Optimal optical orthogonal signature pattern codes with weight three and cross-correlation constraint one

Abstract

Optical orthogonal signature pattern codes (OOSPCs) have attracted wide attention as signature patterns of spatial optical code division multiple access networks. In this paper, an improved upper bound on the size of an (m, n, 3, λ a , 1)-OOSPC with λ a = 2, 3 is established. The exact number of codewords of an optimal (m, n, 3, λ a , 1)-OOSPC is determined for any positive integers m, n ≡ 2 (mod 4) and λ a ∈ {2, 3}.When m and n are coprime, it has been shown in [28] that an (m, n, k, λ a , λ c )-OOSPC is equivalent to a 1-dimensional (mn, k, λ a , λ c )-optical orthogonal code (OOC). See [1-4, 7, 10, 13, 14, 29] and the references therein for more details on OOCs.When m and n are not coprime, various OOSPCs have been constructed via algebraic and combinatorial methods for the case of λ a = λ c (see [5,6,15,[23][24][25][26]28]). We only quote the following result for later use. Theorem 1.1 [24] Θ(m, n, 3, 1) =              J(mn, 3, 1) − 1, if mn ≡ 14, 20 (mod 24), or if mn ≡ 8, 16 (mod 24) and gcd(m, n, 4) = 2, or if mn ≡ 2 (mod 6) and gcd(m, n, 4) = 4; J(mn, 3, 1), otherwise.On the other hand, for the case of λ a = λ c , very little has been done on (m, n, k, λ a , λ c )-OOSPCs with maximum size. Compared with (1.1), an improved upper bound on Θ(m, n, 3, 2, 1) was given by Sawa and Kageyama [26]. That is Θ(m, n, 3, 2, 1) ≤ mn 4 , if mn ≡ 0 (mod 4), mn−1 4 , otherwise.(1.2)And they proved the following theorem. Theorem 1.2 [26] Θ(m, n, 3, 2, 1) = mn−1 4 , if m = n ≡ 1 (mod 4) is a prime and 2 is a primitive root in Z m ; mn−2 4 , if mn ≡ 2 (mod 4).In Section 2, we shall give an equivalent combinatorial description of (m, n, k, λ a , λ c )-OOSPCs by using set-theoretic notation. Section 3 is devoted to improving Sawa and Kageyama's bound (1.2), especially for the case of mn ≡ 0 (mod 4). Throughout this paper, let ξ denote the number of subgroups of order 3 in Z m × Z n , i.e., ξ =    0, if 3 ∤ mn; 1, if 3 | mn and gcd(m, n, 3) = 1; 4, if gcd(m, n, 3) = 3.