Construction of Two Classes of Minimal Binary Linear Codes
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摘要: 线性码在数据存储、信息安全以及秘密共享等领域具有重要的作用。而极小线性码是设计秘密共享方案的首选码，设计极小线性码是当前密码与编码研究的重要内容之一。该文首先选取恰当的布尔函数，研究了函数的Walsh谱值分布，并利用布尔函数的Walsh谱值分布构造了两类极小线性码，确定了码的参数及重量分布。结果表明，所构造的码是不满足Ashikhmin-Barg条件的极小线性码，可用作设计具有良好访问结构的秘密共享方案。Abstract: Linear codes play an important role in data storage, information security and secret sharing. Minimal linear codes are the first choice to design secret sharing schemes, so the design of minimal linear codes is one of the important contents of current cryptosystem and coding theory. In this paper, the Walsh spectrum distribution of the selected Boolean functions is studied, and two kinds of minimal linear codes are obtained by using the Walsh spectrum distribution of the functions, then the weight distribution of the codes are determined. The results show that the constructed codes are minimal linear codes that do not satisfy the Ashikhmin-Barg condition, and can be used to design secret sharing schemes with good access structure.
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Key words:
- Boolean functions /
- Walsh transform /
- Binary linear codes
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表 1 码
${{\boldsymbol{C}}_{\boldsymbol{f}}}$ 的重量分布重量 频数 $ 0 $ $ 1 $ $ {2^{m - 1}} $ ${2^m} - 1 + {2^s}\left({2^t} - 2 - \dfrac{ {s(s + 1)} }{2}\right) + {\varepsilon _1}\left( {\begin{array}{*{20}{c} } s \\ {(2s + 3 \pm k)/4} \end{array} } \right)$ $ {2^{m - 1}} - {2^{t - 1}}A(i) $ $\begin{array}{*{20}{c} } {\left( {\begin{array}{*{20}{c} } s \\ i \end{array} } \right)}&{\left(1 \le i \le s,i \ne \dfrac{ {2s + 3 \pm k} }{4}\right)} \end{array}$ $ {2^{m - 1}} - {2^{t - 1}} $ $ {2^{s - {\text{2}}}}s(s + 1) + {\varepsilon _2}\left( {\begin{array}{*{20}{c}} s \\ {(2s + 3 \pm k)/4} \end{array}} \right) $ $ {2^{m - 1}} + {2^{t - 1}} $ $ {2^s} + {2^{s - {\text{2}}}}s(s + 1) + {\varepsilon _3}\left( {\begin{array}{*{20}{c}} s \\ {(2s + 3 \pm k)/4} \end{array}} \right) $ ${2^{m - 1} } + {2^{t - 1} }\left({2^s} - 1 - \dfrac{ {s(s + 1)} }{2}\right)$ $ 1 $ 
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表 2 码
${{\boldsymbol{C}}_{{{\boldsymbol{f}}_{\bar {\boldsymbol{D}}}}}}$ 的重量分布重量 频数 $ 0 $ $ 1 $ $ {2^{m - 1}} $ $ {2^m} - 1 $ $ {2^{m - 1}} + {2^{t - 1}}A(i) - 1 $ $\begin{array}{*{20}{c} } {\left( {\begin{array}{*{20}{c} } s \\ i \end{array} } \right)}&{ \left(1 \le i \le s,i \ne \dfrac{ {2s + 3 \pm k} }{4}\right)} \end{array}$ $ {2^{m - 1}} - 1 $ ${2^s}\left({2^t} - 2 - \dfrac{ {s(s + 1)} }{2}\right) + {\varepsilon _1}\left( {\begin{array}{*{20}{c} } s \\ {(2s + 3 \pm k)/4} \end{array} } \right)$ $ {2^{m - 1}} + {2^{t - 1}} - 1 $ $ {2^{s - {\text{2}}}}s(s + 1) + {\varepsilon _3}\left( {\begin{array}{*{20}{c}} s \\ {(2s + 3 \pm k)/4} \end{array}} \right) $ $ {2^{m - 1}} - {2^{t - 1}} - 1 $ $ {2^s} + {2^{s - {\text{2}}}}s(s + 1) + {\varepsilon _2}\left( {\begin{array}{*{20}{c}} s \\ {(2s + 3 \pm k)/4} \end{array}} \right) $ ${2^{m - 1} } - {2^{t - 1} }\left({2^s} - 1 - \dfrac{ {s(s + 1)} }{2}\right) - 1$ $ 1 $
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