk ]−1 = ul u−1 k = Γkl (u). Conversely, suppose Γkl (u) = Γkl (v). As follows, two cases are analyzed. C ase 1: Γkl (v) = Γkl (u) = ∞. Thus, vk = uk = 0. Since wt(u) ≥ 2 and wt(v) ≥ 2, the inequalities ul = 0 and vl = 0 hold. Hence, {πkl (u), πkl (v)} is linearly dependent. C ase 2: Γkl (v) = Γkl (u) = ∞. Therefore, vl vk−1 = ul u−1 k . Since vk = 0 and uk = 0, the equalities −1 −1 −1 vk u−1 k πkl (u) = vk uk (uk , ul ) = (vk , vk uk ul ) = (vk , ul uk vk ) = πkl (v)
hold, that is, {πkl (u), πkl (v)} is linearly dependent. The proof is complete. 2 Example 2.5. Choose the vectors u = (1, 1, 1) and v = (0, 0, 1) in F3q . The set {π12 (u), π12 (v)} is linearly dependent while Γ12 (u) = 1 and Γ12 (v) = ∞. This example reveals the importance of the hypotheses wt(u) ≥ 2 and wt(v) ≥ 2 in Proposition 2.4. 2 ∗ The set of the square numbers F q = {γ : γ ∈ Fq } plays a central role for our purpose. Theorem 2.6. Let L be a subset of F q × Fq × Fq . Consider indices such 1 ≤ k < l ≤ 3. If πkl (L) is PWLI, then |Γkl (L)| = |L|.
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Proof. A simple argument shows us that Γkl (L) is a subset of F q , since Fq is a multiplicative subgroup of F∗q . It is clear that |Γkl (L)| ≤ |L| holds. Since L is a subset of F q × Fq × Fq , any vector u ∈ L satisfies wt(u) = 3. By hypothesis, πkl (L) is PWLI, thus Proposition 2.4 implies that the elements in Γkl (L) are pairwise distinct. Hence, |Γkl (L)| ≥ |L|. 2
3. An application to short coverings codes In the same spirit of covering codes, the search of short coverings is generally a hard computational problem, according to [9]. We propose, in this section, a criterion for checking if a subset is in fact a short covering of the space under certain conditions, as Theorem 3.8 will show. In order to achieve this purpose, we need a few preliminary results. 3.1. A criterion for covering a particular vector For a vector u ∈ F3q , consider E(u) = {λu + μej : λ, μ ∈ Fq and 1 ≤ j ≤ 3}.
(3.1)
We say that u is a short covering of v if v ∈ E(u). Acknowledge arbitrary vectors u and v, the standard way of deciding whether v ∈ E(u) consists in finding a solution (a triple (λ, μ, ej )) of the equation v = λu + μej .
(3.2)
Remark 3.1. Let v be a vector in F3q with wt(v) ≤ 1. We claim that any vector u satisfies v ∈ E(u). Indeed, if v has the form v = vj ej for some j, then v = 0u + vj ej holds for any vector u = (u1 , u2 , u3 ). However, the analysis of Eq. (3.2) is not a practical test for the most interesting case in which wt(v) ≥ 2. We present below a more efficient method. Proposition 3.2. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors of F3q with wt(u) ≥ 2 and wt(v) ≥ 2. Consequently, the following statements are equivalent: (i) u is a short covering of v, (ii) there are indices 1 ≤ k < l ≤ 3 such that {πkl (u), πkl (v)} is linearly dependent. Proof. If u is a short covering of v, then there are λ, μ ∈ Fq and a canonical vector ej such that v = λu + μej . Let k < l be distinct indices from j. Note that πkl (v) = λπkl (u), that is, {πkl (u), πkl (v)} is linearly dependent.
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Conversely, suppose that {πkl (u), πkl (v)} is linearly dependent. Thus, there are λ, μ ∈ Fq , with λ = 0 or μ = 0, such that λπkl (u) + μπkl (v) = (0, 0).
(3.3)
It is clear that πkl (u) = (0, 0) and πkl (v) = (0, 0), since wt(u) ≥ 2 and wt(v) ≥ 2. Claim 1 in the proof of Proposition 2.4 states that λ = 0 and μ = 0. Thus, Eq. (3.3) implies πkl (v) = (−μ−1 λ)πkl (u). Let j be the index distinct from both k and l. The scalar τ = vj + μ−1 λuj satisfies v = (−μ−1 λ)u + τ ej , that is, u is a short covering of v. 2 Example 3.3. Acknowledge u = (0, 0, 1) and v = (2, 2, 0) in F33 . The projections π12 (u) = (0, 0) and π12 (v) = (2, 2) produce a linearly dependent set. On the other hand, there is not a solution (λ, μ, ej ) for the equation (2, 2, 0) = λ(0, 0, 1) + μej , thus u is not a short covering of v. This case illustrates the importance of the condition wt(u) ≥ 2 in Proposition 3.2. Corollary 3.4. Consider v = (v1 , v2 , v3 ) as a vector of F3q with wt(v) ≥ 2 and u = (u1 , u2 , u3 ) as a non-zero vector of F3q . The vector u is a short covering of v if and only if there are indices 1 ≤ k < l ≤ 3 such that Γkl (u) = Γkl (v). Proof. For the case in which wt(u) ≥ 2, the result is an immediate consequence of Propositions 2.4 and 3.2. However, it still remains the case in which wt(u) = 1. If u is a short covering of v, then there is an index 1 ≤ j ≤ 3 such that v = λu + βej .
(3.4)
Since wt(v) ≥ 2 and wt(u) = 1, a closer look on Eq. (3.4) reveals that λ = 0, wt(v) = 2. The vectors λu and v have two coincident coordinates, say λ(uk , ul ) = (vk , vl ),
(3.5)
in which uk = 0 or ul = 0 (since wt(u) = 1). If uk = 0, then vk = 0 holds from Eq. (3.5), and consequently, Γkl (u) = Γkl (v) = ∞. Otherwise, ul = 0. Eq. (3.5) implies (λuk , 0) = (vk , vl ), thus vl = 0 and vk = 0 (since wt(v) ≥ 2). Again, by Eq. (3.5), uk = 0 holds. Hence, −1 Γkl (u) = ul u−1 vl λvk−1 = vl vk−1 = Γkl (v). k =λ
Conversely, denote the coordinates of u and v by uk , ul , uj and vk , vl , vj , respectively. Suppose Γkl (v) = Γkl (u) for indices 1 ≤ k < l ≤ 3. Therefore, it should be mentioned that we analyze two cases.
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Case 1: Γkl (v) = Γkl (u) = ∞. Thus, vk = uk = 0. Note that vl = 0, vj = 0 (because wt(v) ≥ 2) and ul = 0 or uj = 0 (because wt(u) = 1). If ul = 0, then uj = 0 and −1 vl u−1 l (uk , ul ) = vl ul (0, ul ) = (0, vl ) = (vk , vl ).
Thus the equality v = vl u−1 l u + vj ej holds. So, v ∈ E(u). Otherwise, uj = 0 and ul = 0. Then, −1 vj u−1 j (uk , uj ) = vj uj (0, uj ) = (0, vj ) = (vk , vj ).
