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极大距离可分纠缠辅助量子纠错码的构造

屈圆月 高健

屈圆月, 高健. 极大距离可分纠缠辅助量子纠错码的构造[J]. 电子与信息学报. doi: 10.11999/JEIT251251
引用本文: 屈圆月, 高健. 极大距离可分纠缠辅助量子纠错码的构造[J]. 电子与信息学报. doi: 10.11999/JEIT251251
QU Yuanyue, GAO Jian. Construction of Entanglement-Assisted Quantum MDS Codes[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT251251
Citation: QU Yuanyue, GAO Jian. Construction of Entanglement-Assisted Quantum MDS Codes[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT251251

极大距离可分纠缠辅助量子纠错码的构造

doi: 10.11999/JEIT251251 cstr: 32379.14.JEIT251251
基金项目: 山东省自然科学基金(ZR2024YQ057),国家自然科学基金项目(12071264)
详细信息
    作者简介:

    屈圆月:女,硕士研究生,研究方向为编码理论及其应用

    高健:男,教授,博士生导师,《电子与信息学报》编委,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

Construction of Entanglement-Assisted Quantum MDS Codes

Funds: The Natural Science Foundation of Shandong Province (ZR2024YQ057), The National Natural Science Foundation of China (12071264)
  • 摘要: 量子通信作为未来信息安全传输的核心技术,在实际应用中高度依赖高效可靠的量子纠错编码方案。纠缠辅助量子纠错码(EAQECCs)通过引入预共享纠缠态,突破了传统量子纠错码对经典纠错码自正交性的条件限制,显著提升了编码设计的灵活性与性能。本文基于有限域上的扭曲Reed–Solomon(TRS)码,系统构造了几类新参数的极大距离可分(MDS)EAQECCs。通过分析TRS码关联的陪集和矩阵的秩,本文确定了EAQECCs所需预共享纠缠态数量的精确条件,借助群的代数结构与陪集和严格证明了构造的EAQECCs具备最优的纠错性能。本文所构造的MDS EAQECCs码长突破了部分文献中所构造MDS EAQECCs码长选择的局限性,拓展了MDS EAQECCs的码长范围。本文的结果不仅深化了人们对MDS EAQECCs代数结构的认知,而且为面向量子通信系统高性能量子纠错码的设计提供了新的理论工具。
  • 表  1  定理3构造的MDS EAQECCs

    q a b $\left[\kern-0.15em\left[ { n,k,d;c} \right]\kern-0.15em\right]_{q} $
    8 9 5 $\left[\kern-0.15em\left[ { 42{,}36{,}7;5} \right]\kern-0.15em\right]_{8} $
    11 12 8 $\left[\kern-0.15em\left[ { 90{,}81{,}10;8} \right]\kern-0.15em\right]_{11} $
    16 17 9 $\left[\kern-0.15em\left[ { 150{,}136{,}13;9} \right]\kern-0.15em\right]_{16} $
    17 18 6 $\left[\kern-0.15em\left[ { 112{,}97{,}12;6} \right]\kern-0.15em\right]_{17} $
    9 10 3 $\left[\kern-0.15em\left[ { 32{,}24{,}7;5} \right]\kern-0.15em\right]_{9} $
    13 7 3 $\left[\kern-0.15em\left[ { 96{,}80{,}11;4} \right]\kern-0.15em\right]_{13} $
    17 18 9 $\left[\kern-0.15em\left[ { 160{,}144{,}14;10} \right]\kern-0.15em\right]_{19} $
    19 20 11 $\left[\kern-0.15em\left[ { 216{,}198{,}16;12} \right]\kern-0.15em\right]_{19} $
    下载: 导出CSV

    表  2  定理4构造的MDS EAQECCs

    q a b $ \left[\kern-0.15em\left[ { n,k,d;c} \right]\kern-0.15em\right]_{q} $
    16 17 12 $ \left[\kern-0.15em\left[ { 195{,}170{,}15;11} \right]\kern-0.15em\right]_{16} $
    17 18 10 $ \left[\kern-0.15em\left[ { 176{,}161{,}14;9} \right]\kern-0.15em\right]_{17} $
    19 20 13 $ \left[\kern-0.15em\left[ { 252{,}234{,}17;12} \right]\kern-0.15em\right]_{19} $
    19 20 14 $ \left[\kern-0.15em\left[ { 270{,}170{,}15;6} \right]\kern-0.15em\right]_{19} $
    19 20 16 $ \left[\kern-0.15em\left[ { 302{,}289{,}18;15} \right]\kern-0.15em\right]_{19} $
    23 24 15 $ \left[\kern-0.15em\left[ { 352{,}330{,}20;14} \right]\kern-0.15em\right]_{23} $
    下载: 导出CSV

    表  3  定理5构造的MDS EAQECCs

    q a b $ \left[\kern-0.15em\left[ { n,k,d;c} \right]\kern-0.15em\right]_{q} $
    9 8 7 $ \left[\kern-0.15em\left[ { 70{,}63{,}8;6} \right]\kern-0.15em\right]_{9} $
    11 10 8 $ \left[\kern-0.15em\left[ { 96{,}88{,}9;7} \right]\kern-0.15em\right]_{11} $
    11 10 9 $ \left[\kern-0.15em\left[ { 108{,}99{,}10;8} \right]\kern-0.15em\right]_{11} $
    13 12 9 $ \left[\kern-0.15em\left[ { 126{,}117{,}10;9} \right]\kern-0.15em\right]_{13} $
    13 12 10 $ \left[\kern-0.15em\left[ { 140{,}128{,}11;10} \right]\kern-0.15em\right]_{13} $
    16 15 11 $ \left[\kern-0.15em\left[ { 187{,}176{,}12;10} \right]\kern-0.15em\right]_{16} $
    16 15 10 $ \left[\kern-0.15em\left[ { 170{,}160{,}11;10} \right]\kern-0.15em\right]_{16} $
    下载: 导出CSV

