On optimal (v,5,2,1) optical orthogonal codes - Semantic Scholar
Des. Codes Cryptogr. (2013) 68:349–371 DOI 10.1007/s10623-012-9654-x
On optimal (v, 5, 2, 1) optical orthogonal codes Marco Buratti · Anita Pasotti · Dianhua Wu
Received: 14 October 2011 / Revised: 8 March 2012 / Accepted: 9 March 2012 / Published online: 5 April 2012 © Springer Science+Business Media, LLC 2012
Abstract The size of a (v, 5, 2, 1) optical orthogonal code (OOC) is shown to be at most v v equal to 12 when v ≡ 11 (mod 132) or v ≡ 154 (mod 924), and at most equal to 12 in all the other cases. Thus a (v, 5, 2, 1)-OOC is naturally said to be optimal when its size reaches the above bound. Many direct and recursive constructions for infinite classes of optimal (v, 5, 2, 1)-OOCs are presented giving, in particular, a very strong indication about the existence of an optimal ( p, 5, 2, 1)-OOC for every prime p ≡ 1 (mod 12). Keywords
Optimal optical orthogonal code · Difference family · Difference matrix
Mathematics Subject Classification
05B30 · 94B25
1 Introduction A (v, k, λa , λc ) optical orthogonal code (OOC for short) can be seen as a collection C = {C1 , . . . , Cs } of k-subsets (codeword-sets) of Zv satisfying the so called autocorrelation and cross-correlation properties:
This is one of several papers published in Designs, Codes and Cryptography comprising the “Special Issue on Finite Geometries”. M. Buratti (B) Dipartimento di Matematica e Informatica, Università di Perugia, Via Vanvitelli 1, 06123 Perugia, Italy e-mail: [email protected] A. Pasotti Dipartimento di Matematica, Facoltà di Ingegneria, Università degli Studi di Brescia, Via Valotti 9, 25133 Brescia, Italy e-mail: [email protected] D. Wu Department of Mathematics, Guangxi Normal University, 541004 Guilin, People’s Republic of China e-mail: [email protected]
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|Ci ∩ (Ci + t)| ≤ λa for 1 ≤ i ≤ s and 1 ≤ t ≤ v − 1;
(1)
|Ci ∩ (C j + t)| ≤ λc for 1 ≤ i < j ≤ s and 0 ≤ t ≤ v − 1.
(2)
The former condition requires that any two distinct translates of a codeword-set share at most λa elements, while the latter requires that any two translates of two distinct codeword-sets share at most λc elements. The integers v and k are the length and the weight of the code. The size of C is the number s of its codeword-sets and C is maximal when every other OOC with the same parameters has size not greater than s. OOCs have been introduced in [17] and they have great importance in the applications [19]; the larger is their size, the greater is their usefulness. A (v, k, λa , λc )-OOC in which λa = λc = λ is briefly denoted by (v, k, λ)-OOC. For a long time the most investigated ones have been the optimal (v, k, 1)-OOCs. The existence problem for an optimal (v, k, 1)-OOC has been solved for k = 3 (see [7]) but to find a complete solution for larger values of k seems to be, at the moment, very difficult (see, e.g., [1,10,14] for some partial answers in the case when k = 4). In the last years several interesting results have been also obtained about optimal (v, k, λ)-OOCs with λ = 2 [3,16,20] and even with λ > 2 [4]. In an earlier paper [28] about (v, k, λa , λc )-OOCs with λa = λc it is shown that the size of a (v, k, λa , 1)-OOC cannot exceed λa v−1 k but this bound is, in general, far from being tight. Several results about (v, k, k − 1, 1)-OOCs (the so called conflict-avoiding codes) have been recently obtained in [23,24] while (v, 4, 2, 1)-OOCs have been deeply investigated in [5,11,25,26]. The present paper deals with (v, 5, 2, 1)-OOCs and it is the natural continuation of [11] and [25]. As one could expect, many results are, in a certain sense, “twin results” of those papers. For instance, it was proved in [11] and [25] that a tight upper bound on the size of a (v, 4, 2, 1)-OOC is the floor or the ceiling of v8 according to the congruence class of v. The twin result here is that a tight upper bound on the size of a (v, 5, 2, 1)-OOC is v the floor or the ceiling of 12 . On the other hand some constructions in this paper are considerably more difficult than their “twins” in [11] and [25]; in various cases we need to invent “ad hoc” tricks to obtain our codes. For instance, we mention the use of some perfect or near-perfect matching of a suitable hypergraph described in the third section to try to obtain an optimal (v, 5, 2, 1)-OOC with v a prime ≡ 1 or 5 (mod 12). We prove that such an optimal code is certainly realizable whenever v ≡ 13 (mod 24) or also when v ≡ 1 (mod 24) provided that 2 is not a 3n -th power modulo v, 3n being the largest power of 3 in v − 1. In all the other unlucky cases we give an algorithm which seems to be always successful to construct the desired optimal code. 2 A bound on the maximum possible size of a (v, 5, 2, 1)-OOC As usual, the list of differences of a given subset C of Zv will be denoted by C. This is the multiset of all differences x − y with (x, y) an ordered pair of distinct elements of C. Also, by C we will denote the set of differences of C, namely the underlying set (or support) of C. C More generally, the list of differences of a set C of subsets of Zv is the multiset C = and, analogously, C =
C∈C
C∈C
C. Saying that C is difference-disjoint we mean that the lists
of differences of every two distinct members of C are disjoint.
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We denote by μ(C) the maximum multiplicity in C, namely, μ(C) = max{μC (i) | 1 ≤ i ≤ v − 1} where μC (i) is the multiplicity of i in C. If δ is the size of C, namely the number of distinct differences appearing in C, we will say that C is of type t (C) = δ. It is obvious that t (C) cannot exceed the size of C so that we have t (C) ≤ k(k − 1) where k is the size of C. Saying that a code C is of type [δ1n 1 , δ2n 2 , . . . , δtn t ], we mean that C has exactly n i codeword-sets of type δi for i = 1, 2, . . . , t. Condition (1) is equivalent to ask that no element of Zv appears in the list of differences of any codeword-set more than λa times: μ(Ci ) ≤ λa for 1 ≤ i ≤ s.
(1 )
Also, in the case that λc = 1, condition (2) is equivalent to ask that C is difference-disjoint, namely that Ci ∩ C j = ∅ for 1 ≤ i < j ≤ s.
(2 )
The set of missing differences of a (v, k, λa , 1)-OOC is the set M of all elements of Zv that are not covered by the lists of differences of its codeword-sets. Following [29] we say that an optical orthogonal code is m-regular in the special case that M is a subgroup of Zv . Note, in particular, that a m-regular (v, k, 1)-OOC is a (v, m, k, 1) relative difference family [9] As already remarked in [11], a codeword-set C of a (v, k, λa , 1)-OOC is of odd type if and only if v is even and v2 ∈ C. Thus a (v, k, λa , 1)-OOC may have at most one codeword-set of odd type otherwise (2 ) would be contradicted. Let C be a codeword-set of type δ of a (v, k, λa , 1)-OOC and set C = {x1 , . . . , xδ }. Denoted by μi the multiplicity of xi in C, we obviously have |C| = k(k −1) = μ1 +· · ·+ μδ . On the other hand we have μi ≤ λa for 1 ≤ i ≤ δ by (1 ) and hence μ1 +· · ·+μδ ≤ δλa . It follows that k(k − 1) ≤ δλa and hence the type of any codeword-set of a (v, k, λa , 1)OOC is at least equal to
k(k−1) λa
. It follows, in particular, that every codeword-set of a
(v, 5, 2, 1)-OOC is of type δ ≥ 10. Let us say that two subsets C and C of Zv are equivalent if we have C = m · C + t for a suitable pair (m, t) with gcd(m, v) = 1. By means of long and tedious calculations we have checked that Zv has a 5-subset C with μ(C) ≤ 2 and |C| = 10 if and only if one of the following two conditions holds: v v 5v 5v v is divisible by 12 and, up to equivalence, C = 0, 12 , 6 , 12 , 6 ; 3v 4v 8v v is divisible by 11 and, up to equivalence, C = 0, 2v 11 , 11 , 11 , 11 .
Hence, by (2 ), a (v, 5, 2, 1)-OOC may have at most two codeword-sets of type 10. They could be two only in the case that v is divisible by 132 = 11 · 12. Further tedious calculations allowed us to see that Zv has a 5-subset C with μ(C) ≤ 2 and |C| = 11 if and only if one of the following two conditions holds:
v is divisible by 4 and, up to equivalence, C = {0, a, −a, v4 , − v4 } for some a ∈ Zv \{ v4 , − v4 , v2 }; v v 3v v v is divisible by 14 and, up to equivalence, C = 0, 14 , − 14 , 14 , 2 . Now note that C = {0, a, −a, b, −b} is a 5-subset of Zv with μ(C) = 2 and |C| = 12 as soon as we have a ∈ {±b, ±2b, ±3b, v4 , v2 ± b} and, symmetrically, b ∈ {±a, ±2a, ±3a, v4 , v 2 ± a}. In this case in fact the set of differences of C is
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C = {±a, ±b, ±2a, ±2b, ±(a − b), ±(a + b)} and we have: μC (±2a) = μC (±2b) = 1;
μC (x) = 2 for x ∈ C\{±2a, ±2b}.
So, for v > 13, we can easily see that Zv has “many” 5-subsets C with μ(C) ≤ 2 and |C| = 12. All the above remarks lead us to the following result. Theorem 2.1 If s is the size of a (v, 5, 2, 1)-OOC, then we have: v 12 for v ≡ 11 (mod 132) or v ≡ 154 (mod 924) s≤ v 12 other wise Proof Let C = {C1 , . . . , Cs } be a (v, 5, 2, 1)-OOC of size s and let δi be the type of Ci so that we have v − 1 ≥ δ1 + · · · + δs . Let v = 12q + r be the Euclidean division of v by v 12 so that q = 12 and 0 ≤ r ≤ 11. By the previous remarks C may have at most two codeword-sets of type 10 and at most one codeword-set of type 11. All other codeword-sets are of type at least 12. Thus we have v − 1 ≥ δ1 + · · · + δs ≥ 12(s − 3) + 31 which gives 12q + r + 4 ≥ 12s. This certainly implies that s never exceeds q + 1. From now on we will assume that s = q + 1. To prove the assertion it is enough to show that either v ≡ 11 (mod 132) or v ≡ 154 (mod 924). 1st case: v is divisible by 12. We have at most two codeword-sets of type 10 and at most one codeword-set of type 11. Thus we have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12(q − 2) + 31 which implies r ≥ 8 against the assumption that r = 0. 2nd case: v is divisible neither by 11 nor by 12. We have at most one codeword-set of type 11 and no codeword-sets of type 10. Thus we have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12q + 11 which implies r ≥ 12, a contradiction. 3rd case: v is odd and divisible by 11 but not by 12. We have at most one codeword-set of type 10 and no codeword-sets of type 11. Thus we have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12q + 10 which gives r ≥ 11 and hence r = 11, i.e., v = 12q + 11. Then, recalling that v is divisible by 11, we necessarily have v ≡ 11 (mod 132). 4th case: v is even and divisible by 11 but not by 12. Here we have at most one codeword-set of type 10 and at most one codeword-set of type 11. If C does not have codeword-sets of type 11, then we have v −1 ≥ δ1 +· · ·+δq+1 ≥ 12q +10 which gives r ≥ 11 and hence r = 11 contradicting that v is even. Thus C actually has a codeword-set of type 11 so that v is divisible either by 14 or by 4. We have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12(q − 1) + 21 which gives r ≥ 10 and hence r = 10 since v is even. Thus v = 12q + 10 besides being divisible by 11 is also divisible by 14 since it is evident that 12q + 10 is not divisible by 4. Using the Chinese Remainder Theorem one can see that the three congruences v ≡ 10 (mod 12), v ≡ 0 (mod 11) and v ≡ 0 (mod 14) simultaneously hold only for v ≡ 154 (mod 924). It is natural to say that a (v, 5, 2, 1)-OOC is optimal when its size s reaches the upper bound given by Theorem 2.1. In particular, we will say that a (v, 5, 2, 1)-OOC is perfect when, besides being optimal, its only missing difference is zero, i.e., when it is optimal and 1-regular at the same time. Note that an optimal (v, 5, 2, 1)-OOC is necessarily perfect in the following cases:
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Optical orthogonal codes Table 1 Optimal (v, 5, 2, 1)-OOCs of small order
353 Length
Code
1 ≤ v ≤ 10 11 12 ≤ v ≤ 23 24 ≤ v ≤ 28 29 ≤ v ≤ 32 33 34 35 36 37 38 39 40 41 42 43 and 44 45 46 and 47 48 ≤ v ≤ 51 52 53 54 55 56 57 58 59 60 61 62
∅ = {1, 3, 4, 5, 9} F11 {C2,3 } ∃ {C2,5 , C8,9 } {C2,8 , C5,9 } {C1,5 , C8,11 } {C2,9 , C5,8 } ∃ {C1,6 , C8,11 , C10,14 } ∃ {C3,4 , C9,14 , C10,12 } ∃ {C1,5 , C7,15 , C9,12 } {C2,15 , C3,8 , C9,10 } {C2,7 , C3,13 , C11,12 } {C1,4 , C6,16 , C7,18 } {C2,7 , C3,15 , C10,11 } ∃ {C6,11 , C7,16 , C13,21 , C24,25 } {C5i ,5i 30 | 1 ≤ i ≤ 4} ∃ {C2,3 , C7,20 , C8,18 , C11,23 } {C2,3 , C7,20 , C9,17 , C12,23 } {C2,3 , C7,24 , C10,22 , C15,23 } {C1,4 , C7,26 , C9,21 , C11,24 } {C2,3 , C8,15 , C10,21 , C12,25 } ∃ {C20i ,20i 11 | 0 ≤ i ≤ 4} ∃
v ≡ 11 (mod 132) necessarily of type 10, 12s−1 ;
v ≡ 154 (mod 924) necessarily of type 10, 11, 12s−2 ; v ≡ 1 (mod 12) but v ≡ 0 (mod 11) necessarily of type 12s . In Table 1 we exhibit an optimal (v, 5, 2, 1)-OOC for any length v ≤ 62 when it exists. Here, and throughout the paper, we agree to briefly denote by C x,y the codewordset {0, x, −x, y, −y}. We point out that for each v ≤ 62 for which the Table 1 indicates the non-existence of a (v, 5, 2, 1)-OOC, there exists a (v, 5, 2, 1)-OOC which fails optimality by missing just one codeword-set. 3 House/dart difference families and (v, 5, 2, 1)-OOCs Optimal OOCs can be often described in terms of certain graph labelings. For instance, in most cases an optimal (v, 4, 2, 1)-OOC can be described in terms of suitable kite labelings as shown in [25]. In this section we will describe optimal (v, 5, 2, 1)-OOCs in terms of suitable house labelings or, equivalently, of dart labelings. The description could appear a little bit artificial but we will see it is quite convenient to briefly explain the recursive constructions.
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Fig. 1 The house graph and the dart graph
Fig. 2 The good house Hx,y
Fig. 3 The good dart Dx,y
Following [27], the graphs H and D depicted in Fig. 1 will be called the house graph and the dart graph, respectively. A house graph with vertices in Zv will be called good if it is of the form shown in Fig. 2 for some elements x and y of Zv with 0 < x < y < v2 . Throughout the paper such a good house graph will be denoted by Hx,y . Sometimes we will also write m · Hx,y in place of Hmx,my . Analogously, we say that a dart graph with vertices in Zv is good if it is of the form shown in Fig. 3 for some elements x and y of Zv and we will denote it by Dx,y . Recall that the list of differences of a simple graph B with vertices in Zv is defined as the multiset B of all possible differences x − y with (x, y) an ordered pair of adjacent vertices of B. We also recall (see [12]) that given a graph and positive integers m and v, a (mv, m, , 1) difference family (or DF for short) is a collection of graphs (blocks) with vertices in Zmv all of which are isomorphic to and whose lists of differences cover exactly once Zmv \{0, v, 2v, . . . , (m −1)v}. In the important case of m = 1 we simply write (v, , 1)DF rather than (v, 1, , 1)-DF. For difference families in the usual sense, corresponding to (v, , 1)-DFs in which is a complete graph, see [2] or [6]. A (v, n, H, 1)-DF or (v, n, D, 1)-DF, where H and D denote the house graph and the dart graph respectively, will be said good if all its blocks are such. Note that if the 5-subset C x,y = {0, x, −x, y, −y} of Zv is of type 12, then its set of differences coincides with the list of differences of the good house Hx,y and also coincides with the list of differences of the good dart Dx,y . We namely have C x,y = Hx,y = Dx,y . It is then obvious that a good (mv, m, H, 1)-DF {Hx1 ,y1 , . . . , Hxt ,yt } is equivalent to the good (mv, m, D, 1)DF {Dx1 ,y1 , . . . , Dxt ,yt } and that it gives a (mv, 5, 2, 1)-OOC whose codeword-sets are C x1 ,y1 , . . . , C xt ,yt . In particular, we have: Proposition 3.1 Every good (mv, m, H, 1)-DF with 1 ≤ m ≤ 9 gives an optimal (mv, 5, 2, 1)-OOC. Proof The (mv, 5, 2, 1)-OOC associated with a (mv, m, H, 1)-DF has size q = mv−m 12 . Thus, having mv = 12q + m and, by assumption, 1 ≤ m ≤ 9, we recognize that q and m
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are quotient and remainder of the Euclidean division of mv by 12. Hence q = mv 12 and the assertion follows from Theorem 2.1 considering that we cannot have neither mv ≡ 11 (mod 132) nor mv ≡ 154 (mod 924) since otherwise we would have mv ≡ 11 or 10 (mod 12) and hence m = 11 or 10 against the assumption 1 ≤ m ≤ 9. 4 On (t, ω)-perfect and (t, ω)-near-perfect subsets of a group In this section we generalize the concepts of a ω-perfect and of a ω-near perfect subset of a group introduced in [14] to those of a (t, ω)-perfect and of a (t, ω)-near perfect subset of a group. In this paper we will need this generalization only for the case of t = 3 but we do not exclude that the general case could be helpful for other purposes in some future works. Let G be a multiplicative abelian group and let ω be an element of G of order o(ω) ≥ t. We define G(t, ω) to be the t-uniform hypergraph whose vertices are the elements of G and whose hyper-edges are all the t-subsets of G of the form {xωi | 0 ≤ i ≤ t − 1} with x ∈ G. Such a hyper-edge will be denoted by [xω0 , xω1 , . . . , xωt−1 ]. Note that using the terminology of [8], G(t, ω) is the t-Cayley hypergraph Cay[G : S] where S is the singleton {ω}. For a given subset X of G, let G(t, ω) − X be the hypergraph obtained from G(t, ω) by deleting all vertices of X and all hyper-edges having non-empty intersection with X . We say that X is (t, ω)-perfect in G if there exists a perfect matching of G(t, ω) − X , that is a set of pairwise disjoint hyper-edges of G(t, ω) − X covering all vertices of G − X . Analogously, we say that X is (t, ω)-near-perfect in G if G(t, ω) − X admits a near-perfect matching, that is a set of pairwise disjoint hyper-edges of G(t, ω) − X covering all but one vertex of G − X . Here is the easy criterion for establishing whether X is (t, ω)-perfect or (t, ω)-near-perfect in G. Proposition 4.1 Let ω be an element of order d of a multiplicative abelian group G, let S be a complete system of representatives for the cosets of the group ω generated by ω, and consider the indicator function of G with respect to ω and S defined by indic ωi s = (i, s) 0 ≤ i ≤ d − 1; s ∈ S. For a given subset X of G and for a given s ∈ S, let indics (X ) = (i s,1 , i s,2 , . . . , i s,s ) be the (possibly empty) increasing sequence of values of i ∈ {0, 1, . . . , d − 1} such that (i, s) is the indicator of a suitable x ∈ X . Then X is (t, ω)-perfect in G if and only if the following two conditions hold: (π1 ) s ≡ d (mod t) for every s ∈ S; (π2 ) i s,k+1 − i s,k ≡ 1 (mod t) for any pair (s, k) with s ∈ S and 0 ≤ k ≤ s − 1. Instead, X is ω-near-perfect in G if and only if the following two conditions hold: (ν1 ) s ≡ d (mod t) for all but exactly one element of S, say s ∗ , for which we have s ∗ ≡ d−1 (mod t); (ν2 ) i s,k+1 − i s,k ≡ 1 (mod t) for any pair (s, k) with s ∈ S and 0 ≤ k ≤ s − 1 with at most one possible exception (s ∗ , k ∗ ) for which one could have i s ∗ ,k ∗ +1 − i s ∗ ,k ∗ ≡ 2 (mod t). The knowledge of indic(X ) immediately allows to determine a perfect or near-perfect matching of G(t, ω) − X when it exists.
