EBK COMPUTER NETWORKING
EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Chapter 8, Problem R6RQ
Program Plan Intro

Symmetric key encryption:

The Symmetric encryption uses a symmetric key, which is a series of numbers and letters. Then this symmetric key is used to encrypt a message as well as decrypt it.

Public key encryption:

A cryptographic system that uses two keys which are: one public key known to everyone and a private key known only to the message recipient.

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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.
R6. Suppose N people want to communicate with each of N-1 other people using symmetric key encryption. All communication between any two people, i and j, is visible to all other people in this group of N, and no other person in this group should be able to decode their communication. How many keys are required in the system as a whole? Now suppose that public key encryption is used. How many keys are required in this case?
An old encryption system uses 20-bit keys. A cryptanalyst, who wants to brute-force attack theencryption system, is working on a computer system with a performance rate N keys per second. How many possible keys will be available in the above encryption system? What will be the maximum number of keys per second (N) that the computer system isworking with, if the amount of time needed to brute-force all the possible keys was 512 milliseconds? Show you detailed calculations.
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