Kolmogorov复杂度中的27个开放问题

Andrei E. Romashchenko, A. Shen, Marius Zimand
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引用次数: 2

摘要

这个公式可以非正式地理解为:第i条消息带来log(1=pi)“信息位”(不管这意味着什么),并且以频率pi出现,因此H是一个随机消息(随机变量的一个样本)提供的预期信息量。此外,我们可以构造一个最优的唯一可解码代码,它平均每条消息需要大约H(确切地说,最多H + 1)位,并且它以大约log(1=pi)位对第i条消息进行编码,遵循对频繁消息使用短码字的自然想法。这很适合上面给出的公式的非正式解读,并且很容易说第i条消息“包含log(1=pi)位的信息”。香农自己也受到了这种诱惑[46,第399页],他写了一篇关于熵估计的文章,认为基础英语和詹姆斯·乔伊斯的书《芬尼根的守魂》是英语文本中高冗余和低冗余的两个极端例子。但是,严格地说,人们只能谈论随机变量的熵,而不能谈论它们的单个值,而且“芬尼根的守灵夜”不是一个随机变量,只是一个特定的字符串。我们能定义单个对象的信息量吗?
本文章由计算机程序翻译,如有差异,请以英文原文为准。
27 Open Problems in Kolmogorov Complexity
This formula can be informally read as follows: the ith messagemi brings us log(1=pi) "bits of information" (whatever this means), and appears with frequency pi, so H is the expected amount of information provided by one random message (one sample of the random variable). Moreover, we can construct an optimal uniquely decodable code that requires about H (at most H + 1, to be exact) bits per message on average, and it encodes the ith message by approximately log(1=pi) bits, following the natural idea to use short codewords for frequent messages. This fits well the informal reading of the formula given above, and it is tempting to say that the ith message "contains log(1=pi) bits of information." Shannon himself succumbed to this temptation [46, p. 399] when he wrote about entropy estimates and considers Basic English and James Joyces's book "Finnegan's Wake" as two extreme examples of high and low redundancy in English texts. But, strictly speaking, one can speak only of entropies of random variables, not of their individual values, and "Finnegan's Wake" is not a random variable, just a specific string. Can we define the amount of information in individual objects?
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