Cryptography In The Information Society by Ryabko Boris; Fionov Andrey;

Author:Ryabko, Boris; Fionov, Andrey;
Language: eng
Format: epub
Publisher: World Scientific Publishing Company
Published: 2021-08-15T00:00:00+00:00

(all operations are modulo 26). Here mi is an odd letter of the message, mi+1 is an even letter, k1, k2 are the key digits, and ci, ci+1 the resulting symbols of ciphertext. For instance, the pair of symbols SE is encrypted under the key k = (3, 5) in the following way:

i.e. SE converts to ZF.

Note that this cipher is decipherable. The decryption algorithm usually called the inverse cipher is written as follows:

Applying the cipher (8.2) to our message under the key (3, 5) we obtain

Here, for clarity, after the first arrow, the intermediate result obtained after the first two operations in (8.2) is shown (this is an “intermingled” but not yet encrypted text). We can see that this cipher hides the frequencies of letters, which complicates the frequency analysis. Of course, the frequencies of the pairs (digrams) are preserved but we can conceal them too if we encrypt the message in blocks of 3 letters, etc.

So, we have considered two ways of improving the cipher: increasing the length of the key and combining symbols into blocks with mixing transformations. Both methods are widely used in real modern ciphers. Now we shall consider one more trick which is also used in practice. For the sake of simplicity, we proceed with the same example.

The cipher (8.2) is more complicated to Eve compared to the cipher (8.1) and gives us the opportunity to consider another scenario of attack. Up to this point we considered only the ciphertext-only attacks. But what will happen if Eve somehow procures the plaintext corresponding to a ciphertext transmitted earlier? Here we have a situation where the known-plaintext attack (see Chap. 1, p. 5) is being used. For instance, Eve has the pair (SEQUENCE, VJTZHSFJ) for the cipher (8.1). Then she immediately computes the secret key k1 = V − S = 3, k2 = J − E = 5 and is able to decrypt all further messages from Alice to Bob. When the cipher (8.2) is used, the pair (SEQUENCE, ZFNJUJJP) does not allow for such an obvious solution, although breaking the cipher (8.2) is simple too. Eve performs the first two operations from (8.2), which do not require the secret key, for the word SEQUENCE, obtains the intermediate word WAKEREGK and using the pair (WAKEREGK, ZFNJUJJP), as in the previous case, finds the key k = 3, 5.

How one can hamper Eve’s actions? The idea is simple. In encryption, we shall use the cipher (8.2) twice. So we obtain:

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