Note that v = vj u−1 j u + vl el , and consequently, v ∈ E(u). Case 2: Γkl (v) = Γkl (u) = ∞. Hence, vl vk−1 = ul u−1 k . Since vk = 0 and uk = 0, we obtain −1 −1 vk u−1 k (uk , ul ) = (vk , vk uk ul ) = (vk , ul uk vk ) = (vk , vl ), −1 that is, v = vk u−1 k u + μej , in which μ = vj − vk uk uj ∈ Fq . Thus, v ∈ E(u). The proof is complete. 2
Example 3.5. Let us return to Example 3.3: u = (0, 0, 1) and v = (2, 2, 0) in F33 . A simple computation shows Γ12 (u) = Γ13 (u) = Γ23 (u) = ∞ but Γ12 (v) = 1 and Γ13 (v) = Γ23 (v) = 0. Thus, Corollary 3.4 states that u is not a short covering of v. Remark 3.6. The equivalence in Corollary 3.4 can be false without the condition wt(v) ≥ 2, as we saw in Example 2.5. Moreover, the corresponding version of Corollary 3.4 for classical coverings is not true. Indeed, (1, 1, 1) is not a classical covering of (2, 2, 2). On the other hand, Γkl (1, 1, 1) = Γkl (2, 2, 2) for all indices 1 ≤ k < l ≤ 3. 3.2. A criterion for covering the whole space According to Eq. (3.1), a subset H is a short covering of F3q if for any v ∈ F3q exists a vector u ∈ H such that v ∈ E(u). We turn our attention to a global perspective. Given a candidate H, in order to check if H is in fact a short covering, we must verify that Eq. (3.2) holds for all vectors of the whole space, which is not a feasible computational procedure. A question naturally arises: how to check it by using a more efficient procedure? As an attempt, we propose here a method to ensure it on the basis of quotients and square numbers. In light of Remark 3.1, we can focus on the following subset of F3q V = {v ∈ F3q : wt(v) ≥ 2}. We assume (0, 0, 0) ∈ / H, without loss of generality.
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Corollary 3.7. Let H be a non-empty subset of F3q . Suppose that for any v in V, there are indices 1 ≤ k < l ≤ 3 such that Γkl (v) ∈ Γkl (H). Then, H is a short covering of F3q . Proof. Let v = (v1 , v2 , v3 ) be an arbitrary vector in F3q . If wt(v) ≤ 1, then any vector in H is a short covering of v, by Remark 3.1. Otherwise, v ∈ V. By hypothesis, there are indices 1 ≤ k < l ≤ 3 such that Γkl (v) ∈ Γkl (H), that is, there is u ∈ H such that Γkl (v) = Γkl (u). Corollary 3.4 states that u is a short covering of v. Therefore, H is a short covering of the whole space F3q . 2 According to Corollary 3.7, in order to verify that H is a short covering of the whole space, it is sufficient to check that each scalar can be generated as a quotient, more specifically, Fq ∪ {∞} ⊂ Γ12 (H) ∪ Γ23 (H) ∪ Γ13 (H). Roughly speaking, the criterion below enhances that only “half of scalars” (the square numbers) are required to be written as quotients. ∗ It is well-known that F q = Fq for any even q. On the other hand, Fq forms a subgroup of index 2 of F∗q whenever q is odd, see [12] for instance. Theorem 3.8. Consider q as an odd prime power. Suppose that a subset H of F3q satisfies the following properties: P1: P2: P3: P4:
F q ⊂ Γ12 (H) ∩ Γ23 (H) ∩ Γ13 (H), 0 ∈ Γ23 (H) ∪ Γ13 (H), ∞ ∈ Γ12 (H) ∪ Γ13 (H), 0 ∈ Γ12 (H) or ∞ ∈ Γ23 (H).
Then, H is a short covering of F3q . Proof. In order to apply Corollary 3.7, we must show that for each v ∈ V, there are indices 1 ≤ k < l ≤ 3 such that Γkl (v) ∈ Γkl (H). Indeed, let v = (v1 , v2 , v3 ) be an arbitrary vector in V. We divide the proof into a few cases, according to the weight of v. Case 1: if wt(v) = 3. By the pigeonhole principle, there are two components, say vk and vl , such that {vk , vl } ⊂ F q either {vk , vl } ⊂ Fq \ (Fq ∪ {0}). It is well-known that Fq is a subgroup of the multiplicative group F∗q , and the product of two non-square numbers is a square number. These facts and the property in (P1) conclude that Γkl (v) = vl vk−1 ∈ F q ⊂ Γkl (H). Case 2: if wt(v) = 2. Here v can be written as one of the forms (x, y, 0), (0, x, y), or (x, 0, y), in which x = 0 and y = 0. Subcase 2.1: Firstly, suppose that v = (x, y, 0). Note that Γ23 (x, y, 0) = 0 and Γ13 (x, y, 0) = 0. From (P2), it follows that Γ23 (x, y, 0) ∈ Γ23 (H) or Γ13 (x, y, 0) ∈ Γ13 (H). Subcase 2.2: v = (0, x, y). We obtain Γ12 (0, x, y) = ∞ and Γ13 (0, x, y) = ∞. The hypothesis in (P3) states that Γ12 (0, x, y) ∈ Γ12 (H) or Γ13 (0, x, y) ∈ Γ13 (H).
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Subcase 2.3: It remains the case v = (x, 0, y). Note that Γ12 (x, 0, y) = 0 and Γ23 (x, 0, y) = ∞. The hypothesis in (P4) assures that Γ12 (x, 0, y) ∈ Γ12 (H) or Γ23 (x, 0, y) ∈ Γ23 (H). Finally, an application of Corollary 3.7 concludes the statement. 2 Example 3.9. Let us illustrate how to obtain a short covering from Theorem 3.8. Select the following subset of F35 H = {(1, 1, 1), (1, 0, 0), (0, 1, 4), (1, 4, 4)}. 0 2 Note that F 5 = {1, 4} = {2 , 2 }. A simple computation shows us that Γ12 (H) = {0, 1, 4, ∞}, Γ23 (H) = {1, 4, ∞}, Γ13 (H) = {0, 1, 4, ∞}. Thus,
F 5 ⊂ Γ12 (H) ∩ Γ23 (H) ∩ Γ31 (H), 0 ∈ Γ23 (H), ∞ ∈ Γ12 (H) and 0 ∈ Γ12 (H). Theorem 3.8 implies that H is a short covering of F35 , and consequently, c(5) ≤ 4 holds. The bound is sharp, according to [9]. The upper bound was previously obtained from the short covering H∗ = {(1, 1, 1), (0, 0, 1), (1, 4, 0), (1, 1, 4)} by using another method. Since 5 ≡ 1 (mod 4), the new short covering in Example 3.9 motives a search for a general construction, as described in the next section. 4. Proof of Theorem 1.1 The following remark will be helpful in outlining our proof. Remark 4.1. It is well-known that the additive group Zq−1 and the multiplicative group F∗q are isomorphic by the function a → ξ a , in which ξ denotes a primite element of F∗q . As usual, the class a ∈ Zq−1 is simply denoted by a, in which 0 ≤ a ≤ q − 1 and the multiplication follows the rule ξ a ξ b = ξ a+b = ξ c , which c = a + b in Zq−1 . We consider 0 2 q−3 F∗q = {ξ 0 , ξ 1 , . . . , ξ q−2 }. Therefore, F } when q is odd. q = {ξ , ξ , . . . , ξ Outline of the construction We present an outline of Theorem’s 1.1 proof. The construction is based on the following steps: Step 1 We initially find a maximal subset L of F q × Fq × Fq regarding the property in which πkl (L) is PWLI for all indices 1 ≤ k < l ≤ 3.
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Step 2 Secondly, we check that L satisfies “almost” all the hypotheses of Theorem 3.8, by using Theorem 2.6. Step 3 We obtain a subset H of F3q by joining suitable vectors to L. Finally, an application of Theorem 3.8 guarantees that H is a short covering of F3q . We have all components to prove the main result. Proof of Theorem 1.1. We initially consider the case in which q ≡ 3 (mod 4). Let ξ be a primitive element of Fq and select the following subset of F q × Fq × Fq L = {(ξ 0 , ξ 0 , ξ 2·0 ), (ξ 0 , ξ 2 , ξ 2·2 ), (ξ 0 , ξ 4 , ξ 2·4 ), . . . , (ξ 0 , ξ q−3 , ξ 2·(q−3) )} = = {(1, 1, 1), (1, ξ 2 , ξ 4 ), (1, ξ 4 , ξ 8 ), . . . , (1, ξ q−3 , ξ 2q−6 )}.