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As an example, consider the group G = U65 of units of Z65 , let ω = 7, and let X = {1, 8, 9, 22, 28, 29, 31, 36, 42, 56, 57, 58}. The element ω has order d = 12 in G and we can take S = {s0 , s1 , s2 , s3 } = {1, 2, 3, 6} as complete system of representatives for the cosets of ω in G. Check that we have: 1 = ω0 s0 ;
8 = ω10 s1 ;
9 = ω5 s1 ;
22 = ω9 s3 ;
28 = ω11 s0 ;
29 = ω9 s1 ;
31 = ω8 s3 ;
36 = ω3 s1 ;
42 = ω s3 ;
56 = ω s1 ;
57 = ω s1 ;
58 = ω7 s0 .
1
11
4
Thus we can write: indics0 (X ) = (0, 7, 11); indics1 (X ) = (3, 4, 5, 9, 10, 11); indics2 (X ) = ();
indics3 (X ) = (1, 8, 9).
We immediately see that conditions (π1 ) and (π2 ) are satisfied with t = 3 and that the unique perfect matching M of G(3, ω) − X is given by {[ω1 s0 , ω2 s0 , ω3 s0 ], [ω4 s0 , ω5 s0 , ω6 s0 ], [ω8 s0 , ω9 s0 , ω10 s0 ], [ω0 s1 , ω1 s1 , ω2 s1 ], [ω6 s1 , ω7 s1 , ω8 s1 ], [ω0 s2 , ω1 s2 , ω2 s2 ], [ω3 s2 , ω4 s2 , ω5 s2 ], [ω6 s2 , ω7 s2 , ω8 s2 ], [ω9 s2 , ω10 s2 , ω11 s2 ], [ω2 s3 , ω3 s3 , ω4 s3 ], [ω5 s3 , ω6 s3 , ω7 s3 ], [ω10 s3 , ω11 s3 , ω0 s3 ]} namely M = {[7, 49, 18], [61, 37, 64], [16, 47, 4], [2, 14, 33], [63, 51, 32], [3, 21, 17], [54, 53, 46], [62, 44, 48], [11, 12, 19], [34, 43, 41], [27, 59, 23], [24, 38, 6]}. As another example, consider the group G = U55 of units of Z55 , let ω = 3, and let X = {1, 8, 16, 18, 28, 29, 38, 43, 48}. The element ω has order d = 20 in G and we can take S = {s0 , s1 } = {1, 2} as complete system of representatives for the cosets of ω in G. Check that we have: 1 = ω0 s0 ; 8 = ω14 s1 ; 16 = ω8 s0 ; 18 = ω2 s1 ; 28 = ω6 s1 ; 29 = ω7 s1 ; 38 = ω13 s0 ; 43 = ω18 s1 ; 48 = ω9 s0 . Here we can write: indics0 (X ) = (0, 8, 9, 13); indics1 (X ) = (2, 6, 7, 14, 18). Conditions (ν1 ) and (ν2 ) are satisfied with t = 3 and we immediately see that G(3, ω) − X has exactly three near-perfect matchings one of which, say M, is given by
10 ω s1 , ω11 s1 , ω12 s1 , ω1 s0 , ω2 s0 , ω3 s0 , ω4 s0 , ω5 s0 , ω6 s0 ,
17 ω3 s1 , ω4 s1 , ω5 s1 , ω s0 , ω18 s0 , ω19 s0 , ω14 s0 , ω15 s0 , ω16 s0 ,
8
11 ω s1 , ω9 s1 , ω10 s1 , ω s1 , ω12 s1 , ω13 s1 , ω15 s1 , ω16 s1 , ω17 s1 ,
19 ω s1 , ω0 s1 , ω1 s1 namely M = {[3, 9, 27], [26, 23, 14], [34, 47, 31], [4, 12, 36], [53, 49, 37], [54, 52, 46], [32, 41, 13], [39, 7, 21], [24, 17, 51], [19, 2, 6]}. The second near-perfect matching of G(3, ω) − X is obtainable from M by multiplying its first two hyper-edges by ω. The third one is obtainable from M by multiplying by ω just the second hyper-edge. As a special trivial case of Proposition 4.1 we have: Proposition 4.2 The empty set is (t, ω)-perfect in G if and only if the order of ω is divisible by t.
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The empty set is (t, ω)-near-perfect in G if and only if G is generated by ω and we have o(G) = o(ω) ≡ 1(mod t). From now on, if L is a subgroup of an abelian group G and S is a complete system of representatives for the cosets of L in G, we will simply say that S is a transversal of G/L. Moreover, given two subsets A and B of a multiplicative group G, we denote by A · B the list of all elements ab with a ∈ A and b ∈ B. The following lemma will be very useful for the construction of some optimal ( p, 5, 2, 1)OOCs with p ≡ 1 (mod 12), p prime. In the lemma, G, ω, S and indic have the same meaning as in Proposition 4.1. Lemma 4.3 Let ω ≤ G 1 ≤ G and let S1 ⊂ S be a transversal of G/G 1 . Assume that X 0 is a subset of G such that indics1 (X 0 ) satisfies properties (π1 ) and (π2 ) for every s1 ∈ S1 while indics (X 0 ) = () for every s ∈ S\S1 . Then, for every transversal S2 of G 1 /ω, we have that the set X := X 0 · S2 is (t, ω)-perfect in G. Proof First observe that S = S1 ·S2 is a transversal of G/ω. Now, let indic be the indicator function with respect to ω and S , and let s be any element of S so that we have s = s1 s2 with s1 ∈ S1 and s2 ∈ S2 . The hypothesis that indics (X 0 ) = () for every s ∈ S \ S1 means that X 0 is entirely contained in ω · S1 . This very easily implies that X ∩ ωs1 s2 = (X 0 ∩ ωs1 )s2 and hence it is clear that indics 1 s2 (X ) = indics1 (X 0 ). So, considering that indics1 (X 0 ) satisfies properties (π1 ) and (π2 ) for every s1 ∈ S1 by assumption, we deduce that indics (X ) satisfies properties (π1 ) and (π2 ) for all elements s ∈ S and hence X is (t, ω)-perfect in G by Proposition 4.1. 5 Constructions for perfect (p, 5, 2, 1)-OOCs with p prime, p ≡ 1 (mod 12) In this section we present a construction for good ( p, H, 1)-DFs with p a prime and hence, necessarily, with p ≡ 1 (mod 12). This will give perfect ( p, 5, 2, 1)-OOCs. Before proceeding, we need to recall some notation and terminology about finite fields. Given a prime p ≡ 1 (mod e), let ρ be a primitive element of Z p , let C e be the set of nonzero e-th powers of Z p and, for any integer i, let Cie := ρ i C e be the coset of C e represented by ρ i in the multiplicative group Z∗p of non-zero elements of Z p . We have Cie = C ej if and only if i ≡ j (mod e) and an e-subset S of Z∗p is a transversal of Z∗p /C e if and only if S has exactly one element in Cie for 0 ≤ i ≤ e − 1. We will need the following application of the Theorem of Weil on multiplicative character sums (see [22], Theorem 5.41) that we obtained in [13]. Theorem 5.1 Given a prime p ≡ 1(mod e), a t-subset B = {b1 , . . . , bt } of Z p , and a t-tuple (β1 , . . . , βt ) of Zte , the existence of an element x ∈ Z p satisfying the t cyclotomic conditions x − bi ∈ Cβei (i = 1, . . . , t) is guaranteed for p > t 2 e2t . As a matter of fact the existence of an element x as in the statement is guaranteed for p greater than a bound Q(e, t) which is better (namely, smaller) than t 2 e2t but whose expression is quite complicated (see also [15]). Theorem 5.2 Let p ≡ 1(mod 12) be a prime, let ε be a primitive 4th root of unity of Z p and set ω := ε + 1. Assume that there exists a set A of good houses of Z p such that A = ±X where X is (3, ω)-perfect in the quotient group G := Z∗p /ε. Then there exists an optimal ( p, 5, 2, 1)-OOC.
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Proof First, consider the good house H1,ε and check that we have H1,ε = 1, ω, ω2 ε.
(1)
By assumption, there exists a perfect matching
M = b, bω, bω2 | b ∈ B
(2)
of G(3, ω) − X so that we have
b, bω, bω2 ε = Z∗p \X ε.
(3)
b∈B
Then, setting B = {b · H1,ε | b ∈ B}, by (1) and (2) we have b, bω, bω2 ε = Z∗p \X ε. (Hb,bε ) = B = b∈B
(4)
b∈B
Now set A+ = A ∪ {ε · H | H ∈ A} and note that we have A+ = X ε. Putting together (3) and (4), we conclude that A+ ∪ B is a good ( p, H, 1)-DF. The assertion then follows from Proposition 3.1. We present an example in which a clever use of Theorem 5.2 and Lemma 4.3 allows us to construct an optimal (1657, 5, 2, 1)-OOC. Example 1 Let p = 1657. Check that ρ = 11 is a primitive element of Z p and hence that ε = ρ ( p−1)/4 = 783 is a primitive 4th root of unity of Z p . Now check that ω = ε + 1 has order d = 23 in G = Z∗p /ε and that S = {ρ j | 0 ≤ j ≤ 17} is a transversal of G/ω. Consider the set of good houses of Z p A0 = H1,ρ , Hω4 ,ω19 ρ , ωH1,ρ , ωHω4 ,ω19 ρ . The list of differences H1,ρ ∪ Hω4 ,ω19 ρ is given by ± 1, ε 3 ω2 , ρ, ε 3 ω2 ρ, εω17 ρ 3 , ω9 ρ 2 , ω4 , ε 3 ω6 , ω19 ρ, ε 3 ω21 ρ, ε 2 ω7 ρ 11 , ε 2 ω5 ρ 10 and hence, if indic is the indicator function of G with respect to ω and S, it is straightforward to see that we have A0 = ±X 0 with indicρ 0 (X 0 ) = (0, 1, 2, 3, 4, 5, 6, 7),
indicρ 1 (X 0 ) = (0, 1, 2, 3, 19, 20, 21, 22),
indicρ 2 (X 0 ) = (9, 10),
indicρ 3 (X 0 ) = (17, 18),
indicρ 10 (X 0 ) = (5, 6),
indicρ 11 (X 0 ) = (7, 8).
We see that indics (X 0 ) satisfies conditions (π1 ) and (π2 ) of Proposition 4.1 with t = 3 for each s ∈ S1 = {ρ 0 , ρ 1 , ρ 2 , ρ 3 , ρ 10 , ρ 11 }. Note that the six-tuple (0, 1, 2, 3, 10, 11) is a complete set of residues modulo 6 so that S1 is a transversal of G/G 1 where G 1 is the subgroup of G between ω and G generated by ρ 6 . Also note that indics (X 0 ) = () for every s ∈ S\S1 . Thus, taking S2 = {ρ 0 , ρ 6 , ρ 12 } as a transversal of G 1 /ω, we have that the set X = X 0 ·S2 is (3, ω)-perfect in G by Lemma 4.3. Of course ±X is the list of differences of the set of good houses A = A0 ∪ ρ 6 · A0 ∪ ρ 12 · A0 , namely of A = ωi ρ 6 j H1,ρ , ωi ρ 6 j Hω4 ,ω19 ρ | i = 0, 1; j = 0, 1, 2 ,
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and hence there exists an optimal ( p, 5, 2, 1)-DF by Theorem 5.2. Following the instructions given in the proof of that theorem one can see that such an optimal code is given by ε h ωi ρ 6 j · C1,ρ , ε h ωi ρ 6 j · Cω4 ,ω19 ρ | h, i = 0, 1; j = 0, 1, 2 ∪ {b · C1,ε | b ∈ B} where B is the union of the following sets ω3i+8 ρ 6 j | 0 ≤ i ≤ 4; 0 ≤ j ≤ 2 ; ω3i+11 ρ 6 j+2 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 ; ω3i+7 ρ 6 j+10 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 ;
ω3i+4 ρ 6 j+1 | 0 ≤ i ≤ 4; 0 ≤ j ≤ 2 ; ω3i+19 ρ 6 j+3 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 ; ω3i+9 ρ 6 j+11 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 .
Theorem 5.3 Let p ≡ 1(mod 12) be a prime, let ε be a primitive 4th root of unity of Z p and let d be the order of ω = ε + 1 in the quotient group G = Z∗p /ε. There exists a perfect ( p, 5, 2, 1)-OOC in each of the following cases: (i) d ≡ 0(mod 3); (ii) d ≡ 1(mod 3) and there exists a ∈
(d−4)/3
{1, a, a − 1} is a transversal of Z∗p /C 3 ; (iii) d ≡ 2(mod 3) and there exists a ∈ is a transversal of Z∗p /C 3 .
k=0
ω3k+2 + 1 ω3k+2 ε + 1 , ω3k+2 − 1 ω3k+2 ε − 1
such that
(d−2)/3
ω3k+1 , ω3k+1 ε such that {1, a −1, a +1}
k=0
e Proof Set e = p−1 4d and note that the group C of e-th powers of Z p is the product of ε and ω. Thus, denoting by ρ a primitive root of Z p , we can write C ej = ε h ωi ρ j | 0 ≤ h ≤ 3; 0 ≤ i ≤ d − 1 for 0 ≤ j ≤ e − 1.
