We address an open problem posed by Chen-Cheng-Qi (IEEE Trans.\ Inf.\ Theory, 2025): can the decoding of binary sum-rank-metric codes $\SR(C_1,C_2)$ with $2\times2$ matrix blocks be reduced entirely to decoding the constituent Hamming-metric codes $C_1$ and $C_2$ without the additional requirement $d_1\ge\tfrac{2}{3}d_{\mathrm{sr}}$ used in their fast decoder? We answer this in the affirmative by exhibiting a simple two-step procedure: first uniquely decode $C_2$, then apply a single error-erasure decoding for $C_1$. This shows that the restrictive hypothesis $d_1\ge\tfrac{2}{3}d_{\mathrm{sr}}$ is theoretically unnecessary. The resulting decoder achieves unique decoding up to $\lfloor (d_{\mathrm{sr}}-1)/2\rfloor$ with overall cost $T_2+T_1$, where $T_2$ and $T_1$ are the complexities of the Hamming decoders for $C_2$ and $C_1$, respectively. We further show that this reduction is asymptotically optimal in a black-box model, as any sum-rank decoder must inherently decode the constituent Hamming codes. For BCH or Goppa instantiations over $\F_4$, the decoder runs in $O(\ell^2)$ time.

翻译：我们解决了Chen-Cheng-Qi（IEEE Trans. Inf. Theory, 2025）提出的一个开放性问题：能否将包含$2\times2$矩阵块的二元和秩度量码$\SR(C_1,C_2)$的译码完全归约为对构成汉明度量码$C_1$和$C_2$的译码，而无需其快速译码器中使用的附加条件$d_1\ge\tfrac{2}{3}d_{\mathrm{sr}}$？我们通过展示一个简单的两步过程对此给出肯定回答：首先唯一译码$C_2$，然后对$C_1$应用单次纠错-删除译码。这表明限制性假设$d_1\ge\tfrac{2}{3}d_{\mathrm{sr}}$在理论上是不必要的。所得译码器可实现达到$\lfloor (d_{\mathrm{sr}}-1)/2\rfloor$的唯一译码，总代价为$T_2+T_1$，其中$T_2$和$T_1$分别是$C_2$和$C_1$的汉明译码器的复杂度。我们进一步证明，在黑盒模型中该归约是渐近最优的，因为任何和秩译码器必须本质上译码构成汉明码。对于定义在$\F_4$上的BCH或Goppa实例，该译码器运行时间为$O(\ell^2)$。