(4.1)
This set L satisfies the following property. Claim A: For any pair of indices 1 ≤ k < l ≤ 3, the set πkl (L) is PWLI. In order to proceed with the proof of Claim A, it should be noted that both sets π12 (L) and π13 (L) are PWLI. Suppose a contradiction in which π23 (L) is not PWLI. Thus, there are x, y ∈ {0, 2, . . . , q − 3} with x = y, such that the vectors (ξ x , ξ 2x ) and (ξ y , ξ 2y ) are linearly dependent. This fact implies that ξ 2x /ξ x = ξ 2y /ξ y , or equivalently, ξ x = ξ y , which is a contradiction, due the fact that x = y. By using Claim A, Theorem 2.6 states that the set L satisfies |Γkl (L)| = |L| = (q − 1)/2 = |F q | for any pair of indices 1 ≤ k < l ≤ 3. Since Γkl (L) ⊂ Fq , the equalities Γ12 (L) = Γ23 (L) = Γ13 (L) = F q hold. It is still necessary to complete L to a short covering H of F3q by joining two vectors H = L ∪ {(1, 0, 0), (0, 1, 1)}. A simple computation reveals that Γ12 (H) = {0, ∞} ∪ F q , Γ23 (H) = {∞} ∪ Fq , and Γ13 (H) = {0, ∞} ∪ Fq .
Thus, Theorem 3.8 states that H is a short covering of F3q . Since |H| = |L| +2 = (q +3)/2, the bound c(q) ≤ (q + 3)/2 holds. It remains the case in which q ≡ 1 (mod 4). Given a primitive element ξ of Fq , we should consider the sets: L1 = {(ξ 0 , ξ 2x , ξ 2·2x ) : 0 ≤ x ≤ (q − 5)/4} = = {(1, 1, 1), (1, ξ 2 , ξ 4 ), . . . , (1, ξ
q−5 2
, ξ q−5 )},
L2 = {(ξ 0 , ξ 2x , ξ 2·(2x+1) ) : (q − 1)/4 ≤ x ≤ (q − 5)/2} = = {(1, ξ
q−1 2
, ξ q+1 ), (1, ξ
q+3 2
, ξ q+5 ), . . . , (1, ξ q−5 , ξ 2q−8 )}.
The union L = L1 ∪ L2 is a subset of F q × Fq × Fq and satisfies the key property:
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Claim B: For any pair of indices 1 ≤ k < l ≤ 3, the set πkl (L) is PWLI. To prove Claim B, note that π12 (L) and π13 (L) are clearly PWLI. Suppose a contradiction that π23 (L) is not PWLI. Thus, there is a pair of vectors (α, β) and (γ, δ) in π23 (L), such that (α, β) = λ(γ, δ) for some λ ∈ Fq . In consequence, we analyze three cases: Case 1: if (α, β) ∈ π23 (L1 ) and (γ, δ) ∈ π23 (L1 ). In this case, there are 0 ≤ x < y ≤ (q − 5)/4, such that (α, β) = (ξ 2x , ξ 2·2x ) and (γ, δ) = (ξ 2y , ξ 2·2y ). The equality (α, β) = λ(γ, δ) can be reformulated as β/α = δ/γ, that is, ξ 4x /ξ 2x = ξ 4y /ξ 2y . Equivalently, ξ 2x = ξ 2y , which is a contradiction due the fact that x < y. Case 2: if (α, β) ∈ π23 (L2 ) and (γ, δ) ∈ π23 (L2 ). This case is analogous to the previous one. There are (q − 1)/4 ≤ x < y ≤ (q − 5)/2, such that (α, β) = (ξ 2x , ξ 2·(2x+1) ) and (γ, δ) = (ξ 2y , ξ 2·(2y+1) ). The equality (α, β) = λ(γ, δ) implies β/α = δ/γ, that is, ξ 4x+2 /ξ 2x = ξ 4y+2 /ξ 2y , or equivalently, ξ 2x+2 = ξ 2y+2 , which is a contradiction due the fact that x < y. Case 3: if (α, β) ∈ π23 (L1 ) but (γ, δ) ∈ π23 (L2 ). In this case, (α, β) = (ξ 2x , ξ 2·2x ) and (γ, δ) = (ξ 2y , ξ 2·(2y+1) ) for some 0 ≤ x ≤ (q−5)/4 and (q − 1)/4 ≤ y ≤ (q − 5)/2. From (α, β) = λ(γ, δ) we obtain that β/α = δ/γ, that is, ξ 4x /ξ 2x = ξ 4y+2 /ξ 2y , or equivalently, ξ 2x = ξ 2y+2 , which is a contradiction due the fact that 0 ≤ 2x ≤ (q − 5)/2 and (q + 3)/2 ≤ 2y + 2 ≤ q − 3. The proof of Claim B is complete. Since L ⊂ F q ×Fq ×Fq and Fq is a multiplicative group, the set L = L1 ∪L2 satisfies Γ12 (L) ∪ Γ23 (L) ∪ Γ13 (L) ⊂ F q .
(4.2)
By applying Claim B, Theorem 2.6 ensures |Γ12 (L)| = |Γ23 (L)| = |Γ13 (L)| = (q − 3)/2.
(4.3)
We need the following statements: Claim 1: ξ q−3 ∈ / Γ12 (L). Indeed, suppose a contradiction that ξ q−3 ∈ Γ12 (L). By the construction of L, there is a vector (1, ξ 2x , ξ 2y ) in L, such that Γ12 (1, ξ 2x , ξ 2y ) = ξ q−3 . Since Γ12 ((1, ξ 2x , ξ 2y )) = ξ 2x , the equality 2x = q − 3 holds, that is, x = (q − 3)/2, thus ξ 2x = ξ q−3 . The last equality is a contradiction because ξ q−3 does not appear in the second coordinate of a vector in L. Claim 2: ξ (q−1)/2 ∈ / Γ23 (L). By simple inspection, we can see that Γ23 (L) = {1, ξ 2 , ξ 4 , . . . , ξ q−5 , ξ (q+3)/2 , ξ (q+7)/2 , . . . , ξ q−3 }. Therefore, ξ (q−1)/2 ∈ / Γ23 (L).
