Answer: D
3DES is also known as triple DES because here we will do DES 3 times. Here we will use 3 keys. Let us consider K1, K2, K3 are 3 Keys.[
C : First Encrypt with K1, Decrypt with K2 and then Encrypt with K3.
therefore, C= EK3[DK2[EK1[P]]]
D : Here, first Decrypt with K3, Encrypt with K2 and then Decrypt with K1.
therefore, P= DK1[EK2[DK3[C]]].
Solution:
The expression which denotes the encryption and decryption of 3DS Algorithm :
d)
C = EK3[DK2[EK1[P]]] and
P = DK1[EK2[DK3[C]]]
Question 1 10 pts Suppose the DES encryption is denoted by C EK P], while the...
Problem 2 (15 pts) In EDE mode of 3DES, encryption of message m with keys k1, k2 and k3 works as follows: C=Ek1(Dk2(Ek3(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)