Question

1. A student recently invented a simple protocol for sending a message securely from A to B by preserving the confidentiality of the message. In his protocol, A and B do not have a common, shared secret/key. Instead, for each message, both A and B generate random numbers, namely R_A and R_B, respectively. The protocol is described as follows, where “->” means that the message is sent over a network (e.g., the Internet). Adversaries have the capability to eavesdrop all messages exchanged between A and B.

1. A -> B: M1       (M1 = M XOR R_A)
2. B -> A: M2       (M2 = M1 XOR R_B)
3. A -> B: M3       (M3 = M2 XOR R_A)

Show how an adversary capable of eavesdropping messages exchanged between A and B can recover the original message M.

Answer 1

B receives message M1 which is XOR of M and random key of A RA

M2 = M1 XOR RB

M1 = M XOR RA

Substituting this value in M2 -> M2 = (M XOR RA) XOR RB

M2 XOR RB = (M XOR RA) XOR (RB XOR RB)

                     = (M XOR RA) XOR 0

                     = M XOR RA

Again using XOR operator on RA, we get

M2 XOR RB XOR RA = M XOR (RA XOR RA) = M XOR 0

And M2 XOR RA = M3, therefore we get M2 XOR RB = M, thus we are able to recover the message M