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Claim 3: ξ q−3 ∈ / Γ13 (L). / Γ13 (L). Indeed, by using an argument similar to Claim 1, we conclude that ξ q−3 ∈ | = (q − 1)/2. A combination of (4.2), (4.3) and Claims 1 to 3 implies Note that |F q q−3 (q−1)/2 q−3 Γ12 (L) = F }, Γ23 (L) = F }, Γ13 (L) = F }. q \ {ξ q \ {ξ q \ {ξ
We now complete L to a short covering of F3q by joining H = L ∪ {(1, 0, 0), (0, 1, ξ (q−1)/2 ), (1, ξ q−3 , ξ q−3 )}. By simple inspection, we can verify that H satisfies the four conditions of Theorem 3.8. Therefore, c(q) ≤ |H| = (q − 3)/2 + 3 = (q + 3)/2. 2 As mentioned in the Introduction, the upper bound c(q) ≤ (q + 3)/2 was obtained for q ≡ 3 (mod 4) in [8]. We proved it once more by using PWLI sets in the first part the proof above. Example 4.2. Let us show that c(11) ≤ 7. Firstly, note that 2 is a primite element of F11 . By the construction given in proof of Theorem 1.1, the correspondent set L is L = {(20 , 20 , 20 ), (20 , 22 , 24 ), (20 , 24 , 28 ), (20 , 26 , 22 ), (20 , 28 , 26 )} = = {(1, 1, 1), (1, 4, 5), (1, 5, 3), (1, 9, 4), (1, 3, 9)}. Thus, a short covering of F311 formed by 7 vectors can be generated from L, H = {(1, 0, 0), (0, 1, 1), (1, 1, 1), (1, 4, 5), (1, 5, 3), (1, 9, 4), (1, 3, 9)}. As a comparative analysis, the set H∗ = {(1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 3, 4), (1, 4, 3), (1, 5, 9), (1, 9, 5)} is a short covering of Z311 too, according to [8, Example 5.2]. Example 4.3. The upper bound c(9) ≤ 6 holds. Let ξ be a primite element of F9 , the construction of Theorem 1.1 is based on the following subset L = L1 ∪L2 of F 9 ×F9 ×F9 , in which L1 = {(ξ 0 , ξ 2x , ξ 2·2x ) : 0 ≤ x ≤ (9 − 5)/4} = {(1, 1, 1), (1, ξ 2 , ξ 4 )}, L2 = {(ξ 0 , ξ 2x , ξ 2·(2x+1) ) : (9 − 1)/4 ≤ x ≤ (9 − 5)/2} = {(1, ξ 4 , ξ 2 )}. Consider H = {(1, 1, 1), (1, ξ 2 , ξ 4 ), (1, ξ 4 , ξ 2 ), (1, 0, 0), (0, 1, ξ 4 ), (1, ξ 6 , ξ 6 )}.
(4.4)
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2 4 6 Note that F 9 = {1, ξ , ξ , ξ }. So, we have Γ12 (H) = {0, ∞} ∪ F 9 , Γ23 (H) = {∞} ∪ F9 and Γ13 (H) = {0, ∞} ∪ F9 .
The subset H in (4.4) satisfies the hypothesis of Theorem 3.8. Thus, H is a short covering of F39 and c(9) ≤ 6 holds, and this improves the previous upper bound c(9) ≤ 7 from [8]. 5. Conclusion As alluded in the introduction of this study, sharp cyclic coverings of a finite module over a finite ring are computed on the basis of PWLI sets. In this work, we explored new connections between PWLI sets and short coverings of the linear space F3q , which allow us to prove c(q) ≤ (q + 3)/2 for any odd prime power q. In particular, the new upper bound improves c(q) ≤ (q + 5)/2 when q ≡ 1 (mod 4). For future research, it would be interesting to find new PWLI sets in a vector space and their connections with short coverings for higher dimensions. New applications of PWLI sets to discrete structures (finite geometries, projective spaces, combinatorial designs) or new relationships with algebraic concepts would be desirable. Acknowledgements The authors are grateful to the anonymous referee for the careful reading and many suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
W.A. Carnielli, On covering and coloring problems for rook domains, Discrete Math. 57 (1985) 9–16. W.A. Carnielli, Hyper-rook domain inequalities, Stud. Appl. Math. 82 (1990) 59–69. G.D. Cohen, I. Honkala, A. Listyn, A. Lobstein, Covering Codes, North-Holland, Amsterdam, 1997. T.J. Dickson, On a covering problem concerning abelian groups, J. London Math. Soc. 2 (1971) 222–232. O.J.N.T.N. dos Santos, E.L. Monte Carmelo, Invariant sets under linear operator and covering codes over modules, Linear Algebra Appl. 444 (2014) 42–52. J.G. Kalbfleisch, R.G. Stanton, A combinatorial problem in matching, J. London Math. Soc. 44 (1969) 60–64. G. Losey, Note of a theorem of Zaremba, J. Combin. Theory 6 (1969) 208–209. A.N. Martinhão, E.L. Monte Carmelo, Short covering codes arising from matchings in weighted graphs, Math. Comp. 82 (2013) 605–616. C. Mendes, E.L. Monte Carmelo, M.V. Poggi de Aragão, Bounds for short covering codes and reactive tabu search, Discrete Appl. Math. 158 (2010) 522–533. astase, P.A. Sissokho, The minimum size of a finite subspace partition, Linear Algebra Appl. E. Nˇ 435 (2011) 1213–1221. I.N. Nakaoka, E.L. Monte Carmelo, O.J.N.T.N. dos Santos, Sharp covering of a module by cyclic submodules, Linear Algebra Appl. 458 (2014) 387–402. J. Rotman, An Introduction to the Theory of Groups, Graduate Texts in Mathematics, vol. 14, Springer, New York, 1999. O. Taussky, J. Todd, Covering theorems for groups, Ann. Soc. Pol. Math. 21 (1948) 303–305.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
New short coverings of F3q from pairwise weakly linearly independent sets A.N. Martinhão a,∗,1 , E.L. Monte Carmelo b,2 a b
Colegiado de Matemática, Universidade Estadual do Paraná, Paranavaí, Brazil Departamento de Matemática, Universidade Estadual de Maringá, Brazil
a r t i c l e
i n f o
Article history: Received 10 May 2017 Accepted 21 December 2018 Available online 28 December 2018 Submitted by R. Brualdi MSC: primary 94B65 secondary 11T71, 51E20, 94B25
a b s t r a c t Pairwise weakly linearly independent (PWLI) sets were introduced in order to compute sharp cyclic coverings of a finite module over a finite ring. In this work, new relationships between PWLI sets and short coverings of the linear space F3q are explored. As a consequence, we investigate the minimum cardinality of a short covering of F3q when q is odd, in order to improve a previous upper bound. © 2018 Elsevier Inc. All rights reserved.
Keywords: Linear independence Short covering code Quotient Square number
1. Introduction Let C be a subset of the three-dimensional space F3q over the finite field Fq , in which q denotes a prime power. We say that C is a covering of F3q , if any vector v in F3q can * Corresponding author. E-mail addresses: [email protected] (A.N. Martinhão), [email protected] (E.L. Monte Carmelo). 1 The author was partially supported by CAPES. 2 The author is partially supported by CNPq/MCT grants: 311703/2016-0. https://doi.org/10.1016/j.laa.2018.12.024 0024-3795/© 2018 Elsevier Inc. All rights reserved.
