# ierra is Alice’s roommate and she has public keys of both (PKA, PKB). She can generate a private and public key for herself. she sends her public key to Alice instead of Bob’s public key. Now whenever Alice sends any message to Bob, she will encrypt the message using (PKs)Sierra’s Public key. Now Sierra can decrypt the message sent by Alice to Bob

Problem 1)
a) Sierra is Alice’s roommate and she has public keys of both (PKA, PKB). She can generate a private and public key for herself. she sends her public key to Alice instead of Bob’s public key. Now whenever Alice sends any message to Bob, she will encrypt the message using (PKs)Sierra’s Public key. Now Sierra can decrypt the message sent by Alice to Bob
b) Alice and Bob can use MAC to share their public keys on the internet. There are two advantage of using MAC
1) An attacker cannot produce a valid MAC for a new message.
2) A receiver can detect whether there is a change in the message or not.
c) True, in IND-CPA probability-based algorithms, randomly apply some ciphertext if the encryption was deterministic, the probability the Sierra wins is ½.
Problem 2)
a) Alice and Bob create one global hash function, in which they are mutually ready for this, and every time they need to pass a dice and key to the hash function.
The hash function has one more capable of rolling a n-dice and outputs the result in the form of the hash value.
They will get the original output when they decrypt the hash value using a key.
Scenario:
-> Alice and Bob generate two key (EK, DK) Encryption key and Decryption key
Alice sends the EK to Bob.
->Bob uses key and n-Dice in the global hash function hash(Dice, EK)
Hash return one hash value (V)
Bob send the generated value(V) to Alice
->Alice Decrypt the V using DK and get a result to decrypt(V, DK)
After getting the result Alice sends the DK to Bob to get the decrypted result
The Xor(DK, EK)=0
In this protocol, everything will be secure, there is a minimum chance of cheating. Alice can’t change the result because after the decryption she has to send DK to Bob and Bob cannot see the outcome without the DK.
If Alice changes the DK to get a different value Bob has performed XOR(EK, DK) if it is not equal then they will abort the protocol.
b) No, all the values in the protocol are hashed so no one can discover the value.
c) JC can work as a host in the Protocol. She manages both parties (EK, DK).
Both parties can use the hash function and send the output to JC, JC will send the decrypted value.
Problem 3)
a)
Erika has the Primary key of both Alice and Bob.
She will receive the message but would not be able to decrypt the message.
● Alice encrypts the message using Bob’s Public key and sends it to him.
● Erica receives the message and encrypts a different message using Bob’s Public key and sends it to Bob.
● Now Bob will receive a different message which is sent by Erica.
● Bob replies to Alice, so he encrypts the message using Alice’s public key.
● Again Erica changes the message and sends it to Alice.
● Erica can’t view the original message.
The second scenario was that Erica received the Public key of Bob and Alice and shared her public key with them.
1. Alice encrypt(M, Erika PK)
2. Erica decrypt(M, Erica SK)
3. Erica changes the Message M to M2
4. Erica encrypt(M2, Bob PK)
5. Bob decrypt (M2, Bob SK)
Bob will receive an edited message.
b)
Erica is trying to do a man-middle-attack. There are many ways to prevent it. Alice and Bob have to share their public key securely, which will reduce the chances of Erica decrypting the message.
So, they can use any Hash-based system or MAC (Message Authentication Code) to share their Public key securely.
c)
Public-key encryption is very slow and not good for a chat-based application, they can use another method for higher security and a private message, their message will be an end to end encrypted.
Alice and Bob can use AES (Advanced encryption standard) for chat-based applications. They can use a message size of 128 bit and a key of 128,256 bit to generate the ciphertext and send it to each other.
Erica can read the message but she has to use Brute force methods or other technique for this
This system is very fast as compared to Public-key encryption.
d)
Erica has the same capability, she can share her public key with both parties.
Alice has APK(Alice public key)
Bob has BPK(Bob public key)
Erica has EPK(Erica public key)
After sharing the public key using the payload, Erica shares her public key with Bob and Alice. Now Alice has EPK, Bob also has EPK and Erica has BPK and APK. So she can decrypt the message and Send the altered message to both parties.
Problem 4)
a) Exceptional access is something where an unknown party enters into another system without informing the owner.
These are some points that show exceptional access reasonably.
● For national security purpose, exceptional access is reasonable
● If parties are indulging in some crime
● The situation may affect the human life
Exceptional access is against the law, Privacy is a human right.
b) We are consuming the product of a company and they are claiming that their system is highly secure but they hide the fact that the system is accessible by the government.
c) No, Because Privacy is one of the most important things in today’s era. The Government is not supposed to listen to our personal conversations until they find that we are doing some malicious activity. Any provider company can’t provide direct access to personal data.
Problem 5)
a)
b) With the help of C1 and C2. Marcus can try to predict the key used in the encryption.
By seeing the message and ciphertext he can visualize the behavior of the key and try to create the same key for the decryption.
c) Change in one bit can corrupt the corresponding plain text
d) False, it is not parallelizable every time we have to create a new ciphertext for the encryption. Then the output of one encryption will be an input of another encryption block, this is a sequential method.
e) True, it is parallelizable, we can decrypt the message parallelly, it means ciphertext decrypted in parts to increase the speed.
f) False, if the attacker determines the key but is not able to determine the initialization vector, and for every encryption, there is a different IV. an attacker can decrypt it.
Problem 6)
a) William, who knows the Primary Key of Zachary and Lilika, so he can encrypt the message using the public key on the other side, without knowing the Secret key(Private key) of Zachary.
William can perform some mathematics-based operation to decrypt the message, operations like prime factorization.
b) Yes, William can tamper the secret message using RSA, RSA is a prime based encryption method where parties select the random prime number, Private and Public key. If Wiliam selects a larger prime number then the new secret is greater than the original secret.
c) It is a public key-based asymmetric encryption technique, where we need two keys: one for encryption and the other for decryption. A forgery attack is a malicious attack on the website or any application. We will use RSA to generate the unique signature and verify it on every access.
Problem 7)
a) False, CTR is a block-based encryption method, CTR uses an initialization vector. If the IV is not random then there is a chance that the attacker can determine the plain text using IND-CPA.
b) We can increase the bit size of Key and encrypt the Plain text more than two times, it will reduce the chances of attacking. We can encrypt the message many times it will increase the level of security

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