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Zigzag Codes: MDS Array Codes With Optimal Rebuilding

机译:之字形代码:具有最佳重建的MDS阵列代码

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摘要

Maximum distance separable (MDS) array codes are widely used in storage systems to protect data against erasures. We address the rebuilding ratio problem, namely, in the case of erasures, what is the fraction of the remaining information that needs to be accessed in order to rebuild exactly the lost information? It is clear that when the number of erasures equals the maximum number of erasures that an MDS code can correct, then the rebuilding ratio is 1 (access all the remaining information). However, the interesting and more practical case is when the number of erasures is smaller than the erasure correcting capability of the code. For example, consider an MDS code that can correct two erasures: What is the smallest amount of information that one needs to access in order to correct a single erasure? Previous work showed that the rebuilding ratio is bounded between 1/2 and 3/4; however, the exact value was left as an open problem. In this paper, we solve this open problem and prove that for the case of a single erasure with a two-erasure correcting code, the rebuilding ratio is 1/2. In general, we construct a new family of r-erasure correcting MDS array codes that has optimal rebuilding ratio of 1/(r) in the case of a single erasure. Our array codes have efficient encoding and decoding algorithms (for the cases r=2 and r=3, they use a finite field of size 3 and 4, respectively) and an optimal update property.
机译:最大距离可分离(MDS)阵列代码广泛用于存储系统中,以防止数据被擦除。我们解决重建率问题,即在擦除的情况下,为了准确地重建丢失的信息,需要访问剩余信息的一部分吗?显然,当擦除次数等于MDS代码可以纠正的最大擦除次数时,则重建比率为1(访问所有剩余信息)。但是,有趣且更实际的情况是擦除次数小于代码的擦除校正能力。例如,考虑一种可以纠正两个擦除的MDS代码:为了纠正一个擦除,一个人需要访问的最少信息量是多少?先前的工作表明,重建比例限制在1/2和3/4之间。但是,确切的值仍然是一个未解决的问题。在本文中,我们解决了这个开放问题,并证明了对于具有两次删除校正码的一次删除,重建比例为1/2。通常,我们构建了一个新的r消除校正MDS阵列代码系列,在单次擦除的情况下,其最佳重建比率为1 /(r)。我们的数组代码具有有效的编码和解码算法(对于r = 2和r = 3的情况,它们分别使用大小为3和4的有限字段)和最佳更新属性。

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