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利用扭曲的Reed-Solomon码构造两类极大距离可分纠缠辅助量子纠错码

潘鑫 高健

潘鑫, 高健. 利用扭曲的Reed-Solomon码构造两类极大距离可分纠缠辅助量子纠错码[J]. 电子与信息学报. doi: 10.11999/JEIT250258
引用本文: 潘鑫, 高健. 利用扭曲的Reed-Solomon码构造两类极大距离可分纠缠辅助量子纠错码[J]. 电子与信息学报. doi: 10.11999/JEIT250258
PAN Xin, GAO Jian. Constructing Two Classes of Maximum Distance Separable Entanglement-Assisted Quantum Error-Correcting Codes by Using Twisted Generalized Reed-Solomon Codes[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT250258
Citation: PAN Xin, GAO Jian. Constructing Two Classes of Maximum Distance Separable Entanglement-Assisted Quantum Error-Correcting Codes by Using Twisted Generalized Reed-Solomon Codes[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT250258

利用扭曲的Reed-Solomon码构造两类极大距离可分纠缠辅助量子纠错码

doi: 10.11999/JEIT250258 cstr: 32379.14.JEIT250258
基金项目: 山东省自然科学基金(ZR2024YQ057, ZR2022MA024),国家自然科学基金(12071264)
详细信息
    作者简介:

    潘鑫:女,硕士生,研究方向为编码理论及其应用

    高健:男,副教授,博士,博士生导师,研究方向为编码理论及其应用

    通讯作者:

    高健 dezhougaojian@163.com

  • 中图分类号: TN911.22

Constructing Two Classes of Maximum Distance Separable Entanglement-Assisted Quantum Error-Correcting Codes by Using Twisted Generalized Reed-Solomon Codes

Funds: The Natural Science Foundation of Shandong Province (ZR2024YQ057, ZR2022MA024), The National Natural Science Foundation of China (12071264)
  • 摘要: 随着量子通信和量子计算技术的飞速发展,高效量子纠错编码技术已成为保障量子系统可靠性的核心需求。传统量子纠错码在参数灵活性和最小距离约束方面存在显著局限性,难以适应复杂量子通信场景中的动态需求。该文基于扭曲的Reed-Solomon(TGRS)码,根据码长中$ i $的奇偶性的不同具体讨论矩阵$ \boldsymbol{G}{\boldsymbol{G}}^{\mathrm{H}} $的秩,进一步通过分析该矩阵的秩确定厄米特正交包的维数,从而得到两类极大距离可分纠缠辅助量子纠错码(MDS EAQECCs)。研究发现,通过特定的扭曲操作,所构造的两类MDS EAQECCs不仅能够灵活调整码长,还能显著提升最小距离,突破了传统理论界限。此外,该文利用扭曲操作将两类MDS EAQECCs提升为最大纠缠态极大距离可分纠缠辅助量子纠错码 (ME-MDS EAQECCs)。该文研究成果不仅为量子纠错码设计提供了更广泛的参数选择,还为动态量子通信场景中的高效纠错提供了理论支撑。
  • 表  1  本文构造的MDS EAQECCs

    构造类 构造参数 参数来源
    1 $ \left[\right[120, 58, 37; 10]{]}_{25} $ 定理3
    $ \left[\right[156, 106, 40; 28]{]}_{27} $ 定理3
    $ \left[\right[288, 194, 73; 50]{]}_{49} $ 定理3
    $ \left[\right[160, 2, 121; 82]{]}_{81} $ 定理4
    $ \left[\right[78, 11, 40; 11]{]}_{27} $ 定理5
    $ \left[\right[72, 22, 40; 28]{]}_{25} $ 定理5
    $ \left[\right[496, 250, 187; 126]{]}_{125} $ 定理6
    $ \left[\right[32, 18, 13; 10]{]}_{9} $ 定理6
    $ \left[\right[32, 20, 13; 12]{]}_{9} $ 定理3的ME提升
    $ \left[\right[78, 39, 40; 39]{]}_{25} $ 定理3的ME提升
    2 $ \left[\right[130, 103, 27; 25]{]}_{25} $ 定理8
    $ \left[\right[30, 23, 7; 5]{]}_{5} $ 定理8
    $ \left[\right[630, 503, 127; 125]{]}_{125} $ 定理8
    $ \left[\right[84, 55, 29; 27]{]}_{27} $ 定理9
    $ \left[\right[12, 7, 5; 3]{]}_{3} $ 定理9
    $ \left[\right[30, 19, 11; 9]{]}_{9} $ 定理9
    $ \left[\right[12, 8, 5; 4]{]}_{3} $ 定理8的ME提升
    $ \left[\right[30, 24, 7; 6]{]}_{5} $ 定理8的ME提升
    下载: 导出CSV

