(1) Find a quote that you like. Make sure that the total number of characters is at least 50 and no more than 100.  (2) Then pick a number that is between 4 and 24 as your additive key.  (3) Create a table of correspondence of 26 alphabets in plaintext and the ciphertext.  (4) Use the table of correspondence to encode the quote you found in Part (1). Post the encrypted message:For example, here is a quote from John Dewey Failure is instructive. The person who really thinks learns quite as much from his failures as from his successes. We will use additive key 17. The table of correspondence between plaintext and ciphertext is as follows.(attached)      Thus the encrypted message is WRZ CLI VZJ ZEJ KIL TKZ MVK YVG VIJ FEN YFI VRC CPK YZE BJC VRI EJH LZK VRJ DLT YWI FDY ZJW RZC LIV JRJ WIF DYZ JJL TTV JJV J (5) Pick an encrypted message from your classmate and try to decrypt it. Remember to use the frequency table because you will not have access to the additive key. Make sure to include all the necessary work including any failed attempts. Do not use any “secret helper” that will “magically” decrypt the message for you. If you get really stuck, ask your classmate which additive key they use.

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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(1) Find a quote that you like. Make sure that the total number of characters is at least 50 and no more than 100. 


(2) Then pick a number that is between 4 and 24 as your additive key. 


(3) Create a table of correspondence of 26 alphabets in plaintext and the ciphertext. 


(4) Use the table of correspondence to encode the quote you found in Part (1).

Post the encrypted message:For example, here is a quote from John Dewey
Failure is instructive. The person who really thinks learns quite as much from his failures as from his successes.
We will use additive key 17. The table of correspondence between plaintext and ciphertext is as follows.(attached) 

 

 

Thus the encrypted message is
WRZ CLI VZJ ZEJ KIL TKZ MVK YVG VIJ FEN YFI VRC CPK YZE BJC VRI EJH
LZK VRJ DLT YWI FDY ZJW RZC LIV JRJ WIF DYZ JJL TTV JJV J
(5) Pick an encrypted message from your classmate and try to decrypt it. Remember to use the frequency table because you will not have access to the additive key. Make sure to include all the necessary work including any failed attempts. Do not use any “secret helper” that will “magically” decrypt
the message for you. If you get really stuck, ask your classmate which additive key they use.

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