Alice and Bob are using the ElGamal cipher with the parameters p = 173 and a = = 2. Alice makes the mistake of using the same ephemeral key for two plaintexts, ₁ and 2. The eavesdropper Eve suspects that x₁ = 169. She sees the two ciphertexts y₁ = 153 and y2 = 135 in transit; these are the encryptions of ₁ and 2, respectively.
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- that the ciphertexts are C = 0xbcc61793955b5eb5 and C2 = 0xdea2c2 fd718d3b77. We also happen to know that M1 A cryptography student has accidentally used the same keystream to encrypt two different messages, M1 and M2. We know Оx805d3а2029c0а386. a) What is M2 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 Ox and use lowercase letters only. Hint: These are all 8 byte numbers and will fit in an unsigned long type on most systems.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).Alice and Bob are using the ElGamal cipher with = : 89 and a = 3. the parameters p Suppose Alice is observed sending public ephemeral key KE 7 and Bob sends public key = 44. = Their private keys are i a) What is the masking key km? = 81 and d = b) What is the encryption of x = 31? 28, respectively.
- Alice and Bob are using the ElGamal cipher with the parameters p = 199 and a = 3. Their parameters are small and we decide to crack the cipher. Suppose Alice is observed sending public ephemeral key k = 103 and Bob sends public key = 136. ß Alice is observed transmitting the ciphertext y = 68. a) What is the masking key km? b) What is the plaintext?In the El Gamal cryptosystem, Alice and Bob use p = 17 and α = 3. Bobchooses his secret to be a = 6, so β = 15. Alice sends the ciphertext (r, t) = (7, 6). Determinethe plaintext m.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=0x42e8ede51496d10cC1=0x42e8ede51496d10c and C2=0xa5ec5666da7419d0C2=0xa5ec5666da7419d0. We also happen to know that M1=0x713dc2f31c1e6c87M1=0x713dc2f31c1e6c87. 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.Let Π = (Gen,Enc,Dec) be a private-key encryption scheme that has indistinguishable en- cryptions in the presence of an eavesdropper. Which of the following encryption schemes are also necessarily secure against an eavesdropper? If you think a scheme is secure, sketch a proof, if not, provide a counterexample. Here, for a bit string s, parity(s) is 1 if the number of 1’s in s is odd, and 0 otherwise. The || symbol stands for concatenation. So, for strings if x = 00 and y = 11, x||y = 0011. (a) Enc1k(m) = 0||Enck(m) (b) Enc2k(m) = Enck(m)||parity(m) (c) Enc3k(m) = Enck(m)||Enck(m) (d) Enc4k(m) = Enck(m)||Enck(m + 1). Here think of m as an integer. (a) Enc1k(m) = 0||Enck(m)(b) Enc2k(m) = Enck(m)||parity(m) (c) Enc3k(m) = Enck(m)||Enck(m)(d) Enc4k(m) = Enck(m)||Enck(m + 1). Here think of m as an integer.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?
- Alice sets up an RSA public/private key, but instead of using two primes, she chooses three primes p, q, and r and she uses n=pqr as her RSA-style modulus. She chooses an encryption exponent e and calculates a decryption exponent d. Encryption and Decryption are defined: C ≡ me mod n and m ≡ Cd mod n where C is the ciphertext corresponding to the message m. Decryption: de ≡ 1 mod φ(n) | Let p = 5, q = 7, r = 3, e = 11, and the decryption exponent d = -13. n = 105 & φ(n) = 48 Q: Alice upgrades to three primes that are each 200 digits long. How many digits does n have?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 01100101 —D ХоXјX2XҙX4X5X6X7 when encrypted by the LFSR produced the ciphertext 10010100 — 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, Pi = 0, po = 1).The operator of a Vigenere encryption machine is bored and encrypts a plaintext consisting of the same letter of the alphabet repeated several hundred times. The key is a seven-letter English word. Eve knows that the key is a word but does not yet know its length. What property of the ciphertext will make Eve suspect that the plaintext is one repeated letter and will allow her to guess that the key length is seven? What will the number of matches be for the different displacements?