Question

Question 2:

1. You are Alice. Bob publishes his ElGamal public key (q, a, ya) = (101, 2, 14). You desire to send the secret message “CALL ME” to Bob. Using the equivalence A = 01, B = 02, and so on up to Z = 26, you encode the message into the number 03 01 12 12 13 05. Regarding each of these two-digit numbers as a plaintext block, compute the message that you will send to Bob using his public key. This requires you to pick a “random” number k; use k = 32.

2. You are Bob. You get a message from Alice. You like Alice a lot, so you are eager to read the message. Use your secret key (101, 2, 10) to decrypt Alice’s message. Notice that you don’t need to know what value of k Alice used in order to do this.

Answer 1

The public key to encrypt message (101,2,14).

Now we calculate the encrypted message for each plaintext block using k=32.

Encrypted message is pair (c1,c2) for each plaintext.

c1=g^kmod p and c2=m*(g^a)^k mod p

Here g=2, p=101 and g^a=14

03 : c1= 2³²mod 101= 68 and c2= 3* (14)³²mod 101 =83 so (68,83)

Using calculation as above:

01 : k=66 (13,87)

12:k=5 (32,89)

12 : k= 45 (41,89)

13:k=17 (75,78)

05: k=67 (26,30)

Now to decrypt the message we have private key 10.

Message decrypted by:

m= c2 * (c1)^-xmod p

here x=10 and (c1)^-x is basically mod inverse p of (c1)^x

Using this message is decrypted as:

(68,83)= mod inverse p of (c1)^x =84 ,m=(84*83)mod 101=3

(13,87)= mod inverse p of (c1)^x =36 ,m=(36*87)mod 101=1

(52,89)= mod inverse p of (c1)^x = 95,m=(95*89)mod 101=12

(41,89)= mod inverse p of (c1)^x = 100,m=(100*89)mod 101=12

(75,78)= mod inverse p of (c1)^x = 17,m=(17*78)mod 101=13

(26,30)= mod inverse p of (c1)^x = 17,m=(17*30)mod 101=5

So message decrypted= 03 01 12 12 13 05