Question

Problem 2 (15 pts) In EDE mode of 3DES, encryption of message m with keys k1, k2 and k3 works as follows: C=E_k1(D_k2(E_k3(m))), where E and D denote the encryption and decryption operation of DES respectively. Given several <m, c> pairs, how can an attacker find the three keys with effort in the order of 2¹¹² instead of 2¹⁶⁸? Describe the attack in detail. (Hint: meet-in-the-middle attack)

Answer 1

The key length of 3Des is 168 bits which is 3 times more than DES whish is of 56 bits only. Making the key complexcity factor to 2¹¹² is increased, but due to the possibilty of meet-in-the-middle attack ,the attacker can make a pair of plain text and chiper text and he can attack the encryption from both sides. the total plain text ecryption key for each level is 2¹¹² and this result is adian compared with the decryption of the cipher text with all 2⁵⁶ possible keys . so overall we have 2¹¹² + 2⁵⁶ = 2¹⁶⁸ when using the brute force method.

MEET - IN - THE - MIDDLE - ATTACK : It is a generic cryptographic attack against encryption schemes which rely on performing multiple encryption operations in sequence. The MITM attack is the primary reason why Double DES is not used and why a Triple DES key (168-bit) can be bruteforced by an attacker with 2⁵⁶ space and 2¹¹² operations.

The meet-in-the-middle attack targets block cipher cryptographic functions. The intruder applies brute force techniques to both the plaintext and ciphertext of a block cipher. He then attempts to encrypt the plaintext according to various keys to achieve an intermediate ciphertext . Simultaneously, he attempts to decrypt the ciphertext according to various keys, seeking a block of intermediate ciphertext that is the same as the one achieved by encrypting the plaintext. If there is a match of intermediate ciphertext, it is highly probable that the key used to encrypt the plaintext and the key used to decrypt the ciphertext are the two encryption keys used for the block cipher.