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applied cryptography - protocols, algorithms, and source code in c

The advantages of this scheme are that the cv can be

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Unformatted text preview: smission. Since a garbled key can mean megabytes of undecryptable ciphertext, this is a problem. All keys should be transmitted with some kind of error detection and correction bits. This way errors in transmission can be easily detected and, if required, the key can be resent. One of the most widely used methods is to encrypt a constant value with the key, and to send the first 2 to 4 bytes of that ciphertext along with the key. At the receiving end, do the same thing. If the encrypted constants match, then the key has been transmitted without error. The chance of an undetected error ranges from one in 216 to one in 232. Key-error Detection during Decryption Sometimes the receiver wants to check if a particular key he has is the correct symmetric decryption key. If the plaintext message is something like ASCII, he can try to decrypt and read the message. If the plaintext is random, there are other tricks. The naïve approach is to attach a verification block: a known header to the plaintext message before encryption. At the receiving end, Bob decrypts the header and verifies that it is correct. This works, but it gives Eve a known plaintext to help cryptanalyze the system. It also makes attacks against short-key ciphers like DES and all exportable ciphers easy. Precalculate the checksum once for each key, then use that checksum to determine the key in any message you intercept after that. This is a feature of any key checksum that doesn’t include random or at least different data in each checksum. It’s very similar in concept to using salt when generating keys from passphrases. Here’s a better way to do this [821]: (1) Generate an IV (not the one used for the message). (2) Use that IV to generate a large block of bits: say, 512. (3) Hash the result. (4) Use the same fixed bits of the hash, say 32, for the key checksum. This gives Eve some information, but very little. If she tries to use the low 32 bits of the final hash value to mount a brute-force attack, she has to do multiple encryptions plus a hash per candidate key; brute-force on the key itself would be quicker. Sh...
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