3- Z intercepts the message sent from A to B. Z then sends B the
message (Z, EB (M), B). B will answer Z according to the
protocol by sending message (B, EZ (M), Z). Z can now decode
EZ (M) to find M.
4- B can fool A by replaying him/her with same nonce N1.
1. A → B: N1
2. B → A: N1
3. A → B: EK[N1]
4. B → A: EK[N1]
5- Two situations arise, the first in which Eve, the attacker, does not send an initial
message to Bob (steps 1 and 2), and the second where she does. In both scenarios,
assume that Eve knows the session key k session , and has intercepted the message at step
5. Begin with the first case. Eve replays the message to Bob. As Bob has not yet received
a new message from Alice (steps 1 and 2), he rejects the message. But if Eve’s message
comes during Alice’s execution of the protocol and after step 2, Bob opens the message
and determines the random number rand 3. Because he kept track of the random number
that he sent to Alice, he recognizes that this was not an attempt to begin a new session,
but instead a replay of an older message (else it could not have been enciphered using the
key he shares with Cathy, k Bob). He therefore knows the message is legitimate but, since
he has already seen the message with that random number in it, that it is a replay.
In the second case, Eve sends message 1 to Bob, who replies with message 2. Eve
immediately sends message 5 to Bob. Bob opens the message, compares rand 3 with the
nonce in the message he just sent Eve (masquerading as Alice), and notes it is different.
So he rejects the message.
7- a. n = 33; (n) = 20; d = 3; C = 26.
b. n = 55; (n) = 40; d = 27; C = 14.
c. n = 77; (n) = 60; d = 53; C = 57.
d. n = 143; (n) = 120; d = 11; C = 106.
e. n = 527; (n) = 480; d = 343; C = 128. For decryption, we have
128343 mod 527 = 128256 12864 12816 1284 1282 1281 mod
527
= 35 256 35 101 47 128 = 2 mod 527
= 2 mod 257
8- 3031
9- If each user wants to communicate with N other users, then each pair of users must
have a shared symmetric key. There are N*(N-1)/2 such pairs and thus there are N*(N-
1)/2 keys. With a public key system, each user has a public key which is known to all,
and a private key (which is secret and only known by the user). There are thus 2N keys
in the public key system.
