


















We consider the problem of communicating over a channel that randomly "tears" the message block into small pieces of different sizes and shuffles them. For the binary torn-paper channel with block length $n$ and pieces of length ${\rm Geometric}(p_n)$, we characterize the capacity as $C = e^{-α}$, where $α= \lim_{n\to\infty} p_n \log n$. Our results show that the case of ${\rm Geometric}(p_n)$-length fragments and the case of deterministic length-$(1/p_n)$ fragments are qualitatively different and, surprisingly, the capacity of the former is larger. Intuitively, this is due to the fact that, in the random fragments case, large fragments are sometimes observed, which boosts the capacity.
此内容由惯性聚合(RSS阅读器)自动聚合整理,仅供阅读参考。 原文来自 — 版权归原作者所有。