Q.14 Encrypt the message "PAY" using Hill cipher with the following key matrix and show the decryption to get the original plain text.
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- ] You receive a message that was encrypted using the RSA system with public key (43, 143), where 43 is the public exponent. A message encrypted using this cryptosystem was captured by an enemy and consisted of the following numbers: 6 82 132 115. Can you decipher what was the original message? (Hint: As a first step, you need to find p and q.)Encrypt the word WEST using p=53 and q=71. Choose e=3. (Map A - Z to 126 and perform the encryption one letter at a time.)b) Find the inverse of the encrypting function f (x) = (5 – x) mod 26,0 < x < 25, and use it to decrypt the message "ZRQUBFNB". -
- You are given the following parameters in RSA cryptosystem: p = 11, q = 13, e = 13 a) Find the integer d. b) List down the public key and private key. You need to encrypt the plaintext message, M = OF. c) Convert this message to value according to ASCII table. To encrypt this message, you can either combine the value, or separate it into the blocks of 2 digits (Hint: it must satisfy the condition M < n). d) Encrypt and decrypt the message OF. e) If the decimal value (not character) of ciphertext is 188267, decrypt this value and convert it to an appropriate message according to the ASCII table.Q3. Given the design of Feistel cipher of DES encryption, assume that the function simply ORs Right block with the key. Show the result of encryption if the original plain text is first two characters of your first name and the key is (D4)16. Plaintexta b с d e fghijklmnopqrstuvwxyz Ciphertext— A|B|C|D E F|G|H|IJKLMNOPQRSTUVWXYZ Value 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 251. Encrypt the message “Take your bag with clothes” using the Hill cipher with the key !3 23 5 <-it is matrixShow your calculations and the result. 2.This problem explores the use of a one-time pad version of the Vigenere cipher. In thisscheme, the key is a stream of random numbers between 0 and 26. For example, if the key is 189 3…, then the first letter of plaintext is encrypted with a shift of 18 letters, the second with ashift of 9 letters, the third with a shift of 3 letters, and so on.Encrypt the plaintext “cryptography” with the key stream10 22 5 4 1 0 2 9 18 16 16 0 3.Does the set of residue classes (mod3) form a group with respect to modular addition?4. Determine -11 mod 95. Determine gcd(65042, 40902)6.Using the extended Euclidean algorithm, find the multiplicative inverse of 1279 mod 9721 Please write the correct answer. Thank you!
- I need help with the following problem: A ciphertext was generated using an affine cipher. The most frequent letter in the ciphertext is “C” and the second most frequent letter is “V”. Break the code. Assume that the most frequent letter in English is “E” and the second most frequent letter is “T”.Encrypt The following correspondence between letters and numbers is used in the Caesar cypher. a b d e f i k lm n o r u V W X y z 1 2 3 4 6 7 8 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Encrypt the animal name zebra with the Caesar cypher with a key of 25. What is the encrypted animal name (enter a string of letters with no spaces)? Save Submit You have used 0 of 1 attempt Decrypt An animal name was encrypted with the scheme described above. The encrypted animal name is azmchbnns . What is the animal name? SaveThere are other modes of block cipher besides the ones (OFB,CFB,CTR). One of these modes is named Plaintext Block Chaining (PBC) Mode. On the encryption side, the following is executed to obtain the nth ciphertext: Cn := Ek(Mn)XOR Mn-1. Suppose that we need to encrypt M1; : : : ;M5 using the PBC mode. Show the explicit formulas to obtain C1; : : : ;C5. What do you need to use for M0? Also, show the steps on the decryption side to obtain M1; : : : ;M5.
- 2. Encrypt the word SHIELD using an alphabetic Caesar shift cipher that starts with a shift of 8 (A to I) and shifts one additional space after each character is encrypted. Original A BC|DE F GHI J KLMN 0PQR STUV WXY Z Maps to I J KLMNOPQRS TUVW XYZ ABCDE FGH Write the encrypted message for the word SHIELD:Answer ALL questions. The ciphertext message below was encrypted using affine transformation C = P + k(mod 26),0 ≤ C ≤ 25. YFXMP CESPZC JTDBF PPYZQX LESPX LETND a) By completing the following table, find the most frequently occurring letter in the ciphertext. Letter Number of Occurrence Question (b) is based on the information in tables below and answer in Question (a). A B C 1 2 3 Numerical 0 equivalent Frequency 7 (in %) Letter Numenical equivalent Frequency 8 (in %) N 13 ABCD E FGHIJKLMNOPQRSTUVWXYZ 1 0 14 7 P 15 3 D 4 Q 16 <1 E 13 13 R 17 8 F 5 3 S 18 6 G 2 T 19 9 H L 3 U 20 3 8 V 21 1 JK 9 <1 W 22 10 <1 LM <1 11 4 X Y 23 24 2 12 3 Z 25 <1 The tables show that the most frequently occurring letters in English text are E, T, N, R, I, O and A, with E occurring substantially more than the other letters. b) Determine what is the letter that represents C and P in affine transformation C = P + k(mod 26),0 ≤ C≤ 25. Then, show that the value of k = 11(mod 26). c) What is the plaintext…Use symmetric ciphers to encrypt message "“promise" and decrypt message "FOG". The representation of characters in modulo 26 is described as follows: Plaintext - a bcdefg|hijk1 mnopar stuvwx y z Ciphertext A BCD|EFG|HIJKL|M|NO|P|Q|R s T|U|VwxY|Z Value 00 01 02|03 04 05 06|07|08|09|10|11 12|13|14|15 16|17 18|19 20 21 22 23 24 25 The mathematical equations for encryption and decryption can be described as follows: Encryption Ea:i→i+kmod 26 Decryption Da : i→i-k mod 26 i represents the messages (plaintext or cipher), k represents a symmetric key. In this case k=20