OpenEuphoria

Re: New encryption thread

• Posted by isaac <isaaca at MINDSPRING.COM> Dec 14, 1998
• 371 views

>isacc wrote:
>> Lets assume he has access to the ciphertext, as we did in your example.
>> The random seed can easily be narrowed down, mainly through the sumup
>> function because the order is insignifigant.  The ciphertext we were given
>> was 27 characters long, but we know the last one is padding, so 26.
>> 26*255 gives us a range of 0-6630, or 6631, possibilities,
>ummmmmmmmmmmmmmmmmmmmmmmmmm no.
>it would be 255 TO THE POWER OF 26...
>255^26...
>and actually, since the max length of a password is 20 characters,
>it's really 255^20...
>
>if you have 2 character, A and B, and 3 positions (length 3 key),
>your possible passwords are:
>AAA
>AAB
>ABA
>ABB
>BAA
>BAB
>BBB
>that is 2^3,  ie: 2 characters with length 3...
>if u have 100 characters (ie:32-132, or approx telnet ascii) and
>a length 20 key it's 100^20 which is 1e40 possibilities for a
>password...

You're describing the number of possible keywords, but many of them will
produce the same seed through you're old sumup().  Try this with the
original sumup:

function sumup(sequence data)
integer sum sum = 0
   for i = 1 to length(data) do
       sum = sum + data[i]
   end for
   return sum
end function

? sumup("elvenbelvedere")
? sumup("uflbenlewceeer")
? sumup("AAAAAAAAAAAAAAAAAAAAAA2")

Try to find a length 26 or less sequence of numbers between 0 and 255 which
add up to less than 26*255 :)

Good luck; I'm actually working on an algorythm of my own (retreivable
data, not hash) which I may post for public critique.