From: Smaug <js4smaug@eden.rutgers.edu>
Date: Wed, 11 Sep 1996 05:33:21 +0800
To: cypherpunks@toad.com
Subject: Allo allo
Message-ID: <CMM-RU.1.5.842366404.js4smaug@er6.rutgers.edu>
MIME-Version: 1.0
Content-Type: text/plain


Just figured i'd drop a hello in the box.

Hopefully at least a few of you have encountered me at some point...
For the rest, I am Si_Druid (si is for Silicon...) and i've been
around too long :)

I consider myself a cyBerpunk by a definition that myself and a group
of other CbP's came up w/ that seemed to fit best.  Whether it is best
or not is a matter of much discrepancy by the hard core "CbP is a
fictional genre" groupies, but its rather irrelevant to the rest of
the populace.

So why am i here?  Well, our ideas on CbP include just about all forms
of survival in this society, and cryptology is probably one of the
leading forms.  I figured I'd better throw my hat in the ring before
it got out of hand and i wound up eating it instead.

I'm not terribly experienced w/ crypts, pgp still annoys me, but 'back
in the day' i used to write some low level crypts that were relatively
impossible to break...
they were also impossible to decrypt, which meant that the only way to
use them was to work forward....

mostly i did password encryption schemes, this one working the best:

string        #a nice set of ascii codes
currkeyletter #letter from string in use
password      #a password from 5 to whatever characters
encrypted     #the password encrypted
char          #the current character of the password we're playing w/
stringchar    #the char # in the string

for char=0 to lengthof(password)

#getting our stringchar
stringchar=char
while stringchar >= lengthof(string) {
  stringchar=stringchar-lengthof(string)
}
currkeyletter=string[stringchar]

#writing the encrypted password
if char = 0   #first letter
encrypted[char]=resolve(password[char],0,password[char+1],string[stringchar]

else if char = lengthof(password) #last letter
encrypted[char]=resolve(password[char],encrypted[char-1],0,string[stringchar])

else
encrypted[char]=resolve(password[char],encrypted[char-1],password[char+1],
       string[stringchar])



end of for

resolve(pw,en,pw2,st){
  result=pw+en+pw2+st
  while result>=250 {
    result=result-250
  }
  return(result)
}


-----
doesnt really matter what language you write it in, the difficulty is
essentialy that the only way to solve it is to test all possible
passwords and strings...  which is...
(lets say a complexity of pw=7, str=6)
52^7 * 250^6 ~=10^26
I think that was somewhere near the number of electrons in the
universe :)

Anyway, just providing y'all with something to toy with....

Si.

Oh what a tangled web we weave, when we practice to deceive
Case in point - www.*.*