Assume that (i) holds. In this case the empty-set is (3, ω)-perfect in G by Proposition 4.2. Thus Theorem 5.2 is successful using as A the empty-set. In this case the presentation of the perfect ( p, 5, 2, 1)-OOC is very easy: d C = ω3i ρ j C1,ε | 0 ≤ i ≤ − 1; 0 ≤ j ≤ e − 1 . 3 Assume that (ii) holds. Set a = ε h 1 ωi1 ρ j1 and a − 1 = ε h 2 ωi2 ρ j2 . It is readily seen that a + 1 ∈ {ω3k+2 (a − 1), εω3k+2 (a − 1)} so that we have a + 1 ∈ {ε h 2 ωi2 +3k+2 ρ j2 , ε h 2 +1 ωi2 +3k+2 ρ j2 }. Note that both ε and ω are cubes of Z∗p so that we have a ∈ C 3j1 and a − 1 ∈ C 3j2 . It follows that {0, j1 , j2 } is a complete set of residues modulo 3 since 1, a and a − 1 lie in pairwise distinct cosets of C 3 by assumption. Let A0 = {H1,a , ω · H1,a } and set A0 = ±X 0 . Observing that ω2 = 2ε and that ε = 1 in the quotient group G, it is easy to check that the following identity holds in G: X 0 = 1, ω, ω2 , ω3 , ωi1 ρ j1 , ωi1 +1 ρ j1 , ωi1 +2 ρ j1 , ωi1 +3 ρ j1 , ωi2 ρ j2 , ωi2 +1 ρ j2 , ωi2 +3k+2 ρ j2 , ωi2 +3k+3 ρ j2 . Taking S = {ρ 0 , ρ 1 , . . . , ρ e−1 } as a transversal of G/ω and denoting by indic the indicator function of G with respect to ω and S, it is straightforward to see that we have:
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indicρ 0 (X 0 ) = (0, 1, 2, 3); indicρ j1 (X 0 ) = (i 1 , i 1 + 1, i 1 + 2, i 1 + 3); indicρ j2 (X 0 ) = (i 2 , i 2 + 1, i 2 + 3k + 2, i 2 + 3k + 3). We see that indics (X 0 ) satisfies conditions (π1 ) and (π2 ) of Proposition 4.1 with t = 3 for each s ∈ S1 = {ρ 0 , ρ j1 , ρ j2 }. Note that S1 is a transversal of G/G 1 where G 1 is the subgroup of G between ω and G generated by ρ 3 , since we have seen that {0, j1 , j2 } is a complete set of residues modulo 3. Also note that indics (X 0 ) = () for every s ∈ S \ S1 . Thus, taking S2 = {ρ 3 j | 0 ≤ j ≤ 3e − 1} as a transversal of G 1 /ω, we have that the set X = X 0 · S2 is (3, ω)-perfect in G by Lemma 4.3. Of course ±X is the list of differences of the set of good houses e A = ρ 3 j · H1,a , ρ 3 j ω · H1,a | 0 ≤ j ≤ − 1 3 and then an optimal ( p, 5, 2, 1)-OOC exists by Theorem 5.2. Now assume that (iii) holds. Set a − 1 = ε h 1 ωi1 ρ j1 and a + 1 = ε h 2 ωi2 ρ j2 . Also here, we can see that {0, j1 , j2 } is a complete set of residues modulo 3 by assumption. Set A0 = {H1,a , ω · H1,a } and A0 = ±X 0 . Note that we have indicρ 0 (X 0 ) = (0, 1, 2, 3, 3k + 1, 3k + 2, 3k + 3, 3k + 4); indicρ j1 (X 0 ) = (i 1 , i 1 + 1); indicρ j2 (X 0 ) = (i 2 , i 2 + 1). Reasoning exactly as in case (ii) one can see that the set of good houses A (also defined as in case (ii)) has list of differences ±X with X that is (3, ω)-perfect in G. Hence the assertion follows from Theorem 5.2. With the same notation as in the above theorem, set p = 3n 4t + 1 with t not divisible by 3. Observing that ω8 = 24 , we can write: n
2 ∈ C 3 ⇐⇒ 24t = 1 ⇐⇒ ω8t = 1 ⇐⇒ ω2t ∈ ε ⇐⇒ d | 2t ⇐⇒ 3 | d. It follows that condition (i) of Theorem 5.3 holds if and only if 2 is not a 3n -th power in Z p . Thus we can state: Corollary 5.4 Let p ≡ 1(mod 12) be a prime, let 3n be the largest power of 3 dividing p −1, and assume that 2 is not a 3n -th power of Z p . Then there exists an optimal ( p, 5, 2, 1)-OOC. Theorem 5.5 There exists a perfect ( p, 5, 2, 1)-OOC for every prime p ≡ 13(mod 24). Proof The assertion is obviously true in the case that 2 is not a cube by Corollary 5.4. Thus, in the following we will assume that p ≡ 13 (mod 24) is a prime and that 2 is a cube of Z p . Recall that 2 is a square in a finite field of order q if and only if q ≡ ±1 (mod 8) by the Quadratic Reciprocity Law [21]. Thus in our case 2 is a non-square of Z p and hence we can claim that 2 ∈ C36 since we have assumed that 2 is a cube. This allows to see that there exists an element x ∈ Z p such that the six-tuple L = {1, 2, x, 2x, x − 1, x + 1} is a transversal of Z∗p /C 6 ; it is enough that x satisfies one of the following cyclotomic conditions: ⎧ 6 ⎨ x ∈ Ci x − 1 ∈ C 6j (i, j, k) ∈ I ⎩ x + 1 ∈ Ck6 with I = {(1, 2, 5), (1, 5, 2), (2, 1, 4), (2, 4, 1), (4, 2, 5), (4, 5, 2), (5, 1, 4), (5, 4, 1)}. The existence of such an element x is assured by Theorem 5.1 for p > 9 · 69 and it has been checked by computer for p < 9 · 69 .
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Table 2 Some optimal ( p, 5, 2, 1)-OOCs with p a prime p
ρ
ε
d
e
a
indic(X 0 )
457 601 1753 1777 1801 2089 2113 2281 2689 2833 4057 4129 4153 4513 4657 4729
13 7 7 5 11 7 5 7 19 5 5 13 5 7 15 17
348 476 713 1002 824 1300 65 710 1142 1357 1857 895 1643 95 1912 1365
19 50 146 74 50 58 11 190 112 118 338 344 346 94 97 197
6 3 3 6 9 9 48 3 6 6 3 3 3 12 12 6
95 54 10 4 4 112 4 11 164 20 32 71 20 34 82 108
0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 70 , 0 0 − 70 , 0 0 − 30 , 0 0 − 70 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 ,
124 − 154 , 65 − 95 340 − 370 , 21 − 31 , 52 − 62 130 − 160 , 1251 − 1261 , 702 − 712 01 − 11 , 302 − 312 431 − 441 , 28 − 38 81 − 111 , 198 − 208 , 338 − 348 01 − 11 , 438 − 538 171 − 201 , 222 − 232 , 1682 − 1692 264 − 274 , 644 − 654 , 375 − 405 41 − 71 , 895 − 905 , 1155 − 1165 100 − 130 , 2051 − 2061 , 92 − 102 730 − 760 , 2681 − 2691 , 762 − 772 41 − 71 , 782 − 792 , 1702 − 1712 502 − 532 , 684 − 694 , 824 − 834 411 − 412 , 428 − 429 , 755 − 785 1150 − 1180 , 101 − 111 , 845 − 855
Now note that {1, −1} · L is the list of differences of the good house H1,x . Hence, if we take a transversal S for the cosets of C 6 /{1, −1}, it is straightforward to see that the set of good houses F = {s · H1,x | s ∈ S} is a good ( p, H, 1)-DF. The assertion follows by Proposition 3.1. In the unlucky case that p ≡ 1 (mod 24) and 2 is a 3n -th power of Z p , it seems that one of the two conditions (ii) and (iii) of Theorem 5.3 almost always holds. Indeed their requirements are very weak and we have checked that p = 1657 is the only prime p ≡ 1 (mod 24) not exceeding 106 for which neither (ii) nor (iii) is satisfied. On the other hand, an optimal (1657, 5, 2, 1)-OOC has been constructed in Example 1. In Table 2 we present an optimal ( p, 5, 2, 1)-OOC obtainable using Theorem 5.3 for each “unlucky prime” p < 10, 000 but p = 1657. The last column gives the indicator function of X 0 with respect to ω = ε + 1 and S = {ρ i | 0 ≤ i ≤ e − 1}. To save space we will simply write indic(ωi ρ j ) = i j rather than indic(ωi ρ j ) = (i, ρ j ). So, the question is whether there are primes p ≡ 1 (mod 24) greater than 106 for which we have d ≡ 1 (mod 3) and condition (ii) fails or d ≡ 2 (mod 3) and conditions (iii) fails. This seems to be the more improbable the more d is “large” considering that the element a has to be found among 2 d3 candidates. Now note that, given d, it is possible to determine the set, say Pd , of primes p ≡ 1 (mod 24) having ω of order d in the quotient group G = Z∗p /ε. In fact, if ω has order d in G, then ωd ∈ ε and hence ω4d ≡ 1 (mod p). Thus, considering that ω4 = −4, we should have (−4)d − 1 ≡ 0 (mod p). This means that if p ∈ Pd , then p is necessarily a divisor of |(−4)d − 1|. Let us determine, for instance, the set P28 . The prime factorization of |(−4)28 − 1| is 428 − 1 = 3 · 5 · 17 · 29 · 43 · 113·127·15790321 and the only prime congruent to 1 (mod 24) appearing in it is the last one. Thus we have P28 = {15790321}.
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We have checked that the conditions of Theorem 5.3 are satisfied by any prime p ∈
101
Pd
d=1
with the only exception of p = 1657 ∈ P23 . Recall, however, once again, that an optimal (1657, 5, 2, 1)-OOC has been constructed in Example 1. In view of all the above remarks it is very reasonable to conjecture that there exists an optimal ( p, 5, 2, 1)-OOC for every prime p ≡ 1 (mod 12). 6 Constructions for optimal (p, 5, 2, 1)-OOCs with p a prime, p ≡ 5 (mod 12) An optimal ( p, 5, 2, 1)-OOC with p ≡ 5 (mod 12) will be said good if every codeword-set of it is of the form C x,y = {0, x, −x, y, −y} and its missing differences are zero and the 4th roots of unity of Z p . For instance, it is readily seen that the singleton {C3,5 } is a good (17, 5, 2, 1)-OOC. It is obvious that every good ( p, 5, 2, 1)-OOC is optimal but the converse is not generally true. We have checked by computer that no good (41, 5, 2, 1)-OOC exists even though, as seen in Table 1, an optimal (41, 5, 2, 1)-OOC exists. Theorem 6.1 Let p ≡ 5(mod 12) be a prime, let ε be a primitive 4th root of unity in Z p and set ω = ε + 1. Assume that there exists a set A of good houses of Z p such that A = ±X where X is (3, ω)-near-perfect in G := Z∗p /ε. Then there exists a good ( p, 5, 2, 1)-OOC. Proof Set A+ = εi H | i = 0, 1; H ∈ A and observe that we have: A+ = X ε. By hypothesis there exists a near-perfect matching
M = b, bω, bω2 | b ∈ B of G(3, ω)− X . Set B = {b · H1,ε | b ∈ B} and observe that we have B =
{b, bω, bω2 }·
b∈B
ε. Thus, by definition of M, we have
B = Z∗p \X ε \yε
for a suitable y. Thus we have (A+ ∪ B) = Z∗p \yε and hence it is obvious that y −1 · A+ ∪ y −1 · B is the set of good houses associated with a good ( p, 5, 2, 1)-OOC Corollary 6.2 If p ≡ 5(mod 12) is a prime and −4 is a primitive 4-th power of Z p , then there exists an optimal ( p, 5, 2, 1) optical orthogonal code. Proof Let ε be a primitive 4th root of unity in Z p and set ω = ε + 1. By assumption, ω4 is a generator of C 4 since we have ω4 = −4. This is equivalent to say that ω is a generator of the quotient group G = Z∗p /ε and hence, by Corollary 4.2, the empty set is (3, ω)-near-perfect in G. The assertion then follows from Theorem 6.1. In particular we have: Corollary 6.3 If p ≡ 5(mod 12) is a prime and optimal ( p, 5, 2, 1)-OOC.
123
p−1 4
is a prime as well, then there exists an
Optical orthogonal codes
363
Proof Here the group C 4 of 4th powers of Z p has prime order p−1 4 and hence −4 certainly is a generator of C 4 since every element of C 4 \ {1} generates C 4 . The assertion then follows from Corollary 6.2. Among the fourtythree primes p ≡ 5 (mod 12) with p < 1000, Corollary 6.2 is not successful only in thirteen cases. The set of these “unlucky” primes is {41, 113, 137, 257, 353, 521, 593, 641, 761, 809, 857, 881, 953}. On the other hand, using Theorem 6.1 we have been able, with the aid of a computer, to find a good ( p, 5, 2, 1)-OOC for each of the above listed primes p with, of course, the exception of p = 41. We present our computer results in Table 3. The variables ρ, ε, d and e have the same Table 3 Some good ( p, 5, 2, 1)-OOCs p
ρ
113
3
ε
d 15
e 7
4
A
indic(X )
(1, 10)
0 0 − 3 0 ; 0 2 − 1 2 , 5 2 − 62 ; 03 − 2 3 , 6 3
137
3
37
17
2
(1, 4)
0 0 − 7 0 ; 0 1 − 1 1 , 8 1 − 91
257
3
16
8
8
(1, 4), (9, 125),
0 0 − 70 ; 01 , 11 ; 02 − 32 ; 13 − 23 ;
(14, 95)
04 − 74 ; 05 − 7 5 ; 16 − 2 6 ; 27 − 37 00 − 30 , 340 − 370 ; 41 − 51 ; 241 − 251
353
3
42
44
2
(1, 11)
521
3
235
65
2
(1, 155)
593
3
77
37
4
(1, 53)
641
3
154
32
5
(1, 4), (16, 23)
761
6
39
95
2
(1, 330)
809
3
318
101
2
(1, 204)
857
3
207
107
2
(1, 4)
00 − 70 ; 01 − 11 ; 181 − 191
881
3
387
110
2
(1, 164)
00 − 30 , 130 − 160 ;
00 − 30 , 310 − 340 ; 441 − 451 , 501 − 511 00 − 30 ; 112 − 142 ; 23 − 33 , 223 − 233 00 − 110 , 180 − 190 ; 01 − 11 ; 232 − 262 ; 263 − 273 ; 64 − 74 00 − 30 , 190 − 220 ; 181 − 191 ; 411 − 421 00 − 30 , 280 − 310 ; 521 − 531 ; 971 − 981
961 − 971 ; 1021 − 1031 953
3
442
17
14
(1, 4), (9, 386),
0 0 − 70 ; 01 − 11 ;
(27, 108), (57, 224),
02 − 32 , 72 − 132 ;
(85,362)
0 3 − 7 3 ; 04 − 14 ; 125 − 135 ; 06 − 76 ; 07 − 17 ; 128 − 138 ; 39 − 49 ; 710 − 1410 ; 1211 − 1311 ; 912 − 1012 ; 1213 − 1513
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meaning as in Section 5. The set A is presented by giving some pairs (y1 , z 1 ), . . . , (yk , z k ); the good houses of A will be Hy1 ,z 1 , ω · Hy1 ,z 1 , . . . , Hyk ,z k , ω · Hyk ,z k . It is also understood that the indicator function of G will be considered with respect to ω and S = {ρ i | 0 ≤ i ≤ e − 1}, agreeing to write indic(ωi ρ j ) = i j rather than indic(ωi ρ j ) = (i, ρ j ). 7 Constructions for perfect (pq, 5, 2, 1)-OOCs with p and q primes, p ≡ q ≡ 5 (mod 12) If p and q are primes such that p ≡ q ≡ 5 (mod 12) and good ( p, 5, 2, 1)- and (q, 5, 2, 1)OOCs exist, there are very good chances to get a perfect ( pq, 5, 2, 1)-OOC. Examine first the case when p and q are distinct. In the following, for a given integer n, we denote by Un the group of units of Zn . Theorem 7.1 Let p and q be distinct primes with p ≡ q ≡ 5(mod 12) and let ε be an element of U pq such that ε 2 = −1. Assume that there exists a set A of good houses of Z pq with A = ±(X ∪ { p, q}) and X ⊂ U pq such that X is (3, ε + 1)-perfect in G := U pq /ε. Also assume that there exists a good ( p, 5, 2, 1)-OOC and a good (q, 5, 2, 1)-OOC. Then there exists a perfect ( pq, 5, 2, 1)-OOC. Proof First observe that ε has order 4 and that we have ε = {1, −1, ε, −ε}. Thus, setting A+ = {εi H | i = 0, 1; H ∈ A}, it is clear that we can write: A+ = X ε ∪ pε ∪ qε. Let ω = ε + 1. By hypothesis there exists a perfect matching
M = b, bω, bω2 | b ∈ B of G(3, ω)− X . Set B = {b · H1,ε | b ∈ B} and observe that we have B =
{b, bω, bω2 }·
b∈B
ε since, also here, we have H1,ε = {1, ω, ω2 }ε. Thus, by definition of M, we have B = U pq \X ε.
(5)
q−5 Let {C yi ,zi | 1 ≤ i ≤ p−5 12 } be a good ( p, 5, 2, 1)-OOC, let {C yi ,z i | 1 ≤ i ≤ 12 } be a good (q, 5, 2, 1)-OOC, and consider the following collection of good houses of Z pq : p−5 q −5 C = q · Hyi ,zi | 1 ≤ i ≤ ∪ p · Hyi ,zi | 1 ≤ i ≤ . 12 12
It is clear that we have
C = p Z pq ∪ q Z pq \ ( pε ∪ qε ∪ {0}) .
(6)
Thus, putting together (5–7) and considering that we have Z pq \ {0} = U pq ( p Z pq \{0}) (q Z pq \{0}),
(7)
we finally see that A+ ∪ B ∪ C is a ( pq, H, 1)-DF and then the assertion follows from Proposition 3.1. We have checked by computer that the above theorem succeeds in constructing a perfect ( pq, 5, 2, 1)-OOC for every pair of distinct primes p and q congruent to 5 (mod 12) and not
123
Optical orthogonal codes
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Table 4 Some perfect ( pq, 5, 2, 1)-OOCs p q
ε
o(ω)
S
17
157
72
{1}
A
29 17
242
208
{1}
53 17
55
176
293
400
{1}
447
182
{1, 5}
568
616
{1}
1404
350
{1, 7}
101 53
10, 44 − 47, 44, 51 − 53, 120 − 121,
432
125 − 128, 147 − 150, 175 − 178
500
1144
{1}
99
32 − 35, 63, 124, 143 − 144, 157 − 160, 164 − 165, 358 − 361, 377 − 380
15
561 − 571 , 761 − 791 , 1281 − 1291 , 1511 , 1551 − 1581 , 1681 − 1711 ; 15 , 1645 − 1675
2
2 − 5, 24, 118 − 119, 147 − 150,
342
454 − 457, 473 − 474, 481 − 484, 494
4
41 − 71 , 801 − 831 , 961 − 991 , 1961 ,
429
2541 − 2571 , 2821 − 2831 , 3351 ; 2887 − 2897
1
818
1300
{1}
101 89
0 − 3, 10 − 11, 147 − 150, 157, 33308 − 309, 517, 683 − 686, 720 − 723
89 53
1471 − 1521 , 1591 ; 1395 − 1435
837
42
89 29
3
221 − 51 , 331 − 341 , 621 − 651 ,
53 29
2 − 3, 19 − 22, 44, 57 − 61, 74 − 77, 87 − 90, 97 − 98
39
101 29
7 172
{1, 5}
89 17
indic(X )
212
2200
{1}
101
1
0 − 3, 10 − 11, 144 − 147, 157,
588
308 − 309, 517, 683 − 686, 720 − 723
4
4 − 9, 802 − 805, 1064 − 1067,
20
1206, 1720, 2165 − 2170
exceeding 101 provided that neither p nor q is 5 or 41. In fact for each of these pairs ( p, q) we have determined a set A of four good houses of Z pq of the form A(y, z) = H p−q , p+q , H ω( p−q) , p+q , Hy,z , Hωy,ωz 2
2
2
2ω
such that A = ±(X ∪ { p, q}) with X a 22-subset of U pq that is (3, ε +1)-perfect in y . The fourth G = U pq /ε. In Table 4 such a set A is presented by giving just the pair z column gives a complete system S of representatives for the cosets of ω in G = U pq /ε and the last column gives the values of the indicator function of G (with respect to ω and S) on the set X . To save space we will write indic(ωi s) = i s rather than indic(ωi s) = (i, s). Of course, in the case when G is generated by ω, namely when S = {1}, we will simply write indic(ωi ) = i. Now examine the case when p = q. Theorem 7.2 Let p ≡ 5(mod 12) be a prime and let ε be an element of U p2 of order 4. Assume that there exists a set A of good houses of Z p2 such that A = ±(X ∪ { p}) with
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Table 5 Some perfect ( p 2 , 5, 2, 1)-OOCs p
ε
o(G)
A
indic(X )
17
38
68
(14, 5, 120)
0, 7 − 15, 22, 35, 45 − 48, 52 − 53, 57 − 60, 67
29
41
203
(1, 12, 68)
0 − 4, 11, 30 − 31, 56 − 59, 63 − 66, 130 − 133, 149 − 150, 202
53
500
689
(1, 37, 475)
0 − 2, 18 − 19, 80, 156 − 159, 460, 515 − 518, 615 − 616, 659, 678 − 681, 688
89
3861
1958
37
231 − 234, 409, 512, 582 − 585, 1714
101
515
2525
97
348, 517 − 520, 1109, 1143 − 1146, 2245
X ⊂ U p2 such that X is (3, ε + 1)-perfect in G := U p2 /ε. Also assume that there exists a good ( p, 5, 2, 1)-OOC. Then there exists a perfect ( p 2 , 5, 2, 1)-OOC. Proof Let M = {[b, bω, bω2 ] | b ∈ B} be a perfect matching of G(3, ω) − X where ω = ε + 1, and let {C yi ,zi | 1 ≤ i ≤ p−5 12 } be a good ( p, 5, 2, 1)-OOC. Reasoning very similarly as in Theorem 7.1, one can see that p−5 εi · H | i = 0, 1; H ∈ A ∪ b · H1,ε | b ∈ B ∪ p · Hyi ,zi | 1 ≤ i ≤ 12 is a ( p 2 , H, 1)-DF and then the assertion follows from Proposition 3.1.