    表  2  与已知$ q $-元MDS EAQECCs的参数对比

    构造类 $ n $ $ d $ $ c $ 限制条件 文献
    1 $ 2\mathrm{\lambda }\left(q-1\right) $ $ d $ $ 2i $ $ i\in \left\{\mathrm{1,2}\right\} $, $ \mathrm{\lambda } $为奇数, $ \mathrm{\lambda }|\left(q-1\right) $, $ 8|\left(q+1\right) $ $ \dfrac{q-1}{2}\left(i-1\right)+4\mathrm{\lambda }+1\le d\le \dfrac{q-1}{2}+2\left(i+1\right)\mathrm{\lambda } $ [30]
    2 $ \dfrac{{q}^{2}-1}{\mathrm{\lambda }} $ $ y+1 $ $ y $ $ 1\le y\le \left\lfloor\dfrac{{q}^{2}-1}{2\mathrm{\lambda }}\right\rfloor $ [31]
    3 $ {q}^{2}+i $ $ d $ 1 $ 2\le d\le 2q+i-1),i\in \{-\mathrm{1,1}\} $, [32]
    $ \dfrac{{q}^{2}-1}{i} $ $ d $ 2 $ \dfrac{q+1}{2}+2\le d\le \dfrac{3}{2}q-\dfrac{1}{2} $
    4 $ {q}^{2}+1 $ $ 2q+2 $ 5 $ q\equiv 3\left(\mathrm{m}\mathrm{o}\mathrm{d}4\right) $ [33]
    $ {q}^{2}+1 $ $ 2\mathrm{\lambda }+2 $ 9 $ q\equiv 3\left(\mathrm{m}\mathrm{o}\mathrm{d}4\right) $, $ q+1\le \mathrm{\lambda }\le 2q-2 $
    $ \dfrac{{q}^{2}-1}{4} $ $ d $ 4 $ q\equiv 3\left(\mathrm{m}\mathrm{o}\mathrm{d}4\right) $, $ \dfrac{3q+7}{4}\le d\le \dfrac{5q+1}{4} $
    5 $ \dfrac{\left(q-1\right)\left(q+1\right)}{8} $ $ d $ 8 $ q+1\le d\le \dfrac{9q-1}{8} $ [34]
    6 $ \dfrac{{q}^{2}-1}{i} $ $ d $ $ i $ $ i\in \{\mathrm{3,5},7\},q+1\le d\le \dfrac{\left(i+1\right)\left(q+1\right)}{j}-1 $ [35]
    $ \dfrac{{q}^{2}-1}{j} $ $ d $ 7 $ d=\dfrac{8\left(q+1\right)}{j}-1 $
    7 $ \dfrac{{q}^{2}+1}{2} $ $ 3q-1 $ 13 $ q\ge 5 $ [36]
    $ 4q- 2 $ 25 $ q\ge 7 $
    $ 6q- 4 $ 61 $ q\ge 11 $
    8 $ i\left(q-1\right) $ $ q+j+1 $ $ q-t+2v+2 $ $ {i}^{2}\equiv 1 \left(\text{mod} p\right) $或$ i\equiv 0 \left(\text{mod} p\right) $, $ i $为奇数,
    $ j=\dfrac{q-3}{2} $, $ v=\left\lfloor\dfrac{\left(j+1\right)t}{q+1}\right\rfloor $, $ t=\mathrm{gcd}\left(i,q+1\right) $
    定理3
    $ i\left(q-1\right) $ $ q+j+1 $ $ q-t+2v+3 $ $ {i}^{2}\equiv 1 \left(\text{mod} p\right) $或$ i\equiv 0 \left(\text{mod} p\right) $, $ i $为偶数,
    $ j=\dfrac{q-3}{2} $, $ v=\left\lfloor\dfrac{\left(j+1\right)t}{q+1}\right\rfloor $, $ t=\mathrm{gcd}\left(i,q+1\right) $
    $ i\left(q-1\right) $ $ q+j+1 $ $ q+j $ $ c=d-1 $ ME提升
    $ 2\left(q-1\right) $ $ q+j+1 $ $ q-t+2v+3 $ $ c=d-1 $,$ i=2 $
    $ 3\left(q-1\right) $ $ q+j+1 $ $ q-t+2v+2 $ $ c=d-1 $,$ i=3 $
    9 $ i\left(q+1\right) $ $ q+2 $ $ q-s+1 $ $ i $为奇数, $ {i}^{2}\equiv 1 \left(\text{mod} p\right) $或$ i\equiv 0 \left(\text{mod} p\right) $,$ 2\le i\le q-2 $, $ s=\mathrm{gcd}\left(i,q-1\right) $ 定理8
    $ i\left(q+1\right) $ $ q+2 $ $ c=d-1 $ ME提升
    $ 3\left(q+1\right) $ $ q+2 $ $ q $ $ c=d-1 $,$ i=3 $
    下载: 导出CSV
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  • 收稿日期:  2025-04-11
  • 修回日期:  2025-07-27
  • 网络出版日期:  2025-08-06

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