Solve the Encryption and Decryption process using Cipher Block Channing (CBC) for the given Plaintext M=1100101111111000. Key= 3214. IV=1011.
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Solve the Encryption and Decryption process using Cipher Block Channing (CBC) for the given
Plaintext M=1100101111111000. Key= 3214. IV=1011.
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- A block cipher has block length 3. When the key is k, the results of encrypting a block with this cipher are given in the table below: m E(m, k) m E(m, k) 000 011 100 101 001 111 101 110 010 001 110 010 011 000 111 100 (i) Decrypt the ciphertext 010011011 given that it was created by using this block cipher in CBC mode, with key k and IV 111. (ii) Decrypt the ciphertext 010011011 given that it was created by using this block cipher in ECB mode, with key k. (iii) Decrypt the ciphertext 010011011 given that it was created by using this block cipher in CTR mode, with key k and initial counter value 100.RSA encryption uses a famous formula for encryption/decryption. Given that N= 12345 , decryption key d = 1961, cipher in numeric form =170826 . what is the decrypted number ?Messages are to be encoded using the RSA method, and the primes chosen are p = 13 and q = 23, so that n = pq = 299. The encryption exponent is e = 13. Thus, the public key is (299, 13). (a) Use the repeated squaring algorithm to find the encrypted form c of the message 84. m = (b) Show that the decryption exponent d (the private key) is 61. (c) Verify that you obtain the original message after decryption. Use the repeated squaring algorithm.
- The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m=4m=4.The plaintext given by 01001011=x0x1x2x3x4x5x6x701001011=x0x1x2x3x4x5x6x7 when encrypted by the LFSR produced the ciphertext 11010110=y0y1y2y3y3y5y6y711010110=y0y1y2y3y3y5y6y7. What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3=0,p2=1,p1=0,p0=1p3=0,p2=1,p1=0,p0=1).The answer above is NOT correct. Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzl. We conduct a known-plaintext attack against an LFSR. Through trial and error we have determined that the number of states is m = 4. The plaintext given by 11100010 — ХоXјX2XҙX4XsX6X7 when encrypted by the LFSR produced the ciphertext 11010110 — Уo У1 У2 Уз Уз У5 У6 Ут. What are the tap bits of the LFSR? Please enter your answer as unspaced binary digits (e.g. 0101 to represent p3 = 0, p2 = 1, p1 = 0, po = 1). 0010Set up an RSA public-key cryptosystem using primes p = 71, q = 31 and a public exponent of 191 by finding the appropriate (positive) private exponent using the extended euclidean algorithm given the table below and identities: ri = ri+1qi+1 + ri+2, si+2 = si − qi+1si+1 and ti+2 = ti − qi+1ti+1 si(a) +ti(b)=ri i ri qi si ti 0 a x 1 0 1 b 0 1 2 3 4 What is the numerical value for n? 2201 What is the numerical value for a? 2100 What is the numerical value for b? 191 What is the numerical value for z? 191 What is the numerical value for q3, S3 and w
- A symmetric block encryption algorithm is shown below. 16-bit blocks of plaintext P are encrypted using a 32-bit key. Encryption is defined as: C = (P EX_OR Ko) + K1 C is the ciphertext, K is the secret key, Ko is the leftmost 16 bits of K, K1 is the rightmost 16 bits of K, EX_OR is bitwise exclusive OR, and + is binary addition. a. The ciphertext C must be the same size as the plaintext P, that is, it must not be larger than 16 bits. How can this be achieved? b. Show the decryption equation. How will the encrypted message be decrypted?Using the concept of Cipher Block Chaining Mode, perform a Simplified DES encryption of a plaintext P=1011111110101011 given K=1111101101 and an Initialization vector 10010011Answer 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…
- 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.Set up an RSA public-key cryptosystem using primes p = 91, q = 73 and a public exponent of 997 by finding the appropriate (positive) private exponent using the extended euclidean algorithm given the table below and identities: ri = ri+1qi+1 + ri+2, si+2 = si − qi+1si+1 and ti+2 = ti − qi+1ti+1 si(a) +ti(b)=ri i ri qi si ti 0 a x 1 0 1 b 0 1 2 3 4 What is the numerical value for n? 6643 What is the numerical value for a? 6480 What is the numerical value for b? 997 What is the numerical value for z? 997 What is the numerical value for q3? 7 What is the numerical value for s3? -2 What is the numerical value for w? 13 is my calculated value right for the one in bold?Encrypt a specific message (see below) using RSA with your encryption key: Let M be the numerical portion of your B-number, minus leading zeroes (for example, B00123456 has numerical portion 123456), plus the constant 3100. 3100 is 515377520732011331036461129765621272702107522001. 1. Compute C=Me (mod n); 2. Compute the “signature” S = Md (mod n) Provide the values C and S. B-number is B00123456