Also here, the aid of a computer allowed us to see that the above theorem succeeds in constructing a perfect ( p 2 , 5, 2, 1)-OOC for every prime p congruent to 5 (mod 12) and not exceeding 101 provided that p is neither 5 or 41, namely for p ∈ {17, 29, 53, 89, 101}. In fact, for each of these primes p we have determined a set A of good houses of Z p2 either of the form A(u) = Hu,u+ p , Hωu,ω−1 (u+ p) or of the form
A(u, v, w) = Hu,u+ p , Hωu,ω−1 (u+ p) , Hv,w , Hωv,ωw
such that A = ±(X ∪ { p}) with X an 11-subset or, respectively, a 23-subset of U p2 that is (3, ε + 1)-perfect in G = U p2 /ε. In Table 5 such a set A is presented by giving just the element u if A = A(u), or the triple (u, v, w) if A = A(u, v, w). In each case ω = ε + 1 generates the group G = U pq /ε whose order is given in the third column. The last column gives the values of the indicator function of G (with respect to ω and S = {1}) on the set X and of course we will write indic(ωi ) = i.
8 Other optimal constructions In this section we present some constructions for optimal (mp, 5, 2, 1)-OOCs where p is a prime and m is a divisor of 12.
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Proposition 8.1 There exists an optimal (2 p, 5, 2, 1)-OOC for any prime p ≡ 1(mod 6) provided that p ≥ 31 and 2 is not a cube of Z p . Proof An application of Theorem 5.1 together with the use of a computer allow us to see that it is possible to find an element a ∈ Z p such that both the triples L 0 = {1, 2, 2a} and L 1 = {a, a − 1, a + 1} are transversals of Z∗p /C 3 . Let us identify Z2 p with Z2 × Z p and consider the good house H(0,1),(1,a) with vertices in Z2 × Z p . It is straightforward to check that H(0,1),(1,a) = {0} × {1, −1} · L 0 ∪ {1} × {1, −1} · L 1 . Thus, if S is a transversal of C 3 /{1, −1}, we see that H = {H(0,s),(1,as) | s ∈ S} is a good (2 p, 2, H, 1)-DF. The assertion then follows from Proposition 3.1. Proposition 8.2 There exists an optimal (3 p, 5, 2, 1)-OOC for any prime p ≡ 5(mod 8). Proof First note that 2 is a non-square of Z p by the Quadratic Reciprocity Law. An application of Theorem 5.1 and the use of a computer allow us to see that for p = 5 it is possible to find an element a ∈ Z p such that the quadruple L = {a, −2a, a − 1, a + 1} is a transversal of Z∗p /C 4 . Let us identify Z3 p with Z3 × Z p and consider the good house H(0,1),(1,a) with vertices in Z3 × Z p . It is straightforward to check that H(0,1),(1,a) = {0} × {±1, ±2} ∪ {1} × L ∪ {2} × (−L). Considering that {±1, ±2} is also a transversal of Z∗p /C 4 , we see that H = {H(0,s),(1,as) |s∈C 4 } is a good (3 p, 3, H, 1)-DF. Then there exists an optimal (3 p, 5, 2, 1)-OOC by Proposition 3.1. Finally recall that {C2,3 } is an optimal (15, 5, 2, 1)-OOC. Proposition 8.3 There exists an optimal (4 p, 5, 2, 1)-OOC for any prime p ≡ 7(mod 12) provided that 2 is not a cube of Z p . Proof An application of Theorem 5.1 together with the use of a computer allow us to see that it possible to find elements a, b, c of Z p such that each of the three sixtuples L 0 = {±1, ±2, ±2c},
L 1 = {a, a − 1, a + 1, b, c − b, −b − c},
L 2 = {±2a, ±2b, ±c}
is a transversal of Z∗p /C 6 . Let us identify Z4 p with Z4 × Z p and consider the good houses H(0,1),(1,a) and H(1,b),(2,c) with vertices in Z4 × Z p . One can easily check that H(0,1),(1,a) ∪ H(1,b),(2,c) =
3
{i} × L i
i=0
where L 3 = −L 1 . It is then easy to see that H = H(0,s),(1,as) , H(1,bs),(2,cs) | s ∈ C 6 is a good (4 p, 4, H, 1)-DF. The assertion then follows from Proposition 3.1.
Proposition 8.4 There exists an optimal (6 p, 5, 2, 1)-OOC for every prime p ≡ 7(mod 8).
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Proof Let a be the first positive integer such that a is a non-square of Z p and consider the good house H(3,a),(2,1) with vertices in Z6 ⊕ Z p Z6 p . One can easily check that H(3,a),(2,1) =
5
{i} × L i
i=0
where L 0 = {2a, −2a};
L 1 = −L 5 = {a − 1, −(a + 1)};
L 2 = −L 4 = {1, −2};
L 3 = {a, −a}.
For p ≡ 7(mod 8), we see that 2 and −1 are a square and a non-square of Z p , respectively, by the Quadratic Reciprocity Law. Observe that a − 1 is a square of Z p by definition of a. Now note that a is necessarily odd otherwise we would have a = 2b with both 2 and b squares by definition of a, contradicting that a is a non-square. Thus we have a + 1 = 2 a+1 2 with a+1 2 an integer smaller than a, namely a square of Z p by definition of a. We see in this way that a + 1 is also a square of Z p . The above observations allow us to see that each L i has a square and a non-square of Z p and hence H = {H(3,as),(2,s) | s ∈ C 2 } is a good (6 p, 6, H, 1)-DF. The assertion then follows from Proposition 3.1.
9 Recursive constructions Let be a simple graph of order k. In [12], generalizing the well known concept of a (v, k, 1) difference matrix [18], the first two authors define a (v, , 1) difference matrix (DM for short) to be a k ×v matrix with elements in Zv whose rows and columns are indexed with the vertices of and the elements of Zv , respectively, in such a way that the difference of any two rows corresponding to two adjacent vertices of is a permutation of Zv . As an example, denoting the vertices of the house graph H with 1, . . . , 5 as in Fig. 1, it is easy to see that ⎞ ⎛ 000 ⎜0 1 2⎟ ⎟ ⎜ ⎟ M =⎜ ⎜0 2 1⎟ ⎝0 0 0⎠ 012 is a (Z3 , H, 1)-DM. Let M be a (v, H, 1)-DM with H as in Fig. 1. We say that M is good if its rows M1 , …, M5 satisfy the following condition: M1 = M2 + M3 = M4 + M5 = (0, 0, . . . , 0). It is clear that the (Z3 , H, 1)-DM displayed above is not good since M4 + M5 is not null. Indeed it is easily seen that a good (Z3 , H, 1)-DM does not exist. On the other hand the following easy result holds. Proposition 9.1 There exists a good (w, H, 1)-DM whenever w is coprime with 6.
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Proof It is straightforward to see that ⎛ 0 0 0 ⎜0 1 2 ⎜ ⎜ 0 −1 −2 ⎜ ⎝0 2 4 0 −2 −4
369
0 3 −3 6 −6
⎞ ... 0 ... w − 1 ⎟ ⎟ . . . −(w − 1) ⎟ ⎟ . . . 2(w − 1) ⎠ . . . −2(w − 1)
is a good (w, H, 1)-DM provided that gcd(w, 6) = 1.
One of the main results of [12] is the construction of a (mvw, mw, , 1)-DF starting from a (mv, m, , 1)-DF and a (w, , 1)-DM. Following the recipe for that construction one can see that if is the house graph H and both the ingredients are good, then the resultant (mvw, mw, H, 1)-DF is also good. Thus we can state: Theorem 9.2 If there exists a good (mv, m, H, 1)-DF and a good (w, H, 1)-DM, then there exists a good (mvw, mw, H, 1)-DF. Now we apply the above theorem for getting optimal (v, 5, 2, 1)-OOCs. Theorem 9.3 Let gcd(w, 6) = 1. Composing a good (mv, m, H, 1)-DF with a (mw, 5, 2, 1)OOC one gets a (mvw, 5, 2, 1)-OOC which is m-regular if the (mw, 5, 2, 1)-OOC is such. If the (mw, 5, 2, 1)-OOC is optimal, the resultant (mvw, 5, 2, 1)-OOC fails optimality by missing just one codeword-set in the exceptional cases listed below, but it is optimal in all the other cases. (i) mw ≡ 11 (mod 12), gcd(mw, 11) = 1 and v ≡ 121 (mod 132); (ii) mw ≡ 10 (mod 12), gcd(mw, 77) = 1 and v ≡ 385 (mod 462); (iii) mw ≡ 70 (mod 84), gcd(mw, 11) = 1 and v ≡ 55 (mod 132); (iv) mw ≡ 22 (mod 132), gcd(mw, 7) = 1 and v ≡ 7 (mod 42). Proof The hypothesis gcd(w, 6) = 1 implies that there exists a (w, H, 1)-DM and hence, applying Theorem 9.2, we can get a (mvw, mw, H, 1)-DF. Let C be the (mvw, 5, 2, 1)-OOC associated with it and note that its size is s = mvw−mw . Thus mvw − mw ≡ 0 (mod 12) so 12 that the Euclidean divisions of mvw and of mw by 12 are of the form mvw = 12q + r and mw = 12q +r with 0 ≤ r ≤ 11. In this way we see that s = q −q . Let C be a (mw, 5, 2, 1)OOC and let φ : x ∈ Zmw −→ vx ∈ Zmvw be the natural monomorphism mapping Zmw into the subgroup of Zmvw of order mw. It is clear that C ∪ φ(C ) is a (mvw, 5, 2, 1)-OOC of size q − q + s where s is the size of C . It is also clear that C ∪ φ(C ) is m-regular if C is such. Now assume that C is optimal, denote by t the size of an optimal (mvw, 5, 2, 1)-OOC, and let V ∗ be the set of all integers v such that either v ≡ 11 (mod 132) or v ≡ 154 (mod 924). By Theorem 2.1 we have q + 1 if mvw ∈ V ∗ q + 1 if mw ∈ V ∗ t= s = ∗ q if mvw ∈ V q if mw ∈ V ∗ It is then clear that the size q − q + s of C ∪ φ(C ) is always equal to t with the only exception that mvw ∈ V ∗ and mw ∈ V ∗ case in which we have q − q + s = t − 1. Thus, if C is optimal, C ∪ φ(C ) is optimal as well unless we have [mvw ∈ V ∗ and mw ∈ V ∗ ] case in which it fails optimality by missing one codeword-set. The assertion then follows observing that the above condition holds exactly when one of (i)-(iv) holds.
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Remark 9.4 The above theorem is the “twin result” of Theorem 5.3 in [25] where, however, there is a minor mistake in the statement. That theorem said that if there exists a m-regular (mv, 4, 2, 1)-OOC of type [8s ] and a (mw, 4, 2, 1)-OOC with gcd(w, 6) = 1, then there exists a (mvw, 4, 2, 1)-OOC that is optimal or m-regular if the (mw, 4, 2, 1)-OOC is such. Whereas the recursive construction itself was correct, there was a mistake in counting the size of the resultant (mvw, 4, 2, 1)-OOC. More precisely, it turns out that the resultant OOC is not optimal in the following cases: (i) mw ≡ 7 (mod 8), gcd(mw, 7) = 1 and v ≡ 49 (mod 56); (ii) mw ≡ 6 (mod 8), gcd(mw, 7) = 1 and v ≡ 21 (mod 28). For example, a (49,4,2,1)-OOC of type [86 ] exists (see Example 5.4 of [25]) and so, according to Theorem 5.3 of [25], a (49 · 23, 4, 2, 1)-OOC should also exist (here m = 1, v = 49 and w = 23). Now note that the OOC with these parameters obtainable via the construction given in that theorem has size 6 · 23 + 2 = 49·23 8 . On the other hand, we have 49 · 23 ≡ 7 (mod 56) so that the size of an optimal (49 · 23, 4, 2, 1)-OOC should be 49·23 8 = 6 · 23 + 3 (see, e.g., Proposition 2.1 in [11]). Thus the existence of an optimal (49 · 23, 4, 2, 1)-OOC is still in doubt. Let us see how Theorem 9.3 and the direct constructions of the previous sections allow us to obtain many recursive constructions for optimal (v, 5, 2, 1)-OOCs. In the following, for a given positive integer k, Wk will denote the set of all integers w such that gcd(w, 6) = 1 and an optimal (kw, 5, 2, 1)-OOC exists. Theorem 9.5 If there exists a good (v, H, 1)-DF with v ≡ 121(mod 132), then there exists an optimal (vw, 5, 2, 1)-OOC for any w ∈ W1 . Observe, for instance, that in view of Corollary 5.4 and Theorem 5.5 the above theorem gives a perfect (v, 5, 2, 1)-OOC for any v of the form v = p1 . . . pt where each pi is a prime such that pi ≡ 13 (mod 24) or pi ≡ 1 (mod 24) and 2 is not a 3n i th power of Z pi where 3n i is the largest power of 3 in pi − 1. By Corollary 6.2 it also gives a (vq, 5, 2, 1)-OOC for any v as above and any prime q ≡ 5 (mod 12) such that −4 is a primitive 4th power of Zq . Theorem 9.6 If p ≡ 1(mod 6), 2 is not a cube of Z p and p ≥ 31 for every prime factor p of v, then there exists an optimal (2vw, 5, 2, 1)-OOC for any w ∈ W2 . Theorem 9.7 If p ≡ 5(mod 8) and p = 5 for every prime factor p of v, then there exists an optimal (3vw, 5, 2, 1)-OOC for any w ∈ W3 . Theorem 9.8 If p ≡ 7(mod 8) and 2 is not a cube of Z p for every prime factor p of v, then there exists an optimal (4v, 5, 2, 1)-OOC for any w ∈ W4 . Theorem 9.9 If p ≡ 7(mod 8) for every prime factor p of v, then there exists an optimal (6vw, 5, 2, 1)-OOC for any w ∈ W6 . Acknowledgments The work of Dianhua Wu was supported in part by NSFC (No. 10961006), Guangxi Science Foundation (No. 0991089) and Program for Excellent Talents in Guangxi Higher Education Institutions.
