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Sequence reconstruction problem for deletion channels: A complete asymptotic solution
Journal of Combinatorial Theory Series A ( IF 0.9 ) Pub Date : 2024-11-27 , DOI: 10.1016/j.jcta.2024.105980
Van Long Phuoc Pham, Keshav Goyal, Han Mao Kiah

Transmit a codeword ▪, that belongs to an (1)-deletion-correcting code of length n, over a t-deletion channel for some 1t<n. Levenshtein (2001) [10], proposed the problem of determining N(n,,t)+1, the minimum number of distinct channel outputs required to uniquely reconstruct ▪. Prior to this work, N(n,,t) is known only when {1,2}. Here, we provide an asymptotically exact solution for all values of and t. Specifically, we show that N(n,,t)=(2)(t)!ntO(nt1). In the special instances: where =t, we show that N(n,,)=(2); and when =3 and t=4, we show that N(n,3,4)20n150. We also provide a conjecture on the exact value of N(n,,t) for all values of n, , and t.

中文翻译:


缺失通道的序列重建问题:一个完整的渐近解



将属于长度为 n 的 (l−1) 删除校正码 的码字 ▪ ,通过 t 删除信道传输约 1≤l≤t<n。Levenshtein (2001) [10] 提出了确定 N(n,l,t)+1 的问题,即唯一重建▪所需的最小不同通道输出数。在这项工作之前,N(n,l,t) 仅在 l∈{1,2} 时已知。在这里,我们提供了 l 和 t 的所有值的渐近精确解。具体来说,我们表明 N(n,l,t)=(2ll)(t−l)!nt−l−O(nt−l−1)。在特殊情况下:其中 l=t,我们表明 N(n,l,l)=(2ll);当 l=3 且 t=4 时,我们表明 N(n,3,4)≤20n−150。我们还提供了 n、l 和 t 的所有值的 N(n,l,t) 的精确值的猜想。
更新日期:2024-11-27
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