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be written as a sum of a vector c ∈ C and a linear combination of at most 1 canonical vector of F3q . In other words, v = c + βej , in which β ∈ Fq and ej denotes a vector of the canonical base {e1 , e2 , e3 }. Kalbfleish and Stanton [6] proved that the minimum cardinality of a covering of F3q is q 2 /2 . A generalization for higher dimensions has been considered a central problem in the combinatorial coding theory since the seminal paper by Tauskky and Todd [13] in 1948. The study of covering codes has been motivated by several applications: data transmission, storage system, data compression and the football pool problem. We refer to [3] for an overview on covering codes. Contributions on similar problems with algebraic constraints appear in literature, as shown in [4] for instance. We deal with the following variant in this work: a subset H of F3q is a short covering of F3q , if any vector v in F3q is written as a liner combination of a vector in H and one in {e1 , e2 , e3 }, that is, v = αu + βej , in which u ∈ H and α, β ∈ Fq . Considering that c(q) denotes the cardinality of a minimum short covering of F3q , the best known bounds are
⎧ ⎪ ⎪ ⎨(q + 3)/2 if q ≡ 3 (mod 4) q+1 ≤ c(q) ≤ (q + 5)/2 if q ≡ 1 (mod 4) ⎪ 2 ⎪ ⎩3(q + 4)/4 if q is even,
(1.1)
according to [8] and their references. Another closely related problem that is worth mentioning is: given a finite commutative ring A with identity, An denotes the A-module in the usual way. A subset M of An is called a cyclic covering of An if An is the union of the cyclic submodules [u] = Au = {au : a ∈ A}, in which u ∈ M. For that purpose, we illustrate the simplest case: the minimum cardinality of a cyclic covering of Fnq is (q n −1)/(q −1), which induces a vector partition of the space into lines. Vector partitions of spaces, in a more general context, are explored in [10]. A lot of tools from linear algebra have been investigated to construct good upper bounds on covering codes and their variants: perfect linear code, MDS code and the matrix method. See [1–3,5,7] for instance. In particular, a set of vectors is pairwise weakly linearly independent (PWLI) if none of the vectors is a scalar multiple of another one. PWLI sets were introduced in [11] in order to compute sharp cyclic covering of a finite module over a finite ring. Since PWLI sets produce cyclic coverings, we can ask how PWLI sets and short coverings are related. The contributions of this work focus on this question for the case in which A = Fq and n = 3. As the first goal, new connections between PWLI sets and short coverings are explored. Then, we present in Theorem 3.8 a criterion for checking if a subset of F3q is in fact a short covering. As a numerical application, this criterion
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enables us to compute the following refinement of Eq. (1.1), in order to improve the previous upper bound when q ≡ 1 (mod 4). Theorem 1.1. If q is an odd prime power, then c(q) ≤ (q + 3)/2. The paper is organized as the following: in Section 2 we explore new relationships between PWLI sets and certain quotients. The problem of finding minimum short coverings of F3q is discussed in Section 3. Finally, Theorem 1.1 is proved in Section 4 through a construction from suitable PWLI sets. 2. Tools: pairwise weakly linearly independent sets and quotients In this section we briefly describe the main tools adopted in this work. / P. We say that P is pairwise weakly Definition 2.1. Let P be a subset of Fnq such that 0 ∈ linearly independent (PWLI) if it satisfies the property: for every pair of distinct vectors u and v in P, the inequality u = λv holds for any λ ∈ Fq . As usual, a PWLI set P is maximal in a subset W of Fnq if P ⊆ W and P ∪{v} do not satisfy the property above for every v ∈ W \ P. For instance, the subset {(1, x) : x ∈ Fq } ∪ {(0, 1)} is a maximal PWLI in F2q . We refer to [11] for more details on PWLI sets. Notation 2.2. Consider u = (u1 , u2 , u3 ) as a vector in F3q . For indices 1 ≤ k < l ≤ 3, the quotient is denoted by
Γkl (u) =
ul u−1 k
if uk = 0
∞
if uk = 0,
in which ∞ denotes a symbol. More generally, the quotients generated by a set L are represented by Γkl (L) = {Γkl (u) : u ∈ L}. This way of defining a quotient has been successfully employed in constructions of remarkable discrete structures: projective geometries, combinatorial designs and Mathieu groups. See the book [12] for instance. Notation 2.3. The three projections from F3q to F2q (x, y, z) → (x, y), (x, y, z) → (x, z), and (x, y, z) → (y, z) are denoted by π12 , π13 , and π23 , respectively. The weight of a vector v, denoted by wt(v), is the number of non-zero coordinates of v.
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In this section, we discuss relationships between PWLI sets and quotients. Proposition 2.4. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors of F3q such that wt(u) ≥ 2 and wt(v) ≥ 2. Given indices 1 ≤ k < l ≤ 3, the set {πkl (u), πkl (v)} is linearly dependent if and only if Γkl (u) = Γkl (v). Proof. Note that πkl (u) = (0, 0) and πkl (v) = (0, 0) because wt(u) ≥ 2 and wt(v) ≥ 2. Suppose that {πkl (u), πkl (v)} is linearly dependent. There are scalars λ, μ in Fq , with λ = 0 or μ = 0, such that λπkl (u) + μπkl (v) = (0, 0).
(2.1)
Claim 1: We claim that λ = 0 and μ = 0. Indeed, suppose the contradiction that λ = 0 and μ = 0. Then, Eq. (2.1) implies the absurd μπkl (v) = (0, 0) (because πkl (v) = (0, 0)). The case in which λ = 0 and μ = 0 follows analogously. Eq. (2.1) and Claim 1 yield (vk , vl ) = πkl (v) = (−μ−1 λ)πkl (u) = (−μ−1 λ)(uk , ul ).
(2.2)
We analyze two cases: if vk = 0, then Eq. (2.2) yields uk = 0, and consequently, Γkl (u) = ∞ = Γkl (v), otherwise, vk = 0. Hence, Eq. (2.2) assures that uk = 0. The following equalities follow from Eq. (2.2), Γkl (v) = vl vk−1 = (−μ−1 λ)ul [(−μ−1 λ)uk ]−1 = ul u−1 k = Γkl (u). Conversely, suppose Γkl (u) = Γkl (v). As follows, two cases are analyzed. C ase 1: Γkl (v) = Γkl (u) = ∞. Thus, vk = uk = 0. Since wt(u) ≥ 2 and wt(v) ≥ 2, the inequalities ul = 0 and vl = 0 hold. Hence, {πkl (u), πkl (v)} is linearly dependent. C ase 2: Γkl (v) = Γkl (u) = ∞. Therefore, vl vk−1 = ul u−1 k . Since vk = 0 and uk = 0, the equalities −1 −1 −1 vk u−1 k πkl (u) = vk uk (uk , ul ) = (vk , vk uk ul ) = (vk , ul uk vk ) = πkl (v)
hold, that is, {πkl (u), πkl (v)} is linearly dependent. The proof is complete. 2 Example 2.5. Choose the vectors u = (1, 1, 1) and v = (0, 0, 1) in F3q . The set {π12 (u), π12 (v)} is linearly dependent while Γ12 (u) = 1 and Γ12 (v) = ∞. This example reveals the importance of the hypotheses wt(u) ≥ 2 and wt(v) ≥ 2 in Proposition 2.4. 2 ∗ The set of the square numbers F q = {γ : γ ∈ Fq } plays a central role for our purpose. Theorem 2.6. Let L be a subset of F q × Fq × Fq . Consider indices such 1 ≤ k < l ≤ 3. If πkl (L) is PWLI, then |Γkl (L)| = |L|.
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Proof. A simple argument shows us that Γkl (L) is a subset of F q , since Fq is a multiplicative subgroup of F∗q . It is clear that |Γkl (L)| ≤ |L| holds. Since L is a subset of F q × Fq × Fq , any vector u ∈ L satisfies wt(u) = 3. By hypothesis, πkl (L) is PWLI, thus Proposition 2.4 implies that the elements in Γkl (L) are pairwise distinct. Hence, |Γkl (L)| ≥ |L|. 2
3. An application to short coverings codes In the same spirit of covering codes, the search of short coverings is generally a hard computational problem, according to [9]. We propose, in this section, a criterion for checking if a subset is in fact a short covering of the space under certain conditions, as Theorem 3.8 will show. In order to achieve this purpose, we need a few preliminary results. 3.1. A criterion for covering a particular vector For a vector u ∈ F3q , consider E(u) = {λu + μej : λ, μ ∈ Fq and 1 ≤ j ≤ 3}.
(3.1)
We say that u is a short covering of v if v ∈ E(u). Acknowledge arbitrary vectors u and v, the standard way of deciding whether v ∈ E(u) consists in finding a solution (a triple (λ, μ, ej )) of the equation v = λu + μej .
