Apr 1999 SC223: Computer Security (Question 2)

(a)	(i) What is a block cipher?	[1]
	A block cipher is an technique of encryption the plain text in character of more than 2, such as having to encrypt 2 or 3 or more character at a time.
	(ii) Name one example of a block cipher.	[1]
	Transposition cipher
	(iii) Name one advantage and one disadvantage of block ciphers.	[2]
	One advantage of block cipher is that it is difficult to cryptanalysis thereby not easy for hacker to know the plain text. One disadvantage of block cipher is that it may generate error due to the fact that the plain text are encrypted in block and it is difficult to find out the errors.
(b)	Define the following terms.
	(i) Monoalphabetic cipher.	[2]
	Monoalphabetic cipher is an encryption technique that uses the shifting of the original alphabet to a different order but the position of the original alphabet remain. In another word, which mean replacing the original fixed alphabet which another alphabet but position remains. Example: n= 3 C = E(pi) = pi+3 APPLE -> DSSOH
	(ii) Polyalphabetic cipher.	[2]
	Polyalphabetic cipher is an encryption technique that allows replacing of the original alphabet replaced with more than one alphabet.
(c)	A Caesar cipher was used to produce the following ciphertext message. AOPZ PZ AOL LUK VM AOL WYVISLT Use the most efficient cryptanalysis you can to deduce the plaintext message.	[4]
	The plain text for the cipher text message is THIS IS THE END OF THE PROBLEM The answer was deduce by shifting every aplabet by positive 19 which mean n = +19. In the ciphertext message given, there is a high frequency usage of A, O, I , E and this has given the biggest hint to its hacker. From this analysis, the above answer can be deduce.
(d)	(i) Perform encryption and decryption using the RSA algorithm, where p = 5, q = 11, e = 3, and M = 9. Show each step in detail.	[7]
	RSA algorithm p = 5 q= 11 e = 3 m= 9 n=pq m = (p-1)(q-1) = 5(11) = (5-1) (11-1) = 55 = 40 Given that e=3 where e is between 3 to m with no common factor to m(40). (DE) = (? m) + 1 273 = (2 40) + 1 81 = 81 D=27 C =P^E mod n (encrypted) = 9³ mod 55 =14 P =C^D mod n = 14²⁷ mod 55 (decrypted) = 9
	(ii) In the RSA algorithm, n is usually chosen to be the product of two large prime numbers. Why is this?	[1]
	Because such a number is difficult to factorize.