In the RSA public-key encryption scheme, each user has a public key, e, and a private key, d. Suppose Bob leaks his private key. Rather than generating a new modulus, he decides to generate a new public and a new private key. Is this safe?
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- 5. Use RSA public-key encryption to decrypt the message 111 using the private keys n = 133 and d = 5.Bob's public key is (55,3). Alice wants to send Bob the "message" 6. What is the cyphertext (encrypted version) of her message in the RSA cryptosystem? (Show the arithmetic, and you can use a calculator.)Exercise 1 (The RSA basics ( Alice forms a public key for the "textbook" RSA public-key encryption and signature schemes, choosing N = 1073 and e = 715. 1. Encrypt the message M = 3 for Alice. 2. What is Alice's private key, and how does she use it for decrypting the received ciphertext? 3. Generate a signature on the message M = 1234567 using the following simple hash function H: {0, 1} ZN such that H(M)= 3M + 5 (mod N). 4. Verify that the signature you generated in the previous step is a valid signature.
- Question 2: 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. 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.Alice sets up an RSA public/private key, but instead of using two primes, she chooses three primes p, q, and r and she uses n=pqr as her RSA-style modulus. She chooses an encryption exponent e and calculates a decryption exponent d. Encryption and Decryption are defined: C ≡ me mod n and m ≡ Cd mod n where C is the ciphertext corresponding to the message m. Decryption: de ≡ 1 mod φ(n) | Let p = 5, q = 7, r = 3, e = 11, and the decryption exponent d = -13. n = 105 & φ(n) = 48 Q: Alice upgrades to three primes that are each 200 digits long. How many digits does n have?t) Alice and Bob are using the ElGamal cipher with the parameters p 173 and a = 2. = Alice makes the mistake of using the same ephemeral key for two plaintexts, ₁ and 2. The eavesdropper Eve suspects that x₁ 169. She sees the two ciphertexts y₁ = 153 and y2 = 135 in transit; these are the encryptions of ₁ and 2, respectively. a) What is the masking key k? = b) What is the plaintext ₂ ?
- Let's assume that Bob and Alice share a secret key in a symmetric key system. Bob wishes tobegin a conversation with Alice, and sends her a message that essentially says "let's talk". Aliceis suspicious that the person contacting her might not be Bob, so Alice generates a single-userandom number (a nonce) and sends it to Bob. Bob encrypts the nonce with the secret key heshares with Alice and sends it back. Alice decrypts this message and retrieves the same nonceshe sent to Bob. Which of the following is true, provided the secret key K has not been compromised?3. Alice publishes her RSA public key: modulus n = 27455269 and exponent e = 191. (a) Bob wants to send Alice the message m = 27453912. What ciphertext does Bob send to Alice? (b) Alice knows that her modulus factors into a product of two primes, one of which is p = 5011. Find a decryption exponent d for Alice. (c) Alice receives the ciphertext c= 5329179 from Bob. Decrypt the message.Suppose that Alice and Bob communicate using ElGamal cipher and f (p. 9. Z) is common public values. Bob generates his private key d ER Z and then computes the corresponding and public public key y=g" (mod p). To save time, Bob uses the same number r each time he encrypts a plaintext message m (ie., r is a fixed nonce of Bob, and it is not randomly generated each time encryption is performed). Assume that Alice compute the ciphertext for the message m as (cc) = (g mod p, mxy mod p). and for the message m as (1,2)=(g" mod p, xy' mod p). Show how an adversary who possesses a plaintext-ciphertext pair (m. (c.ca)) can decrypt (1, 2) without knowing the private key d of Bob.
- Question 1 Study the scenario and complete the question(s) that follow: Ceasar Cipher The Caesar Cipher technique is one of the earliest and simplest method of encryption technique. It's simply a type of substitution cipher, i.e., each letter of a given text is replaced by a letter some fixed number of positions down the alphabet. For example with a shift of 1, A would be replaced by B, B would become C, and so on. The method is apparently named after Julius Caesar, who apparently used it to communicate with his officials. 1.1 Write a Ceasar cipher algorithm in such a way that a character D is changed to N. Derive the encryption of the other characters accordingly. 1.2 Based on your algorithm, what will be the encrypted code of the message “my mother is not home". End of Question 1Suppose your RSA public key is PK: {n, e} = {13861, 37}. Your friend sends you a ciphertext C = 9908. But unfortunately you have forgotten your private key, now you have to crack it yourself.a) Write down a possible condition of factors p and q.p= q=b) What is your private key SK: {d}?d=c) What is the plaintext of your friend’s message?The plaintext M =d) Suppose the plaintext M is a 12-digit number consisting of a prefix “19” and 1234567890. What is the corresponding ciphertext? Since M is greater than n, you only need to encrypt four digits at a time. Ignore redundant zeros. e.g., 0001 = 1. The result should contain three integers.The ciphertext C0 =RSA (Rivest–Shamir–Adleman) and AES (Advanced Encryption Standard) are two different types of ciphers. You are now given two large primes, p and q, a cryptographic hash function H(x).a) DESCRIBE the RSA algorithm. In your description, you should make clear how RSA keys are generated, how a message is encrypted and how a cipher text is decrypted.