Cryptography and security CSC8419

Cryptography and security CSC8419 - 2. Try to decrypt the...

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University of Southern Queensland 10 Cryptography and security CSC8419 Assignment 1 Tejas Patel
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Cryptography and security CSC8419 ID: W0094835 Name: Tejas K Patel Page 2 Assignment 1: 26 th March 2009 Task 1: (8/100) According to Stallings (2006, p36), ‘The Caesar cipher involves replacing each latter of the alphabet with the latter stan ding three places’ For each plain text latter p, substitute the ciphertext letter C 2 C = E(3,P) = ( P + 3) mod 26 Valu e 0 1 2 3 4 5 6 7 8 9 1 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 2 2 3 2 4 2 5 Plai n a B c D e f g h I j k l m n o p q r s t u v w x y z ciph er D E F G H I J K L M N O P Q R S T U V W X Y Z A B C 1. Try to encrypt the following plaintext with Caesar Cipher. The lazy dog catches no fleas. Encrypt above plain text using key 3 as follow: WKH ODCB GRJ FDWFKHV QR IOHDV
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Unformatted text preview: 2. Try to decrypt the following ciphertext decrypted by Caesar Cipher L DP KDSSB WR OHDUQ P= D(k, C) = (C - k) mod 26 So the decrypted text is: I am happy to learn. Cryptography and security CSC8419 ID: W0094835 Name: Tejas K Patel Page 3 Task 2: (8/100) Given p= 5 & q = 7 and (n) = (p-1)*(q-1), find the inverse of 17 mod (n), and demonstrate the inverse is correct. (n) = (p-1)*(q-1) = (5-1)*(7-1) = (4)*(6) = 24 24 = 1*17+7 17 = 2*7+3 7 = 2*3+1 So, 1 = 7-2*3 = 7-2*(17-2*7) = 5*7-2*17 = 5*(24-17)-2*17 = 5*24-7*17 (mod 24) = -7*17 mod 24 = 17*17 mod 24 So, here inverse is correct. The Inverse of 17 mod 24 is 17. Reference : Stallings, W 2006, Cryptography and network security principles and practices, 4 th edition, Pearson Education International, new jersey....
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This note was uploaded on 04/27/2010 for the course CSC 8419 at University of Sydney.

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Cryptography and security CSC8419 - 2. Try to decrypt the...

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