Question

Suppose that X = (x1, . . . , xn) and X0 = (x01 , . . . , x 0n ) are two sequences of n plaintext blocks. Define same(X, X0 ) = max{j : xi = x 0i for all i ≤ j}. Suppose X and X0 are encrypted in CBC or CFB mode using the same key and the same IV. Show that it is easy for an adversary to compute same(X, X0 ).

Answer 1

In CBC or CFB mode, an adversary can easily compute the value of same(X, X0) by observing or performing an XOR operation on the ciphertext blocks.

This value represents the index up to which the two sequences have the same intermediate ciphertext blocks.

Step-by-step explanation:

In CBC or CFB mode, each plaintext block is XORed with the previous ciphertext block before encryption.

This means that the decryption of a ciphertext block depends on the ciphertext block that preceded it. If the same key and IV are used for encrypting two sequences of plaintext blocks, the same XOR operations will be performed, resulting in the same intermediate ciphertext blocks.

The maximum value of j, denoted as same(X, X0), represents the index up to which the two sequences have the same intermediate ciphertext blocks.

By observing the ciphertext or performing an XOR operation between the two ciphertexts, an adversary can easily determine which ciphertext blocks have equal values and calculate same(X, X0).

For example, if X = (A, B, C, D, E) and X0 = (A, C, C, D, E), the intermediate ciphertext blocks would be (IV xor A), (A xor B), (B xor C), (C xor D), and (D xor E).

By comparing these values, it is apparent that same(X, X0) = 2, indicating that the first two blocks of the sequences are identical.