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四元Preparata码是汉明度量下的准完全覆盖码

Quaternary Preparata codes are quasi-perfect covering codes in the Hamming metric

Tamal Maharaj

arXiv 2610.03760首次发表:更新:

发表机构

Ramakrishna Mission Vivekananda Educational and Research Institute(罗摩克里希那传教士维韦卡南达教育研究机构)

机构由 AI 辅助整理,请以论文原文为准。

AI 中文总结

该论文证明四元Preparata码在汉明度量下对奇数长度参数为准完全覆盖码,覆盖半径为2、最小距离为4,并改进覆盖数上界,如$K_4(8,2)\le 256$。

AI 中文摘要

Hammons、Kumar、Calderbank、Sloane和Solé提出的四元Preparata码是长度为$2^m$、包含$4^{2^m-m-1}$个码字的$\mathbb{Z}_4$-线性码。它们通常通过其二元Gray像进行研究,即在Lee度量下。我们转而将它们视为在4字母字母表上、采用汉明度量的码。我们证明,对于奇数$m$,它们的覆盖半径为2、最小距离为4,因此它们是准完全的。对于偶数$m$,相同的构造具有覆盖半径3。证明是在Galois环$\mathrm{GR}(4,m)$中使用Teichmüller代表元进行的简短计算,同时给出了陪集重量分布。这些码的覆盖密度趋于$9/8$。此前已知的最佳无限族四元码(覆盖半径为2)的覆盖密度约为1.5。因此,对于所有$n\ge 2^m$且$m$为奇数,有$K_4(n,2)\le 4^{n-m-1}$,这改进了在$2^m$以上一定长度范围内的最佳可用界。对于$m=3$,该码是八元码,给出$K_4(8,2)\le 256$;之前的界是2005年的352,最佳下界为251。我们还给出一个四元码,表明$K_4(6,2)\le 48$,而之前的界是52,并给出了对更大字母表的影响。

英文摘要

The quaternary Preparata codes of Hammons, Kumar, Calderbank, Sloane and Solé are $\mathbb{Z}_4$-linear codes of length $2^m$ with $4^{2^m-m-1}$ codewords. They are usually studied through their binary Gray images, that is, in the Lee metric. We consider them instead as codes over a 4-letter alphabet with the Hamming metric. We show that for odd $m$ they have covering radius 2 and minimum distance 4, so they are quasi-perfect. For even $m$ the same construction has covering radius 3. The proof is a short computation with Teichmüller representatives in the Galois ring $\mathrm{GR}(4,m)$, and it also gives the coset weight distribution. The covering density of these codes tends to $9/8$. The best previously known infinite families of quaternary codes with covering radius 2 have covering density about 1.5. Consequently $K_4(n,2)\le 4^{n-m-1}$ for all $n\ge 2^m$ and odd $m$, which improves the best available bounds for a range of lengths above each $2^m$. For $m=3$ the code is the octacode, and it gives $K_4(8,2)\le 256$; the previous bound was 352, from 2005, and the best lower bound is 251. We also give a quaternary code showing $K_4(6,2)\le 48$, where the previous bound was 52, together with consequences for larger alphabets.

论文原文

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