Cryptography - Sage

For each of these we can extend our model for the English language to strings of length n.How well do you think each of these model the English language?4.2 Rate and RedundancyLet X be a discrete probability space. We define the rate of X to ber(X) =H(X)log 2 (|X|) ,and the redundancy to be 1 − r(X). The redundancy in a language derives from thestructures such as character frequency distributions, digram frequency distributions (theprobabilities of ordered, adjacent character pairs), and more generally n-gram frequencydistributions. Global structures of a natural language such as vocabulary and grammarrules determine yet more structure, adding to the redundancy of the language.4.3 Conditional ProbabilityWe would now like to have a concept of conditional probability for cryptosystems. Let Ebe a cryptosystem, M a plaintext space, K a key space, and C a ciphertext space. For asymmetric key system the space of plaintext and ciphertext coincide, but the probabilitydistributions on them may differ in the context of the cryptosystem.We use P for both the probability function on the plaintext space M and on K. We cannow define a probability function on C relative to the cryptosystem E:P (y) = ∑ K∈KP (K) ∑ P (x).x∈ME K (x)=yWe can now define P (x, y), for x ∈ M and y ∈ C to be the probability that the pair(x, y) appears as a plaintext–ciphertext pair. Assuming the independence of plaintext andkey spaces, we can define this probability as:P (x, y) = ∑ P (K)P (x).K∈KE K (x)=yx and y are said to be independent if P (x, y) = P (x)P (y). For ciphertext y and plaintextx, define the conditional probability P (y|x) by⎧P (x, y) ⎪⎨ if P (x) ≠ 0P (x)P (y|x) =⎪⎩0 if P (x) = 04.2. Rate and Redundancy 33