Re: TIB: Encrytion...
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Re: TIB: Encrytion...
Christopher Robin MacDougald wrote:
>
> Encrytion now that is fun. I had a good friend in High School write a
> recursive encrytion program based off the case sensitive password... and
> like all good encrytion programs it didn't store the password in the
> file. You know you would save memory if you didn't put the password in
> but rather decrypted using the users input that way if they enter the
> wrong password it just won't work, they have to try again until it was
> of real/integer type! That would work BTW is this for the 92?
>
> Happy Encrypting,
> Christopher
All of my work was done the 83. My favorite encryption/decryption
program was the one that used a pattern of dots as a picture. My
program workied like this. The calculator would start in the top left
corner and work down. Each column represented a letter and each row
was the next letter in the message. Theata was a space, and since A=1,
B=2, C=3...it worked out perfectly that a space was represented in
column 0. Since the 83 screen is 62 x 94 pixels, this allows for three
columns each of which has 62 letters. Do the math - 27*3 is 81, 94-81
leaves 13 unused columns. To optimize the number of letters that can
be encrypted these last 13 columns are used in the same way, except
everything is reversed. In those last 13 columns the letter is
represented by the row and each letter is the next column. Anyway,
this allows another 26 letters. If you notice there is a chunk up in
the upper right that is unused. This is where the password and length
of message are recorded. Now, you might be thinking that this would be
fairly easy to figure out. At least I did. To get around this problem,
each row is shifted over by a random number. In fact every 14 rows a
pattern repeats. I know it sounds confusing, but think about it like
this.
:randInt(1,7,14->L1 --say it comes up with {3,6,3,2,4,2.....}--
Now, each number in list one is the amount that the letter is shifted.
Take, for example, my name; JODY. In numbers, this is {10,15,4,25}.
However, before displaying the pattern of dots, these numbers will be
shifted. The 10 will shift over 3, and become 13 (the letter M), the
15 will shift over 6, becoming U,...get the idea? After you run out of
numbers in L1, it starts back over at the beginning with the numbers
3,6,3,2...If the person trying to decipher the picture doesn't input
the correct password than the decryption program will not take into
account these extra shifts, and thus show only gibberish. My name
would look like this; MUG_. If the message is short then you can
imagine what the picture would look like; a small pattern of dots in
the upper left and a small pattern in the upper right. This would be a
pretty quick giveaway that one of those patterns is part of the
encryption. So, after the actual message is encoded the calc continues
with a random pattern of dots until the whole screen is full. In the
upper right, along with the encryption pattern, is also a record of
how long the message actually is. This way when the decryption program
is running, it stops at the end of the message and doesn't show all
the fill. I guess that was more than anyone wanted to know, but
perhaps someone would like to take this idea and improve on it? If you
do, tell me what you did so I can see for myself.
Jody Snider
jody1@alaska.net
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