3 ### Cryptography Exercise1. **Affine Cipher Decryption Task** - An intercepted encrypted message reveals through frequency analysis that 'R' is the most common letter and 'X' is the least in the ciphertext. - **Ciphertext**: SMRER TL CJ LQJJC - Task: Use the frequency analysis information and the affine cipher method to decrypt the message.2. **Vigenère Cipher Decryption Task** - **Ciphertext**: MS. RZ MS QWC. KKMAI LA WS WZH. - **Keyword**: JEDI - Task: Determine the plaintext message using the Vigenère cipher and the given keyword.3. **RSA Decryption Task** - **Ciphertext**: 1259 0854 1647 1139 - **Public Key**: (2021,97) - Task: Decrypt the message using the RSA algorithm and the provided public key.4. **ElGamal Encryption Task** - Using the ElGamal cryptosystem, create and encrypt a message that is at least eight alphabetic characters long. - **Block Requirement**: Convert the plaintext into two-digit blocks. - **Public Key**: (89,14,8) - Task: Use a unique ephemeral key (j) to encrypt each block.5. **Vernam Cipher and Baudot Code Task** - **Ciphertext Lines**: - 01011 11010 00110 10110 00101 01100 10101 10011 01111 10011 - 00001 01101 10110 00001 11000 01000 00000 01101 10101 10011 - Task: Determine the plaintext message using the first line of Chapter 2 in the textbook as the key.6. **Additional Task** - **Ciphertext Specified**: Implicit from the context, potentially one of the above tasks requires further deciphering.### Explanation of Graphs or DiagramsThere are no graphs or diagrams in the provided tasks. Each task focuses purely on textual cryptographic challenges.

Here, N = 16 Then possible φ(n) = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 But the No. of co-primes to n = 1, 3, 5, 7, 9, 11, 13, 15 Hence, φ(n) = 8 We are given 2 public keys (2021, 97). Here, e = 2021 and n = 97 We have to calculate d (private key) Hence, d * e mod φ(n) = 1 => d * 2021 mod 8 = 1 Hence, d = 5 5 * 2021 = 10105 mod 8 = 1Now, for decryption, we have the formula. M = Cd mod n M(1) = 15 mod 97 = 1M(2) = 25 mod 97 = 32M(3) = 55 mod 97 = 21M(4) = 95 mod 97 = 73M(5) = 05 mod 97 = 0M(6) = 85 mod 97 = 79M(7) = 55 mod 97 = 21M(8) = 45 mod 97 = 54M(9) = 15 mod 97 = 1M(10) = 65 mod 97 = 16M(11) = 45 mod 97 = 54M(12) = 75 mod 97 = 26M(13) = 15 mod 97 = 1M(14) = 15 mod 97 = 1M(15) = 35 mod 97 = 49M(16) = 95 mod 97 = 73Hence, the Plain Text is : M = 1322173 0732154 1165426 114973 Hope it helps !!