(3.2)
Remark 3.1. Let v be a vector in F3q with wt(v) ≤ 1. We claim that any vector u satisfies v ∈ E(u). Indeed, if v has the form v = vj ej for some j, then v = 0u + vj ej holds for any vector u = (u1 , u2 , u3 ). However, the analysis of Eq. (3.2) is not a practical test for the most interesting case in which wt(v) ≥ 2. We present below a more efficient method. Proposition 3.2. Let u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) be vectors of F3q with wt(u) ≥ 2 and wt(v) ≥ 2. Consequently, the following statements are equivalent: (i) u is a short covering of v, (ii) there are indices 1 ≤ k < l ≤ 3 such that {πkl (u), πkl (v)} is linearly dependent. Proof. If u is a short covering of v, then there are λ, μ ∈ Fq and a canonical vector ej such that v = λu + μej . Let k < l be distinct indices from j. Note that πkl (v) = λπkl (u), that is, {πkl (u), πkl (v)} is linearly dependent.
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Conversely, suppose that {πkl (u), πkl (v)} is linearly dependent. Thus, there are λ, μ ∈ Fq , with λ = 0 or μ = 0, such that λπkl (u) + μπkl (v) = (0, 0).
(3.3)
It is clear that πkl (u) = (0, 0) and πkl (v) = (0, 0), since wt(u) ≥ 2 and wt(v) ≥ 2. Claim 1 in the proof of Proposition 2.4 states that λ = 0 and μ = 0. Thus, Eq. (3.3) implies πkl (v) = (−μ−1 λ)πkl (u). Let j be the index distinct from both k and l. The scalar τ = vj + μ−1 λuj satisfies v = (−μ−1 λ)u + τ ej , that is, u is a short covering of v. 2 Example 3.3. Acknowledge u = (0, 0, 1) and v = (2, 2, 0) in F33 . The projections π12 (u) = (0, 0) and π12 (v) = (2, 2) produce a linearly dependent set. On the other hand, there is not a solution (λ, μ, ej ) for the equation (2, 2, 0) = λ(0, 0, 1) + μej , thus u is not a short covering of v. This case illustrates the importance of the condition wt(u) ≥ 2 in Proposition 3.2. Corollary 3.4. Consider v = (v1 , v2 , v3 ) as a vector of F3q with wt(v) ≥ 2 and u = (u1 , u2 , u3 ) as a non-zero vector of F3q . The vector u is a short covering of v if and only if there are indices 1 ≤ k < l ≤ 3 such that Γkl (u) = Γkl (v). Proof. For the case in which wt(u) ≥ 2, the result is an immediate consequence of Propositions 2.4 and 3.2. However, it still remains the case in which wt(u) = 1. If u is a short covering of v, then there is an index 1 ≤ j ≤ 3 such that v = λu + βej .
(3.4)
Since wt(v) ≥ 2 and wt(u) = 1, a closer look on Eq. (3.4) reveals that λ = 0, wt(v) = 2. The vectors λu and v have two coincident coordinates, say λ(uk , ul ) = (vk , vl ),
(3.5)
in which uk = 0 or ul = 0 (since wt(u) = 1). If uk = 0, then vk = 0 holds from Eq. (3.5), and consequently, Γkl (u) = Γkl (v) = ∞. Otherwise, ul = 0. Eq. (3.5) implies (λuk , 0) = (vk , vl ), thus vl = 0 and vk = 0 (since wt(v) ≥ 2). Again, by Eq. (3.5), uk = 0 holds. Hence, −1 Γkl (u) = ul u−1 vl λvk−1 = vl vk−1 = Γkl (v). k =λ
Conversely, denote the coordinates of u and v by uk , ul , uj and vk , vl , vj , respectively. Suppose Γkl (v) = Γkl (u) for indices 1 ≤ k < l ≤ 3. Therefore, it should be mentioned that we analyze two cases.
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Case 1: Γkl (v) = Γkl (u) = ∞. Thus, vk = uk = 0. Note that vl = 0, vj = 0 (because wt(v) ≥ 2) and ul = 0 or uj = 0 (because wt(u) = 1). If ul = 0, then uj = 0 and −1 vl u−1 l (uk , ul ) = vl ul (0, ul ) = (0, vl ) = (vk , vl ).
Thus the equality v = vl u−1 l u + vj ej holds. So, v ∈ E(u). Otherwise, uj = 0 and ul = 0. Then, −1 vj u−1 j (uk , uj ) = vj uj (0, uj ) = (0, vj ) = (vk , vj ).
Note that v = vj u−1 j u + vl el , and consequently, v ∈ E(u). Case 2: Γkl (v) = Γkl (u) = ∞. Hence, vl vk−1 = ul u−1 k . Since vk = 0 and uk = 0, we obtain −1 −1 vk u−1 k (uk , ul ) = (vk , vk uk ul ) = (vk , ul uk vk ) = (vk , vl ), −1 that is, v = vk u−1 k u + μej , in which μ = vj − vk uk uj ∈ Fq . Thus, v ∈ E(u). The proof is complete. 2
Example 3.5. Let us return to Example 3.3: u = (0, 0, 1) and v = (2, 2, 0) in F33 . A simple computation shows Γ12 (u) = Γ13 (u) = Γ23 (u) = ∞ but Γ12 (v) = 1 and Γ13 (v) = Γ23 (v) = 0. Thus, Corollary 3.4 states that u is not a short covering of v. Remark 3.6. The equivalence in Corollary 3.4 can be false without the condition wt(v) ≥ 2, as we saw in Example 2.5. Moreover, the corresponding version of Corollary 3.4 for classical coverings is not true. Indeed, (1, 1, 1) is not a classical covering of (2, 2, 2). On the other hand, Γkl (1, 1, 1) = Γkl (2, 2, 2) for all indices 1 ≤ k < l ≤ 3. 3.2. A criterion for covering the whole space According to Eq. (3.1), a subset H is a short covering of F3q if for any v ∈ F3q exists a vector u ∈ H such that v ∈ E(u). We turn our attention to a global perspective. Given a candidate H, in order to check if H is in fact a short covering, we must verify that Eq. (3.2) holds for all vectors of the whole space, which is not a feasible computational procedure. A question naturally arises: how to check it by using a more efficient procedure? As an attempt, we propose here a method to ensure it on the basis of quotients and square numbers. In light of Remark 3.1, we can focus on the following subset of F3q V = {v ∈ F3q : wt(v) ≥ 2}. We assume (0, 0, 0) ∈ / H, without loss of generality.
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Corollary 3.7. Let H be a non-empty subset of F3q . Suppose that for any v in V, there are indices 1 ≤ k < l ≤ 3 such that Γkl (v) ∈ Γkl (H). Then, H is a short covering of F3q . Proof. Let v = (v1 , v2 , v3 ) be an arbitrary vector in F3q . If wt(v) ≤ 1, then any vector in H is a short covering of v, by Remark 3.1. Otherwise, v ∈ V. By hypothesis, there are indices 1 ≤ k < l ≤ 3 such that Γkl (v) ∈ Γkl (H), that is, there is u ∈ H such that Γkl (v) = Γkl (u). Corollary 3.4 states that u is a short covering of v. Therefore, H is a short covering of the whole space F3q . 2 According to Corollary 3.7, in order to verify that H is a short covering of the whole space, it is sufficient to check that each scalar can be generated as a quotient, more specifically, Fq ∪ {∞} ⊂ Γ12 (H) ∪ Γ23 (H) ∪ Γ13 (H). Roughly speaking, the criterion below enhances that only “half of scalars” (the square numbers) are required to be written as quotients. ∗ It is well-known that F q = Fq for any even q. On the other hand, Fq forms a subgroup of index 2 of F∗q whenever q is odd, see [12] for instance. Theorem 3.8. Consider q as an odd prime power. Suppose that a subset H of F3q satisfies the following properties: P1: P2: P3: P4:
F q ⊂ Γ12 (H) ∩ Γ23 (H) ∩ Γ13 (H), 0 ∈ Γ23 (H) ∪ Γ13 (H), ∞ ∈ Γ12 (H) ∪ Γ13 (H), 0 ∈ Γ12 (H) or ∞ ∈ Γ23 (H).
