We have the following encryption scheme for message space . Gen will uniformly choose the key k from the key space . Enc(k, m) returns and Dec(k, c) returns .
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A: The answer of the following is given below.
We have the following encryption scheme for message space .
Gen will uniformly choose the key k from the key space .
Enc(k, m) returns and Dec(k, c) returns .
Is EAV secure (computationally indistinguishable)?
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- 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).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 11000101 — ХоXјX2XҙХ4X5X6X7 when encrypted by the LFSR produced the ciphertext 10010000 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).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). 0010
- 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).Consider the following encryption scheme for a block cipher (using the notationfrequently used in class).y1 = ek(x1 ⊕ IV), y2 = ek(x2), y3 = ek(x3), etc.a. Suppose IV is a block consisting entirely of zeros. Which mode ofoperation does the described mode reduce to?b. Describe the corresponding decryption scheme.c. Can encryption and/or decryption be run in parallel for this scheme?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.
- 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.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?
- (b) Suppose we have public key encryption system with E the encryption function and Dr the decryption function, where Er is publicly known for a key k and Dx is secret. Suppose further that the composition of the operations E and Dµ is com- mutative: Er o Dr = Dx o Ex. Propose a simple way to implement authenticating messages using components of this encryption system. Answer: to send a signed message x we compute and send:Design a cryptographic solution for implementing the dinning cryptographers protocol. You should consider the following: The secret messages shared between any two agents in the protocol are communicated over public channels only. The announcements made by each agent must be traceable in a trustworthy way to that agent. Your solution should be a new protocol that addresses the above two requirements.Suppose the two prime numbers p = 23 and q = 37 are used for an RSA encryption scheme. What are the values of the connection key n, the encryption e and the decryption key d, given that d < e? e = d =