Two-Step Decoding of Binary $2\times2$ Sum-Rank-Metric Codes
Hao Wu, Bocong Chen, Guanghui Zhang, Hongwei Liu
TL;DR
The paper resolves an open problem on decoding binary sum-rank-metric codes with $2\times2$ blocks by removing the previously required $d_1\ge\tfrac{2}{3}d_{ ext{sr}}$ constraint. It shows a simple two-step decoding that first decodes $C_2$ and then decodes $C_1$ with erasures, under the assumption $d_2\ge d_{ ext{sr}}$, achieving unique decoding up to $\left\lfloor\frac{d_{ ext{sr}}-1}{2}\right\rfloor$ with cost $T_2+T_1$. The approach is asymptotically optimal in a black-box model and preserves quadratic-time decoding for BCH or Goppa codes over $\mathbb{F}_4$, while expanding the feasible design region in the $(d_1,d_2)$-plane. This yields greater flexibility in code design and tightens the connection between sum-rank Decoding and constituent Hamming-code decoders, with practical impact for multishot network coding and related applications.
Abstract
We resolve an open problem posed by Chen--Cheng--Qi (IEEE Trans.\ Inf.\ Theory, 2025): can 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 $d_1\ge\tfrac{2}{3}d_{\mathrm{sr}}$ that underlies 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 of $C_1$.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.