Then, H is a short covering of F3q . Proof. In order to apply Corollary 3.7, we must show that for each v ∈ V, there are indices 1 ≤ k < l ≤ 3 such that Γkl (v) ∈ Γkl (H). Indeed, let v = (v1 , v2 , v3 ) be an arbitrary vector in V. We divide the proof into a few cases, according to the weight of v. Case 1: if wt(v) = 3. By the pigeonhole principle, there are two components, say vk and vl , such that {vk , vl } ⊂ F q either {vk , vl } ⊂ Fq \ (Fq ∪ {0}). It is well-known that Fq is a subgroup of the multiplicative group F∗q , and the product of two non-square numbers is a square number. These facts and the property in (P1) conclude that Γkl (v) = vl vk−1 ∈ F q ⊂ Γkl (H). Case 2: if wt(v) = 2. Here v can be written as one of the forms (x, y, 0), (0, x, y), or (x, 0, y), in which x = 0 and y = 0. Subcase 2.1: Firstly, suppose that v = (x, y, 0). Note that Γ23 (x, y, 0) = 0 and Γ13 (x, y, 0) = 0. From (P2), it follows that Γ23 (x, y, 0) ∈ Γ23 (H) or Γ13 (x, y, 0) ∈ Γ13 (H). Subcase 2.2: v = (0, x, y). We obtain Γ12 (0, x, y) = ∞ and Γ13 (0, x, y) = ∞. The hypothesis in (P3) states that Γ12 (0, x, y) ∈ Γ12 (H) or Γ13 (0, x, y) ∈ Γ13 (H).
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Subcase 2.3: It remains the case v = (x, 0, y). Note that Γ12 (x, 0, y) = 0 and Γ23 (x, 0, y) = ∞. The hypothesis in (P4) assures that Γ12 (x, 0, y) ∈ Γ12 (H) or Γ23 (x, 0, y) ∈ Γ23 (H). Finally, an application of Corollary 3.7 concludes the statement. 2 Example 3.9. Let us illustrate how to obtain a short covering from Theorem 3.8. Select the following subset of F35 H = {(1, 1, 1), (1, 0, 0), (0, 1, 4), (1, 4, 4)}. 0 2 Note that F 5 = {1, 4} = {2 , 2 }. A simple computation shows us that Γ12 (H) = {0, 1, 4, ∞}, Γ23 (H) = {1, 4, ∞}, Γ13 (H) = {0, 1, 4, ∞}. Thus,
F 5 ⊂ Γ12 (H) ∩ Γ23 (H) ∩ Γ31 (H), 0 ∈ Γ23 (H), ∞ ∈ Γ12 (H) and 0 ∈ Γ12 (H). Theorem 3.8 implies that H is a short covering of F35 , and consequently, c(5) ≤ 4 holds. The bound is sharp, according to [9]. The upper bound was previously obtained from the short covering H∗ = {(1, 1, 1), (0, 0, 1), (1, 4, 0), (1, 1, 4)} by using another method. Since 5 ≡ 1 (mod 4), the new short covering in Example 3.9 motives a search for a general construction, as described in the next section. 4. Proof of Theorem 1.1 The following remark will be helpful in outlining our proof. Remark 4.1. It is well-known that the additive group Zq−1 and the multiplicative group F∗q are isomorphic by the function a → ξ a , in which ξ denotes a primite element of F∗q . As usual, the class a ∈ Zq−1 is simply denoted by a, in which 0 ≤ a ≤ q − 1 and the multiplication follows the rule ξ a ξ b = ξ a+b = ξ c , which c = a + b in Zq−1 . We consider 0 2 q−3 F∗q = {ξ 0 , ξ 1 , . . . , ξ q−2 }. Therefore, F } when q is odd. q = {ξ , ξ , . . . , ξ Outline of the construction We present an outline of Theorem’s 1.1 proof. The construction is based on the following steps: Step 1 We initially find a maximal subset L of F q × Fq × Fq regarding the property in which πkl (L) is PWLI for all indices 1 ≤ k < l ≤ 3.
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Step 2 Secondly, we check that L satisfies “almost” all the hypotheses of Theorem 3.8, by using Theorem 2.6. Step 3 We obtain a subset H of F3q by joining suitable vectors to L. Finally, an application of Theorem 3.8 guarantees that H is a short covering of F3q . We have all components to prove the main result. Proof of Theorem 1.1. We initially consider the case in which q ≡ 3 (mod 4). Let ξ be a primitive element of Fq and select the following subset of F q × Fq × Fq L = {(ξ 0 , ξ 0 , ξ 2·0 ), (ξ 0 , ξ 2 , ξ 2·2 ), (ξ 0 , ξ 4 , ξ 2·4 ), . . . , (ξ 0 , ξ q−3 , ξ 2·(q−3) )} = = {(1, 1, 1), (1, ξ 2 , ξ 4 ), (1, ξ 4 , ξ 8 ), . . . , (1, ξ q−3 , ξ 2q−6 )}.
(4.1)
This set L satisfies the following property. Claim A: For any pair of indices 1 ≤ k < l ≤ 3, the set πkl (L) is PWLI. In order to proceed with the proof of Claim A, it should be noted that both sets π12 (L) and π13 (L) are PWLI. Suppose a contradiction in which π23 (L) is not PWLI. Thus, there are x, y ∈ {0, 2, . . . , q − 3} with x = y, such that the vectors (ξ x , ξ 2x ) and (ξ y , ξ 2y ) are linearly dependent. This fact implies that ξ 2x /ξ x = ξ 2y /ξ y , or equivalently, ξ x = ξ y , which is a contradiction, due the fact that x = y. By using Claim A, Theorem 2.6 states that the set L satisfies |Γkl (L)| = |L| = (q − 1)/2 = |F q | for any pair of indices 1 ≤ k < l ≤ 3. Since Γkl (L) ⊂ Fq , the equalities Γ12 (L) = Γ23 (L) = Γ13 (L) = F q hold. It is still necessary to complete L to a short covering H of F3q by joining two vectors H = L ∪ {(1, 0, 0), (0, 1, 1)}. A simple computation reveals that Γ12 (H) = {0, ∞} ∪ F q , Γ23 (H) = {∞} ∪ Fq , and Γ13 (H) = {0, ∞} ∪ Fq .
Thus, Theorem 3.8 states that H is a short covering of F3q . Since |H| = |L| +2 = (q +3)/2, the bound c(q) ≤ (q + 3)/2 holds. It remains the case in which q ≡ 1 (mod 4). Given a primitive element ξ of Fq , we should consider the sets: L1 = {(ξ 0 , ξ 2x , ξ 2·2x ) : 0 ≤ x ≤ (q − 5)/4} = = {(1, 1, 1), (1, ξ 2 , ξ 4 ), . . . , (1, ξ
q−5 2
, ξ q−5 )},
L2 = {(ξ 0 , ξ 2x , ξ 2·(2x+1) ) : (q − 1)/4 ≤ x ≤ (q − 5)/2} = = {(1, ξ
q−1 2
, ξ q+1 ), (1, ξ
q+3 2
, ξ q+5 ), . . . , (1, ξ q−5 , ξ 2q−8 )}.
