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首页> 外文期刊>IEEE Transactions on Pattern Analysis and Machine Intelligence >Breaking substitution cyphers using stochastic automata
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Breaking substitution cyphers using stochastic automata

机译:使用随机自动机打破替代密码

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摘要

Let Lambda be a finite plaintext alphabet and V be a cypher alphabet with the same cardinality as Lambda . In all one-to-one substitution cyphers, there exists the property that each element in V maps onto exactly one element in Lambda and vice versa. This mapping of V onto Lambda is represented by a function T*, which maps any v in V onto some lambda in Lambda (i.e., T*(v)= lambda ). The problem of learning the mapping of T* (or its inverse (T*)/sup -1/) by processing a sequence of cypher text is discussed. The fastest reported method to achieve this is a relaxation scheme that utilizes the statistical information contained in the unigrams and trigrams of the plaintext language. A new learning automaton solution to the problem called the cypher learning automaton (CLA) is given. The proposed scheme is fast, and the advantages of the scheme in terms of time and space requirements over the relaxation method have been listed. Simulation results comparing both cypher-breaking techniques are presented.
机译:令Lambda为有限的纯文本字母,V为与Lambda具有相同基数的密文字母。在所有一对一的替换密码中,存在以下属性:V中的每个元素都精确映射到Lambda中的一个元素,反之亦然。 V到Lambda的映射由函数T *表示,该函数将V中的任何v映射到Lambda中的某个lambda(即T *(v)= lambda)。讨论了通过处理密文序列来学习T *(或其反(T *)/ sup -1 /)映射的问题。达到此目的的最快的报告方法是一种松弛方案,该方案利用了明文语言的单字组和三字组中包含的统计信息。给出了一种新的针对该问题的学习自动机解决方案,称为密码学习自动机(CLA)。所提出的方案是快速的,并且在时间和空间要求方面该方案相对于松弛方法具有优点。仿真结果比较了两种密码破解技术。

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