This paper addresses fundamental challenges in two-dimensional error correction by constructing optimal codes for \emph{criss-cross deletions}. We consider an $ n \times n $ array $\boldsymbol{X}$ over a $ q $-ary alphabet $\Sigma_q := \{0, 1, \ldots, q-1\}$ that is subject to a \emph{$(t_r, t_c)$-criss-cross deletion}, which involves the simultaneous removal of $ t_r $ rows and $ t_c $ columns. A code $\mathcal{C} \subseteq \Sigma_q^{n \times n}$ is defined as a \emph{$(t_r,t_c)$-criss-cross deletion correcting code} if it can successfully correct these deletions. We derive a sphere-packing type lower bound and a Gilbert-Varshamov type upper bound on the redundancy of optimal codes. Our results indicate that the optimal redundancy for a $(t_r, t_c)$-criss-cross deletion correcting code lies between $(t_r + t_c)n\log q + (t_r + t_c)\log n + O_{q,t_r,t_c}(1)$ and $(t_r + t_c)n\log q + 2(t_r + t_c)\log n + O_{q,t_r,t_c}(1)$, where the logarithm is on base two, and $O_{q,t_r,t_c}(1)$ is a constant that depends solely on $q$, $t_r$, and $t_c$. For the case of $(1,1)$-criss-cross deletions, we propose two families of constructions that achieve $2n\log q + 2\log n + O_q(1)$ bits of redundancy. This redundancy is optimal up to an additive constant term $O_q(1)$, which depends solely on $q$. One family is designed for non-binary alphabets, while the other addresses arbitrary alphabets. For the case of $(t_r, t_c)$-criss-cross deletions, we provide a strategy to derive optimal codes when both unidirectional deletions occur consecutively. We propose decoding algorithms with a time complexity of $O(n^2)$ for our codes, which are optimal for two-dimensional scenarios.
翻译:本文通过构建针对\emph{十字交叉删除}的最优码,解决了二维纠错中的基本挑战。我们考虑定义在$q$元字母表$\Sigma_q := \{0, 1, \ldots, q-1\}$上的一个$n \times n$阵列$\boldsymbol{X}$,该阵列受到\emph{$(t_r, t_c)$-十字交叉删除}的影响,即同时删除$t_r$行和$t_c$列。若一个码$\mathcal{C} \subseteq \Sigma_q^{n \times n}$能够成功纠正这些删除,则其被定义为一个\emph{$(t_r,t_c)$-十字交叉删除纠错码}。我们推导了最优码冗余度的球包型下界和吉尔伯特-瓦尔沙莫夫型上界。我们的结果表明,$(t_r, t_c)$-十字交叉删除纠错码的最优冗余度介于$(t_r + t_c)n\log q + (t_r + t_c)\log n + O_{q,t_r,t_c}(1)$与$(t_r + t_c)n\log q + 2(t_r + t_c)\log n + O_{q,t_r,t_c}(1)$之间,其中对数以二为底,$O_{q,t_r,t_c}(1)$是一个仅依赖于$q$、$t_r$和$t_c$的常数。针对$(1,1)$-十字交叉删除的情况,我们提出了两个构造族,实现了$2n\log q + 2\log n + O_q(1)$比特的冗余度。该冗余度在仅依赖于$q$的加法常数项$O_q(1)$范围内是最优的。其中一个构造族针对非二进制字母表设计,而另一个则适用于任意字母表。针对$(t_r, t_c)$-十字交叉删除的情况,我们提供了一种在单向删除连续发生时推导最优码的策略。我们为所提出的码设计了时间复杂度为$O(n^2)$的解码算法,该复杂度在二维场景下是最优的。