> OK, fine, but the point is that if you find a star and a time (and
> whatever other set of parameters you need --- sampling rate, bits per
> sample, wavelength filters, etc etc etc) that decrypts a ciphertext
> into something sensible, you can be sure that it's the right sensible
> thing. That's not the case with one-time pads --- given any
> ciphertext-plaintext pair of the same length, you can find a pad that
> will encrypt the plaintext as the ciphertext.
And if I happen across a random bitstring that just happens to decrypt my ciphertext, well,
fantastic. :-) Proves what, exactly? In either case, I've just proved that I've decrypted
the ciphertext. (:makes spinning motion in air with index finger, ;-)
That doesn't help me compromise future messages.
The point is, that with a large enough set of randomness sources (and high enough bit rates)
and enough time, you've got a problem that's much more an approx. of the one-time pad problem
than the n-length key problem.
> This is because the one-time pad has a keyspace as large as the number
> of possible plaintexts.
And the total size of the possible streams generated by a vast number of bit sources across a
large quantity of time?
jb
This archive was generated by hypermail 2b29 : Fri Apr 27 2001 - 23:18:43 PDT