Let's have an encryption CHALLENGE!

[Agamemnus]

From now on, can we just stick to:
key%, ciphertext% (output), and plaintext% (input) for ease of use?

oracle:
First, I'd get analyze the ciphertext string and find the average repetition length of numbers after zero. (which will be 100% len.temp$ - 7).

Then I'd remove the extras.

Then, I'd divide the string into numbers according to average.repetition.length% + 7 (or somethin like that).

Then I'd subtract 9999 from each number. Then, I'd do the same thing as I did before.

*OR*
I can even just factor a bunch of numbers and get the factor that appears in all of them. If I reach a number that has two factors and one of those factors didn't appear before, the other one must be the key.

Then I divide by the key.

*OR*
I go through every number from 1 to 256 and divide each number by that. As soon as I get a number that doesn't divide evenly (gets a remainder), I skip that test key and go on to the next. The one that passes is the one that I will use to divide each number in the number array and get the answer.

--------------------------------------------

Neo:

I changed the variables in your code:

Code:

FUNCTION KT.ENC.DifCipher$(plaintext$ AS STRING, key AS INTEGER)

'a completely new ciphering method -> Dif Ciphering
'by Neo Deus Ex Machina

'I thought of this ciphering, it's completely new ;) lol
'It uses the difference with a sequent character from the base character

base.char$ = MID$(plaintext$, 1, 1)
ans$ = base.char$
cur.key% = 1

FOR i = 2 TO LEN(plaintext$)

IF cur.key% = key THEN
cur.key% = cur.key% + 1
base.char$ = mid$(plaintext$, i, 1)
ans$ = ans$ + base.char$
ELSE
new.char$ = MID$(plaintext$, i, 1)
diff% = ASC(new.char$) - ASC(base.char$)
IF diff% >= 0 THEN ans$ = ans$ + CHR$(diff%) ELSE ans$ = ans$ + CHR$(256 + diff%)
END IF

NEXT i

KT.ENC.DifCipher$ = ans$
END FUNCTION

I'm sure it's breakable but I can't figure out how, yet.