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Cryptography and Architectures for Computer Security Exam Code: 095947 (old 090959), A.Y. 2015–2016, Semester: 2 Prof. G. Pelosi F ebruary 9th, 2017 – Exam Session Name: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Surname: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Student ID: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Signature: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Time: 2h:30’. Use of textbooks, notes, or Internet connected devices (including smart- phones) is not allowed. The usage of simple calculators is allowed. Prior to turn in your paper, write your name on any additional sheet and sign it. Question 1 [2 pts] In order to accelerate the computation of the RSA encryption function, it is popular to use e=3 as a public exponent. Although RSA is considered to be a secure public-key cryptosystem, both the implementations and wrong parameter configurations of RSA can nullify the security of the system. Consider an RSA cryptosystem with 2048-bit modulus, public exponent e=3 and assume each plaintext message to be encoded as a binary string with at most 128-bit length. Explain why this is completely insecure. Solution: employing 128-bit plaintext messages, every ciphertext c≡nm3 will be computed as an integer number which is smaller than the modulus n. Indeed, c will be encoded with only 128· 3=384 bits, thus no modular reduction will occur during the computation. A simple way to reverse the effect of the encryption transformation, without knowing the secret key, is computing the plaintext value m as c1/3. Question 2 [3 pts] (a) What is the main advantage of the One Time Pad cipher and why is it hard to use in practice? (b) How do we…