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References 1. Abel R.J.R., Buratti M.: Some progress on (v, 4, 1) difference families and optical orthogonal codes. J. Comb. Theory Ser. A 106, 59–75 (2004). 2. Abel R.J.R., Buratti M.: Difference Families. In: Colbourn C.J., Dinitz J.H. (eds.) The CRC Handbook of Combinatorial Designs, 2nd edn., pp. 392–409. Chapman & Hall/CRC Press, Boca Raton (2006). 3. Alderson T.L., Mellinger K.E.: Families of optimal OOCs with λ = 2. IEEE Trans. Inf. Theory 54, 3722–3724 (2008). 4. Alderson T.L., Mellinger K.E.: Geometric constructions of optimal optical orthogonal codes. Adv. Math. Commun. 2, 451–467 (2008). 5. Baicheva T., Topalova S.: Optimal (v, 4, 2, 1) optical orthogonal codes with small parameters. J. Comb. Des. 20 142–160 (2012). 6. Beth T., Jungnickel D., Lenz H.: Design Theory. Cambridge University Press, Cambridge (1999). 7. Brickell E.F., Wei V.: Optical orthogonal codes and cyclic block designs. Congr. Numer. 58, 175–182 (1987). 8. Buratti M.: Cayley, Marty and Schreier hypergraphs. Abh. Math. Sem. Univ. Hamburg 64, 151–162 (1994). 9. Buratti M.: Recursive constructions for difference matrices and relative difference families. J. Comb. Des. 6, 165–182 (1998). 10. Buratti M.: Cyclic designs with block size 4 and related optimal optical orthogonal codes. Des. Codes Cryptogr. 26, 111–125 (2002). 11. Buratti M., Momihara K., Pasotti A.: New results on optimal (v, 4, 2, 1) optical orthogonal codes. Des. Codes Cryptogr. 58, 89–109 (2011). 12. Buratti M., Pasotti A.: Graph decompositions with the use of difference matrices. Bull. Inst. Comb. Appl. 47, 23–32 (2006). 13. Buratti M., Pasotti A.: Combinatorial designs and the theorem of Weil on multiplicative character sums. Finite Fields Appl. 15, 332–344 (2009). 14. Buratti M., Pasotti A.: Further progress on difference families with block size 4 or 5. Des. Codes Cryptogr. 56, 1–20 (2010). 15. Chang Y.X., Ji L.: Optimal (4up, 5, 1) optical orthogonal codes. J. Comb. Des. 12, 346–361 (2004). 16. Chu W., Colbourn C.J.: Recursive constructions for optimal (n, 4, 2)-OOCs. J. Comb. Des. 12, 333–345 (2004). 17. Chung F.R.K., Salehi J.A., Wei V.K.: Optical orthogonal codes: design, analysis and applications. IEEE Trans. Inf. Theory 35, 595–604 (1989). 18. Colbourn C.J.: Difference Matrices. In: Colbourn C.J., Dinitz J.H. (eds.) The CRC Handbook of Combinatorial Designs, 2nd edn., pp. 392–409. Chapman & Hall/CRC Press, Boca Raton (2006). 19. Colbourn C.J., Dinitz J.H., Stinson D.R.: Applications of Combinatorial Designs to Communications, Cryptography, and Networking. In: Lamb J.D., Preece D.A. (eds.) Surveys in Combinatorics, pp. 37–100. Cambridge University Press, London (1999). 20. Feng T., Chang Y., Ji L.: Constructions for strictly cyclic 3-designs and applications to optimal OOCs with λ = 2. J. Comb. Theory Ser. A 115, 1527–1551 (2008). 21. Lemmermeyer F., Reciprocity Laws: From Euler to Eisenstein. Springer, Heidelberg (2000). 22. Lidl R., Neiderreiter H.: Finite Fields. Encyclopedia of Mathematics, vol 20. Cambridge University Press, Cambridge (1983). 23. Mishima M., Fu H.L., Uruno S.: Optimal conflict-avoiding codes of length n ≡ 0 (mod 16) and weight 3. Des. Codes Cryptogr. 52, 275–291 (2009). 24. Momihara K.: On cyclic 2(k − 1)-support (n, k)k−1 difference families. Finite Fields Appl. 15, 415–427 (2009). 25. Momihara K., Buratti M.: Bounds and constructions of optimal (n, 4, 2, 1) optical orthogonal codes. IEEE Trans. Inf. Theory 55, 514–523 (2009). 26. Wang X., Chang Y.: Further results on optimal (v, 4, 2, 1)-OOCs. Discret. Math. 312, 331–340 (2012). 27. West D.: Introduction to Graph Theory, 2nd edn. Prentice Hall, Upper Saddle River (2001). 28. Yang G.C., Fuja T.E.: Optical orthogonal codes with unequal auto- and cross-correlation constraints. IEEE Trans. Inf. Theory 41, 96–106 (1995). 29. Yin J.: Some combinatorial constructions for optical orthogonal codes. Discret. Math. 185, 201–219 (1998).
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On optimal (v, 5, 2, 1) optical orthogonal codes Marco Buratti · Anita Pasotti · Dianhua Wu
Received: 14 October 2011 / Revised: 8 March 2012 / Accepted: 9 March 2012 / Published online: 5 April 2012 © Springer Science+Business Media, LLC 2012
Abstract The size of a (v, 5, 2, 1) optical orthogonal code (OOC) is shown to be at most v v equal to 12 when v ≡ 11 (mod 132) or v ≡ 154 (mod 924), and at most equal to 12 in all the other cases. Thus a (v, 5, 2, 1)-OOC is naturally said to be optimal when its size reaches the above bound. Many direct and recursive constructions for infinite classes of optimal (v, 5, 2, 1)-OOCs are presented giving, in particular, a very strong indication about the existence of an optimal ( p, 5, 2, 1)-OOC for every prime p ≡ 1 (mod 12). Keywords
Optimal optical orthogonal code · Difference family · Difference matrix
Mathematics Subject Classification
05B30 · 94B25
1 Introduction A (v, k, λa , λc ) optical orthogonal code (OOC for short) can be seen as a collection C = {C1 , . . . , Cs } of k-subsets (codeword-sets) of Zv satisfying the so called autocorrelation and cross-correlation properties:
This is one of several papers published in Designs, Codes and Cryptography comprising the “Special Issue on Finite Geometries”. M. Buratti (B) Dipartimento di Matematica e Informatica, Università di Perugia, Via Vanvitelli 1, 06123 Perugia, Italy e-mail: [email protected] A. Pasotti Dipartimento di Matematica, Facoltà di Ingegneria, Università degli Studi di Brescia, Via Valotti 9, 25133 Brescia, Italy e-mail: [email protected] D. Wu Department of Mathematics, Guangxi Normal University, 541004 Guilin, People’s Republic of China e-mail: [email protected]
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|Ci ∩ (Ci + t)| ≤ λa for 1 ≤ i ≤ s and 1 ≤ t ≤ v − 1;
(1)
|Ci ∩ (C j + t)| ≤ λc for 1 ≤ i < j ≤ s and 0 ≤ t ≤ v − 1.
(2)
The former condition requires that any two distinct translates of a codeword-set share at most λa elements, while the latter requires that any two translates of two distinct codeword-sets share at most λc elements. The integers v and k are the length and the weight of the code. The size of C is the number s of its codeword-sets and C is maximal when every other OOC with the same parameters has size not greater than s. OOCs have been introduced in [17] and they have great importance in the applications [19]; the larger is their size, the greater is their usefulness. A (v, k, λa , λc )-OOC in which λa = λc = λ is briefly denoted by (v, k, λ)-OOC. For a long time the most investigated ones have been the optimal (v, k, 1)-OOCs. The existence problem for an optimal (v, k, 1)-OOC has been solved for k = 3 (see [7]) but to find a complete solution for larger values of k seems to be, at the moment, very difficult (see, e.g., [1,10,14] for some partial answers in the case when k = 4). In the last years several interesting results have been also obtained about optimal (v, k, λ)-OOCs with λ = 2 [3,16,20] and even with λ > 2 [4]. In an earlier paper [28] about (v, k, λa , λc )-OOCs with λa = λc it is shown that the size of a (v, k, λa , 1)-OOC cannot exceed λa v−1 k but this bound is, in general, far from being tight. Several results about (v, k, k − 1, 1)-OOCs (the so called conflict-avoiding codes) have been recently obtained in [23,24] while (v, 4, 2, 1)-OOCs have been deeply investigated in [5,11,25,26]. The present paper deals with (v, 5, 2, 1)-OOCs and it is the natural continuation of [11] and [25]. As one could expect, many results are, in a certain sense, “twin results” of those papers. For instance, it was proved in [11] and [25] that a tight upper bound on the size of a (v, 4, 2, 1)-OOC is the floor or the ceiling of v8 according to the congruence class of v. The twin result here is that a tight upper bound on the size of a (v, 5, 2, 1)-OOC is v the floor or the ceiling of 12 . On the other hand some constructions in this paper are considerably more difficult than their “twins” in [11] and [25]; in various cases we need to invent “ad hoc” tricks to obtain our codes. For instance, we mention the use of some perfect or near-perfect matching of a suitable hypergraph described in the third section to try to obtain an optimal (v, 5, 2, 1)-OOC with v a prime ≡ 1 or 5 (mod 12). We prove that such an optimal code is certainly realizable whenever v ≡ 13 (mod 24) or also when v ≡ 1 (mod 24) provided that 2 is not a 3n -th power modulo v, 3n being the largest power of 3 in v − 1. In all the other unlucky cases we give an algorithm which seems to be always successful to construct the desired optimal code. 2 A bound on the maximum possible size of a (v, 5, 2, 1)-OOC As usual, the list of differences of a given subset C of Zv will be denoted by C. This is the multiset of all differences x − y with (x, y) an ordered pair of distinct elements of C. Also, by C we will denote the set of differences of C, namely the underlying set (or support) of C. C More generally, the list of differences of a set C of subsets of Zv is the multiset C = and, analogously, C =
C∈C
C∈C
C. Saying that C is difference-disjoint we mean that the lists
of differences of every two distinct members of C are disjoint.
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We denote by μ(C) the maximum multiplicity in C, namely, μ(C) = max{μC (i) | 1 ≤ i ≤ v − 1} where μC (i) is the multiplicity of i in C. If δ is the size of C, namely the number of distinct differences appearing in C, we will say that C is of type t (C) = δ. It is obvious that t (C) cannot exceed the size of C so that we have t (C) ≤ k(k − 1) where k is the size of C. Saying that a code C is of type [δ1n 1 , δ2n 2 , . . . , δtn t ], we mean that C has exactly n i codeword-sets of type δi for i = 1, 2, . . . , t. Condition (1) is equivalent to ask that no element of Zv appears in the list of differences of any codeword-set more than λa times: μ(Ci ) ≤ λa for 1 ≤ i ≤ s.
(1 )
Also, in the case that λc = 1, condition (2) is equivalent to ask that C is difference-disjoint, namely that Ci ∩ C j = ∅ for 1 ≤ i < j ≤ s.
(2 )
The set of missing differences of a (v, k, λa , 1)-OOC is the set M of all elements of Zv that are not covered by the lists of differences of its codeword-sets. Following [29] we say that an optical orthogonal code is m-regular in the special case that M is a subgroup of Zv . Note, in particular, that a m-regular (v, k, 1)-OOC is a (v, m, k, 1) relative difference family [9] As already remarked in [11], a codeword-set C of a (v, k, λa , 1)-OOC is of odd type if and only if v is even and v2 ∈ C. Thus a (v, k, λa , 1)-OOC may have at most one codeword-set of odd type otherwise (2 ) would be contradicted. Let C be a codeword-set of type δ of a (v, k, λa , 1)-OOC and set C = {x1 , . . . , xδ }. Denoted by μi the multiplicity of xi in C, we obviously have |C| = k(k −1) = μ1 +· · ·+ μδ . On the other hand we have μi ≤ λa for 1 ≤ i ≤ δ by (1 ) and hence μ1 +· · ·+μδ ≤ δλa . It follows that k(k − 1) ≤ δλa and hence the type of any codeword-set of a (v, k, λa , 1)OOC is at least equal to
k(k−1) λa
. It follows, in particular, that every codeword-set of a
(v, 5, 2, 1)-OOC is of type δ ≥ 10. Let us say that two subsets C and C of Zv are equivalent if we have C = m · C + t for a suitable pair (m, t) with gcd(m, v) = 1. By means of long and tedious calculations we have checked that Zv has a 5-subset C with μ(C) ≤ 2 and |C| = 10 if and only if one of the following two conditions holds: v v 5v 5v v is divisible by 12 and, up to equivalence, C = 0, 12 , 6 , 12 , 6 ; 3v 4v 8v v is divisible by 11 and, up to equivalence, C = 0, 2v 11 , 11 , 11 , 11 .
Hence, by (2 ), a (v, 5, 2, 1)-OOC may have at most two codeword-sets of type 10. They could be two only in the case that v is divisible by 132 = 11 · 12. Further tedious calculations allowed us to see that Zv has a 5-subset C with μ(C) ≤ 2 and |C| = 11 if and only if one of the following two conditions holds:
v is divisible by 4 and, up to equivalence, C = {0, a, −a, v4 , − v4 } for some a ∈ Zv \{ v4 , − v4 , v2 }; v v 3v v v is divisible by 14 and, up to equivalence, C = 0, 14 , − 14 , 14 , 2 . Now note that C = {0, a, −a, b, −b} is a 5-subset of Zv with μ(C) = 2 and |C| = 12 as soon as we have a ∈ {±b, ±2b, ±3b, v4 , v2 ± b} and, symmetrically, b ∈ {±a, ±2a, ±3a, v4 , v 2 ± a}. In this case in fact the set of differences of C is
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C = {±a, ±b, ±2a, ±2b, ±(a − b), ±(a + b)} and we have: μC (±2a) = μC (±2b) = 1;
μC (x) = 2 for x ∈ C\{±2a, ±2b}.
So, for v > 13, we can easily see that Zv has “many” 5-subsets C with μ(C) ≤ 2 and |C| = 12. All the above remarks lead us to the following result. Theorem 2.1 If s is the size of a (v, 5, 2, 1)-OOC, then we have: v 12 for v ≡ 11 (mod 132) or v ≡ 154 (mod 924) s≤ v 12 other wise Proof Let C = {C1 , . . . , Cs } be a (v, 5, 2, 1)-OOC of size s and let δi be the type of Ci so that we have v − 1 ≥ δ1 + · · · + δs . Let v = 12q + r be the Euclidean division of v by v 12 so that q = 12 and 0 ≤ r ≤ 11. By the previous remarks C may have at most two codeword-sets of type 10 and at most one codeword-set of type 11. All other codeword-sets are of type at least 12. Thus we have v − 1 ≥ δ1 + · · · + δs ≥ 12(s − 3) + 31 which gives 12q + r + 4 ≥ 12s. This certainly implies that s never exceeds q + 1. From now on we will assume that s = q + 1. To prove the assertion it is enough to show that either v ≡ 11 (mod 132) or v ≡ 154 (mod 924). 1st case: v is divisible by 12. We have at most two codeword-sets of type 10 and at most one codeword-set of type 11. Thus we have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12(q − 2) + 31 which implies r ≥ 8 against the assumption that r = 0. 2nd case: v is divisible neither by 11 nor by 12. We have at most one codeword-set of type 11 and no codeword-sets of type 10. Thus we have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12q + 11 which implies r ≥ 12, a contradiction. 3rd case: v is odd and divisible by 11 but not by 12. We have at most one codeword-set of type 10 and no codeword-sets of type 11. Thus we have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12q + 10 which gives r ≥ 11 and hence r = 11, i.e., v = 12q + 11. Then, recalling that v is divisible by 11, we necessarily have v ≡ 11 (mod 132). 4th case: v is even and divisible by 11 but not by 12. Here we have at most one codeword-set of type 10 and at most one codeword-set of type 11. If C does not have codeword-sets of type 11, then we have v −1 ≥ δ1 +· · ·+δq+1 ≥ 12q +10 which gives r ≥ 11 and hence r = 11 contradicting that v is even. Thus C actually has a codeword-set of type 11 so that v is divisible either by 14 or by 4. We have v − 1 ≥ δ1 + · · · + δq+1 ≥ 12(q − 1) + 21 which gives r ≥ 10 and hence r = 10 since v is even. Thus v = 12q + 10 besides being divisible by 11 is also divisible by 14 since it is evident that 12q + 10 is not divisible by 4. Using the Chinese Remainder Theorem one can see that the three congruences v ≡ 10 (mod 12), v ≡ 0 (mod 11) and v ≡ 0 (mod 14) simultaneously hold only for v ≡ 154 (mod 924). It is natural to say that a (v, 5, 2, 1)-OOC is optimal when its size s reaches the upper bound given by Theorem 2.1. In particular, we will say that a (v, 5, 2, 1)-OOC is perfect when, besides being optimal, its only missing difference is zero, i.e., when it is optimal and 1-regular at the same time. Note that an optimal (v, 5, 2, 1)-OOC is necessarily perfect in the following cases:
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353 Length
Code
1 ≤ v ≤ 10 11 12 ≤ v ≤ 23 24 ≤ v ≤ 28 29 ≤ v ≤ 32 33 34 35 36 37 38 39 40 41 42 43 and 44 45 46 and 47 48 ≤ v ≤ 51 52 53 54 55 56 57 58 59 60 61 62
∅ = {1, 3, 4, 5, 9} F11 {C2,3 } ∃ {C2,5 , C8,9 } {C2,8 , C5,9 } {C1,5 , C8,11 } {C2,9 , C5,8 } ∃ {C1,6 , C8,11 , C10,14 } ∃ {C3,4 , C9,14 , C10,12 } ∃ {C1,5 , C7,15 , C9,12 } {C2,15 , C3,8 , C9,10 } {C2,7 , C3,13 , C11,12 } {C1,4 , C6,16 , C7,18 } {C2,7 , C3,15 , C10,11 } ∃ {C6,11 , C7,16 , C13,21 , C24,25 } {C5i ,5i 30 | 1 ≤ i ≤ 4} ∃ {C2,3 , C7,20 , C8,18 , C11,23 } {C2,3 , C7,20 , C9,17 , C12,23 } {C2,3 , C7,24 , C10,22 , C15,23 } {C1,4 , C7,26 , C9,21 , C11,24 } {C2,3 , C8,15 , C10,21 , C12,25 } ∃ {C20i ,20i 11 | 0 ≤ i ≤ 4} ∃
v ≡ 11 (mod 132) necessarily of type 10, 12s−1 ;
v ≡ 154 (mod 924) necessarily of type 10, 11, 12s−2 ; v ≡ 1 (mod 12) but v ≡ 0 (mod 11) necessarily of type 12s . In Table 1 we exhibit an optimal (v, 5, 2, 1)-OOC for any length v ≤ 62 when it exists. Here, and throughout the paper, we agree to briefly denote by C x,y the codewordset {0, x, −x, y, −y}. We point out that for each v ≤ 62 for which the Table 1 indicates the non-existence of a (v, 5, 2, 1)-OOC, there exists a (v, 5, 2, 1)-OOC which fails optimality by missing just one codeword-set. 3 House/dart difference families and (v, 5, 2, 1)-OOCs Optimal OOCs can be often described in terms of certain graph labelings. For instance, in most cases an optimal (v, 4, 2, 1)-OOC can be described in terms of suitable kite labelings as shown in [25]. In this section we will describe optimal (v, 5, 2, 1)-OOCs in terms of suitable house labelings or, equivalently, of dart labelings. The description could appear a little bit artificial but we will see it is quite convenient to briefly explain the recursive constructions.
