Bob's Knapsack public-key is (45, 87, 143,472, 75, 5, 57, 97). Encrypt the message x=179 for sending to Bob. Select one: O a. 145 O b. 265 O c. 193 O d. 257 e. 814
Q: A message authentication code, MAC, takes as input a key K and message M and outputs a tag T. In…
A: The answer is given in step 2.
Q: Alice and Bob agree to use the prime p = 1373 and the base g = 2 for communications using the…
A: The ciphertext is been decrypted below
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Q: message along with the encrypted symmetric key. Bob (Receiver): Decrypts the symmetric key using his…
A: Public Key: A public key is a cryptographic key that is used in asymmetric encryption systems. It is…
Q: im not sure whats wrong
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A: The Answer for the following question is given in next step.
Q: Bob's Knapsack public-key is (45, 4, 189, 366, 75, 45, 57, 13). Encrypt the message x=233 for…
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A: We need to find the plaintext from given key stream and cipher text.
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Q: Alice and Bob use the ElGamal scheme with a common prime q= 131 and a primitive root a = 6. Let…
A: The Answer is in Below Steps
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Q: A.Given p=25, q=9, and e=7, generate the public key (n,e) and the private key (n,d) using the RSA…
A: Solution: Given, A.Given p=25, q=9, and e=7, generate the public key (n,e) and the private key…
Q: 3. Decrypt the following RSA message. Public key: (n, E) = (91,23) Ciphertext: 76 14 58 73
A: Here, we are going to decrypt the given ciphertext. We are given a public key. With the help of this…
Q: Alice and Bob use the ElGamal scheme with a common prime q= 131 and a primitive root a = 6. Let…
A: The answer is
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- Alice and Bob agree to use the prime p = 1373 and the base g = 2 for communications using the Elgamal public key cryptosystem. Bob chooses b = 716 as his private key, so his public key is B ≡ 2716 ≡ 469 (mod 1373). Alice encrypts the message m = 583 using the random element k = 877. What is the ciphertext (c1, c2) that Alice sends to Bob?A.Given p=25, q=9, and e=7, generate the public key (n,e) and the private key (n,d) using the RSA Key generation algorithm. (Use Excel for computations) b. Given Bob's public key of (85, 7) and private key of (85, 55), show how Alice can encrypt m=3 to send to Bob. Show the ciphertext. c. Given Bob's public key of (85, 7) and private key of (85, 55), show how Bob can decrypt the ciphertext y=2 received from Alice. Show the plaintext. d. Using the fast exponentiation method, determine 530 mod 47. Show your work.dont post copied or existing one skip skip if dont know
- Alice and Bod have decided to use a symmetric encryption algorithm. They have some assumptions about their messages:- Messages only contain capital letters (i.e. A to Z)- The length of their shared key must be greater than or equal to the length of the plaintext- They assign each letter a number as follows: (A,0), (B,1), (C,2), (D,3),…, (Z,25)Their algorithm combines the key and the message using modular addition. The numerical values of corresponding message and key letters are added together, modulo 26. For example, if the plain text is “HELLO” and the key is “SECRET” then the encrypted message is calculated as following:Since the length of the plaintext is 5, we just need the first 5 letters of the key (i.e. “SECRE”), then for each letter, we should add corresponding letters in both the plaintext and the key modulo 26.Plaintext: H (7) E (4) L (11) L (11) O (14)Key: S (18) E (4) C (2) R (17) E(4)Cipher: Z (25) I (8) N(13) C(2) S (18) Write a program in Python, C/C++ or JavaScript to…You witness Alice and Bob agree on a secret key using the Diffie-Hellman key exchange. Alice andBob chose p = 97 and g = 5. Alice send Bob the number 3 and Bob sent Alice the number 7. Bruteforce crack their code: What is the secret key that Alice and Bob agreed upon? What were theirsecret keys?Let's encrypt a message using RSA: Choose p = 7 and q = 11, and then select e=13. a.Compute d b.Select plaintext message x=7. Produce the ciphertext y using the fast exponentiation algorithm. c. Decrypt the ciphertext (y) to verify that the initial plaintext (x) is produced. Again, please use the fast exponentiation algorithm.