Date: Sep 4, 2013 2:19 PM
Subject: Re: Simply Decrypting
"Ar " <firstname.lastname@example.org> wrote in message <email@example.com>...
> Trying to solve to problem, but have no idea where to start from.
If you are stuck,
(1) Forget MATLAB, first try solving it with just pencil and paper.
(2) Then use MATLAB to encode exactly what you did step-by-step.
(3) If you get stuck at a specific step, ask us a SPECIFIC question.
> One of the simplest methods to encrypt a message is by using a key that defines
> how many positions the letters in the text is replaced.
> If you encrypt a text with a key 3, all letters in the original text moves down the
> alphabet by 3 positions.
> If you decrypt a text with a key 3, all letters in the decrypted text moves up the
> alphabet by 3 positions (i.e d a ; f c; etc )
> if you encrypt the text ?trial? with a key of 3, it will become ?wuldo?
> if you decrypt the text ?wuldo? with a key of 3, it becomes ?trial?
> (a) Write an M-file to can decrypt the following message ?brx pdqdjhg wr gr
> lw.? using a key of 3.
> (b) Write an M-file that can load the text file ?encrypted_text.txt?
> i. Only alpha-characters have been encrypted. Punctuations and white
> spaces have not been altered and as such in your codes, you should not
> decrypt any punctuations or white spaces.
> ii. The alphabet has 26 letters. So your encryption key should always be
> between 1 and 26.
> iii. You might use MATLAB inbuilt function ?find?
> iv. See more hints in the lab4presentation file on moodle
What are those hints? There must be clues in there.
But we don't know them. My hints:
How are you going to handle negative numbers?
For example, if 'a' == 1, then what is the result of:
'a' - 3 = ?