Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Question
Chapter 31.7, Problem 3E
Program Plan Intro
To prove that the RSA is a multiplicative for the below given expression,
And, to calculate the probability of encrypting a message with high success rate.
Expert Solution & Answer
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Check out a sample textbook solutionStudents have asked these similar questions
One of the one-way functions used in public key cryptography is the discrete logarithm. Computing r = gº mod p from
g, e, and p is easy. But given only r, g and p, recovering e is hard.
Suppose p
1597,
2 and r =
128.
What is the smallest positive integer e such that r = ge mod p?
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.
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.
Chapter 31 Solutions
Introduction to Algorithms
Ch. 31.1 - Prob. 1ECh. 31.1 - Prob. 2ECh. 31.1 - Prob. 3ECh. 31.1 - Prob. 4ECh. 31.1 - Prob. 5ECh. 31.1 - Prob. 6ECh. 31.1 - Prob. 7ECh. 31.1 - Prob. 8ECh. 31.1 - Prob. 9ECh. 31.1 - Prob. 10E
Ch. 31.1 - Prob. 11ECh. 31.1 - Prob. 12ECh. 31.1 - Prob. 13ECh. 31.2 - Prob. 1ECh. 31.2 - Prob. 2ECh. 31.2 - Prob. 3ECh. 31.2 - Prob. 4ECh. 31.2 - Prob. 5ECh. 31.2 - Prob. 6ECh. 31.2 - Prob. 7ECh. 31.2 - Prob. 8ECh. 31.2 - Prob. 9ECh. 31.3 - Prob. 1ECh. 31.3 - Prob. 2ECh. 31.3 - Prob. 3ECh. 31.3 - Prob. 4ECh. 31.3 - Prob. 5ECh. 31.4 - Prob. 1ECh. 31.4 - Prob. 2ECh. 31.4 - Prob. 3ECh. 31.4 - Prob. 4ECh. 31.5 - Prob. 1ECh. 31.5 - Prob. 2ECh. 31.5 - Prob. 3ECh. 31.5 - Prob. 4ECh. 31.6 - Prob. 1ECh. 31.6 - Prob. 2ECh. 31.6 - Prob. 3ECh. 31.7 - Prob. 1ECh. 31.7 - Prob. 2ECh. 31.7 - Prob. 3ECh. 31.8 - Prob. 1ECh. 31.8 - Prob. 2ECh. 31.8 - Prob. 3ECh. 31.9 - Prob. 1ECh. 31.9 - Prob. 2ECh. 31.9 - Prob. 3ECh. 31.9 - Prob. 4ECh. 31 - Prob. 1PCh. 31 - Prob. 2PCh. 31 - Prob. 3PCh. 31 - Prob. 4P
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