Maximum-likelihood (ML) decoding for arbitrary block codes remains fundamentally hard, with worst-case time complexity-measured by the total number of multiplications-being no better than straightforward exhaustive search, which requires $q^{k} n$ operations for an $[n,k]_q$ code. This paper introduces a simple, code-agnostic framework that reduces the worst-case complexity by a factor of $n$, down to $q^{k}$ operations, a highly desirable reduction in practice. The result holds for both linear and nonlinear block codes over general memoryless channels and under both hard-decision and soft-decision decoding. It naturally extends to intersymbol-interference (ISI) channels and ML list decoding with only a negligible increase in complexity. Our core insight is that, upon receipt of each sequence at the receiver, the conditional probability of that sequence for each codeword in the codebook (i.e., the \emph{likelihood}) can be expressed as the inner product of two carefully constructed vectors -- the first depending on the received sequence, and the second on that codeword itself. As a result, evaluating the likelihoods for all codewords in the codebook reduces to a single vector-matrix multiplication, and ML decoding (MLD) becomes the simple task of picking the maximum entry in the resulting vector. The only non-trivial cost lies in the vector-matrix product. However, our matrix construction allows the use of the Mailman algorithm to reduce this cost. This time reduction is achieved at the cost of high space complexity, requiring $\mathcal{O}(q^{k+1} n)$ space to store the pre-computed codebook matrix.

翻译：对于任意分组码的最大似然解码，其根本困难性依然存在，其最坏情况时间复杂度——以乘法总次数衡量——并不优于朴素的穷举搜索，后者对于$[n,k]_q$码需要$q^{k} n$次运算。本文提出了一种简单、与码无关的框架，将最坏情况复杂度降低了$n$倍，降至$q^{k}$次运算，这在实践中是极为理想的降幅。该结果适用于一般无记忆信道上的线性和非线性分组码，且同时适用于硬判决与软判决解码。它自然地扩展到符号间干扰信道和最大似然列表解码，且复杂度仅可忽略地增加。我们的核心洞见在于：接收端收到每个序列后，码本中每个码字对该序列的条件概率（即似然）可表示为两个精心构造的向量的内积——第一个向量依赖于接收序列，第二个向量依赖于码字本身。因此，计算码本中所有码字的似然值简化为一次向量-矩阵乘法，而最大似然解码则简化为选取结果向量中最大元素的简单任务。唯一的非平凡开销在于向量-矩阵乘积运算。然而，我们的矩阵构造允许利用Mailman算法来降低这一开销。这种时间复杂度的降低是以高空间复杂度为代价的，需要$\mathcal{O}(q^{k+1} n)$的空间来存储预计算的码本矩阵。