用vasil 'ev码构造二权或三权二进制线性码

Pub Date : 2021-01-01 DOI:10.4134/JKMS.J190429
J. Hyun, Jaeseon Kim
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引用次数: 0

摘要

。线性码的生成矩阵的列向量的集合D称为线性码的定义集合。本文研究了从定义集构造少权(主要是二权或三权)线性码的问题。很容易看出,当我们取一个定义集为线性码的非零码字时,我们得到一个一权码。因此,我们必须从非线性码中选择一个定义集来获得二权码或三权码,并且我们面临着构造的码包含多个权值的问题。为了克服这个困难,我们采用以下形式的线性编码:设D是fn2的一个子集,W (p = 1)。V)是f2的一个子空间。F n 2)。我们定义线性码C D (W;V),定义集合D,限定为W,V受C D (W;V) = {(s + u·x) x∈D∗| s∈W,u∈V}。取D为长度为n = 2m - 1(m≥3)的Vasil 'ev码,并选择合适的W,得到两权码或三权码。对于D是Vasil 'ev代码的补充,我们做同样的工作。构造的低权重代码共享一些很好的属性。其中一些是最优的,因为它们达到了Griesmer界或Grey-Rankin界。它们中的大多数都是最小的代码,反过来在秘密共享方案中有应用。最后,我们得到了不成立Ashikhmin和Barg充分条件的无穷极小码族。
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CONSTRUCTION OF TWO- OR THREE-WEIGHT BINARY LINEAR CODES FROM VASIL'EV CODES
. The set D of column vectors of a generator matrix of a linear code is called a defining set of the linear code. In this paper we consider the problem of constructing few-weight (mainly two- or three-weight) linear codes from defining sets. It can be easily seen that we obtain an one- weight code when we take a defining set to be the nonzero codewords of a linear code. Therefore we have to choose a defining set from a non-linear code to obtain two- or three-weight codes, and we face the problem that the constructed code contains many weights. To overcome this difficulty, we employ the linear codes of the following form: Let D be a subset of F n 2 , and W (resp. V ) be a subspace of F 2 (resp. F n 2 ). We define the linear code C D ( W ; V ) with defining set D and restricted to W,V by C D ( W ; V ) = { ( s + u · x ) x ∈ D ∗ | s ∈ W,u ∈ V } . We obtain two- or three-weight codes by taking D to be a Vasil’ev code of length n = 2 m − 1( m ≥ 3) and a suitable choices of W . We do the same job for D being the complement of a Vasil’ev code. The constructed few-weight codes share some nice properties. Some of them are optimal in the sense that they attain either the Griesmer bound or the Grey-Rankin bound. Most of them are minimal codes which, in turn, have an application in secret sharing schemes. Finally we obtain an infinite family of minimal codes for which the sufficient condition of Ashikhmin and Barg does not hold.
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