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Fig. 1 The house graph and the dart graph
Fig. 2 The good house Hx,y
Fig. 3 The good dart Dx,y
Following [27], the graphs H and D depicted in Fig. 1 will be called the house graph and the dart graph, respectively. A house graph with vertices in Zv will be called good if it is of the form shown in Fig. 2 for some elements x and y of Zv with 0 < x < y < v2 . Throughout the paper such a good house graph will be denoted by Hx,y . Sometimes we will also write m · Hx,y in place of Hmx,my . Analogously, we say that a dart graph with vertices in Zv is good if it is of the form shown in Fig. 3 for some elements x and y of Zv and we will denote it by Dx,y . Recall that the list of differences of a simple graph B with vertices in Zv is defined as the multiset B of all possible differences x − y with (x, y) an ordered pair of adjacent vertices of B. We also recall (see [12]) that given a graph and positive integers m and v, a (mv, m, , 1) difference family (or DF for short) is a collection of graphs (blocks) with vertices in Zmv all of which are isomorphic to and whose lists of differences cover exactly once Zmv \{0, v, 2v, . . . , (m −1)v}. In the important case of m = 1 we simply write (v, , 1)DF rather than (v, 1, , 1)-DF. For difference families in the usual sense, corresponding to (v, , 1)-DFs in which is a complete graph, see [2] or [6]. A (v, n, H, 1)-DF or (v, n, D, 1)-DF, where H and D denote the house graph and the dart graph respectively, will be said good if all its blocks are such. Note that if the 5-subset C x,y = {0, x, −x, y, −y} of Zv is of type 12, then its set of differences coincides with the list of differences of the good house Hx,y and also coincides with the list of differences of the good dart Dx,y . We namely have C x,y = Hx,y = Dx,y . It is then obvious that a good (mv, m, H, 1)-DF {Hx1 ,y1 , . . . , Hxt ,yt } is equivalent to the good (mv, m, D, 1)DF {Dx1 ,y1 , . . . , Dxt ,yt } and that it gives a (mv, 5, 2, 1)-OOC whose codeword-sets are C x1 ,y1 , . . . , C xt ,yt . In particular, we have: Proposition 3.1 Every good (mv, m, H, 1)-DF with 1 ≤ m ≤ 9 gives an optimal (mv, 5, 2, 1)-OOC. Proof The (mv, 5, 2, 1)-OOC associated with a (mv, m, H, 1)-DF has size q = mv−m 12 . Thus, having mv = 12q + m and, by assumption, 1 ≤ m ≤ 9, we recognize that q and m
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are quotient and remainder of the Euclidean division of mv by 12. Hence q = mv 12 and the assertion follows from Theorem 2.1 considering that we cannot have neither mv ≡ 11 (mod 132) nor mv ≡ 154 (mod 924) since otherwise we would have mv ≡ 11 or 10 (mod 12) and hence m = 11 or 10 against the assumption 1 ≤ m ≤ 9. 4 On (t, ω)-perfect and (t, ω)-near-perfect subsets of a group In this section we generalize the concepts of a ω-perfect and of a ω-near perfect subset of a group introduced in [14] to those of a (t, ω)-perfect and of a (t, ω)-near perfect subset of a group. In this paper we will need this generalization only for the case of t = 3 but we do not exclude that the general case could be helpful for other purposes in some future works. Let G be a multiplicative abelian group and let ω be an element of G of order o(ω) ≥ t. We define G(t, ω) to be the t-uniform hypergraph whose vertices are the elements of G and whose hyper-edges are all the t-subsets of G of the form {xωi | 0 ≤ i ≤ t − 1} with x ∈ G. Such a hyper-edge will be denoted by [xω0 , xω1 , . . . , xωt−1 ]. Note that using the terminology of [8], G(t, ω) is the t-Cayley hypergraph Cay[G : S] where S is the singleton {ω}. For a given subset X of G, let G(t, ω) − X be the hypergraph obtained from G(t, ω) by deleting all vertices of X and all hyper-edges having non-empty intersection with X . We say that X is (t, ω)-perfect in G if there exists a perfect matching of G(t, ω) − X , that is a set of pairwise disjoint hyper-edges of G(t, ω) − X covering all vertices of G − X . Analogously, we say that X is (t, ω)-near-perfect in G if G(t, ω) − X admits a near-perfect matching, that is a set of pairwise disjoint hyper-edges of G(t, ω) − X covering all but one vertex of G − X . Here is the easy criterion for establishing whether X is (t, ω)-perfect or (t, ω)-near-perfect in G. Proposition 4.1 Let ω be an element of order d of a multiplicative abelian group G, let S be a complete system of representatives for the cosets of the group ω generated by ω, and consider the indicator function of G with respect to ω and S defined by indic ωi s = (i, s) 0 ≤ i ≤ d − 1; s ∈ S. For a given subset X of G and for a given s ∈ S, let indics (X ) = (i s,1 , i s,2 , . . . , i s,s ) be the (possibly empty) increasing sequence of values of i ∈ {0, 1, . . . , d − 1} such that (i, s) is the indicator of a suitable x ∈ X . Then X is (t, ω)-perfect in G if and only if the following two conditions hold: (π1 ) s ≡ d (mod t) for every s ∈ S; (π2 ) i s,k+1 − i s,k ≡ 1 (mod t) for any pair (s, k) with s ∈ S and 0 ≤ k ≤ s − 1. Instead, X is ω-near-perfect in G if and only if the following two conditions hold: (ν1 ) s ≡ d (mod t) for all but exactly one element of S, say s ∗ , for which we have s ∗ ≡ d−1 (mod t); (ν2 ) i s,k+1 − i s,k ≡ 1 (mod t) for any pair (s, k) with s ∈ S and 0 ≤ k ≤ s − 1 with at most one possible exception (s ∗ , k ∗ ) for which one could have i s ∗ ,k ∗ +1 − i s ∗ ,k ∗ ≡ 2 (mod t). The knowledge of indic(X ) immediately allows to determine a perfect or near-perfect matching of G(t, ω) − X when it exists.
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As an example, consider the group G = U65 of units of Z65 , let ω = 7, and let X = {1, 8, 9, 22, 28, 29, 31, 36, 42, 56, 57, 58}. The element ω has order d = 12 in G and we can take S = {s0 , s1 , s2 , s3 } = {1, 2, 3, 6} as complete system of representatives for the cosets of ω in G. Check that we have: 1 = ω0 s0 ;
8 = ω10 s1 ;
9 = ω5 s1 ;
22 = ω9 s3 ;
28 = ω11 s0 ;
29 = ω9 s1 ;
31 = ω8 s3 ;
36 = ω3 s1 ;
42 = ω s3 ;
56 = ω s1 ;
57 = ω s1 ;
58 = ω7 s0 .
1
11
4
Thus we can write: indics0 (X ) = (0, 7, 11); indics1 (X ) = (3, 4, 5, 9, 10, 11); indics2 (X ) = ();
indics3 (X ) = (1, 8, 9).
We immediately see that conditions (π1 ) and (π2 ) are satisfied with t = 3 and that the unique perfect matching M of G(3, ω) − X is given by {[ω1 s0 , ω2 s0 , ω3 s0 ], [ω4 s0 , ω5 s0 , ω6 s0 ], [ω8 s0 , ω9 s0 , ω10 s0 ], [ω0 s1 , ω1 s1 , ω2 s1 ], [ω6 s1 , ω7 s1 , ω8 s1 ], [ω0 s2 , ω1 s2 , ω2 s2 ], [ω3 s2 , ω4 s2 , ω5 s2 ], [ω6 s2 , ω7 s2 , ω8 s2 ], [ω9 s2 , ω10 s2 , ω11 s2 ], [ω2 s3 , ω3 s3 , ω4 s3 ], [ω5 s3 , ω6 s3 , ω7 s3 ], [ω10 s3 , ω11 s3 , ω0 s3 ]} namely M = {[7, 49, 18], [61, 37, 64], [16, 47, 4], [2, 14, 33], [63, 51, 32], [3, 21, 17], [54, 53, 46], [62, 44, 48], [11, 12, 19], [34, 43, 41], [27, 59, 23], [24, 38, 6]}. As another example, consider the group G = U55 of units of Z55 , let ω = 3, and let X = {1, 8, 16, 18, 28, 29, 38, 43, 48}. The element ω has order d = 20 in G and we can take S = {s0 , s1 } = {1, 2} as complete system of representatives for the cosets of ω in G. Check that we have: 1 = ω0 s0 ; 8 = ω14 s1 ; 16 = ω8 s0 ; 18 = ω2 s1 ; 28 = ω6 s1 ; 29 = ω7 s1 ; 38 = ω13 s0 ; 43 = ω18 s1 ; 48 = ω9 s0 . Here we can write: indics0 (X ) = (0, 8, 9, 13); indics1 (X ) = (2, 6, 7, 14, 18). Conditions (ν1 ) and (ν2 ) are satisfied with t = 3 and we immediately see that G(3, ω) − X has exactly three near-perfect matchings one of which, say M, is given by
10 ω s1 , ω11 s1 , ω12 s1 , ω1 s0 , ω2 s0 , ω3 s0 , ω4 s0 , ω5 s0 , ω6 s0 ,
17 ω3 s1 , ω4 s1 , ω5 s1 , ω s0 , ω18 s0 , ω19 s0 , ω14 s0 , ω15 s0 , ω16 s0 ,
8
11 ω s1 , ω9 s1 , ω10 s1 , ω s1 , ω12 s1 , ω13 s1 , ω15 s1 , ω16 s1 , ω17 s1 ,
19 ω s1 , ω0 s1 , ω1 s1 namely M = {[3, 9, 27], [26, 23, 14], [34, 47, 31], [4, 12, 36], [53, 49, 37], [54, 52, 46], [32, 41, 13], [39, 7, 21], [24, 17, 51], [19, 2, 6]}. The second near-perfect matching of G(3, ω) − X is obtainable from M by multiplying its first two hyper-edges by ω. The third one is obtainable from M by multiplying by ω just the second hyper-edge. As a special trivial case of Proposition 4.1 we have: Proposition 4.2 The empty set is (t, ω)-perfect in G if and only if the order of ω is divisible by t.
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The empty set is (t, ω)-near-perfect in G if and only if G is generated by ω and we have o(G) = o(ω) ≡ 1(mod t). From now on, if L is a subgroup of an abelian group G and S is a complete system of representatives for the cosets of L in G, we will simply say that S is a transversal of G/L. Moreover, given two subsets A and B of a multiplicative group G, we denote by A · B the list of all elements ab with a ∈ A and b ∈ B. The following lemma will be very useful for the construction of some optimal ( p, 5, 2, 1)OOCs with p ≡ 1 (mod 12), p prime. In the lemma, G, ω, S and indic have the same meaning as in Proposition 4.1. Lemma 4.3 Let ω ≤ G 1 ≤ G and let S1 ⊂ S be a transversal of G/G 1 . Assume that X 0 is a subset of G such that indics1 (X 0 ) satisfies properties (π1 ) and (π2 ) for every s1 ∈ S1 while indics (X 0 ) = () for every s ∈ S\S1 . Then, for every transversal S2 of G 1 /ω, we have that the set X := X 0 · S2 is (t, ω)-perfect in G. Proof First observe that S = S1 ·S2 is a transversal of G/ω. Now, let indic be the indicator function with respect to ω and S , and let s be any element of S so that we have s = s1 s2 with s1 ∈ S1 and s2 ∈ S2 . The hypothesis that indics (X 0 ) = () for every s ∈ S \ S1 means that X 0 is entirely contained in ω · S1 . This very easily implies that X ∩ ωs1 s2 = (X 0 ∩ ωs1 )s2 and hence it is clear that indics 1 s2 (X ) = indics1 (X 0 ). So, considering that indics1 (X 0 ) satisfies properties (π1 ) and (π2 ) for every s1 ∈ S1 by assumption, we deduce that indics (X ) satisfies properties (π1 ) and (π2 ) for all elements s ∈ S and hence X is (t, ω)-perfect in G by Proposition 4.1. 5 Constructions for perfect (p, 5, 2, 1)-OOCs with p prime, p ≡ 1 (mod 12) In this section we present a construction for good ( p, H, 1)-DFs with p a prime and hence, necessarily, with p ≡ 1 (mod 12). This will give perfect ( p, 5, 2, 1)-OOCs. Before proceeding, we need to recall some notation and terminology about finite fields. Given a prime p ≡ 1 (mod e), let ρ be a primitive element of Z p , let C e be the set of nonzero e-th powers of Z p and, for any integer i, let Cie := ρ i C e be the coset of C e represented by ρ i in the multiplicative group Z∗p of non-zero elements of Z p . We have Cie = C ej if and only if i ≡ j (mod e) and an e-subset S of Z∗p is a transversal of Z∗p /C e if and only if S has exactly one element in Cie for 0 ≤ i ≤ e − 1. We will need the following application of the Theorem of Weil on multiplicative character sums (see [22], Theorem 5.41) that we obtained in [13]. Theorem 5.1 Given a prime p ≡ 1(mod e), a t-subset B = {b1 , . . . , bt } of Z p , and a t-tuple (β1 , . . . , βt ) of Zte , the existence of an element x ∈ Z p satisfying the t cyclotomic conditions x − bi ∈ Cβei (i = 1, . . . , t) is guaranteed for p > t 2 e2t . As a matter of fact the existence of an element x as in the statement is guaranteed for p greater than a bound Q(e, t) which is better (namely, smaller) than t 2 e2t but whose expression is quite complicated (see also [15]). Theorem 5.2 Let p ≡ 1(mod 12) be a prime, let ε be a primitive 4th root of unity of Z p and set ω := ε + 1. Assume that there exists a set A of good houses of Z p such that A = ±X where X is (3, ω)-perfect in the quotient group G := Z∗p /ε. Then there exists an optimal ( p, 5, 2, 1)-OOC.
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Proof First, consider the good house H1,ε and check that we have H1,ε = 1, ω, ω2 ε.
(1)
By assumption, there exists a perfect matching
M = b, bω, bω2 | b ∈ B
(2)
of G(3, ω) − X so that we have
b, bω, bω2 ε = Z∗p \X ε.
(3)
b∈B
Then, setting B = {b · H1,ε | b ∈ B}, by (1) and (2) we have b, bω, bω2 ε = Z∗p \X ε. (Hb,bε ) = B = b∈B
(4)
b∈B
Now set A+ = A ∪ {ε · H | H ∈ A} and note that we have A+ = X ε. Putting together (3) and (4), we conclude that A+ ∪ B is a good ( p, H, 1)-DF. The assertion then follows from Proposition 3.1. We present an example in which a clever use of Theorem 5.2 and Lemma 4.3 allows us to construct an optimal (1657, 5, 2, 1)-OOC. Example 1 Let p = 1657. Check that ρ = 11 is a primitive element of Z p and hence that ε = ρ ( p−1)/4 = 783 is a primitive 4th root of unity of Z p . Now check that ω = ε + 1 has order d = 23 in G = Z∗p /ε and that S = {ρ j | 0 ≤ j ≤ 17} is a transversal of G/ω. Consider the set of good houses of Z p A0 = H1,ρ , Hω4 ,ω19 ρ , ωH1,ρ , ωHω4 ,ω19 ρ . The list of differences H1,ρ ∪ Hω4 ,ω19 ρ is given by ± 1, ε 3 ω2 , ρ, ε 3 ω2 ρ, εω17 ρ 3 , ω9 ρ 2 , ω4 , ε 3 ω6 , ω19 ρ, ε 3 ω21 ρ, ε 2 ω7 ρ 11 , ε 2 ω5 ρ 10 and hence, if indic is the indicator function of G with respect to ω and S, it is straightforward to see that we have A0 = ±X 0 with indicρ 0 (X 0 ) = (0, 1, 2, 3, 4, 5, 6, 7),
indicρ 1 (X 0 ) = (0, 1, 2, 3, 19, 20, 21, 22),
indicρ 2 (X 0 ) = (9, 10),
indicρ 3 (X 0 ) = (17, 18),
indicρ 10 (X 0 ) = (5, 6),
indicρ 11 (X 0 ) = (7, 8).
We see that indics (X 0 ) satisfies conditions (π1 ) and (π2 ) of Proposition 4.1 with t = 3 for each s ∈ S1 = {ρ 0 , ρ 1 , ρ 2 , ρ 3 , ρ 10 , ρ 11 }. Note that the six-tuple (0, 1, 2, 3, 10, 11) is a complete set of residues modulo 6 so that S1 is a transversal of G/G 1 where G 1 is the subgroup of G between ω and G generated by ρ 6 . Also note that indics (X 0 ) = () for every s ∈ S\S1 . Thus, taking S2 = {ρ 0 , ρ 6 , ρ 12 } as a transversal of G 1 /ω, we have that the set X = X 0 ·S2 is (3, ω)-perfect in G by Lemma 4.3. Of course ±X is the list of differences of the set of good houses A = A0 ∪ ρ 6 · A0 ∪ ρ 12 · A0 , namely of A = ωi ρ 6 j H1,ρ , ωi ρ 6 j Hω4 ,ω19 ρ | i = 0, 1; j = 0, 1, 2 ,
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and hence there exists an optimal ( p, 5, 2, 1)-DF by Theorem 5.2. Following the instructions given in the proof of that theorem one can see that such an optimal code is given by ε h ωi ρ 6 j · C1,ρ , ε h ωi ρ 6 j · Cω4 ,ω19 ρ | h, i = 0, 1; j = 0, 1, 2 ∪ {b · C1,ε | b ∈ B} where B is the union of the following sets ω3i+8 ρ 6 j | 0 ≤ i ≤ 4; 0 ≤ j ≤ 2 ; ω3i+11 ρ 6 j+2 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 ; ω3i+7 ρ 6 j+10 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 ;
ω3i+4 ρ 6 j+1 | 0 ≤ i ≤ 4; 0 ≤ j ≤ 2 ; ω3i+19 ρ 6 j+3 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 ; ω3i+9 ρ 6 j+11 | 0 ≤ i ≤ 6; 0 ≤ j ≤ 2 .
Theorem 5.3 Let p ≡ 1(mod 12) be a prime, let ε be a primitive 4th root of unity of Z p and let d be the order of ω = ε + 1 in the quotient group G = Z∗p /ε. There exists a perfect ( p, 5, 2, 1)-OOC in each of the following cases: (i) d ≡ 0(mod 3); (ii) d ≡ 1(mod 3) and there exists a ∈
(d−4)/3
{1, a, a − 1} is a transversal of Z∗p /C 3 ; (iii) d ≡ 2(mod 3) and there exists a ∈ is a transversal of Z∗p /C 3 .
k=0
ω3k+2 + 1 ω3k+2 ε + 1 , ω3k+2 − 1 ω3k+2 ε − 1
such that
(d−2)/3
ω3k+1 , ω3k+1 ε such that {1, a −1, a +1}
k=0
e Proof Set e = p−1 4d and note that the group C of e-th powers of Z p is the product of ε and ω. Thus, denoting by ρ a primitive root of Z p , we can write C ej = ε h ωi ρ j | 0 ≤ h ≤ 3; 0 ≤ i ≤ d − 1 for 0 ≤ j ≤ e − 1.