    表  4  一些已知的MDS EAQECCs

    参数 限制条件 参考文献
    $ \left[\kern-0.15em\left[ { n=\dfrac{{q}^{2}-1}{2}+\dfrac{{q}^{2}-1}{2},n-2d+c+2,d;c} \right]\kern-0.15em\right]_{q} $ $ q> 3,b\mid \left(q+1\right),b\equiv 2\left(mod4\right),2\leq d\leq \dfrac{3\left(q+1\right)}{4}+\dfrac{q+1}{2b},c=\dfrac{b}{2}+1 $ [13]
    $ \left[\kern-0.15em\left[ { n=\dfrac{{q}^{2}+1}{a},n-2d+4k-2,d;4k-1} \right]\kern-0.15em\right]_{q} $ $ a=4{h}^{2}+{(4h+1)}^{2},q=(2t-1)a-10h-2,h\geq 2,t\in {\mathbb{Z}}^{+};2f(1)-4t-4\leq $
    $d\leq 2f(2)\text{偶数} ,k=2\mathrm{或}2f(k-1)+2\leq d\leq 2f(k)\text{偶数},k=3{,}4 $
    [14]
    $ \left[\kern-0.15em\left[ { n=\dfrac{{q}^{2}+1}{a},n-2d+4k-1,d;1} \right]\kern-0.15em\right]_{q} $ $ a={h}^{2}+{(3h+1)}^{2},,q=2ta-1-h-3,h\geq 2,t\in {\mathbb{Z}}^{+};2\leq d\leq 2f(1)\text{偶数} $ [14]
    $ \left[\kern-0.15em\left[ { n=\dfrac{{q}^{2}+1}{a},n-2d+4k-1,d;1+4\left(k-1\right)} \right]\kern-0.15em\right]_{q} $ $ a={h}^{2}+{(3h+1)}^{2},q=2ta-1-h-3,h\geq 2,t\in {\mathbb{Z}}^{+};$
    $2f(k-1)+2\leq d\leq 2f(k)\text{偶数},k=2{,}3 $
    [14]
    $ \left[\kern-0.15em\left[ { n=\dfrac{t(p-l)}{{p}^{e-l}-1},k-h,n-k+1;n-k-h} \right]\kern-0.15em\right]_{q} $ $ {q=p}^{e},2(e-l)\mid e,1\leq t\leq {p}^{e-l}-1{,}1\leq k\leq \left\lfloor \dfrac{{p}^{e}+n}{{p}^{e}+1}\right\rfloor ,0\leq h\leq k-1 $ [15]
    $ \left[\kern-0.15em\left[ { \begin{matrix}n=\dfrac{({t}_{1}+1)({q}^{2}-1)}{{h}_{1}}+\dfrac{(2{t}_{2}+2)({q}^{2}-1)}{{h}_{2}},\\ n-2k+c,k+1;c\\ \end{matrix}} \right]\kern-0.15em\right]_{q} $ $ {h}_{1},{h}_{2}\text{偶数},{K}_{1}=\dfrac{({t}_{1}+1)({q}^{2}-1)}{{h}_{1}}+\dfrac{q-1}{2},{K}_{2}=(\dfrac{{h}_{2}}{2}+{t}_{2}+1)\cdot \dfrac{q+1}{{h}_{2}}-2, $
    $ 0\leq {t}_{1}\leq \dfrac{{h}_{1}-1}{2},0\leq {t}_{2}\leq \dfrac{{h}_{2}-3}{4},min\{{K}_{1},{K}_{2}\}\leq k\leq max\{{K}_{1},{K}_{2}\} $
    [16]
    $ \left[\kern-0.15em\left[ { n=\dfrac{{q}^{2}-1}{t},n-2d+t+2,d;t} \right]\kern-0.15em\right]_{q} $ $ q\text{奇数},t\mid (q+1),t\geq 3\text{奇数},\dfrac{(t-1)(q+1)}{t}+2\leq d\leq \dfrac{(t-1)(q+1)}{t}-2 $ [17]
    $ \left[\kern-0.15em\left[ { n=\dfrac{b\left({q}^{2}-1\right)}{a}+\dfrac{{q}^{2}-1}{a},n-2d+c+3,d;c} \right]\kern-0.15em\right]_{q} $ $ a\geq 3,a\mid (q+1),b\leq min\{a-3,q-3\},2\leq d\leq \dfrac{a+b+1}{2}\cdot \dfrac{q+1}{a},$
    $c=b\mathrm{或}b+1 $
    定理3
    $ \left[\kern-0.15em\left[ { n=\dfrac{b\left({q}^{2}-1\right)}{a}+\dfrac{{q}^{2}-1}{a},n-2d+c+3,d;c} \right]\kern-0.15em\right]_{q} $ $ a\geq 3,a\mid (q+1),b\leq min\{a-3,q-3\},2\leq d\leq \dfrac{a+b+1}{2}\cdot \dfrac{q+1}{a}, $
    $ c=b-1\mathrm{或}b\mathrm{或}b+1 $
    定理4
    $ \left[\kern-0.15em\left[ { n=\dfrac{b\left({q}^{2}-1\right)}{a},n-2d+c+3,d;c} \right]\kern-0.15em\right]_{q} $ $ b\leq a,a\mid (q-1),2\leq d\leq \dfrac{b(q+1)}{a}+1,c=b-1\mathrm{或}b $ 定理5
    下载: 导出CSV
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  • 收稿日期:  2025-11-26
  • 修回日期:  2026-03-09
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  • 网络出版日期:  2026-03-26

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