Citadel Encryption

ID:188202
 
Jun 13 2004, 12:16 pm
In a few minutes I wrote a simple encryption system, and I want to know whether or not to expand on it. Here is an encrypted sentence:

242049214418291746121313511356164220131359213822451746205620 671946185212 2616341825135313461222141913 502257176022 571928142012 54165319311521173414

Each block of numbers represents a word. They are all real words. The numbers follow a definite pattern, so once you find that it will work the whole way through. If you can decrypt it, you get a cookie.

Good luck.
Jun 14 2004, 4:31 am
Well, it seemed pretty effective. No one I talked to came close to figuring it out, so I'll explain how it works:

Each letter consists of four digits, which was easy to see just by finding the GCF. The first letter in the earlier code was "2420".

Take the last two digits of that letter (20) and subtract 2 (18). Now subtract that number from the first two letters (24-18=6) and divide the difference by two (3). That'll be the letter (C).

I plan to expand on this quite a bit, mainly by making it alpha-numeric. You can go ahead and decrypt that if you feel bored, that trick will work the whole way through.
Jun 14 2004, 4:34 am
In response to HavenMaster
HavenMaster wrote:
Well, it seemed pretty effective. No one I talked to came close to figuring it out, so I'll explain how it works:

I told you already, I knew it was 4 digits per letter and that in each, the pairs were significant. I said that either addition or subtraction of those pairs was involved.

Each letter consists of four digits, which was easy to see just by finding the GCF. The first letter in the earlier code was "2420".

Take the last two digits of that letter (20) and subtract 2 (18). Now subtract that number from the first two letters (24-18=6) and divide the difference by two (3). That'll be the letter (C).

That seems pretty darn close to me. I didn't take the extra time to bother cracking it, but I came so close in my guess that I could have if I'd had the interest.

Lummox JR
Jun 14 2004, 1:16 pm
In response to HavenMaster
I got bored and made an open sourced decripter to get that thing done with.

I learned new things from making it. :)

The source is here: http://developer.byond.com/hub/UnknownPerson/ HavenMaster_Decripter

Get the answer to the encryption from this.
Jun 14 2004, 1:39 pm
"CONGRATULATIONS YOU FIGURED OUT THE TRICK"

Buahaha!! THE COOKIE IS MYNE!!
Jun 14 2004, 1:48 pm
In response to Shun Di
You wouldn't of gotten it if I didn't make that. -_-
Jun 14 2004, 1:52 pm
In response to Unknown Person
That was pretty inefficient. Here is a much easier way to do it:
var/list/alphabet=list("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")
var/decryptedstring
mob/verb/Decrypt(t as text)
    t=ckey(t)
    if(!(length(t)%4))
        for(var/i=1,i<=length(t)/4,i++)
            if(i==1)
                var/currentletter=copytext(t,i,i+4)
                var/lastnum=text2num(copytext(currentletter,3,5))
                var/letter=text2num(copytext(currentletter,1,3))
                lastnum-=2
                letter-=lastnum
                letter=letter/2
                previous=1
                previous2=5
                decryptedstring+=alphabet[letter]
            else
                var/currentletter=copytext(t,previous+4,previous2+4)
                var/lastnum=text2num(copytext(currentletter,3,5))
                var/letter=text2num(copytext(currentletter,1,3))
                lastnum-=2
                letter-=lastnum
                letter=letter/2
                previous+=4
                previous2+=5
                decryptedstring+=alphabet[letter]
        world<<decryptedstring
        decryptedstring=null
Jun 14 2004, 1:54 pm
In response to HavenMaster
HavenMaster wrote:
That was pretty inefficient. Here is a much easier way to do it:
> var/list/alphabet=list("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")
> var/decryptedstring
> mob/verb/Decrypt(t as text)
>   t=ckey(t)
>   if(!(length(t)%4))
>       for(var/i=1,i<=length(t)/4,i++)
>           if(i==1)
>               var/currentletter=copytext(t,i,i+4)
>               var/lastnum=text2num(copytext(currentletter,3,5))
>               var/letter=text2num(copytext(currentletter,1,3))
>               lastnum-=2
>               letter-=lastnum
>               letter=letter/2
>               previous=1
>               previous2=5
>               decryptedstring+=alphabet[letter]
>           else
>               var/currentletter=copytext(t,previous+4,previous2+4)
>               var/lastnum=text2num(copytext(currentletter,3,5))
>               var/letter=text2num(copytext(currentletter,1,3))
>               lastnum-=2
>               letter-=lastnum
>               letter=letter/2
>               previous+=4
>               previous2+=5
>               decryptedstring+=alphabet[letter]
>       world<<decryptedstring
>       decryptedstring=null
>


...

...

...

I forgot all about the ckey() proc...

Doh to me.