In the paper, the Levenshtein's sequence reconstruction problem is considered in the case where at most $t$ substitution errors occur in each of the $N$ channels and the decoder outputs a list of length $\mathcal{L}$. Moreover, it is assumed that the transmitted words are chosen from an $e$-error-correcting code $C \ (\subseteq \{0,1\}^n)$. Previously, when $t = e+\ell$ and the length $n$ of the transmitted word is large enough, the numbers of required channels are determined for $\mathcal{L} =1, 2 \text{ and } \ell+1$. Here we determine the exact number of channels in the cases $\mathcal{L} = 3, 4, \ldots, \ell$. Furthermore, with the aid of covering codes, we also consider the list sizes in the cases where the length $n$ is rather small (improving previously known results). After that we study how much we can decrease the number of required channels when we use list-decoding codes. Finally, the majority algorithm is discussed for decoding in a probabilistic set-up; in particular, we show that with high probability a decoder based on it is verifiably successful, i.e., the output word of the decoder can be verified to be the transmitted one.
翻译:在本文中,Levenshtein的序列重建问题在以下情况下得到考虑:在美元频道和解码器输出的每个频道中,最多出现美元替代错误,最多出现美元替换错误,每个频道和解码器输出都有一个长度为$mathcal{L}美元的清单。此外,我们假设所传送的字词是从美元-error更正代码$C\ (\\ subseqeq $0,1 ⁇ n) 中选择的。之前,当美元= e ⁇ ell$ 和传送单词的长度足够大时,当所传送的频道数量为$\ mathcal{L} = 1, 2\ text{ 和}\\ ell+1$。我们在这里确定所传送的字词的确切数量是 $\ mathcal{L} = 3, 4,\ ldots,\ ell$。此外,在覆盖代码的帮助下,我们还会考虑所传送的单词长度相当小(改进了以前已知的结果) 之后, 我们研究一下, 当我们使用一个加密的解码的路径的概率会减少多少, 当我们最后使用高的解码时, 时, 显示一个加密的解码的路径的概率的路径的概率是显示一个特定的路径的概率。