The union L = L1 ∪ L2 is a subset of F q × Fq × Fq and satisfies the key property:
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Claim B: For any pair of indices 1 ≤ k < l ≤ 3, the set πkl (L) is PWLI. To prove Claim B, note that π12 (L) and π13 (L) are clearly PWLI. Suppose a contradiction that π23 (L) is not PWLI. Thus, there is a pair of vectors (α, β) and (γ, δ) in π23 (L), such that (α, β) = λ(γ, δ) for some λ ∈ Fq . In consequence, we analyze three cases: Case 1: if (α, β) ∈ π23 (L1 ) and (γ, δ) ∈ π23 (L1 ). In this case, there are 0 ≤ x < y ≤ (q − 5)/4, such that (α, β) = (ξ 2x , ξ 2·2x ) and (γ, δ) = (ξ 2y , ξ 2·2y ). The equality (α, β) = λ(γ, δ) can be reformulated as β/α = δ/γ, that is, ξ 4x /ξ 2x = ξ 4y /ξ 2y . Equivalently, ξ 2x = ξ 2y , which is a contradiction due the fact that x < y. Case 2: if (α, β) ∈ π23 (L2 ) and (γ, δ) ∈ π23 (L2 ). This case is analogous to the previous one. There are (q − 1)/4 ≤ x < y ≤ (q − 5)/2, such that (α, β) = (ξ 2x , ξ 2·(2x+1) ) and (γ, δ) = (ξ 2y , ξ 2·(2y+1) ). The equality (α, β) = λ(γ, δ) implies β/α = δ/γ, that is, ξ 4x+2 /ξ 2x = ξ 4y+2 /ξ 2y , or equivalently, ξ 2x+2 = ξ 2y+2 , which is a contradiction due the fact that x < y. Case 3: if (α, β) ∈ π23 (L1 ) but (γ, δ) ∈ π23 (L2 ). In this case, (α, β) = (ξ 2x , ξ 2·2x ) and (γ, δ) = (ξ 2y , ξ 2·(2y+1) ) for some 0 ≤ x ≤ (q−5)/4 and (q − 1)/4 ≤ y ≤ (q − 5)/2. From (α, β) = λ(γ, δ) we obtain that β/α = δ/γ, that is, ξ 4x /ξ 2x = ξ 4y+2 /ξ 2y , or equivalently, ξ 2x = ξ 2y+2 , which is a contradiction due the fact that 0 ≤ 2x ≤ (q − 5)/2 and (q + 3)/2 ≤ 2y + 2 ≤ q − 3. The proof of Claim B is complete. Since L ⊂ F q ×Fq ×Fq and Fq is a multiplicative group, the set L = L1 ∪L2 satisfies Γ12 (L) ∪ Γ23 (L) ∪ Γ13 (L) ⊂ F q .
(4.2)
By applying Claim B, Theorem 2.6 ensures |Γ12 (L)| = |Γ23 (L)| = |Γ13 (L)| = (q − 3)/2.
(4.3)
We need the following statements: Claim 1: ξ q−3 ∈ / Γ12 (L). Indeed, suppose a contradiction that ξ q−3 ∈ Γ12 (L). By the construction of L, there is a vector (1, ξ 2x , ξ 2y ) in L, such that Γ12 (1, ξ 2x , ξ 2y ) = ξ q−3 . Since Γ12 ((1, ξ 2x , ξ 2y )) = ξ 2x , the equality 2x = q − 3 holds, that is, x = (q − 3)/2, thus ξ 2x = ξ q−3 . The last equality is a contradiction because ξ q−3 does not appear in the second coordinate of a vector in L. Claim 2: ξ (q−1)/2 ∈ / Γ23 (L). By simple inspection, we can see that Γ23 (L) = {1, ξ 2 , ξ 4 , . . . , ξ q−5 , ξ (q+3)/2 , ξ (q+7)/2 , . . . , ξ q−3 }. Therefore, ξ (q−1)/2 ∈ / Γ23 (L).
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Claim 3: ξ q−3 ∈ / Γ13 (L). / Γ13 (L). Indeed, by using an argument similar to Claim 1, we conclude that ξ q−3 ∈ | = (q − 1)/2. A combination of (4.2), (4.3) and Claims 1 to 3 implies Note that |F q q−3 (q−1)/2 q−3 Γ12 (L) = F }, Γ23 (L) = F }, Γ13 (L) = F }. q \ {ξ q \ {ξ q \ {ξ
We now complete L to a short covering of F3q by joining H = L ∪ {(1, 0, 0), (0, 1, ξ (q−1)/2 ), (1, ξ q−3 , ξ q−3 )}. By simple inspection, we can verify that H satisfies the four conditions of Theorem 3.8. Therefore, c(q) ≤ |H| = (q − 3)/2 + 3 = (q + 3)/2. 2 As mentioned in the Introduction, the upper bound c(q) ≤ (q + 3)/2 was obtained for q ≡ 3 (mod 4) in [8]. We proved it once more by using PWLI sets in the first part the proof above. Example 4.2. Let us show that c(11) ≤ 7. Firstly, note that 2 is a primite element of F11 . By the construction given in proof of Theorem 1.1, the correspondent set L is L = {(20 , 20 , 20 ), (20 , 22 , 24 ), (20 , 24 , 28 ), (20 , 26 , 22 ), (20 , 28 , 26 )} = = {(1, 1, 1), (1, 4, 5), (1, 5, 3), (1, 9, 4), (1, 3, 9)}. Thus, a short covering of F311 formed by 7 vectors can be generated from L, H = {(1, 0, 0), (0, 1, 1), (1, 1, 1), (1, 4, 5), (1, 5, 3), (1, 9, 4), (1, 3, 9)}. As a comparative analysis, the set H∗ = {(1, 1, 0), (1, 0, 1), (0, 1, 1), (1, 3, 4), (1, 4, 3), (1, 5, 9), (1, 9, 5)} is a short covering of Z311 too, according to [8, Example 5.2]. Example 4.3. The upper bound c(9) ≤ 6 holds. Let ξ be a primite element of F9 , the construction of Theorem 1.1 is based on the following subset L = L1 ∪L2 of F 9 ×F9 ×F9 , in which L1 = {(ξ 0 , ξ 2x , ξ 2·2x ) : 0 ≤ x ≤ (9 − 5)/4} = {(1, 1, 1), (1, ξ 2 , ξ 4 )}, L2 = {(ξ 0 , ξ 2x , ξ 2·(2x+1) ) : (9 − 1)/4 ≤ x ≤ (9 − 5)/2} = {(1, ξ 4 , ξ 2 )}. Consider H = {(1, 1, 1), (1, ξ 2 , ξ 4 ), (1, ξ 4 , ξ 2 ), (1, 0, 0), (0, 1, ξ 4 ), (1, ξ 6 , ξ 6 )}.
(4.4)
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2 4 6 Note that F 9 = {1, ξ , ξ , ξ }. So, we have Γ12 (H) = {0, ∞} ∪ F 9 , Γ23 (H) = {∞} ∪ F9 and Γ13 (H) = {0, ∞} ∪ F9 .
The subset H in (4.4) satisfies the hypothesis of Theorem 3.8. Thus, H is a short covering of F39 and c(9) ≤ 6 holds, and this improves the previous upper bound c(9) ≤ 7 from [8]. 5. Conclusion As alluded in the introduction of this study, sharp cyclic coverings of a finite module over a finite ring are computed on the basis of PWLI sets. In this work, we explored new connections between PWLI sets and short coverings of the linear space F3q , which allow us to prove c(q) ≤ (q + 3)/2 for any odd prime power q. In particular, the new upper bound improves c(q) ≤ (q + 5)/2 when q ≡ 1 (mod 4). For future research, it would be interesting to find new PWLI sets in a vector space and their connections with short coverings for higher dimensions. New applications of PWLI sets to discrete structures (finite geometries, projective spaces, combinatorial designs) or new relationships with algebraic concepts would be desirable. Acknowledgements The authors are grateful to the anonymous referee for the careful reading and many suggestions. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]
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