Construction of optimal locally recoverable codes and connection with hypergraph

November 22, 2018 Β· Declared Dead Β· πŸ› International Colloquium on Automata, Languages and Programming

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Authors Chaoping Xing, Chen Yuan arXiv ID 1811.09142 Category cs.IT: Information Theory Cross-listed math.CO Citations 38 Venue International Colloquium on Automata, Languages and Programming Last Checked 6 months ago
Abstract
Recently, it was discovered by several authors that a $q$-ary optimal locally recoverable code, i.e., a locally recoverable code archiving the Singleton-type bound, can have length much bigger than $q+1$. This is quite different from the classical $q$-ary MDS codes where it is conjectured that the code length is upper bounded by $q+1$ (or $q+2$ for some special case). This discovery inspired some recent studies on length of an optimal locally recoverable code. It was shown in \cite{LXY} that a $q$-ary optimal locally recoverable code is unbounded for $d=3,4$. Soon after, it was proved that a $q$-ary optimal locally recoverable code with distance $d$ and locality $r$ can have length $Ξ©_{d,r}(q^{1 + 1/\lfloor(d-3)/2\rfloor})$. Recently, an explicit construction of $q$-ary optimal locally recoverable codes for distance $d=5,6$ was given in \cite{J18} and \cite{BCGLP}. In this paper, we further investigate construction optimal locally recoverable codes along the line of using parity-check matrices. Inspired by classical Reed-Solomon codes and \cite{J18}, we equip parity-check matrices with the Vandermond structure. It is turns out that a parity-check matrix with the Vandermond structure produces an optimal locally recoverable code must obey certain disjoint property for subsets of $\mathbb{F}_q$. To our surprise, this disjoint condition is equivalent to a well-studied problem in extremal graph theory. With the help of extremal graph theory, we succeed to improve all of the best known results in \cite{GXY} for $d\geq 7$. In addition, for $d=6$, we are able to remove the constraint required in \cite{J18} that $q$ is even.
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