User login
Poll
Steganography
History of Cryptography
Basics of Cryptography
Downloads
- Advanced-Encryption-Standard.pdf [PDF, 383.41 KB]
- Developing-the-Advanced-encryption-stander.pdf [PDF, 1.02 MB]
- Rijndael.pdf [PDF, 1019.09 KB]
- Rijndael_Flash_Animation.exe [X-MSDOWNLOAD, 506.02 KB]
Language Redundancy & Unicity Distance
* human languages are highly redundant
o e.g., th lrd s m shphrd shll nt wnt
* Claude Shannon derived several important results about the information content of languages in 1949
* entropy of a message H(X) is related to the number of bits of information needed to encode a message X
o cannot exceed log_(2)n bits for n possible messages
* the rate of langauge for messages of length k denotes the average number of bits in each character
D = F(H(M),k)
* rate of English is about 3.2 bits/letter
* distinguish information context and redundancy
* Shannon defined the unicity distance of a cipher to give a quantatative measure of:
o the security of a cipher (must not be too small)
o the amount of ciphertext N needed to break it
N = F(H(K),D)
where H(K) is entropy (amount of info) of the key, and is D the rate of the language used (eg 3.2)
For polynomial based monoalphabetic substitution ciphers have:
N = F(H(K),D) = F(log_(2)26,3.2) = 1.5
hence only need 2 letters to break. For general monoalphabetic substitution ciphers have
N = F(H(K),D) = F(log_(2)n!,D) = F(log_(2)26!,3.2) = F(26 log_(2)F(26,e),3.2) = 27.6
hence only need 27 or 28 letters to break. For polyalphabetic substitution ciphers, if have s possible keys for each simple subs, and d keys used, then
N = F(H(K),D) = F(log_(2)s^(d),D) = F(d log_(2)26,3.2) = 1.5d
hence need 1.5 times the number of separate substitutions used letters to break the cipher but first need to determine just how many alphabets were used
* Kasiski method
* index of coincidence
l0ck.com data encryption
