For the given parameters 'P' = 3 and 'Q' = 19 find the value of 'e' and 'd' using RSA algorithm and encrypt message 'M' = 6.
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- Using C programming language: A Transposition Cipher A very simple transposition cipher encrypt(S, N) can be described by the following rules: If the length of S is 1 or 2, then encrypt(S, N) is S. If S is a string of N characters s1 s2 s3... sN and k = N/2, then encrypt(S)= encrypt(sk sk−1... s2 s1 ,K)+ encrypt(sN sN−1... sk+1 ,N - K) where + indicates string concatenation. For example, encrypt("Ok", 2) = "Ok" and encrypt("12345678", 8) = "34127856". Write a program to implement this cipher. The input is a file that is guaranteed to have less then 2048 characters. Code structure might look like this: #define MAX_SIZE 2048char text_buffer[MAX_SIZE];int main(){ // read file into text_buffer encrypt(text_buffer, n); // print out text_buffer return 0;}In an RSA system, the public key (n,e) of a given user is (323, 11). 1. What is the value of the exponent in the private key (n, d), of this user? 2. Suppose you want to send this user the message m = 45, write down the expression to generate the ciphertext for this message. 3. Suppose that your ciphertext, c, is 5, write down the expression to generate the plaintext matching this ciphertext. 4. Can you encrypt the message m=322 with this public key? O a. Yes O b. NoSuppose the parameters for an instance of the RSA cryptosystem are N = 26671, e = 3, and d 17563. = Decrypt the message y Decrypt the message y = 6666. = 1576.
- IH A cryptography student has accidentally used the same keystream to encrypt two different messages, M1 and M2. We know that the ciphertexts are C¡ = 0x2e38ac20ƒe51b68c and C2 = 0x638cd0bla9114b4d. We also happen to know that M1 = 0x2b1d510e6a67ae98. a) What is M2 in hexadecimal? Please give your answer a leading Ox and use lowercase letters only. b) What is the keystream? Please give your answer a leading Ox and use lowercase letters only.Given is the following string of ciphertext which was encrypted with substitution cipher. asvphgyt The encryption rule is given as C= (M + K) mod 26, where C is the ciphertext, M is the plaintext and K is the key. We assume that the plaintext is in English. You know that the first plaintext letter is a w. Decrypt the message.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
- A cryptography student has accidentally used the same keystream to encrypt two different messages, M1M1 and M2M2. We know that the ciphertexts are C1=0x4ab397cf20db1ce0C1=0x4ab397cf20db1ce0 and C2=0x2f7fded503744af5C2=0x2f7fded503744af5. We also happen to know that M1=0xbc66dac8daeee8e2M1=0xbc66dac8daeee8e2. a) What is M2M2 in hexadecimal? Please give your answer a leading 0x and use lowercase letters only. b) What is the keystream? Please give your answer a leading 0x and use lowercase letters only. Hint: These are all 8 byte numbers and will fit in an unsigned long type on most systems.Two aspirants talking to each other use the RSA algorithm to encrypt their messages. They encrypt the message character by character. The value of p, q and d are 5, 17 and 13 respectively, where p, q and d are their integers having usual meaning in the RSA algorithm. Identify the sum of integers in cipher text for corresponding characters in plain text: "IIT". Assume that corresponding cipher characters are placed in their corresponding plain text character places. Also each character is converted to ASCII value before applying RSA (ASCII value of A, B, C.... and so on are 1, 2, 3,..., respectively).Encrypt the animal name marmot using the letter to number correspondence abc de f 8hijk 1 mnopar stuvw x y 2 e 1 2 3 4 5 6 7 8 9 18 11 12 13 14 15 16 17 18 19 28 21 22 23 24 25 and the encryption function e :n + (n +1) mod 26 The answer will be a nonsense word (no numbers).
- 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…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.Suppose that the encryption function mod 26, and , encrypt the letter “F” by “A”. Use the function to decrypt the message “ZSVXO”