Given the following Diffie-Hellman scheme, is an attack possible? Alice to Bob-[Tm Alice'lasce mod p Bob to Alice-gmod p Lisencryption with specified PUBLIC key Hisencryption with specified PRIVATE key.
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Q: Show the encryption using Affine Cipher with a = 20 and B=6 as plaintext letter g and r yield…
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Q: Alice and Bob use the ElGamal scheme with a common prime q=131 and a primitive root a=6. Let Bob's…
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Q: Identify the type of cryptanalysis attack - Knows Encryption Algorithm + Cipher text + one or more…
A: Identify the type of cryptanalysis attack - Knows Encryption Algorithm + Cipher text + one or more…
Q: Note: The notation from this problem is from Understanding Cryptography by Paar and Pelzi. We…
A: Explanation: The plain text is given by 01110111 the ciphertext is given by 00001101. 01110111…
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Q: Alice and Bob use the ElGamal scheme with a common prime q= 131 and a primitive root a = 6. Let…
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Q: Alice and Bob use the ElGamal scheme with a common prime q=131 and a primitive root a=6. Let Bob's…
A: Key generation Common prime q=131 primary root = 6 Bob's public key =3 Random integer chosen k=…
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Q: Suppose Alice has a public key with modulus 815153 and encryption exponent 91. Eve intercepts a…
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- Alice uses RSA to send a key to Bob for use in encryption of future messages. The key is the word "CINEMA" and the encoding from letters to numbers is done using the ASCII: A = 65... Z = 90, space = 99. The public parameters (n, e) = (9379, 11) are shared. Encrypt Alice's key word, using 2-letter blocks. Please show work.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.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?
- Let's assume that Bob and Alice share a secret key in a symmetric key system. Bob wishes tobegin a conversation with Alice, and sends her a message that essentially says "let's talk". Aliceis suspicious that the person contacting her might not be Bob, so Alice generates a single-userandom number (a nonce) and sends it to Bob. Bob encrypts the nonce with the secret key heshares with Alice and sends it back. Alice decrypts this message and retrieves the same nonceshe sent to Bob. Which of the following is true, provided the secret key K has not been compromised?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 01110010 = 0X1X2X3X4X5X6X7 when encrypted by the LFSR produced the ciphertext 11000011 = YOY1Y2Y3Y3Y5Y6Y7- 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).Consider an RSA key set for Alice with p = 23, q = 17, n = 391 and e = 15.a) Her public key is (e, n) = (15, 391). Is her private key (d, n) = (47, 391)? Justffy your answer. b) Suppose Bob wants to encrypt a message 90 for Alice using RSA keys for confidenVality. Whatis the corresponding cipher text?Justify your answer.
- 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. Please type answer note write by hend.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.Consider an RSA key set for Alice with p = 23, q = 17, n = 391 and e = 15.a) Her public key is (e, n) = (15, 391). Is her private key (d, n) = (47, 391)? JusVfy your answer. b) Suppose Bob wants to encrypt a message 90 for Alice using RSA keys for confidenVality. Whatis the corresponding cipher text?Justify your answer.
- 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). 00103. Alice publishes her RSA public key: modulus n = 27455269 and exponent e = 191. (a) Bob wants to send Alice the message m = 27453912. What ciphertext does Bob send to Alice? (b) Alice knows that her modulus factors into a product of two primes, one of which is p = 5011. Find a decryption exponent d for Alice. (c) Alice receives the ciphertext c= 5329179 from Bob. Decrypt the message.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).