Assume that (i) holds. In this case the empty-set is (3, ω)-perfect in G by Proposition 4.2. Thus Theorem 5.2 is successful using as A the empty-set. In this case the presentation of the perfect ( p, 5, 2, 1)-OOC is very easy: d C = ω3i ρ j C1,ε | 0 ≤ i ≤ − 1; 0 ≤ j ≤ e − 1 . 3 Assume that (ii) holds. Set a = ε h 1 ωi1 ρ j1 and a − 1 = ε h 2 ωi2 ρ j2 . It is readily seen that a + 1 ∈ {ω3k+2 (a − 1), εω3k+2 (a − 1)} so that we have a + 1 ∈ {ε h 2 ωi2 +3k+2 ρ j2 , ε h 2 +1 ωi2 +3k+2 ρ j2 }. Note that both ε and ω are cubes of Z∗p so that we have a ∈ C 3j1 and a − 1 ∈ C 3j2 . It follows that {0, j1 , j2 } is a complete set of residues modulo 3 since 1, a and a − 1 lie in pairwise distinct cosets of C 3 by assumption. Let A0 = {H1,a , ω · H1,a } and set A0 = ±X 0 . Observing that ω2 = 2ε and that ε = 1 in the quotient group G, it is easy to check that the following identity holds in G: X 0 = 1, ω, ω2 , ω3 , ωi1 ρ j1 , ωi1 +1 ρ j1 , ωi1 +2 ρ j1 , ωi1 +3 ρ j1 , ωi2 ρ j2 , ωi2 +1 ρ j2 , ωi2 +3k+2 ρ j2 , ωi2 +3k+3 ρ j2 . Taking S = {ρ 0 , ρ 1 , . . . , ρ e−1 } as a transversal of G/ω and denoting by indic the indicator function of G with respect to ω and S, it is straightforward to see that we have:
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indicρ 0 (X 0 ) = (0, 1, 2, 3); indicρ j1 (X 0 ) = (i 1 , i 1 + 1, i 1 + 2, i 1 + 3); indicρ j2 (X 0 ) = (i 2 , i 2 + 1, i 2 + 3k + 2, i 2 + 3k + 3). We see that indics (X 0 ) satisfies conditions (π1 ) and (π2 ) of Proposition 4.1 with t = 3 for each s ∈ S1 = {ρ 0 , ρ j1 , ρ j2 }. Note that S1 is a transversal of G/G 1 where G 1 is the subgroup of G between ω and G generated by ρ 3 , since we have seen that {0, j1 , j2 } is a complete set of residues modulo 3. Also note that indics (X 0 ) = () for every s ∈ S \ S1 . Thus, taking S2 = {ρ 3 j | 0 ≤ j ≤ 3e − 1} as a transversal of G 1 /ω, we have that the set X = X 0 · S2 is (3, ω)-perfect in G by Lemma 4.3. Of course ±X is the list of differences of the set of good houses e A = ρ 3 j · H1,a , ρ 3 j ω · H1,a | 0 ≤ j ≤ − 1 3 and then an optimal ( p, 5, 2, 1)-OOC exists by Theorem 5.2. Now assume that (iii) holds. Set a − 1 = ε h 1 ωi1 ρ j1 and a + 1 = ε h 2 ωi2 ρ j2 . Also here, we can see that {0, j1 , j2 } is a complete set of residues modulo 3 by assumption. Set A0 = {H1,a , ω · H1,a } and A0 = ±X 0 . Note that we have indicρ 0 (X 0 ) = (0, 1, 2, 3, 3k + 1, 3k + 2, 3k + 3, 3k + 4); indicρ j1 (X 0 ) = (i 1 , i 1 + 1); indicρ j2 (X 0 ) = (i 2 , i 2 + 1). Reasoning exactly as in case (ii) one can see that the set of good houses A (also defined as in case (ii)) has list of differences ±X with X that is (3, ω)-perfect in G. Hence the assertion follows from Theorem 5.2. With the same notation as in the above theorem, set p = 3n 4t + 1 with t not divisible by 3. Observing that ω8 = 24 , we can write: n
2 ∈ C 3 ⇐⇒ 24t = 1 ⇐⇒ ω8t = 1 ⇐⇒ ω2t ∈ ε ⇐⇒ d | 2t ⇐⇒ 3 | d. It follows that condition (i) of Theorem 5.3 holds if and only if 2 is not a 3n -th power in Z p . Thus we can state: Corollary 5.4 Let p ≡ 1(mod 12) be a prime, let 3n be the largest power of 3 dividing p −1, and assume that 2 is not a 3n -th power of Z p . Then there exists an optimal ( p, 5, 2, 1)-OOC. Theorem 5.5 There exists a perfect ( p, 5, 2, 1)-OOC for every prime p ≡ 13(mod 24). Proof The assertion is obviously true in the case that 2 is not a cube by Corollary 5.4. Thus, in the following we will assume that p ≡ 13 (mod 24) is a prime and that 2 is a cube of Z p . Recall that 2 is a square in a finite field of order q if and only if q ≡ ±1 (mod 8) by the Quadratic Reciprocity Law [21]. Thus in our case 2 is a non-square of Z p and hence we can claim that 2 ∈ C36 since we have assumed that 2 is a cube. This allows to see that there exists an element x ∈ Z p such that the six-tuple L = {1, 2, x, 2x, x − 1, x + 1} is a transversal of Z∗p /C 6 ; it is enough that x satisfies one of the following cyclotomic conditions: ⎧ 6 ⎨ x ∈ Ci x − 1 ∈ C 6j (i, j, k) ∈ I ⎩ x + 1 ∈ Ck6 with I = {(1, 2, 5), (1, 5, 2), (2, 1, 4), (2, 4, 1), (4, 2, 5), (4, 5, 2), (5, 1, 4), (5, 4, 1)}. The existence of such an element x is assured by Theorem 5.1 for p > 9 · 69 and it has been checked by computer for p < 9 · 69 .
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Table 2 Some optimal ( p, 5, 2, 1)-OOCs with p a prime p
ρ
ε
d
e
a
indic(X 0 )
457 601 1753 1777 1801 2089 2113 2281 2689 2833 4057 4129 4153 4513 4657 4729
13 7 7 5 11 7 5 7 19 5 5 13 5 7 15 17
348 476 713 1002 824 1300 65 710 1142 1357 1857 895 1643 95 1912 1365
19 50 146 74 50 58 11 190 112 118 338 344 346 94 97 197
6 3 3 6 9 9 48 3 6 6 3 3 3 12 12 6
95 54 10 4 4 112 4 11 164 20 32 71 20 34 82 108
0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 70 , 0 0 − 70 , 0 0 − 30 , 0 0 − 70 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 , 0 0 − 30 ,
124 − 154 , 65 − 95 340 − 370 , 21 − 31 , 52 − 62 130 − 160 , 1251 − 1261 , 702 − 712 01 − 11 , 302 − 312 431 − 441 , 28 − 38 81 − 111 , 198 − 208 , 338 − 348 01 − 11 , 438 − 538 171 − 201 , 222 − 232 , 1682 − 1692 264 − 274 , 644 − 654 , 375 − 405 41 − 71 , 895 − 905 , 1155 − 1165 100 − 130 , 2051 − 2061 , 92 − 102 730 − 760 , 2681 − 2691 , 762 − 772 41 − 71 , 782 − 792 , 1702 − 1712 502 − 532 , 684 − 694 , 824 − 834 411 − 412 , 428 − 429 , 755 − 785 1150 − 1180 , 101 − 111 , 845 − 855
Now note that {1, −1} · L is the list of differences of the good house H1,x . Hence, if we take a transversal S for the cosets of C 6 /{1, −1}, it is straightforward to see that the set of good houses F = {s · H1,x | s ∈ S} is a good ( p, H, 1)-DF. The assertion follows by Proposition 3.1. In the unlucky case that p ≡ 1 (mod 24) and 2 is a 3n -th power of Z p , it seems that one of the two conditions (ii) and (iii) of Theorem 5.3 almost always holds. Indeed their requirements are very weak and we have checked that p = 1657 is the only prime p ≡ 1 (mod 24) not exceeding 106 for which neither (ii) nor (iii) is satisfied. On the other hand, an optimal (1657, 5, 2, 1)-OOC has been constructed in Example 1. In Table 2 we present an optimal ( p, 5, 2, 1)-OOC obtainable using Theorem 5.3 for each “unlucky prime” p < 10, 000 but p = 1657. The last column gives the indicator function of X 0 with respect to ω = ε + 1 and S = {ρ i | 0 ≤ i ≤ e − 1}. To save space we will simply write indic(ωi ρ j ) = i j rather than indic(ωi ρ j ) = (i, ρ j ). So, the question is whether there are primes p ≡ 1 (mod 24) greater than 106 for which we have d ≡ 1 (mod 3) and condition (ii) fails or d ≡ 2 (mod 3) and conditions (iii) fails. This seems to be the more improbable the more d is “large” considering that the element a has to be found among 2 d3 candidates. Now note that, given d, it is possible to determine the set, say Pd , of primes p ≡ 1 (mod 24) having ω of order d in the quotient group G = Z∗p /ε. In fact, if ω has order d in G, then ωd ∈ ε and hence ω4d ≡ 1 (mod p). Thus, considering that ω4 = −4, we should have (−4)d − 1 ≡ 0 (mod p). This means that if p ∈ Pd , then p is necessarily a divisor of |(−4)d − 1|. Let us determine, for instance, the set P28 . The prime factorization of |(−4)28 − 1| is 428 − 1 = 3 · 5 · 17 · 29 · 43 · 113·127·15790321 and the only prime congruent to 1 (mod 24) appearing in it is the last one. Thus we have P28 = {15790321}.
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We have checked that the conditions of Theorem 5.3 are satisfied by any prime p ∈
101
Pd
d=1
with the only exception of p = 1657 ∈ P23 . Recall, however, once again, that an optimal (1657, 5, 2, 1)-OOC has been constructed in Example 1. In view of all the above remarks it is very reasonable to conjecture that there exists an optimal ( p, 5, 2, 1)-OOC for every prime p ≡ 1 (mod 12). 6 Constructions for optimal (p, 5, 2, 1)-OOCs with p a prime, p ≡ 5 (mod 12) An optimal ( p, 5, 2, 1)-OOC with p ≡ 5 (mod 12) will be said good if every codeword-set of it is of the form C x,y = {0, x, −x, y, −y} and its missing differences are zero and the 4th roots of unity of Z p . For instance, it is readily seen that the singleton {C3,5 } is a good (17, 5, 2, 1)-OOC. It is obvious that every good ( p, 5, 2, 1)-OOC is optimal but the converse is not generally true. We have checked by computer that no good (41, 5, 2, 1)-OOC exists even though, as seen in Table 1, an optimal (41, 5, 2, 1)-OOC exists. Theorem 6.1 Let p ≡ 5(mod 12) be a prime, let ε be a primitive 4th root of unity in Z p and set ω = ε + 1. Assume that there exists a set A of good houses of Z p such that A = ±X where X is (3, ω)-near-perfect in G := Z∗p /ε. Then there exists a good ( p, 5, 2, 1)-OOC. Proof Set A+ = εi H | i = 0, 1; H ∈ A and observe that we have: A+ = X ε. By hypothesis there exists a near-perfect matching
M = b, bω, bω2 | b ∈ B of G(3, ω)− X . Set B = {b · H1,ε | b ∈ B} and observe that we have B =
{b, bω, bω2 }·
b∈B
ε. Thus, by definition of M, we have
B = Z∗p \X ε \yε
for a suitable y. Thus we have (A+ ∪ B) = Z∗p \yε and hence it is obvious that y −1 · A+ ∪ y −1 · B is the set of good houses associated with a good ( p, 5, 2, 1)-OOC Corollary 6.2 If p ≡ 5(mod 12) is a prime and −4 is a primitive 4-th power of Z p , then there exists an optimal ( p, 5, 2, 1) optical orthogonal code. Proof Let ε be a primitive 4th root of unity in Z p and set ω = ε + 1. By assumption, ω4 is a generator of C 4 since we have ω4 = −4. This is equivalent to say that ω is a generator of the quotient group G = Z∗p /ε and hence, by Corollary 4.2, the empty set is (3, ω)-near-perfect in G. The assertion then follows from Theorem 6.1. In particular we have: Corollary 6.3 If p ≡ 5(mod 12) is a prime and optimal ( p, 5, 2, 1)-OOC.
123
p−1 4
is a prime as well, then there exists an
Optical orthogonal codes
363
Proof Here the group C 4 of 4th powers of Z p has prime order p−1 4 and hence −4 certainly is a generator of C 4 since every element of C 4 \ {1} generates C 4 . The assertion then follows from Corollary 6.2. Among the fourtythree primes p ≡ 5 (mod 12) with p < 1000, Corollary 6.2 is not successful only in thirteen cases. The set of these “unlucky” primes is {41, 113, 137, 257, 353, 521, 593, 641, 761, 809, 857, 881, 953}. On the other hand, using Theorem 6.1 we have been able, with the aid of a computer, to find a good ( p, 5, 2, 1)-OOC for each of the above listed primes p with, of course, the exception of p = 41. We present our computer results in Table 3. The variables ρ, ε, d and e have the same Table 3 Some good ( p, 5, 2, 1)-OOCs p
ρ
113
3
ε
d 15
e 7
4
A
indic(X )
(1, 10)
0 0 − 3 0 ; 0 2 − 1 2 , 5 2 − 62 ; 03 − 2 3 , 6 3
137
3
37
17
2
(1, 4)
0 0 − 7 0 ; 0 1 − 1 1 , 8 1 − 91
257
3
16
8
8
(1, 4), (9, 125),
0 0 − 70 ; 01 , 11 ; 02 − 32 ; 13 − 23 ;
(14, 95)
04 − 74 ; 05 − 7 5 ; 16 − 2 6 ; 27 − 37 00 − 30 , 340 − 370 ; 41 − 51 ; 241 − 251
353
3
42
44
2
(1, 11)
521
3
235
65
2
(1, 155)
593
3
77
37
4
(1, 53)
641
3
154
32
5
(1, 4), (16, 23)
761
6
39
95
2
(1, 330)
809
3
318
101
2
(1, 204)
857
3
207
107
2
(1, 4)
00 − 70 ; 01 − 11 ; 181 − 191
881
3
387
110
2
(1, 164)
00 − 30 , 130 − 160 ;
00 − 30 , 310 − 340 ; 441 − 451 , 501 − 511 00 − 30 ; 112 − 142 ; 23 − 33 , 223 − 233 00 − 110 , 180 − 190 ; 01 − 11 ; 232 − 262 ; 263 − 273 ; 64 − 74 00 − 30 , 190 − 220 ; 181 − 191 ; 411 − 421 00 − 30 , 280 − 310 ; 521 − 531 ; 971 − 981
961 − 971 ; 1021 − 1031 953
3
442
17
14
(1, 4), (9, 386),
0 0 − 70 ; 01 − 11 ;
(27, 108), (57, 224),
02 − 32 , 72 − 132 ;
(85,362)
0 3 − 7 3 ; 04 − 14 ; 125 − 135 ; 06 − 76 ; 07 − 17 ; 128 − 138 ; 39 − 49 ; 710 − 1410 ; 1211 − 1311 ; 912 − 1012 ; 1213 − 1513
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meaning as in Section 5. The set A is presented by giving some pairs (y1 , z 1 ), . . . , (yk , z k ); the good houses of A will be Hy1 ,z 1 , ω · Hy1 ,z 1 , . . . , Hyk ,z k , ω · Hyk ,z k . It is also understood that the indicator function of G will be considered with respect to ω and S = {ρ i | 0 ≤ i ≤ e − 1}, agreeing to write indic(ωi ρ j ) = i j rather than indic(ωi ρ j ) = (i, ρ j ). 7 Constructions for perfect (pq, 5, 2, 1)-OOCs with p and q primes, p ≡ q ≡ 5 (mod 12) If p and q are primes such that p ≡ q ≡ 5 (mod 12) and good ( p, 5, 2, 1)- and (q, 5, 2, 1)OOCs exist, there are very good chances to get a perfect ( pq, 5, 2, 1)-OOC. Examine first the case when p and q are distinct. In the following, for a given integer n, we denote by Un the group of units of Zn . Theorem 7.1 Let p and q be distinct primes with p ≡ q ≡ 5(mod 12) and let ε be an element of U pq such that ε 2 = −1. Assume that there exists a set A of good houses of Z pq with A = ±(X ∪ { p, q}) and X ⊂ U pq such that X is (3, ε + 1)-perfect in G := U pq /ε. Also assume that there exists a good ( p, 5, 2, 1)-OOC and a good (q, 5, 2, 1)-OOC. Then there exists a perfect ( pq, 5, 2, 1)-OOC. Proof First observe that ε has order 4 and that we have ε = {1, −1, ε, −ε}. Thus, setting A+ = {εi H | i = 0, 1; H ∈ A}, it is clear that we can write: A+ = X ε ∪ pε ∪ qε. Let ω = ε + 1. By hypothesis there exists a perfect matching
M = b, bω, bω2 | b ∈ B of G(3, ω)− X . Set B = {b · H1,ε | b ∈ B} and observe that we have B =
{b, bω, bω2 }·
b∈B
ε since, also here, we have H1,ε = {1, ω, ω2 }ε. Thus, by definition of M, we have B = U pq \X ε.
(5)
q−5 Let {C yi ,zi | 1 ≤ i ≤ p−5 12 } be a good ( p, 5, 2, 1)-OOC, let {C yi ,z i | 1 ≤ i ≤ 12 } be a good (q, 5, 2, 1)-OOC, and consider the following collection of good houses of Z pq : p−5 q −5 C = q · Hyi ,zi | 1 ≤ i ≤ ∪ p · Hyi ,zi | 1 ≤ i ≤ . 12 12
It is clear that we have
C = p Z pq ∪ q Z pq \ ( pε ∪ qε ∪ {0}) .
