一对一编码及其抗冗余性

W. Szpankowski
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引用次数: 8

摘要

一对一代码是“一次性”代码,它为源符号分配一个不同的码字,不一定是前缀代码(更一般地说,是唯一可解码的)。例如,当存在“消息结束”通道符号时,就会出现这种代码。有趣的是,正如Wyner在1972年证明的那样,对于这样的代码,平均代码长度可以小于源熵。减少多少?我们称这种差异为反冗余。多年来,许多作者已经表明,反冗余可以与源熵的负对数一样大。然而,据我们所知,并不存在精确的估计。本文考虑由二进制无记忆源生成的长度为n的块码,并证明了平均抗冗余为-(1/2)log2n + C + F(n) + o(1),其中C为常数,如果log2(1 - p)/p为无理数(其中p为生成“0”的概率),则F(n) = o,否则F(ri)是随码长增加的波动函数。这个相对简单的发现需要结合相当复杂的分析工具,如伯努利和的精确计算、鞍点法和模为1的序列分布理论
本文章由计算机程序翻译,如有差异,请以英文原文为准。
One-to-one code and its anti-redundancy
One-to-one codes are "one shot" codes that assign a distinct codeword to source symbols and are not necessarily prefix codes (more generally, uniquely decodable). For example, such codes arise when there exists an "end of message" channel symbol. Interestingly, as Wyner proved in 1972, for such codes the average code length can be smaller than the source entropy. By how much? We call this difference the anti-redundancy. Various authors over the years have shown that the anti-redundancy can be as big as minus the logarithm of the source entropy. However, to the best of our knowledge precise estimates do not exist. In this note, we consider a block code of length n generated by a binary memoryless source, and prove that the average anti-redundancy is -(1/2)log2n + C + F(n) + o(1) where C is a constant and either F(n) = O if log2(1 - p)/p is irrational (where p is the probability of generating a "0") or otherwise F(ri) is a fluctuating function as the code length increases. This relatively simple finding requires a combination of quite sophisticated analytic tools such as precise evaluation of Bernoulli sums, the saddle point method, and theory of distribution of sequences modulo 1
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