(6)
Thus, putting together (5–7) and considering that we have Z pq \ {0} = U pq ( p Z pq \{0}) (q Z pq \{0}),
(7)
we finally see that A+ ∪ B ∪ C is a ( pq, H, 1)-DF and then the assertion follows from Proposition 3.1. We have checked by computer that the above theorem succeeds in constructing a perfect ( pq, 5, 2, 1)-OOC for every pair of distinct primes p and q congruent to 5 (mod 12) and not
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365
Table 4 Some perfect ( pq, 5, 2, 1)-OOCs p q
ε
o(ω)
S
17
157
72
{1}
A
29 17
242
208
{1}
53 17
55
176
293
400
{1}
447
182
{1, 5}
568
616
{1}
1404
350
{1, 7}
101 53
10, 44 − 47, 44, 51 − 53, 120 − 121,
432
125 − 128, 147 − 150, 175 − 178
500
1144
{1}
99
32 − 35, 63, 124, 143 − 144, 157 − 160, 164 − 165, 358 − 361, 377 − 380
15
561 − 571 , 761 − 791 , 1281 − 1291 , 1511 , 1551 − 1581 , 1681 − 1711 ; 15 , 1645 − 1675
2
2 − 5, 24, 118 − 119, 147 − 150,
342
454 − 457, 473 − 474, 481 − 484, 494
4
41 − 71 , 801 − 831 , 961 − 991 , 1961 ,
429
2541 − 2571 , 2821 − 2831 , 3351 ; 2887 − 2897
1
818
1300
{1}
101 89
0 − 3, 10 − 11, 147 − 150, 157, 33308 − 309, 517, 683 − 686, 720 − 723
89 53
1471 − 1521 , 1591 ; 1395 − 1435
837
42
89 29
3
221 − 51 , 331 − 341 , 621 − 651 ,
53 29
2 − 3, 19 − 22, 44, 57 − 61, 74 − 77, 87 − 90, 97 − 98
39
101 29
7 172
{1, 5}
89 17
indic(X )
212
2200
{1}
101
1
0 − 3, 10 − 11, 144 − 147, 157,
588
308 − 309, 517, 683 − 686, 720 − 723
4
4 − 9, 802 − 805, 1064 − 1067,
20
1206, 1720, 2165 − 2170
exceeding 101 provided that neither p nor q is 5 or 41. In fact for each of these pairs ( p, q) we have determined a set A of four good houses of Z pq of the form A(y, z) = H p−q , p+q , H ω( p−q) , p+q , Hy,z , Hωy,ωz 2
2
2
2ω
such that A = ±(X ∪ { p, q}) with X a 22-subset of U pq that is (3, ε +1)-perfect in y . The fourth G = U pq /ε. In Table 4 such a set A is presented by giving just the pair z column gives a complete system S of representatives for the cosets of ω in G = U pq /ε and the last column gives the values of the indicator function of G (with respect to ω and S) on the set X . To save space we will write indic(ωi s) = i s rather than indic(ωi s) = (i, s). Of course, in the case when G is generated by ω, namely when S = {1}, we will simply write indic(ωi ) = i. Now examine the case when p = q. Theorem 7.2 Let p ≡ 5(mod 12) be a prime and let ε be an element of U p2 of order 4. Assume that there exists a set A of good houses of Z p2 such that A = ±(X ∪ { p}) with
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Table 5 Some perfect ( p 2 , 5, 2, 1)-OOCs p
ε
o(G)
A
indic(X )
17
38
68
(14, 5, 120)
0, 7 − 15, 22, 35, 45 − 48, 52 − 53, 57 − 60, 67
29
41
203
(1, 12, 68)
0 − 4, 11, 30 − 31, 56 − 59, 63 − 66, 130 − 133, 149 − 150, 202
53
500
689
(1, 37, 475)
0 − 2, 18 − 19, 80, 156 − 159, 460, 515 − 518, 615 − 616, 659, 678 − 681, 688
89
3861
1958
37
231 − 234, 409, 512, 582 − 585, 1714
101
515
2525
97
348, 517 − 520, 1109, 1143 − 1146, 2245
X ⊂ U p2 such that X is (3, ε + 1)-perfect in G := U p2 /ε. Also assume that there exists a good ( p, 5, 2, 1)-OOC. Then there exists a perfect ( p 2 , 5, 2, 1)-OOC. Proof Let M = {[b, bω, bω2 ] | b ∈ B} be a perfect matching of G(3, ω) − X where ω = ε + 1, and let {C yi ,zi | 1 ≤ i ≤ p−5 12 } be a good ( p, 5, 2, 1)-OOC. Reasoning very similarly as in Theorem 7.1, one can see that p−5 εi · H | i = 0, 1; H ∈ A ∪ b · H1,ε | b ∈ B ∪ p · Hyi ,zi | 1 ≤ i ≤ 12 is a ( p 2 , H, 1)-DF and then the assertion follows from Proposition 3.1.
Also here, the aid of a computer allowed us to see that the above theorem succeeds in constructing a perfect ( p 2 , 5, 2, 1)-OOC for every prime p congruent to 5 (mod 12) and not exceeding 101 provided that p is neither 5 or 41, namely for p ∈ {17, 29, 53, 89, 101}. In fact, for each of these primes p we have determined a set A of good houses of Z p2 either of the form A(u) = Hu,u+ p , Hωu,ω−1 (u+ p) or of the form
A(u, v, w) = Hu,u+ p , Hωu,ω−1 (u+ p) , Hv,w , Hωv,ωw
such that A = ±(X ∪ { p}) with X an 11-subset or, respectively, a 23-subset of U p2 that is (3, ε + 1)-perfect in G = U p2 /ε. In Table 5 such a set A is presented by giving just the element u if A = A(u), or the triple (u, v, w) if A = A(u, v, w). In each case ω = ε + 1 generates the group G = U pq /ε whose order is given in the third column. The last column gives the values of the indicator function of G (with respect to ω and S = {1}) on the set X and of course we will write indic(ωi ) = i.
8 Other optimal constructions In this section we present some constructions for optimal (mp, 5, 2, 1)-OOCs where p is a prime and m is a divisor of 12.
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Proposition 8.1 There exists an optimal (2 p, 5, 2, 1)-OOC for any prime p ≡ 1(mod 6) provided that p ≥ 31 and 2 is not a cube of Z p . Proof An application of Theorem 5.1 together with the use of a computer allow us to see that it is possible to find an element a ∈ Z p such that both the triples L 0 = {1, 2, 2a} and L 1 = {a, a − 1, a + 1} are transversals of Z∗p /C 3 . Let us identify Z2 p with Z2 × Z p and consider the good house H(0,1),(1,a) with vertices in Z2 × Z p . It is straightforward to check that H(0,1),(1,a) = {0} × {1, −1} · L 0 ∪ {1} × {1, −1} · L 1 . Thus, if S is a transversal of C 3 /{1, −1}, we see that H = {H(0,s),(1,as) | s ∈ S} is a good (2 p, 2, H, 1)-DF. The assertion then follows from Proposition 3.1. Proposition 8.2 There exists an optimal (3 p, 5, 2, 1)-OOC for any prime p ≡ 5(mod 8). Proof First note that 2 is a non-square of Z p by the Quadratic Reciprocity Law. An application of Theorem 5.1 and the use of a computer allow us to see that for p = 5 it is possible to find an element a ∈ Z p such that the quadruple L = {a, −2a, a − 1, a + 1} is a transversal of Z∗p /C 4 . Let us identify Z3 p with Z3 × Z p and consider the good house H(0,1),(1,a) with vertices in Z3 × Z p . It is straightforward to check that H(0,1),(1,a) = {0} × {±1, ±2} ∪ {1} × L ∪ {2} × (−L). Considering that {±1, ±2} is also a transversal of Z∗p /C 4 , we see that H = {H(0,s),(1,as) |s∈C 4 } is a good (3 p, 3, H, 1)-DF. Then there exists an optimal (3 p, 5, 2, 1)-OOC by Proposition 3.1. Finally recall that {C2,3 } is an optimal (15, 5, 2, 1)-OOC. Proposition 8.3 There exists an optimal (4 p, 5, 2, 1)-OOC for any prime p ≡ 7(mod 12) provided that 2 is not a cube of Z p . Proof An application of Theorem 5.1 together with the use of a computer allow us to see that it possible to find elements a, b, c of Z p such that each of the three sixtuples L 0 = {±1, ±2, ±2c},
L 1 = {a, a − 1, a + 1, b, c − b, −b − c},
L 2 = {±2a, ±2b, ±c}
is a transversal of Z∗p /C 6 . Let us identify Z4 p with Z4 × Z p and consider the good houses H(0,1),(1,a) and H(1,b),(2,c) with vertices in Z4 × Z p . One can easily check that H(0,1),(1,a) ∪ H(1,b),(2,c) =
3
{i} × L i
i=0
where L 3 = −L 1 . It is then easy to see that H = H(0,s),(1,as) , H(1,bs),(2,cs) | s ∈ C 6 is a good (4 p, 4, H, 1)-DF. The assertion then follows from Proposition 3.1.
Proposition 8.4 There exists an optimal (6 p, 5, 2, 1)-OOC for every prime p ≡ 7(mod 8).
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Proof Let a be the first positive integer such that a is a non-square of Z p and consider the good house H(3,a),(2,1) with vertices in Z6 ⊕ Z p Z6 p . One can easily check that H(3,a),(2,1) =
5
{i} × L i
i=0
where L 0 = {2a, −2a};
L 1 = −L 5 = {a − 1, −(a + 1)};
L 2 = −L 4 = {1, −2};
L 3 = {a, −a}.
For p ≡ 7(mod 8), we see that 2 and −1 are a square and a non-square of Z p , respectively, by the Quadratic Reciprocity Law. Observe that a − 1 is a square of Z p by definition of a. Now note that a is necessarily odd otherwise we would have a = 2b with both 2 and b squares by definition of a, contradicting that a is a non-square. Thus we have a + 1 = 2 a+1 2 with a+1 2 an integer smaller than a, namely a square of Z p by definition of a. We see in this way that a + 1 is also a square of Z p . The above observations allow us to see that each L i has a square and a non-square of Z p and hence H = {H(3,as),(2,s) | s ∈ C 2 } is a good (6 p, 6, H, 1)-DF. The assertion then follows from Proposition 3.1.
9 Recursive constructions Let be a simple graph of order k. In [12], generalizing the well known concept of a (v, k, 1) difference matrix [18], the first two authors define a (v, , 1) difference matrix (DM for short) to be a k ×v matrix with elements in Zv whose rows and columns are indexed with the vertices of and the elements of Zv , respectively, in such a way that the difference of any two rows corresponding to two adjacent vertices of is a permutation of Zv . As an example, denoting the vertices of the house graph H with 1, . . . , 5 as in Fig. 1, it is easy to see that ⎞ ⎛ 000 ⎜0 1 2⎟ ⎟ ⎜ ⎟ M =⎜ ⎜0 2 1⎟ ⎝0 0 0⎠ 012 is a (Z3 , H, 1)-DM. Let M be a (v, H, 1)-DM with H as in Fig. 1. We say that M is good if its rows M1 , …, M5 satisfy the following condition: M1 = M2 + M3 = M4 + M5 = (0, 0, . . . , 0). It is clear that the (Z3 , H, 1)-DM displayed above is not good since M4 + M5 is not null. Indeed it is easily seen that a good (Z3 , H, 1)-DM does not exist. On the other hand the following easy result holds. Proposition 9.1 There exists a good (w, H, 1)-DM whenever w is coprime with 6.
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Proof It is straightforward to see that ⎛ 0 0 0 ⎜0 1 2 ⎜ ⎜ 0 −1 −2 ⎜ ⎝0 2 4 0 −2 −4
369
0 3 −3 6 −6
⎞ ... 0 ... w − 1 ⎟ ⎟ . . . −(w − 1) ⎟ ⎟ . . . 2(w − 1) ⎠ . . . −2(w − 1)
is a good (w, H, 1)-DM provided that gcd(w, 6) = 1.
One of the main results of [12] is the construction of a (mvw, mw, , 1)-DF starting from a (mv, m, , 1)-DF and a (w, , 1)-DM. Following the recipe for that construction one can see that if is the house graph H and both the ingredients are good, then the resultant (mvw, mw, H, 1)-DF is also good. Thus we can state: Theorem 9.2 If there exists a good (mv, m, H, 1)-DF and a good (w, H, 1)-DM, then there exists a good (mvw, mw, H, 1)-DF. Now we apply the above theorem for getting optimal (v, 5, 2, 1)-OOCs. Theorem 9.3 Let gcd(w, 6) = 1. Composing a good (mv, m, H, 1)-DF with a (mw, 5, 2, 1)OOC one gets a (mvw, 5, 2, 1)-OOC which is m-regular if the (mw, 5, 2, 1)-OOC is such. If the (mw, 5, 2, 1)-OOC is optimal, the resultant (mvw, 5, 2, 1)-OOC fails optimality by missing just one codeword-set in the exceptional cases listed below, but it is optimal in all the other cases. (i) mw ≡ 11 (mod 12), gcd(mw, 11) = 1 and v ≡ 121 (mod 132); (ii) mw ≡ 10 (mod 12), gcd(mw, 77) = 1 and v ≡ 385 (mod 462); (iii) mw ≡ 70 (mod 84), gcd(mw, 11) = 1 and v ≡ 55 (mod 132); (iv) mw ≡ 22 (mod 132), gcd(mw, 7) = 1 and v ≡ 7 (mod 42). Proof The hypothesis gcd(w, 6) = 1 implies that there exists a (w, H, 1)-DM and hence, applying Theorem 9.2, we can get a (mvw, mw, H, 1)-DF. Let C be the (mvw, 5, 2, 1)-OOC associated with it and note that its size is s = mvw−mw . Thus mvw − mw ≡ 0 (mod 12) so 12 that the Euclidean divisions of mvw and of mw by 12 are of the form mvw = 12q + r and mw = 12q +r with 0 ≤ r ≤ 11. In this way we see that s = q −q . Let C be a (mw, 5, 2, 1)OOC and let φ : x ∈ Zmw −→ vx ∈ Zmvw be the natural monomorphism mapping Zmw into the subgroup of Zmvw of order mw. It is clear that C ∪ φ(C ) is a (mvw, 5, 2, 1)-OOC of size q − q + s where s is the size of C . It is also clear that C ∪ φ(C ) is m-regular if C is such. Now assume that C is optimal, denote by t the size of an optimal (mvw, 5, 2, 1)-OOC, and let V ∗ be the set of all integers v such that either v ≡ 11 (mod 132) or v ≡ 154 (mod 924). By Theorem 2.1 we have q + 1 if mvw ∈ V ∗ q + 1 if mw ∈ V ∗ t= s = ∗ q if mvw ∈ V q if mw ∈ V ∗ It is then clear that the size q − q + s of C ∪ φ(C ) is always equal to t with the only exception that mvw ∈ V ∗ and mw ∈ V ∗ case in which we have q − q + s = t − 1. Thus, if C is optimal, C ∪ φ(C ) is optimal as well unless we have [mvw ∈ V ∗ and mw ∈ V ∗ ] case in which it fails optimality by missing one codeword-set. The assertion then follows observing that the above condition holds exactly when one of (i)-(iv) holds.
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Remark 9.4 The above theorem is the “twin result” of Theorem 5.3 in [25] where, however, there is a minor mistake in the statement. That theorem said that if there exists a m-regular (mv, 4, 2, 1)-OOC of type [8s ] and a (mw, 4, 2, 1)-OOC with gcd(w, 6) = 1, then there exists a (mvw, 4, 2, 1)-OOC that is optimal or m-regular if the (mw, 4, 2, 1)-OOC is such. Whereas the recursive construction itself was correct, there was a mistake in counting the size of the resultant (mvw, 4, 2, 1)-OOC. More precisely, it turns out that the resultant OOC is not optimal in the following cases: (i) mw ≡ 7 (mod 8), gcd(mw, 7) = 1 and v ≡ 49 (mod 56); (ii) mw ≡ 6 (mod 8), gcd(mw, 7) = 1 and v ≡ 21 (mod 28). For example, a (49,4,2,1)-OOC of type [86 ] exists (see Example 5.4 of [25]) and so, according to Theorem 5.3 of [25], a (49 · 23, 4, 2, 1)-OOC should also exist (here m = 1, v = 49 and w = 23). Now note that the OOC with these parameters obtainable via the construction given in that theorem has size 6 · 23 + 2 = 49·23 8 . On the other hand, we have 49 · 23 ≡ 7 (mod 56) so that the size of an optimal (49 · 23, 4, 2, 1)-OOC should be 49·23 8 = 6 · 23 + 3 (see, e.g., Proposition 2.1 in [11]). Thus the existence of an optimal (49 · 23, 4, 2, 1)-OOC is still in doubt. Let us see how Theorem 9.3 and the direct constructions of the previous sections allow us to obtain many recursive constructions for optimal (v, 5, 2, 1)-OOCs. In the following, for a given positive integer k, Wk will denote the set of all integers w such that gcd(w, 6) = 1 and an optimal (kw, 5, 2, 1)-OOC exists. Theorem 9.5 If there exists a good (v, H, 1)-DF with v ≡ 121(mod 132), then there exists an optimal (vw, 5, 2, 1)-OOC for any w ∈ W1 . Observe, for instance, that in view of Corollary 5.4 and Theorem 5.5 the above theorem gives a perfect (v, 5, 2, 1)-OOC for any v of the form v = p1 . . . pt where each pi is a prime such that pi ≡ 13 (mod 24) or pi ≡ 1 (mod 24) and 2 is not a 3n i th power of Z pi where 3n i is the largest power of 3 in pi − 1. By Corollary 6.2 it also gives a (vq, 5, 2, 1)-OOC for any v as above and any prime q ≡ 5 (mod 12) such that −4 is a primitive 4th power of Zq . Theorem 9.6 If p ≡ 1(mod 6), 2 is not a cube of Z p and p ≥ 31 for every prime factor p of v, then there exists an optimal (2vw, 5, 2, 1)-OOC for any w ∈ W2 . Theorem 9.7 If p ≡ 5(mod 8) and p = 5 for every prime factor p of v, then there exists an optimal (3vw, 5, 2, 1)-OOC for any w ∈ W3 . Theorem 9.8 If p ≡ 7(mod 8) and 2 is not a cube of Z p for every prime factor p of v, then there exists an optimal (4v, 5, 2, 1)-OOC for any w ∈ W4 . Theorem 9.9 If p ≡ 7(mod 8) for every prime factor p of v, then there exists an optimal (6vw, 5, 2, 1)-OOC for any w ∈ W6 . Acknowledgments The work of Dianhua Wu was supported in part by NSFC (No. 10961006), Guangxi Science Foundation (No. 0991089) and Program for Excellent Talents in Guangxi Higher Education Institutions.
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